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Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 12 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 12 Thermodynamics

PSEB 11th Class Physics Guide Thermodynamics Textbook Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10⁴ J/g?
Solution:
Water is flowing at a rate of 3.0 liter/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
Rise in temperature, ΔT = T₂ -T₁
= 77-27 = 50°C

Heat of combustion = 4 x 10⁴ J/g
Specific heat of water, C = 4.2 J g^-1 °C^-1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mC ΔT
= 3000 x 4.2 x 50
= 6.3 x 10⁵ J/min
∴ Rate of consumption = \(\frac{6.3 \times 10^{5}}{4 \times 10^{4}}\) = 15.75 g/min

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol^-1K^-1).
Solution:
Mass of nitrogen, m = 2.0 x 10^-2 kg = 20g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂,M =28
Universal gas constant, R = 8.3 J mol^-1K^-1
Number of moles, n = \(\frac{m}{M}\)
= \(\frac{2.0 \times 10^{-2} \times 10^{3}}{28}\) = 0.714
Molar specific heat at constant pressure for nitrogen,
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) x 8.3
= 29.05J mol^-1 K^-1

The total amount of heat to be supplied is given by the relation
ΔQ = nC_pΔT
= 0.714 x 29.05 x 45 = 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J

Question 3.
Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂) / 2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂) / 2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution:
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4 For an adiabatic process,
we have
P₁V₁^γ = P₂V₂^γ
The final volume is compressed to half of its initial volume.

Hence, the pressure increases by a factor of 2.639.

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:
The work done (W) on the system while the gas changes from state A to state Bis 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
ΔQ = 0
ΔW = -22.3 J (since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW

where, ΔU = Change in the internal energy of the gas .
ΔU = ΔQ – ΔW = -(-22.3 J)
ΔU = +22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is
ΔQ =9.35cal = 9.35 x 4.19 = 39.1765J
Heat absorbed, ΔQ = ΔU + ΔW
ΔW = ΔQ – ΔU = 39.1765 – 22.3 – = 16.8765J
Therefore, 16.88 J of work is done by the system.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock.
A contains a gas at standard temperature and pressure.
B is completely evacuated. The entire system is thermally insulated.
The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P -V – T surface?
Solution:
(a) 0.5 atm
The volume available to the gas is doused as soon as the stopcock between cylinders A and B is opened? Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5atm.

(b) Zero
The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(c) Zero
Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) No
The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Work done by the steam engine per minute, W = 5.4 x 10⁸ J
Heat supplied from the boiler, H = 3.6 x 10⁹ J
Efficiency of the engine = \(\frac{\text { Output energy }}{\text { Input energy }}\)
∴ η = \(\frac{W}{H}=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
Hence, the percentage efficiency of the engine is 15%.
Amount of heat wasted= 3.6 x 10⁹ – 5.4 x 10⁸
= 30.6 x 10⁸ = 3.06 x 10⁹ J
Therefore, the amount of heat wasted per minute is 3.06 x 10⁹J.

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
Heat is supplied to the system at a rate of 100 W.
∴ Heat supplied, ΔQ = 100 J/s
The system performs at a rate of 75 J/s.
∴ Work done, ΔW = 75 J/s

From the first law of thermodynamics, we have
ΔQ = ΔU + ΔW
where ΔU = Rate of change in internal energy
ΔU = ΔQ – ΔW = 100 – 75 = 25 J/s = 25W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure given below.

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Solution:
Total work done by the gas from D to E to F = Area of ΔDEF
Area of ΔDEF = \(\frac{1}{2}\) DF x EF
where, DF = Change in pressure
=600 N/m²
= 300N/m² = 300N/m²
FE = Change in volume
5.0 m³ – 2.0 m³ = 3.0m³
Area of ADEF = \(\frac{1}{2} \) x 300 x 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, Calculate the coefficient of performance.
Solution:
Temperature inside the refrigerator, T₁ = 9°C = 273 + 9 = 282 K
Room temperature, T₂ = 36°C = 273+36
Coefficient of performance = \(\frac{T_{1}}{T_{2}-T_{1}}\)
= \(\frac{282}{309-282}=\frac{282}{27}\)
309-282 = 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 11 Thermal Properties of Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

PSEB 11th Class Physics Guide Thermal Properties of Matter Textbook Questions and Answers

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Solution:
Kelvin and Celsius’s scales are related as
T_C =T_K -273.15
Celsius and Fahrenheit’s scales are related as …(i)
T_F = \(\frac{9}{5}\) T_C +32 ………………………….. (ii)
For neon:
T_K = 24.57K
∴ T_G = 24.57 – 273.15 = -284.58°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-248.58) + 32
=-415.44° F

For carbon dioxide:
T_K = 216.55K
∴ T_C =216.55-273.15 = -56.60°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-56.60) + 32 = -69.88°C

Question 2.
Two absolute scales A and B have triple points of water defined to be 200A and 350B.
What is the relation between T_Aand T_B?
Solution:
Triple point of water on absolute scale A, T₁ = 200 A
Triple point of water on absolute scale B, T₂ = 350 B
Triple point of water on Kelvin scale, T_K = 273.15K
The temperature 273.15K on Kelvein scale is equivalent to 200 A on absolute scale A.
T₁ =T_K
200 A = 273.15 K .
∴ A= \(\frac{273.15}{200}\)

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
T₂=T_K
350 B = 273.15
∴ B = \(\frac{273.15}{350}\)
T_A is triple point of water on scale A.
T_B is triple point of water on scale B.
∴ \(\frac{273.15}{200} \times T_{A}=\frac{273.15}{350} \times T_{B}\)
T_A = \(\frac{200}{350} T_{B}\)
T_A = \(\frac{4}{7} T_{B}\)
Therefore, the ratioT_A:T_B is given as 4 : 7.

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R₀[l + α (T-T₀)]
The resistance is 101.6Ω at the triple-point of water 273.16 K, and 165.50 at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4Ω?
Solution:
It is given that:
R = R₀[l + α (T-T₀)] ………………………….(i)
where R₀ and T₀ are the initial resistance and temperature respectively R and T are the final resistance and temperature respectively α is a constant At the triple point of water, T₀ = 273.16 K
Resistance of lead, R₀ =101.6 Ω

At normal melting point of lead, T = 600.5 K
Resistance of lead, R = 165.5 Ω
Substituting these values in equation (i), we get:
R=R₀[l + α (T-T₀)]
165.5 = 101.6[1 + α (600.5-273.16)]
1.629 = 1+α (327.34)
∴ α = \(\frac{0.629}{327.34} \) = 1.92 x10^-3K^-1
For resistance, R₁ = 123.4Ω
R₁ =R₀[l + α (T-T₀)]

where, T is the temperature when the resistance of lead is 123.4Ω
123.4 =101.6[1 +1.92 x 10^-3(T-273.16)]
1.214 =1+1.92 x 10^-3(T- 273.16)
\(\frac{0.214}{1.92 \times 10^{-3}}\) = T -273.16
111.46 = T-273.16
⇒ T =111.46 +273.16
∴ T = 384.62 K

Question 4.
Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c = T – 273.15
Why do we have 273.15 in this relation, and not 273.16?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Solution:
(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

(c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K. Hence, absolute temperature (Kelvin scafe) T, is related to temperature t_c, on Celsius scale as:
t_c =T -273.15

(d) Let T_F be the temperature on Fahrenheit scale and T_K be the temperature on absolute scale. Both the temperatures can be related as
\(\frac{T_{F}-32}{180}=\frac{T_{K}-273.15}{100}\) ………………………….. (i)
Let T_F1 be the temperature on Fahrenheit scale and T_K1 be the temperature on absolute scale. Both the temperatures can be relates as
\(\frac{T_{F 1}-32}{180}=\frac{T_{K 1}-273.15}{100}\) ……………………….. (ii)
It is given that:
T_K1-T_K= 1K

Subtracting equation (i) from equation (ii), we get
\(\frac{T_{F 1}-T_{F}}{180}=\frac{T_{K 1}-T_{K}}{100}\)
\(T_{F 1}-T_{F}=\frac{1 \times 180}{100}=\frac{9}{5}\)
Triple point of water = 273.16 K
∴ Triple point of water on absolute scale = 273.16 x \( \frac{9}{5}\) = 491.69

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Temperature	Pressure thermometer A	Pressure thermometer B
Triple-point of water	1.250 x 105 Pa	0.200 x 105 Pa
Normal melting point of sulphur	1.797 x 105 Pa	0.287 x 105 Pa

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Solution:
Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, P_A = 1.250 x 10⁵ Pa
Let T₁ be the normal melting point of sulphur.
At this temperature, pressure in thermometer A,P₁ = 1.797 x 10⁵ Pa
According to Charles’ law, we have the relation
\(\frac{P_{A}}{T}=\frac{P_{1}}{T_{1}}\)
∴ T₁ = \(\frac{P_{1} T}{P_{A}}=\frac{1.797 \times 10^{5} \times 273.16}{1.250 \times 10^{5}}\)
= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.
At triple point 237.16 K, the pressure in thermometer B,
P_B =0.200 x 10⁵ Pa
At temperature T₂, the pressure in thermometer B, P₂ = 0.287 x 10⁵ Pa
According to Charles’ law, we can write the relation
\(\frac{P_{B}}{T}=\frac{P_{2}}{T_{2}}\)
\(\frac{0.200 \times 10^{5}}{273.16}=\frac{0.287 \times 10^{5}}{T_{2}}\)
∴ T₂ = \(\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \times 273.16\) = 391.98 K
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low-pressure conditions. At low pressure, these gases behave as perfect ideal gases.

Question 6.
A steel tape lm long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cmon a hot day when the temperature is 45°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel = 1.20 x 10^-5 K^-1.
Solution:
Length of the steel tape at temperature T = 27°C, l = 1 m = 100 cm
At temperature T₁ = 45°C,
the length of the steel rod, l₁ = 63 cm
Coefficient of linear expansion of steel, α = 1.20 x 10^-5K^-1

Let l₂ be the actual length of the steel rod and l’ be the length of the steel tape at 45°C.
l’ =l+αl(T₁ -T)
∴ l’ = 100 +1.20 x 10^-5x 100(45 -27)
= 100,0216 cm

Hence, the actual length of the steel rod measured by the steel tape at 45° C can be calculated as
l₂ = \(\frac{100.0216}{100} \times 63\) = 63.0136cm
Therefore, the actual length of the rod at 45°C is 63.0136 cm. Its length at 27.0 °C is 63.0 cm

Question 7.
A large steel wheel is to be fitted onto a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm, and the diameter of the central hole in the wheel is 8.69cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 x 10^-5 K^-1.
Solution:
The given temperature, T = 27°C can be written in Kelvin as
27 + 273 =300K
Outer diameter of the steel shaft at T, d₁ = 8.70 cm
Diameter of the central hole in the wheel at T, d₂ = 8.69 cm
Coefficient of linear expansion of steel, a steel = 1.20 x 10^-5 K ^-1
After the shaft is cooled using ‘dry ice’, its temperature becomes T₁.
The wheel will slip on the shaft if the change in diameter,
Δd = 8.69-8.70 =-0.01 cm

Temperature T₁; can be calculated from the relation
Δd = d₁α_steel(T₁ -T)
-0.01 =8.70 x 1.20 x 10^-5(T₁ -300)
(T₁ – 300) = \(\frac{-0.01}{8.70 \times 1.20 \times 10^{-5}}\)
(T₁ -300) = -95.78
∴ T₁ = 300 – 95.78 = 204.22 K
= (204.22-273)°C
= -68.78°C ≈ 69°C
Therefore, the wheel will slip on the shaft when the temperature of the shaft is -69°C.

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = 1.70 x 10^-5 K ^-1.
solution:
Initial temperature, T₁ = 27.0°C
Diameter of the hole at T₁, d₁ = 4.24 cm
Final temperature, T₂ = 227°C
Let, diameter of the hole at T₂=d₂
Coefficient of linear expansion of copper, α_Cu = 1.70 x 10^-5 K^-1
For coefficient of superficial expansion β, and change in temperature ΔT, we have the relation:
\(\frac{\text { Change in area }(\Delta \mathrm{A})}{\text { Original area }(\mathrm{A})}=\beta \Delta \mathrm{T}\)

Change in diameter = d₂ – d₁ = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 x 10^-2 cm.

Question 9.
A brass wire 1.8m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of-39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x 10^-5K^-1; Young’s modulus of brass = 0.91 x 10¹¹ Pa.
Solution:
Initial temperature, T₁ = 27°C
Length of the brass wire at T₁,l = 1.8 m
Final temperature, T₂ = -39 °C
Diameter of the wire, d = 2.0 mm = 2 x 10^-3 m
Let, tension developed in the wire = F
Coefficient of linear expansion of brass, a = 2.0 x 10^-5 K^-1

Young’s modulus of brass, Y = 0.91 x 10¹¹Pa
Young’s modulus is given by the relation ‘
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
ΔL = \(\frac{F \times L}{A \times Y}\) ……………………………….. (i)
where, F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation
ΔL = αL(T₂ -T₁) …………………………… (ii)

Equating equations (i) and (ii), we get

= -3.8 x 10² N
(The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is 3.8 x 10² N.

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1, steel = 1.2 x 10^-5K^-1).
Solution:
Initial temperature, T₁ = 40°C
Final temperature, T₂ = 250 °C
Change in temperature, ΔT = T₂ – T₁ = 210°C
Length of the brass rod at T₁,l₁ = 50 cm
Diameter of the brass rod at T₂, d₁ = 3.0 mm
Length of the steel rod at T₂,l₂ = 50 cm
Diameter of the steel rod at T₂,d₂ = 3.0 mm
Coefficient of linear expansion of brass, α₁ = 2.0 x 10^-5 K^-1
Coefficient of linear expansion of steel, α₂ = 1.2 x 10^-5 K^-1

For the expansion in the brass rod, we have
\(\frac{\text { Change in length }\left(\Delta l_{1}\right)}{\text { Original length }\left(l_{1}\right)}=\alpha_{1} \Delta T\)
∴ Δl₁ = 50 x (2.1 x 10^-5)X210
= 0.2205cm
For the expansion in the steel rod, we have Change in length
\(\frac{\text { Change in length }\left(\Delta l_{2}\right)}{\text { Original length }\left(l_{2}\right)}=\alpha_{2} \Delta T\)
∴ Δl₂ =50 x (1.2 x 10^-5)x 210
= 0.126 cm

Total change in the lengths of brass and steel,
Δl = Δl₁ + Δl₂
=0.2205 + 0.126
= 0.346 cm
Total change in the length of the combined rod = 0.346 cm Since the rod expands freely from both ends, no thermal stress is developed at the junction.

Question 11.
The coefficient of volume expansion of glycerin is 49 x 10^-5K^-1.
What is the fractional change in its density for a 30°C rise in temperature?
Solution:
Coefficient of volume expansion of glycerin, α_v = 49 x 10^-5 K^-1
Rise in temperature, ΔT = 30°C
Fractional change in its volume = \(\frac{\Delta V}{V}\)

where, m = Mass of glycerine
PT₁ = Initial density at T₁
PT₂ = Final density at T₂
\(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}=\alpha_{\mathrm{V}} \Delta T\)
Where, \(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}\) = Fractional change in the density
∴ Fractional change in the density of glycerin
= 49 x 10^-5 x 30 = 1.47 x 10^-2

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 Jg^-1K^-1.
Solution:
Power of the drilling machine, P =10 kW = 10 x 10³ W
Mass of the aluminum block, m = 8.0 kg = 8 x 10³ g
Time for which the machine is used, t = 2.5min = 2.5x 60 = 150 s
Specific heat of aluminium, C = 0.91 J g^-1K^-1
Rise in the temperature of the block after drilling = ΔT

Total energy of the drilling machine = Pt
= 10 x 10³ x 150 = 1.5 x 10⁶ J

It is given that only 50% of the power is useful.

Useful energy, ΔQ = \(\frac{50}{100} \times 1.5 \times 10^{6}\) = 7.5×10⁵J
But ΔQ = mCΔT
∴ ΔT = \(\frac{\Delta Q}{m C}=\frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}\)
=103°C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg^-1K^-1; heat of fusion of water = 335Jg^-1).
Solution:
Mass of the copper block, m = 2.5kg = 2500 g
Rise in the temperature of the copper block, Δθ = 500 °C
Specific heat of copper, C = 0.39 J g^-1 °C^-1
The heat of fusion of water, L = 335 J g^-1

The maximum heat the copper block can lose, Q = mCΔθ
= 2500×0.39×500 =487500J
Let m₁ g be the amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice, Q = m₁L
∴ m₁ = \(\frac{Q}{L}=\frac{487500}{335}\) =1455.22g
Hence, the maximum amount of ice that can melt is 1.45 kg.

Question 14.
In an experiment on the specific heat of a metal, a 0.20kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40° C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Solution:
Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T₁ = 150°C
Final temperature of the metal, T₂ = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025kg = 25g
Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C,
150×1 =150g
Fall in the temperature of the metal,
ΔT =T₁ -T₂ =150-40 =110°C
Specific heat of water, C_w = 4.186 J/g/K
Specific heat of metal = C
Heat lost by the metal, Q = mCΔT …………………………….. (i)

Rise in the temperature of the water and calorimeter system,
ΔT’ = 40 -27 = 13°C
Heat gained by the water and calorimeter system,
ΔQ’ = m₁C_wΔT’
= (M + m’)C_wΔT’ ……………………………… (ii)

Heat lost by the metal = Heat gained by the water and colourimeter system
mCΔT =(M + m’)C_wΔT’
200 xC x 110 = (150+25) x 4.186 x 13
∴ C = \(\frac{175 \times 4.186 \times 13}{110 \times 200} \) = 0.43 Jg^-1K^-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat (Cv) (cal mol-1K-1)
Hydrogen	4.87
Nitrogen	4.97
Oxygen	5.02
Nitric oxide	4.99
Carbon monoxide	5.01
Chlorine	6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Solution:
The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion). Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion.

Hence, the molar specific heat of diatomic gases is more than that of monatomic gases. If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = \(\frac{5}{2} R=\frac{5}{2} \times 1.98\) = 4.95 cal mol^-1K^-1 With the exception of chlorine, all the observations in the given table agree with(\(\frac{5}{2} R\)) This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?
(b)What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at
(a) -70°C under 1 atm,
(b) -60°C under 10 atm,
(c) 15°C under 56 atm?
Solution:
The P-T phase diagram for CO₂ is shown in the following figure.

(a) C is the triple point of the CO₂ phase diagram. This means that at the temperature and pressure corresponding to this point (i.e., at -56.6°C and 5.11 atm), the solid, liquid, and vaporous phases of CO₂ co-exist in equilibrium.
(b) The fusion and boiling point of CO₂ decrease with a decrease in pressure.
(c) The critical temperature and critical pressure of CO₂ are 31.1°C and 73 atm respectively.
Even if it is compressed to a pressure greater than 73 atm, CO₂ will not liquefy above the critical temperature.
(d) It can be concluded from the P-T phase diagram of CO₂ that:
(i) CO₂ is gaseous at -70 °C, under 1 atm pressure
(ii) CO₂ is solid at -60 °C, under 10 atm pressure
(iii) CO₂ is liquid at 15°C, under 56 atm pressure.

Question 17.
Answer the following questions based on the P-T phase diagram of CO₂:
(a) CO₂ at 1 atm pressure and temperature -60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when C0₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
The P-T phase diagram for CO₂ is shown in the following figure:

Solution:
(a) No
Explanation:
At 1 atm pressure and at -60°C, CO₂ lies to the left of -56.6°C (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, CO₂ condenses into the solid-state directly, without going through the liquid state.
(b) It condenses to solid directly.
Explanation:
At 4 atm pressure, C0₂ lies below 5.11 atm (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid-state directly, without passing through the liquid state.

(c) The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.
Explanation:
When the temperature of a mass of solid CO₂ (at 10 atm pressure and at -65° C) is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

(d) It departs from ideal gas behaviour as pressure increases.
Explanation:
If CO₂ is heated to 70 °C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70 °C is higher than the critical temperature of CO₂. It will remain in the vapour state but will depart from its ideal behaviour as pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrin (i. e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Solution:
Initial temperature of the body of the child, T₁ = 101°F
Final temperature of the body of the child, T₂ = 98°F
Change in temperature, ΔT = \(\left[(101-98) \times \frac{5}{9}\right] \)°c
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 x 10³ g

Specific heat of the human body = Specific heat of water = C
= 1000 cal/kg/°C
Latent heat of evaporation of water, L = 580 cal g^-1
The heat lost by the child is given as:
ΔQ = mCΔT
= 30 x 1000 x (101-98)x \(\frac{5}{9}\) = 50000cal

Let m₁ be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is given by
ΔQ = m₁L
∴ m₁ = \(\frac{\Delta Q}{L}=\frac{50000}{580}\) = 86.2g
∴ Average rate of extra evaporation caused by the drug = \(\frac{m_{1}}{t}\)
= \(\frac{86.2}{20}\) = 4.3 g/mm.

Question 19.
A ‘thermal’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and coefficient of thermal conductivity of thermal is 0.01 J s^-1m^-1 K^-1.
[Heat of fusion of water = 335 x 10³ Jkg ^-1 ]
Solution:
Side of the given cubical icebox, s = 30 cm = 0.3 m
Thickness of the icebox, l = 5.0 cm = 0.05 m
Mass of ice kept in the icebox, m = 4 kg
Time gap, t=6h = 6x 60 x 60 s
Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermacole,
K =0.01 Js^-1 m^-1 K^-1
Heat of fusion of water, L = 335 x 10³ J kg^-1
Let m be the total amount of ice that melts in 6 h.
The amount of heat lost by the food,
Q = \(\frac{K A(T-0) t}{l}\)

where, A = Surface area of the box = 6s² =6 x (0.3)² = 0.54 m²
Q = \(\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}\) = 104976 J
But Q=m’L
∴ m’ = \(\frac{Q}{L}=\frac{104976}{335 \times 10^{3}}\) = 0.313 kg
Mass of ice left = 4-0.313 = 3.687kg .
Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 20.
A brass boiler has a base area of 0.15m² and a thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js^-1m^-1 K^-1; Heat of vaporisation of water = 2256 x 10³ Jkg^-1.
Solution:
Base area of the boiler, A = 0.15 m²
Thickness of the boiler,l = 1.0 cm = 0.01 m
Roiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s^-1 m^-1 K^-1
Heat of vaporisation, L = 2256 x10³ J kg^-1
The amount of heat flowing into water through the brass base of the boiler is given by
Q = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\) ………………………. (i)
where, T₁ = Temperature of the flame in contact with the boiler
T₂ = Boiling point of water = 100 °C
Heat required for boiling the water
Q = mL …………………………………. (ii)

Equating equations (i) and (ii), we get
∴ mL = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\)
T₁ -T₂= \(\frac{m L l}{K A t}\)
= \(\frac{6 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60}\) = 137.98°C
T₁ -T₂ = 137.98°C
∴ T₁=137.98+100 = 237.98°C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

Question 21.
Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day.
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
Solution:
(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.

(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler. Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool. •
Thus, a brass tumbler feels colder than a wooden tray on a chilly day.

(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open. Black body radiation equation is given by
E = σ(T⁴ -T₀⁴) where, E – Energy radiation
T = Temperature of optical pyrometer
T₀ = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E =σT⁴

(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be .trapped. All the heat would be radiated back from earth’s surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20 C.
Solution:
According to Newton’s law of cooling, the rate of cooling cc difference in temperature.
Here, average of 80 °C and 50 °C, T = \(\frac{T_{1}+T_{2}}{2}=\frac{80+50}{2}\) = 65°C
Temperature of surroundings, T₀ = 20 °C .
∴Difference, ΔT = T – T₀ = 65 – 20 = 45°C
Under these conditions, the body cools 30 °C in time 5 minutes.
∴ 
or \(\frac{30}{5}\) = k x 45 ……………………………….. (i)
The average of 60°C and 30 °C is 45°C which is 25°C (45 – 20) above the room temperature and the body cools by 30 °C (60 – 30) in a time t (say).
∴\(\frac{30}{t}\) = k x 25 ……………………………….. (ii)

where k is same for this situation as for the original.
Dividing eq. (i) from eq. (ii), we get
\(\frac{30 / 5}{30 / t}=\frac{k \times 45}{k \times 25}\)
or \(\frac{t}{5}=\frac{9}{5}\)
or t = 9 min

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 10 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

PSEB 11th Class Physics Guide Mechanical Properties of Fluids Textbook Questions and Answers

Question 1.
Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer:
(a) The pressure of a liquid is given by the relation
P =hρg
where, P = Pressure
h = Height of the liquid column
ρ = Density of the liquid ‘ .
g = Acceleration due to the gravity

It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, jthe blood pressure at the feet is more than it is at the brain.

(b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.

(c) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

Question 2.
Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Solution:
(a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (0), as shown in the given figure.

S_la, S_sa, and S_sl are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i. e.,
cos θ = \(\frac{S_{s a}-S_{s l}}{S_{l a}}\)
The angle of contact 0, is obtuse if S_sa < S_la (as in the case of mercury on glass). This angle is acute if Ss < Sa (as in the case of water on glass).

(b) Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops. On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.

(c) Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

(d) Water with detergent dissolved in it has small angles of contact (0). This is because for a small 0, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (0). If 0 is small, then cos 0 will be large and the rise of the detergent water in the cloth will be fast.

(e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally…with temperatures (increases/ decreases)
(b) Viscosity of gases …………………. with temperature, whereas viscosity of liquids ………………………… with temperature (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to …………………………… while for fluids it is proportional to ………………………………… (shear strain/rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ………………………….. speed for turbulence for an actual plane (greater /smaller)
Solution:
(a) decreases
The surface tension of a liquid is inversely proportional to temperature.

(b) increases; decreases
Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.

(c) shear strain; rate of shear strain
With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

(d) conservation of mass/Bernoulli’s principle
For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bemoulli’s principle.

(e) greater
For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds numbers are associated with the motions of the two planes. ,

Question 4.
Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory
Answer:
(a) When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.

(b) According to the equation of continuity,
Area x Velocity = Constant
For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.

(c) The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.

(d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity,
Area x Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.

(e) A spinning cricket ball has two simultaneous motions-rotatory and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Solution:
Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = \(\frac{d}{2}\) = 0.005 m
Area of the heel = πr²
= 3.14 x (0.005)²
= 7.85 x 10^-5 m²

Force exerted by the heel on the floor,
F = mg
= 50 x 9.8 = 490 N
Pressure exerted by the heel on the floor,
P = \(\frac{\text { Force }}{\text { Area }}\)
= \(\frac{490}{7.85 \times 10^{-5}}\) = 6.24 x 10⁶Nm^-2
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 x 10⁶Nm^-2 .

Question 6.
Torieelli’s barometer used mercury. Pascal duplicated it using French wine of density 984kg m3. Determine the height of the wine column for normal atmospheric pressure.
Solution:
Density of mercury, ρ₁ = 13.6 x 10³ kg / m³
Height of the mercury column, h₁ = 0.76 m
Density of French wine, ρ₂ = 984 kg / m³
Height of the French wine column = h₂
Acceleration due to gravity, g = 9.8 m / s²

The pressure in both the columns is equal, i. e.,
Pressure in the mercury column = Pressure in the French wine column
ρ₁h₁g = ρ₂h₂g
h₂ = \(\frac{\rho_{1} h_{1}}{\rho_{2}}\)
= \(\frac{13.6 \times 10^{3} \times 0.76}{984}\)
= 10.5m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to bet roughly 3 km, and ignore ocean currents.
Solution:
Yes The maximum allowable stress for the structure, P = 10⁹Pa
Depth of the ocean, d = 3 km = 3 x 10³ m
Density of water, ρ = 10³ kg / m³
Acceleration due to gravity, g = 9.8 m / s²

The pressure exerted because of the sea water at depth, d = ρdg
= 3 x 10³ x 10³ x 9.8 = 2.94 x 10⁷ Pa
The maximum allowable stress for the structure (10⁹ Pa) is greater than the pressure of the seawater (2.94 x 10⁷ Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425cm2. What maximum pressure would the smaller piston have to bear?
Solution:
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm² = 425x 10^-4m²
The maximum force exerted by the load, F = mg
= 3000 x 9.8 = 29400N
The maximum pressure exerted on the load-carrying piston, P = \(\frac{F}{A}\)
= \(\frac{29400}{425 \times 10^{-4}}\)
= 6.917 x 10⁵ Pa
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 x 10⁵Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
Solution:
The given system of water, mercury, and methylated spirit is shown as follows:

Height of the spirit column, h₁ = 12.5cm = 0.125m
Height of the water column, h₂ = 10 cm = 0.1 m
P₀ = Atmospheric pressure
ρ₁ = Density of spirit
ρ₂ = Density of water
Pressure at point B = P₀ + h₁ρ₁g
Pressure at point D = P₀ + h₂ρ₂g
Pressure at points B and D is the same.

Therefore, the specific gravity of spirit is 0.8.

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Solution:
Height of the water column, h₁ =10+15 = 25cm
Height of the spirit column, h₂ = 12.5 +15 = 27.5cm
Density of water, ρ₁ = 1 g cm^-3
Density of spirit, ρ₂ = 0.8 g cm^-3
Density of mercury = 13.6 g cm^-3

Let h be the difference between the levels of mercury in the two arms. Pressure exerted by height h, of the mercury column:
= hρg = h x 13.6g ……………………………….. (i)
Difference between the pressures exerted by water and spirit
h₁ρ₁g – h₂ρ₁g
= g (25 x 1 – 27.5 x 0.8) = 3g ……………………………….. (ii)
Equating equations (i) and (ii), we get
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm .
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
No Explanation: Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamlined flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
No Explanation: It does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 x 10^-3 kgs^-1, what is the pressure difference between the two ends of the tube?(Density of glycerine = 1.3 x 10³ kg m ^-3 and viscosity of glycerine = 0.83Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
Solution:
Length of the horizontal tube, l = 1.5m
Radius of the tube, r = 1 cm = 0.01 m
Diameter of the tube, d = 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 x 10 kgs .
M = 4.0 x 10^-3 kgs^-1
Density of glycerine, ρ = 1.3 x 10^-3 kg m^-3
Viscosity of glycerine, η = 0.83Pas
Volume of glycerine flowing per sec,
V = \(\frac{M}{\rho}=\frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}}\)
= 3.08 x 10^-6 m³ s^-1
According to Poisevelle’s formula, we have the relation for the rate of flow,
V = \(\frac{\pi p r^{4}}{8 \eta l} \)
where, p is the pressure difference between the two ends of the tube
∴ p = \(\frac{V 8 \eta l}{\pi r^{4}}\)
= \(\frac{3.08 \times 10^{-6} \times 8 \times 0.83 \times 1.5}{3.14 \times(0.01)^{4}} \)
= 9.8 x 10² Pa
Reynold’s number is given by the relation,
R = \(\frac{4 \rho V}{\pi d \eta}=\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{-6}}{3.14 \times(0.02) \times 0.83}\)
= 0.3
Reynold’s number is about 0.3. Hence, the flow is laminar.

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5m²? Take the density of air to be 1.3 kg m^-3.
Solution:
Speed of wind on the upper surface of the wing, V₁ = 70 m/s
Speed of wind on the lower surface of the wing, V₂ = 63 m/s
Area of the wing, A = 2.5 m²
Density of air, ρ = 1.3 kg m
According to Bernoulli’s theorem, we have the relation:

where, P₁ = Pressure on the upper surface of the wing
P₂ = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P₂ – P₁ )A

Therefore, the lift on the wing of the aeroplane is 1.51 x 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Solution:
Figure (a) is incorrect.
Take the case given in figure (b).

where, A₁ = Area of pipe 1
A₂ = Area of pipe 2
V₁ = Speed of the fluid in pipe 1
V₂ = Speed of the fluid in pipe 2
From the law of continuity, we have
A₁V₁ = A₂V₂
When the area of a cross-section in the middle of the venturi meter is small, the speed of the flow of liquid through this part is more. According to Bernoulli’s principle, if speed is more, then pressure is less. Pressure is directly proportional to height. Hence, the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes?
Solution:
Area of cross-section of the spray pump. A = 8 cm² = 8 x 10^-4 m²
number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 x 10^-3 m
Radius of each hole,r = d/2 = 0.5 x 10^-3 m
Area of cross-section of each hole, a = πr² = π(0.5 x 10^-3)²m²
Total area of 40 holes, A₂ = n x a
= 40 x 3.14 x (0.5 x 10^-3)² m²
= 31.41 x 10^-6 m²

Speed of flow of liquid inside the tube, V₁ = 1.5 m/min = 0.025m/s
Speed of ejection of liquid through the holes = V₂
According to the law of continuity, we have A₁V₁ = A₂V₂
V₂ = \(\frac{A_{1} V_{1}}{A_{2}}=\frac{8 \times 10^{-4} \times 0.025}{31.41 \times 10^{-6}}\)
= 0.636 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.636 m/s.

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Solution:
The weight that the soap film supports, W = 1.5 x 10^-2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴ Total length = 2l = 2 x 0.3 = 0.6 m
Surface tension, T = \(\frac{\text { Force or Weight }}{2 l} \)
= \(\frac{1.5 \times 10^{-2}}{0.6}\) =  2.5 x10^-2  N/m
Therefore, the surface tension of the film is 2.5 x10^-2Nm^-1.

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 x 10^-2 N.
What is the weight supported by a film of the same liquid at the same temperature in fig. (b) and (c)? Explain your answer physically.

Solution:
Take case (a):
The length of the liquid film supported by the weight, l = 40 cm = 0.4 m
The weight supported by the film, W = 4.5 x 10^-2 N
A liquid film has two free surfaces.
∴ Surface tension = \(\frac{W}{2 l}=\frac{4.5 \times 10^{-2}}{2 \times 0.4}\) = 5.625 x 10^-2 Nm^-1
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625x 10 ~2Nm-1.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 x 10^-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 x 10^-1 Nm^-1. The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.
Solution:
Radius of the mercury drop, r = 3.00 mm = 3 x 10^-3 m
Surface tension of mercury, T = 4.65 x 10^-1 N m^-1
Atmospheric pressure, P₀ = 1.01 x 10⁵ Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= \(\frac{2 T}{r}+P_{0}\)

Excess pressure = \(\frac{2 T}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\) = 310 Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of ‘ radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 x 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 10⁵ Pa).
Solution:
Soap bubble is of radius, r = 5.00 mm = 5 x 10^-3 m
Surface tension of the soap solution, T = 2.50 x 10^-2 Nm^-1
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 x 10³ kg/m³
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 x 10^-3 m
1 atmospheric pressure = 1.01 x 10⁵Pa

Acceleration due to gravity, g = 9.8 m/s²
Hence, the excess pressure inside the soap bubble is given by the relation
P = \(\frac{4 T}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation
P’ = \(\frac{2 T}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
=10 Pa

Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble =Atmospheric pressure + hρg + P’
= 1.01 x 10⁵ + 0.4 x 1.2 x 10³ x 9.8 + 10 ,
= 1.057 x 10⁵ Pa = 1.06 x 10⁵ Pa
Therefore, the pressure inside the air bubble is 1.06 x 10⁵ Pa.

Additional Exercises

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Solution:
Base area of the given tank, A = 1.0 m²
Area of the hinged door, a = 20 cm² = 20 x 10^-4 m²
Density of water, ρ₁ = 10³ kg/m³
Density of acid, ρ₂ = 1.7 x 10³ kg/m³
Height of the water column, h₁ = 4 m
Height of the acid column, h₂ = 4 m
Acceleration due to gravity, g = 9.8 m/s²
Pressure due to water is given as
P₁ =h₁ρ₁g = 4 x 10³ x 9.8 = 3.92 x 10⁴Pa
Pressure due to acid is given as, P₂ = h₂ρ₂g
= 4 x 1.7 x10³ x 9.8
= 6.664 x 10⁴ Pa

Pressure difference between the water and acid columns,
ΔP=P₂– P₁
= 6.664 x 10⁴ -3.92 x10⁴
= 2.744 x10⁴ Pa
Hence, the force exerted on the door = ΔP x a
= 2.744 x 10⁴ x 20 x 10^-4 = 54.88N
Therefore, the force necessary to keep the door closed is 54.88N.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in figure (a). When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Solution:
(a) For figure (a)
Atmospheric pressure, P₀ = 76 cm of Hg
The difference between the levels of mercury in the two limbs gives gauge pressure Hence, gauge pressure is 20 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76+20 =96 cm of Hg

For figure (b)
Difference between the levels of mercury in the two limbs = -18 cm Hence, gauge pressure is -18 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm-18cm = 58 cm

(b) 13.6 cm of water is poured into the right limb of figure (b).
Relative density of mercury = 13.6
Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury. Let h be the difference between the levels of mercury in the two limbs. The pressure in the right limb is given as,
P_R = Atmospheric pressure + 1 cm of Hg
= 76+1 = 77 cm of Hg …………………………. (i)
The mercury column will rise in the left limb.
Hence, pressure in the left limb,
P_L = 58 + h ……………………………. (ii)
Equating equations (i) and (ii), we get
77 = 58 + h
h = 19 cm
Hence, the difference between the levels of mercury in the two limbs will be 19 cm.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Answer:
Yes.
Two vessels having the same base area have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components. When these vertical components are added, the total force on one vessel comes out to be greater than that on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000Pa. At what height must the blood container be placed so that blood may just enter the vein? [Take the density of whole blood = 1.06 x 10³ kg m^-3 ].
Solution:
Given, gauge pressure, P = 2000 Pa
Density of whole blood, p = 1.06 x 10³ kg m^-3
Acceleration due to gravity, g = 9.8 m/s²
Height of the blood container = h
Pressure of the blood container, P = hρg
h = \(\frac{P}{\rho g}=\frac{2000}{1.06 \times 10^{3} \times 9.8}\)
= 0.1925 m
The blood may enter the vein if the blood container is kept at a height greater than 0.1925m, i. e., about 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy,
(a) What is the largest average velocity of blood flow in an artery of diameter 2 x 10^-3 m if the flow must remain laminar?
(b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Solution:
(a) Diameter of the artery, d = 2×10^-3 m
Viscosity of blood, η = 2.084 x 10^-3 Pas
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given as
V_avg = \(\frac{N_{R} \eta}{\rho d}\)
= \(\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}\)
= 1.966 m/s
Therefore, the largest average velocity of blood is 1.966 m/s
(b) Yes, as the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

Question 26.
(a) What is the largest average velocity of blood flow in an artery of radius 2 x 10^-3 m if the flow must remain laminar?
(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 x 10^-3 Pas).
Solution:
(a) Radius of the artery, r = 2 x 10^-3 m
Diameter of the artery, d=2 x 2x 10^-3 m = 4 x 10^-3m
Viscosity of blood, η = 2.084 x 10^-3 Pa s
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given by the relation
V_Avg = \(\frac{N_{R} \eta}{\rho d}=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 4 \times 10^{-3}}\)
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.98,3 m/s.

(b) Flow rate is given by the relation
R = πr² V_avg
= 3.14 x (2 x 10^-3)² x 0.983
= 1.235 x 10^-5m³s^-1
Therefore, the corresponding flow rate is 1.235 x 10^-5m³s^-1.

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1kg m^-3).
Solution:
The area of the wings of the plane, A = 2 x 25 = 50 m²
Speed of air over the lower wing,
V₁ = 180 km/h = 180 x \(\frac{5}{18}\) m/s = 50 m/s
Speed of air over the upper wing,
V₂ = 234 km/h = 234 x \(\frac{5}{18}\) m/s = 65 m/s
Density of air, ρ = 1 kg m^-3
Pressure of air over the lower wing = P₁
Pressure of air over the upper wing = P₂
The upward force on the plane can be obtained using Bernoulli’s equation as

The upward force (F) on the plane can be calculated as

Using Newton’s force equation, we can obtain the mass (m) of the plane as
F = mg
m = \(\frac{43125}{9.8}\)
= 4400.51 kg ≈ 4400 kg
Hence, the mass of the plane is about 4400 kg.

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-5 m and density 1.2 x 10^-5 kg m?
Take the viscosity of air at the temperature of the experiment to be 1.8x 10⁵ Pas. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Solution:
Terminal speed = 5.8cm/s; Viscous force = 3.9 x 10^-10 N
Radius of the given uncharged drop, r = 2.0 x 10^-5 m
Density of the uncharged drop, ρ = 1.2 x 10³ kg m^-3
Viscosity of air, η = 1.8 x 10^-5 Pa s
Density of air (ρ₀) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s²
Terminal velocity (ν) is given by the relation
ν = \(\frac{2 r^{2} \times\left(\rho-\rho_{0}\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^{2}\left(1.2 \times 10^{3}-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.807 x 10^-2ms^-1
= 5.8 cm s^-1
Hence, the terminal speed of the drop is 5.8 cms^-1.
The viscous force on the drop is given by:
F = 6πηrν
∴ F = 6 x 3.14 x 1.8 x 10^-5 x 2.0 x 10^-5 x 5.8 x 10^-2
= 3.9 x 10^-10N
Hence, the viscous force on the drop is 3.9 x 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm^-1. Density of mercury = 13.6 x 10³ kgm^-3.
Solution:
Angle of contact between mercury and soda-lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 x 10^-3 m
Surface tension of mercury at the given temperature, T = 0.465N m^-1
Density of mercury, ρ = 13.6 x 10³ kg/m³
Dip in the height of mercury = h

Acceleration due to gravity, g = 9.8 m/s²
Surface tension is related with the angle of contact and the dip in the height as
T = \(\frac{h \rho g r}{2 \cos \theta}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\)

= -5.34 mm
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm

Question 30.
Two narrow bores of diameters 3.0mm and 6.0mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube?
Surface tension of water at the temperature of the experiment is 7.3 x 10^-2Nm^-1.
Take the angle of contact to be zero and density of water to be 1.0x 10³ kg m^-3 (g = 9.8ms^-2).
Solution:
Diameter of the first bore, d₁ = 3.0 mm = 3 x 10^-3 m
Hence, the radius of the first bore, r₁ = \(\frac{d_{1}}{2}\) =1.5 x 10^-3m
Diameter of the second bore, d₂ =6.0 mm
Hence, the radius of the second bore, r₂ = \(\frac{d_{2}}{2} \) = 3 x 10^-3 m
Surface tension of water, T = 7.3 x 10^-2 N m^-1
Angle of contact between the bore surface and water, θ=0
Density of water, ρ = 1.0 x 10³ kg/m^-3
Acceleration due to gravity, g =9.8 m/s²
Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively.

These heights are given by the relations
h₁ = \(\frac{2 T \cos \theta}{r_{1} \rho g}\) …………………..(i)
h₂ = \(\frac{2 T \cos \theta}{r_{2} \rho g}\) …………………… (ii)
The difference between the levels of water in the two limbs of the tube can be calculated as
= h₁ – h₂
= \(\frac{2 T \cos \theta}{r_{1} \rho g}-\frac{2 T \cos \theta}{r_{2} \rho g}\)

= 4.966 x 10^-3m = 4.97 mm
Hence, the difference between levels of water in the two bores is 4.97 mm.

Question 31.
(a) It is known that density p of air decreases with height y as ρ = ρ_oe^-y/yo
where ρ_o = 1.25kg m^-3 is the density at sea level, and a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.

(b) A large He balloon of volume 1425m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y₀ =8000m and ρ_He = 018 kg m^-3]
Solution:
Volume of the balloon, V = 1425m³
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s²
y_o =8000m
ρ_He =0.18kgm^-3
ρ_o =1.25kg/m³

Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as
ρ = ρ₀e^-y/yo
\(\frac{\rho}{\rho_{0}}=e^{-y / y_{0}}\) …………………………… (i)

This density variation is called the law of atmospheres.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i. e.,

where, k is the constant of proportionality
Height changes from 0 to y, while density changes from ρ_o to ρ).
Integrating the sides between these limits, we get

Comparing equations (i) and (ii) we get
y₀ = \(\frac{1}{k}\)
k = \(\frac{1}{y_{0}}\) ……………………………………. (iii)

From equations (ii) and (iii), we get
ρ = ρ₀e^-y/yo
(b) Density,
ρ = \(\frac{\text { Mass }}{\text { Volume }}\)

= 0.46 kg/m³
From equations (ii) arid (iii), we can obtain y as
ρ = ρ₀e^-y/yo
log e\(\frac{\rho}{\rho_{0}}=-\frac{y}{y_{0}}\)
∴ y =-8000 x log_e \(\frac{0.46}{1.25}\)
=-8000 x-1=8000m8 km
Hence, the balloon will rise to a height of 8 km.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 5 Laws of Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 5 Laws of Motion

PSEB 11th Class Physics Guide Laws of Motion Textbook Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
(a) Zero net force
The rain drop is falling with a constant speed. Hence, its acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.

(b) Zero net force
The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

(c) Zero net force
The kite is stationary in the sky, i. e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.

(d) Zero net force
The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.

(e) Zero net force
The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.
Solution:
Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:
F = m × a
Where,
F = Net force
m = Mass of the pebble = 0.05 kg
a = g =10 m/s²
F =0.05 × 10 =0.5 N
The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
(d) lying on the floor of a train which is accelerating with 1 ms^-2, the stone being at rest relative to the train.
Neglect air resistance throughout.
Solution:
(a) Mass of the stone, m = 0.1 kg
Acceleration of the stone, a = g = 10 m/s²
As per Newton’s second law of motion, the net force acting on the stone,
F = ma = mg = 0.1 × 10 = 1 N
Acceleration due to gravity always acts in the downward direction.
The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(b) The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

(c) It is given that the train is accelerating at the rate of 1 m/s² .
Therefore, the net force acting on the stone, F’ = ma = 0.1 × 1 = 0.1 N
This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F’, stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.
F = mg = 1 N
This force acts vertically downward.

(d) The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
Acceleration of the train, a = 0.1 m/s²
The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:
F = ma
= 0.1 × 1 = 0.1 N

Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed υ the net force on the
particle (directed towards the centre) is:
(i) T,
(ii) T – \(\frac{m v^{2}}{l}\),
(iii) T + \(\frac{m v^{2}}{l}\),
(iv) 0
T is the tension in the string. [Choose the correct alternative].
Solution:
(i) When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced , in the string. Hence, in the given case, the net force on the particle is the tension T, i. e.,
F = T = \(\frac{m v^{2}}{l}\)
where F is the net force acting on the particle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Solution:
Retarding force, F = -50 N
Mass of the body, m = 20 kg
Initial velocity of the body, u = 15 m/s
Final velocity of the body, υ = 0
Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:
F = ma
-50 = 20 × a
∴ a = \(\frac{-50}{20}\) = -2.5 m/s²
20
Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:
υ = u + at
t = \(\frac{-u}{a}=\frac{-15}{-2.5}\) = 6s

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, υ = 3.5 m/s Time,
Time t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
υ = u + at
∴ a = \(\frac{v-u}{t}\)
= \(\frac{3.5-2}{25}=\frac{1.5}{25}\) = 0.06 m/s²
As per Newton’s second law of motion, force is given as:
F = ma
= 3 × 0.06 = 0.18 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Solution:
Mass of the body, m = 5 kg
The given situation can be represented as follows:

The resultant of two forces is given as:
R = \(\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}\) = 10N
θ is the angle made by R with the force of 8 N
∴ θ = tan^-1 (\(\frac{-6}{8}\)) = -36.87°
The negative sign indicates that 0 is in the clockwise direction with respect to the force of magnitude 8 N.
Hence, the magnitude of the acceleration is 2 m/s2, at an angle of 37° with a force of 8 N.
As per Newton’s second law of motion, the acceleration (a) of the body is given as :
F = ma
a = \(\frac{F}{m}=\frac{10}{5}\) = 2m/s²
Hence, the magnitude of the acceleration is 2 m/s², at an angle of 37° with a force of 8 N.

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child.
What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:
Initial speed of the three-wheeler, u = 36 km/h
Final speed of the three-wheeler, υ = 10 m/s
Time, t = 4s
Mass of the three-wheeler, m = 400 kg
Mass of the driver, = m’ = 65 kg
Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:
= u + at
a = \(\frac{v-u}{t}=\frac{0-10}{4}\) = -2.5 m/s²
The negative sign indicates that the velocity of the three-wheeler is decreasing with time.
Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:
F = Ma
= 465 × (-2.5) = -1162.5 N
= -1.2 × 10³ N
The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Solution:
Mass of the rocket, m = 20,000 kg
Initial acceleration, a = 5 m/s²
Acceleration due to gravity, g = 10 m/s²
Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:
F – mg = ma
F = m(g + a)
= 20000 × (10 + 5)
= 20000 × 15 = 3 × 10⁵ N

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s^-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be t = 0, and predict its position at t = -5 s, 25 s, 100 s.
Solution:
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = -8.0 N
Acceleration produced in the body, a = \(\frac{F}{m}=\frac{-8.0}{0.40}\) = -20 m/s²
At t = -5 s
Acceleration, a’ = 0 and u = 10 m/s
s = ut + \(\frac{1}{2}\) a’t²
= 10 × (-5) + 0
= -50 m

At t = 258
Acceleration, a” = -20 m/s²
and u = 10 m/s
s’ =ut + \(\frac{1}{2}\) a” t²
= 10 × 25 + \(\frac{1}{2}\) × (-20) × (25)²
= 250 – 6250 = -6000 m

At t = 100 s
For 0 ≤ t ≤ 30 s
a = -20 m/s²
u = 10 m/s
s₁ = ut + \(\frac{1}{2}\) a”t²
= 10 × 30 + \(\frac{1}{2}\) × (-20) × (30)²
= 300 – 9000
= -8700 m
For 30 < t ≤ 100 s
As per the first equation of motion, for t = 30 s, final velocity is given as:
υ = u + at
= 10 + (-20) × 30 =-590 m/s
Velocity of the body after 30 s = -590 m/s
Distance travelled in time interval from t = 30 s to t =100 s
s₂ = υt
= -590 × 70 = -41300 m
.’.Total distance, s” = s₁ + s₂ = -8700 – 41300 = -50000 m = -50 km

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Solution:
(a) Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s²
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
υ = u + at
= 0 + 2 × 10 =20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (υ_x ) of velocity, in the absence of air resistance, remains unchanged, i.e.,
υ_x = 20 m/s
The vertical component (υ_y) of velocity of the stone is given by the first equation of motion as :
υ_y = u + a_yδt
where, δt = 11 – 10 = 1 s
and a_y = g = 10 m/s²
υ_y = 0 + 10 × 1 =10 m/s
The resultant Velocity (υ) of the stone is given as:

υ = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{20^{2}+10^{2}}=\sqrt{400+100}\)
= \(\sqrt{500}\) = 22.36 m/s

Hence, velocity is 22.36 mIs, at an angle of 26.57° with the motion of the truck.

b) Let θ be the angle made by the resultant velocity with the horizontal component of velocity, υ_x
∴ tanθ = (\(\frac{v_{y}}{v_{x}}\))
θ = tan^-1(\(\frac{10}{20}\))
= tan^-1 (0.5)
= 26.57°
When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s² and it acts vertically downward.

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the String is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Solution:
(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 ms^-1,
(b) downwards with a uniform acceleration of 5 m s^-2,
(c) upwards with a uniform acceleration of 5ms^-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Solution:
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
where, ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
∴ R = mg
= 70 × 10= 700 N
∴ Reading on the weighing scale = \(\frac{700}{g}=\frac{700}{10}\) 70 kg

(b) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s² downward
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= 70 (10 – 5) = 70 × 5
= 350 N
Reading on the weighing scale = \(\frac{350}{g}=\frac{350}{10}\) = 35 kg

(c) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s² upward
Using,Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5)
= 70 × 15 = 1050 N
∴ Reading on the weighing scale = \(\frac{1050}{g}=\frac{1050}{10}\) = 105 kg

(d) When the lift moves freely under gravity, acceleration a = g
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= m(g – g) = 0
∴ Reading on the weighing scale = \(\frac{0}{g}\) = 0 kg
The man will be in a state of weightlessness.

Question 14.
Following figure shows the position-time graph of a particle of mass 4 kg. What is the (a) Force on the particle for t< 0, t > 4 s, 0< t< 4s? (b) impulse at f = 0 and f = 4s? (Consider one-dimensional motion only).

Solution:
(a) For t < 0 It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero. For t > 4 s
It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4
It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0
Impulse = Change in momentum
= mυ – mu
Mass of the particle, m = 4 kg
Initial velocity of the particle, u = 0
Final velocity of the particle, υ = \(\frac{3}{4}\) m/s
∴ Impulse = (\(\frac{3}{4}\) – 0) = 3 kg m/s
At t = 4s
Initial velocity of the particle, u = \(\frac{3}{4}\) m/s
Final velocity of the particle, υ = 0
∴ Impulse = 4(0 – \(\frac{3}{4}\)) = -3 kg m/s

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B, along the direction of string. What is the tension in the string in each case?
Solution:
Horizontal force, F = 600 N
Mass of body A, m₁ = 10 kg
Mass of body B, m₂ = 20 kg
Total mass of the system, m = m₁ + m₂ = 30 kg
Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as :
F = ma
∴ a = \(\frac{F}{m}=\frac{600}{30}\) = 20 m/s²
When force F is applied on body A:

The equation of motion can be written as:
F – T = m₁a
∴ T = F – m₁a
= 600 – 10 × 20 =400 N
When force F is applied on body B:

The equation of motion can be written as:
F – T = m₂a
T = F – m₂a
∴ T =600 – 20 × 20 = 200 N

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Solution:
The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, m₁ = 8 kg
Larger mass, m₂ = 12 kg
Tension in the string = T
Mass m₂, owing to its weight, moves downward with acceleration a and mass m₁ moves upward.
Applying Newton’s second law of motion to the system of each mass:
For mass m₁:
The equation of motion can be written as:
T – m₁g = ma ……………. (i)

For mass m₂:
m₂g – T = m₂ a ………………. (ii)
Adding equations (i) and (ii),we get:
(m₂ – m₁)g = (m₁ + m₂)a
∴ a = [Latex](\frac{m_{2}-m_{1}}{m_{1}+m_{2}}[/Latex]) g
= (\(\frac{12-8}{12+8}\)) × 10 = \(\frac{4}{20}\) × 10 = 2m/s²
Therefore, the acceleration of the masses is 2 m/s² .
Substituting the value of a in equation (ii), we get:

Therefore, the tension in the string is 96 N.

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
Let m, m₁ and m₂ be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) = 0
Let υ₁ and υ₂ be the respective velocities of the daughter nuclei having masses m₁ and m₂.
Total linear momentum of the system after disintegration
= m₁ υ₁ + m₂υ₂
According to the law of conservation of momentum,
Total initial momentum = Total final momentum
0 = m₁υ₁+ m₂υ₂
υ₁ = \(\frac{-m_{2} v_{2}}{m_{1}}\)
Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each hall due to the other?
Solution:
Mass of each ball = 0.05 kg
Initial velocity of each ball = 6 m/s
Magnitude of the initial momentum of each ball, p_i = 0.3 kg m/s
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball, p_f = -0.3 kg m/s
Impulse imparted to each ball = Change in the momentum of the system
= P_f – P_i
= -0.3 -0.3 = -0.6 kg m/s
The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s^-1, what is the recoil speed of the gun?
Solution:
Mass of the gun, M = 100 kg
Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, υ = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mυ – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum,
Final momentum = Initial momentum
mυ – MV = 0
∴ V = \(\frac{m v}{M}\)
= \(\frac{0.020 \times 80}{100}\) = 0.016 M/S
= 1.6 cm/s

Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Solution:
The given situation can be represented as shown in the following figure.

where,
AO = Incident path of the ball
OB = Path followed by the ball after deflection
∠AOB = Angle between the incident and deflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
Initial and final velocities of the ball = υ
Horizontal component of the initial velocity = υcosθ along RO
Vertical component of the initial velocity = υ sinθ along PO
Horizontal component of the final velocity = υ cosθ along OS
Vertical component of the final velocity = υ sinθ along OP
The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.
∴ Impulse imparted to the ball
= Change in the linear momentum of the ball
= m υcosθ – (-mυ cosθ)
= 2mυ cosθ
Mass of the ball, m = 0.15 kg
Velocity of the ball, υ = 54 km/h = 54 × \(\frac{5}{18}\) m/s = 15 m/s
∴ Impulse = 2 × 0.15 × 15 cos 22.5° = 4.5 × 0.9239 = 4.16 kg m/s

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Number of revolution per second, n = \(\frac{40}{60}=\frac{2}{3}\) rps

Angular velocity, ω = \(\frac{v}{r}\) = 2πn ……………… (i)
The centripetal force for the stone is provided by the tension T, in the string, i.e.,

Therefore, the maximum speed of the stone is 34.64 m/s.

Question 22.
If, in question 21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Solution:
(b) When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Question 23.
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Answer:
(a) In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.

(b) When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

(c) While pulling a lawn mower, a force at ah angle θ is applied on it, as shown in the following figure.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower.
On the other hand, while pushing a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

In this case, the vertical component of the applied fords acts in the direction of the weight of the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.

(d) According to Newton’s second law of motion, we have the equation of motion:
F = ma = m\(\frac{\Delta v}{\Delta t}\) ……………. (i)
where,
F = Stopping force experienced by the cricketer as he catches the ball m = Mass of the ball
∆t = Time of impact of the ball with the hand It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,
f ∝ \(\frac{1}{\Delta t}\) ………….. (ii)
Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.
While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (∆t). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.
Additional Exercises

Question 24.
Figure below shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Solution:
A ball rebounding between two walls located between at x = 0 and x = 2 cm; after every 2 s, the bah receives an impulse of magnitude 0.08 × 10^-2 kg-m/s from the walls.
The given graph shows that a body changes its direction of motion after every 2 s.
Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x – t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.
Mass of the ball, m = 0.04 kg
The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:
u = \(\frac{(2-0) \times 10^{-2}}{(2-0)}\) = 10^-2 m/s
Velocity of the ball before collision, u = 10^-2 m/s
Velocity of the ball after collision, υ = -10^-2 m/s
(Here, the negative sign arises as the ball reverses its direction of motion.) Magnitude of impulse = Change in momentum
= | mυ – mu | = 10.04 (υ – u) |
= | 0.04 (-10^-2 – 10 ^-2 ) |
= 0.08 × 10^-2 kg-m/s
= 8 × 10^-4 kg-ms^-1

Question 25.
Figure below shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

Solution:
Mass of the man, m = 65 kg
Acceleration of the belt, a = 1 m/s²
Coefficient of static friction, μ = 0.2
The net force F, acting on the man is given by Newton’s second law of motion as:
F_net = ma = 65 × 1 – 65 N
The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force f_s, exerted by the belt, i. e.,
F’_net = f_s
ma’ = μmg
∴ a’ =0.2 × 10 = 2 m/s²
Therefore, the maximum acceleration of the belt up to which the man can stand stationary is 2 m/s².

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

	Lowest Point	Highest Point
(a)	mg – T1	mg + T2
(b)	mg + T1	mg – T1
(c)	mg + T1 – (mυ12)/R	mg – T2 + (mυ 12) / R
(d)	mg – T1 – (mυ12 )/ R	mg + T2 + (mυ12 ) / R

T₁ and υ ₁ denote the tension and speed at the lowest point. T₂ and υ₂denote corresponding values at the highest point.
Solution:
(a) The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law. of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,
F_net = mg – T₁ ……………….. (i)
where, υ₁ = Velocity at the lowest point
The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:
F_net = mg + T₂ ……………… (ii)
where, υ ₂ = Velocity at the highest point
It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are (mg – T₁ ) and (mg + T₂) respectively.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force oh the helicopter due to the surrounding air.
Solution:
Mass of the helicopter, m_h = 1000 kg
Mass of the crew and passengers, m_p = 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s²

(a) Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:
R – m_pg = m_pa
or R = m_p(g + a)
= 300 (10 + 15) = 300 × 25
= 7500 N
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:
R’ – mg = ma
or R’ = m(g + a)
= 1300 (10 + 15) = 1300 × 25
= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Question 28.
A stream of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:
Speed of the water stream, υ = 15 m/s
Cross-sectional area of the tube, A = 10^-2 m²
Volume of water coming out from the pipe per second,
V = Aυ = 15 × 10^-2 m³/s
Density of water, ρ = 10³ kg/m³
Mass of water flowing out through the pipe per second = ρ × V =150 kg/s The water strikes the wall and does not rebound. Therefore, the force , exerted by the water on the wall is given by Newton’s second law of motion as:
F = Rate of change of momentum = \(\frac{\Delta P}{\Delta t}=\frac{m v}{t}\)
= 150 × 15 = 2250 N

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7th coin by the eighth coin,
(c) the reaction of the 6th coin on the 7th coin.
Solution:
(a) Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3 mg
Hence, the force exerted on the 7th coin by the three coins on its top is 3 mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3 mg
Hence, the force exerted on the 7th coin by the eighth coin is 3 mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th and 10th) on its top.
Therefore, the total downward force experienced by the 6th coin is 4 mg.
As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th coin is of magnitude 4 mg. This force acts in the upward direction.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:
Speed of the aircraft, υ = 720 km/h = 720 × \(\frac{5}{18}\) = 200 m/s
Acceleration due to gravity, g = 10 m/s²
Angle of banking, θ = 15°
For radius r, of the loop, we have the relation:
tan0 =\(\frac{v^{2}}{r g}\)
r = \(\frac{v^{2}}{g \tan \theta}=\frac{200 \times 200}{10 \times \tan 15^{\circ}}=\frac{4000}{0.268}\)
= 14925.37 m = 14.92 km

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Solution:
Radius of the circular track, r = 30 m
Speed of the train, υ = 54 km/h = 15 m/s
Mass of the train, m = 10⁶ kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail.
The angle of banking 0, is related to the radius (r) and speed (υ) by the relation:
tanθ = \(\frac{v^{2}}{r g}=\frac{(15)^{2}}{30 \times 10}=\frac{225}{300}\)
θ = tan^-1 (0.75) = 36.87°
Therefore, the angle of banking is about 36.87°.

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure below. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Solution:
Mass of the block, m = 25 kg
Mass of the man, M = 50 kg
Acceleration due to gravity, g = 10 m/s²
Force applied on the block, F =25 × 10 = 250 N
Weight of the man, W = 50 × 10 = 500 N

Case (a): When the man lifts the block directly
In this case, the man applies a force in the upward direction. This increases his apparent weight.
.’. Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley
In this case, the man applies a force in the downward direction. This decreases his apparent weight.
Action on the floor by the man = 500 – 250 = 250 N
If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Question 33.
A monkey of mass 40 kg climbs on a rope (see figure), which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 m s^-2
(b) climbs down with an acceleration of 4 m s^-2
(c) climbs up with a uniform speed of 5 m s^-1
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).

Solution:
Case (a)
Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, T_max = 600 N
Acceleration of the monkey, a = 6 m/s² upward
Using Newton’s second law of motion, we can write the equation of motion as:
T – mg = ma
∴ T = m(g +a)
= 40(10 + 6)
= 640 N
Since T > T_max, the rope will break in this case.

Case (b)
Acceleration of the monkey, a = 4 m/s² downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg – T = ma
∴ T = m(g – a)
= 40(10 – 4)
= 240 N
Since T < T_max, the rope will not break in this case.

Case (c)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e.,a = 0.
Using Newton’s second law of motion, we can write the equation of motion as:
T – mg = ma
T- mg = 0
∴ T = mg
= 40 × 10
= 400 N
Since T < T_max, the rope will not break in this case.

Case (d)
When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e.,a = g
Using Newton’s second law of motion, we can write the equation of motion as:
mg – T = mg
∴ T = m(g – g) = 0
Since T < T_max, the rope will not break in this case.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (see figure). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between (μ_s and μ_k

Solution:
Mass of body A, m_A = 5 kg
Mass of body B, m_B =10 kg ,
Applied force, F = 200 N
Coefficient of friction, μ_s = 0.15

(a) The force of friction is given by the relation:
f_s = μ(m_A + m_B)g
= 0.15(5 + 10) × 10
= 1.5 × 15 = 22.5 N leftward
Net force acting on the partition = 200 – 22.5 = 177.5 N rightward
As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.
Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:
f_A = μ m_Ag
= 0.15 × 5 × 10 = 7.5 N leftward
Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward
As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i. e., 192.5 N acting leftward.
When the wall is removed, the two bodies will move in the direction of the applied force.
Net force acting on the moving system = 177.5 N
The equation of motion for the system of acceleration a, can be written as: Net force = (m_A + m_B)a
Net force
∴ a = \(\frac{\text { Net force }}{m_{A}+m_{B}}\)
= \(\frac{177.5}{5+10}=\frac{177.5}{15}\) = 11.83 m/s2
Net force causing mass A to move:
F_A =m_Aa = 5 × 11.83 = 59.15N
Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N
This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i, e., 133.35 N, acting opposite to the direction of motion.

Question 35.
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Solution:
Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleiation of the trolley, a = 0.5 m/s²

(a) As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 × 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = μ mg = 0.18 × 15 × 10 = 27 N
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in figure below. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Solution:
Mass of the box, m = 40 kg
Coefficient of friction, μ = 0.15
Initial velocity, u = 0
Acceleration, a = 2 m/s²
Distance of the box from the end of the truck, s’ = 5 m
As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:
F = ma – 40 × 2 = 80 N
As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction /, acting between the box and the floor of the truck. This force is given by:
f = μmg = 0.15 × 40 × 10 = 60 N
∴ Net force acting on the block:
F_net = 80 – 60 = 20 N backward
The backward acceleration produced in the box is given by:
a_back = \(\frac{F_{\text {net }}}{m}=\frac{20}{40}\) = 0.5m/s²
Using the second equation of motion, time t can be calculated as :
s’ =ut + \(\frac{1}{2}\)a_backt²
5 = 0 + \(\frac{1}{2}\) × 0.5 × t²
∴ t = \(\sqrt{20}\) s
Hence, the box will fall from the truck after \(\sqrt{20}\) s from start.
The distance s, travelled by the truck in \(\sqrt{20}\) s is given by the relation :
s = ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2 × (\(\sqrt{20}\) )² = 20 m

Question 37.
A disc revolves with a speed of 33\(\frac{1}{3}\) rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
Solution:
Mass of each coin = m
Radius of the disc, r = 15 cm = 0.15 m
Frequency of revolution, v = 33 \(\frac{1}{3}\) rev/min = \(\) rev/s
Coefficient of friction, μ = 0.15
In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:
Radius of revolution, r’ = 4 cm = 0.04 m
Angular frequency, ω = 2πv = 2 × \(\frac{22}{7}\) × \(\frac{5}{9}\) = 3.49 s^-1
Frictional force, f = μ mg = 0.15 × m × 10 = 1.5m N
Centripetal force on the coin:
F_cent = mr’ω²
= m × 0.04 × (3.49)²
= 0.49 m N
Since f > F_cent, the coin will revolve along with the record.

Coin placed at 14 cm:
Radius, r” = 14 cm = 0.14 m
Angular frequency, ω = 3.49 s^-1
Frictional force, f’ = 1.5 m N
Centripetal force is given as:
F_cent = mr”ω²
= m × 0.14 × (3.49)² = 1.7m N
Since f < _cent, the coin will slip from the surface of the record.

Question 38.
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?
Solution:
In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (F_N) and the force due to gravity (F_g = mg).
The equation of motion for the centripetal acceleration ac, can be written as :
F_net – ma_c
F_N + F_g = ma_c
F_N + mg = \(\frac{m v^{2}}{r}\)
Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (υ_min),
F_N = 0
mg = \(\frac{m v_{\min }^{2}}{r}\)
∴ υ_min = \(\frac{r}{\sqrt{r g}}=\sqrt{25 \times 10}\) = 15.8 m/s

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Solution:
Mass of the man, m = 70 kg .
Radius of the drum, r = 3 m
Coefficient of friction, μ = 0.15
Frequency of rotation, v = 200 rev/mm = \(\frac{200}{60}=\frac{10}{3}\) rev/s
The necessary centripetal force required for the rotation of the man is provided by the normal force (F_N).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force (f = μF_N) acting upward.
Hence, the man will not fall until:
mg< f
mg< μF_N = μmrω
g < μ rω²
ω = \(\sqrt{\frac{g}{\mu r}}\)
The minimum angular speed is given as:
ω_min = \(\sqrt{\frac{g}{\mu r}}\)
= \(\sqrt{\frac{10}{0.15 \times 3}}\) = 4.71 rad s^-1

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ \(\sqrt{g / R}\). What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = \(\sqrt{2 g / R}\)?  Neglect friction.
Solution:
Let the radius vector joining the bead with the centre makes an angle 0, with the vertical downward direction.

OP = R = Radius of the circle
N = Normal reaction
The respective vertical and horizontal equations of forces can be written as:
mg = N cosθ ……………. (i)
mlω²= N sinθ ………….. (ii)
In Δ OPQ, we have:
sinθ = \(\frac{l}{R}\)
l = R sinθ ………….. (iii)
Substituting equation (iii) in equation (ii), we get:
m (R sinθ) ω² = N sinθ
mR ω² = N ……………. (iv)
Substituting equation (iv) in equation (i), we get:
mg = mRω²cosθ
cosθ = \(\frac{g}{R \omega^{2}}\) …………….. (v)
Since cosθ ≤ 1, the bead will remain at its lowermost point for \(\frac{g}{R \omega^{2}}\) ≤ 1,
i.e for ω ≤ \(\sqrt{\frac{g}{R}}\)
For ω = \(\sqrt{\frac{2 g}{R}}\) or ω² = \(\frac{2 g}{R}\) …………….. (vi)
On equating equations (v) and (vi), we get:
\(\frac{2 g}{R}=\frac{g}{R \cos \theta}\)
cosθ = \(\frac{1}{2}\)
∴ θ = cos^-1 (0.5) = 60°

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 4 Motion in a Plane Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 4 Motion in a Plane

PSEB 11th Class Physics Guide Motion in a Plane Textbook Questions and Answers

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector:
volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:
Scalar quantities: Volume, mass, speed, density, number of moles, angular frequency.
Vector quantities: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Work and current are scalar quantities.

Question 3.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Impulse is only a vector quantity in the given quantities.

Question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars,
(b) adding a scalar to a vector of the same dimensions,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.
Answer:
(a) Not meaningful
Explanation: Adding any two scalars is not meaningful because only the scalars of same dimensions can be added.

(b) Not meaningful
Explanation: The addition of a vector quantity with a scalar quantity is not meaningful.

(c) Meaningful
Explanation: A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.

(d) Meaningful
Explanation: A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.

(e) Not meaningful
Explanation: Adding any two vectors is not meaningful because only vectors of same dimensions can be added.

(f) Meaningful
Explanation: A component of a vector can be added to the same vector as they both have the same dimensions.

Question 5.
Read each statement below carefully and state with reasons, if it is true or false:
(a) The magnitude of a vector is always a scalar,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the displacement vector of a particle,
(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
(e) Three Vectors not lying in a plane can never add up to give a null vector.
Answer:
(a) True
Explanation: The magnitude of a vector is a number. Hence, it is a scalar.

(b) False
Explanation: Each component of a vector is also a vector.

(c) False
Explanation: Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

(d) True
Explanation: It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) True
Explanation: Three vectors, which do not he in a plane, cannot be represented by the sides of a triangle taken in the same order.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
(a) \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
(b) \(|\vec{a}+\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
(c) \(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
(d) \(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
When does the equality sign above apply?
Solution:
(a) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

\(|\overrightarrow{O M}|=|\vec{a}|\) ……………. (i)
\(|\overrightarrow{M N}|=|\overrightarrow{O P}|=|\vec{b}|\) ……………. (ii)
\(|\overrightarrow{O N}|=|\vec{a}+\vec{b}|\) ……………. (iii)
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ∆ OMN, we have:
ON < (OM + MN)
\(|\vec{a}+\vec{b}|<|\vec{a}|+|\vec{b}|\) ………….. (iv)
If the two vectors \(\vec{a}\) and \(\vec{b}\) act along a straight line in the same direction, then we can write: \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\) …………… (v)
Combining equations (iv) and (v), we get: \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|\) (b) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:
\(|\overrightarrow{O M}|=|\vec{a}|\) …………….. (i)
\(|\overrightarrow{M N}|=|\overrightarrow{O P}|=|\vec{b}|\) …………….. (ii)
\(|\overrightarrow{O N}|=|\vec{a}+\vec{b}|\) …………….. (iii)
In a triangle, each side is smaller than the sum of the-other two sides. Therefore, in ∆ OMN, we have: ON + MN > OM
ON + OM > MN
\(|\overrightarrow{O N}|>|\overrightarrow{O M}-\overrightarrow{O P}|\) (∵ OP = MN)
\(|\vec{a}+\vec{b}|>\| \vec{a}|-| \vec{b}||\) ……………….. (iv)

If the two vectors \(\vec{a}\) and \(\vec{b}\) act along a straight line in the same direction, then we can write:
\(|\vec{a}+\vec{b}|=\|\vec{a}|-| \vec{b}\|\) ……………. (v)
Combining equations (iv) and (v), we get:
\(|\vec{a}+\vec{b}| \geq \| \vec{a}|-| \vec{b}||\)

(c) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:
\(|\overrightarrow{O R}|=|\overrightarrow{P S}|=|-\vec{b}|\) ……………… (i)
\(|\overrightarrow{O P}|=|\vec{a}|\) …………….. (ii)
\(|\overrightarrow{O S}|=|\vec{a}-\vec{b}|\) …………….. (iii)
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ∆ OPS, we have:
OS < OP + PS
\(|\vec{a}-\vec{b}|<|\vec{a}|+|-\vec{b}|\)
\(|\vec{a}-\vec{b}|<|\vec{a}|+|\vec{b}|\) …………… (iv)
If the two vectors act in a straight line but in opposite directions, then we can write:
\(|\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}|\) …………….. (v)
Combining equations (iv) and (v), we get
\(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)

(d) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

The following relations can be written for the given parallelogram. OS + PS > OP …………… (i)
OS > OP – PS ……………. (ii)
\(|\vec{a}-\vec{b}|>|\vec{a}|-|\vec{b}|\) ……………….. (iii)
The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:
\(\|\vec{a}-\vec{b}\|>\|\vec{a}|-| \vec{b}\|\)
\(|\vec{a}-\vec{b}|>|| \vec{a}|-| \vec{b}||\) ………………. (iv)
If the two vectors act in a straight line but in the opposite directions, then we can write:
\(|\vec{a}-\vec{b}|=\| \vec{a}|-| \vec{b}||\) …………….(v)
Combining equations (iv) and (v), we get
\(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)

Question 7.
Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0, which of the following statements are correct:
(a) \(\vec{a}, \vec{b}, \vec{c}\) and \(\overrightarrow{\boldsymbol{d}}\) must each be a null vector,
(b) The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\),
(c) The magnitude of a can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\),
(d) \(\vec{b}+\vec{c}\) must lie in the plane of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) if \([latex]\overrightarrow{\boldsymbol{a}}\)[/latex] and \(\overrightarrow{\boldsymbol{d}}\) are not collinear, and in the line of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{d}}\), if they are collinear?
Solution:
(a) Incorrect
In order to make \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

(b) Correct
\(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}+\vec{c}=-(\vec{b}+\vec{d})\)
Taking modulus on both the sides, we get:
\(|\vec{a}+\vec{c}|=|-(\vec{b}+\vec{d})|=|\vec{b}+\vec{d}|\)
Hence, the magnitude of (\(\vec{a}+\vec{c}\)) is the same as the magnitude of (\(\vec{b}+\vec{d}\)).

(c) Correct \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}=(\vec{b}+\vec{c}+\vec{d})\)

Taking modulus on both sides, we get
\(|\vec{a}|=|\vec{b}+\vec{c}+\vec{d}|\)
\(|\vec{a}| \leq|\vec{a}|+|\vec{b}|+|\vec{c}|\) ………………. (i)

Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).

(d) Correct For \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}+(\vec{b}+\vec{c})+\vec{d}\) = 0
The resultant sum of the three vectors \(\vec{a},(\vec{b}+\vec{c})\) and \(\vec{d}\) can be zero
only if (\(\vec{b}+\vec{c}\)) lie in a plane containing a and d, assuming that these
three vectors are represented by the three sides of a triangle.

If a and d are collinear/ then it implies that the vector (\(\vec{b}+\vec{c}\) ) is in the line of \(\vec{a}\) and \(\vec{d}\). This implication holds only then the vector sum of all the vectors will be zero.

Question 8.
Three girls skating on a circular ice ground of radius 200m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in figure below. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skate?

Solution:
Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground.
Radius of the ground = 200 m
Diameter of the ground = 2 × 200 = 400 m
Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.

Question 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in figure below. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Solution:
(a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.

(b) Average velocity is given by the relation;

Since the net displacement of the cyclist is zero, his average velocity will also be zero.

(c) Average speed of the cyclist is given by the relation

Question 10.
On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Solution:
The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

Let the motorist start from point P.
The motorist takes the third turn at S.
∴ Magnitude of displacement PS = PV + VS = 500 + 500 = 1000 m
Total path length = PQ + QR +RS = 500 + 500 + 500 = 1500 m
The motorist takes the sixth turn at point P, which is the starting point.
∴ Magnitude of displacement = 0
Total path length = PQ + QR + RS + ST +TU + UP
= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m
The motorist takes the eight turn at point R
∴ Magnitude of displacement = PR

= 866.03 m
If it is inclined at an angle β from the direction of PQ, then

or β = 30°
Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.
Total path length = Circumference of the hexagon + PQ + QR
= 6 × 500 + 500 + 500 = 4000 m
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given

Turn	Magnitude of dispalcement (m)	Total path length (m)
Third	1000	1500
Sixth	0	3000
Eighth	866.03; 30°	4000

Comparison of the magnitude of displacement with the total path length in each case:

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
Solution:
Total distance travelled = 23 km
Total time taken = 28 min = \(\frac{28}{60}\) h

(b) Distance between the hotel and the station =10 km = Displacement of the taxi
∴ Average velocity = \(\frac{\frac{10}{28}}{\frac{60}{60}}\) = 21.43 km/ h

Therefore, the two physical quantities (average speed and average velocity) are not equal.

Question 12.
Rain is falling vertically with a speed of 30 m s^-1. A woman rides a bicycle with a speed of 10 ms^-1 in the north to south direction. What is the direction in which she should hold her umbrella?
Solution:
The described situation is shown in the given figure.

Here, υ_c = Velocity of the cyclist
υ_r = Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.
υ = υ_r + (-υ_c)
= 30 + (-10) = 20 m/s
tanθ = \(\frac{v_{c}}{v_{r}}=\frac{10}{30}\)
θ = tan^-1 (\(\frac{1}{3}\))
= tan^-1 (0.333) ≈ 18°
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other hank?
Solution:

Speed of the man, υ_m = 4 km/h
Width of the river = 1 km

= \(\frac{1}{4}\) h = \(\frac{1}{4}\) × 60 = 15 min
Speed of the river, υ_r = 3 km/h
Distance covered with flow of the river = υ_r × t
= 3 × \(\frac{1}{4}\) = \(\frac{3}{4}\) km
= \(\frac{3}{4}\) × 1000 = 750 m

Question 14.
In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
Velocity of the boat, υ_b = 51 km/h
Velocity of the wind, υ_w = 72 km/h
The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (υ_wb) of the wind with respect to the boat.

∴ β = tan^-1 (1.0038) = 45.18°
Angle with respect to the east direction = 45.18° – 45° = 0.18°
Hence, the flag will flutter almost due east.

Question 15.
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s 1 can go without hitting the ceiling of the hall?
Solution:
Speed of the ball, u = 40 m/s
Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle 0, is given by the relation:

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Solution:
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = \(\frac{u^{2} \sin 2 \theta}{g}\)
100 = \(\frac{u^{2}}{g}\) sin90°
\(\frac{u^{2}}{g}\) = 100 ……………… (i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = -g
Using the third equation of motion:
υ² – u² = -2gH
H = \(\frac{1}{2}\) × \(\frac{u^{2}}{g}\) = \(\frac{1}{2}\) × 100 = 50 m

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Solution:
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s

= 9.91 m/s²
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Question 18.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:
Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft,υ = 900 km/h = 900 × \(\frac{5}{18}\) = 250 m/s
Centripetal acceleration, a_c = \(\frac{v^{2}}{r}\)
= \(\frac{(250)^{2}}{1000}\) = 62.5 m/s²
Acceleration due to gravity, g = 9.8 m/s²
\(\frac{a_{c}}{g}=\frac{62.5}{9.8}\) = 6.38
a_c = 6.38 g

Question 19.
Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Answer:
(a) False
Reason: The net acceleration of a particle in circular motion is not always directed along the radius of the circle towards the centre. It happens only in the case of uniform circular motion.

(b) True
Reason: At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.

(c) True
Reason: In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Question 20.
The position of a particle is given by
r̂ = 3.0 t î – 2.0 t²ĵ + 4.0k̂ m
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the υ and a of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 3.0 s?
Solution:

= – tan^-1 (2.667)
= -69.45°
The negative sign indicates that the direction of velocity is below the x-axis.

Question 21.
A particle starts from the origin at t 0 s with a velocity of 10.0 ĵ m/s and moves in the x – y plane with a constant acceleration of (8.0î + 2.0ĵ) ms^-2.
(a) At what time is the x – coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time?
Solution:

Question 22.
î and ĵ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors î + ĵ and î – ĵ? What are the components of a vector \(\overrightarrow{\boldsymbol{A}}\) = 2î + 3ĵ along the directions of î + ĵ and î – ĵ? [You may use graphical method]
Solution:
Consider a vector \(\vec{P}\), given as:
\(\vec{P}\) = î + ĵ
P_xî +P_y ĵ = î + ĵ
On comparing the components on both sides, we get:
P_x = P_y = 1
\(|\vec{P}|=\sqrt{P_{x}^{2}+P_{y}^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}\) …………… (i)
Hence, the magnitude of the vector î + ĵ is √2.
Let 0 be the angle made by the vector \(\), with the x-axis, as shwon in the following figure.

Hence, the vector î + ĵ makes an angle of 45° with the x-axis.
Let \(\vec{Q}\) = î – ĵ

Question 23.
For any arbitrary motion in space, which of the following relations are true:

(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)
Solution:
(b) and (e)
(a) It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
(b) The arbitrary motion of the particle can be represented by this equation.
(c) The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.
(d) The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space.
(e) The arbitrary motion of the particle can be represented by this equation.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
Answer:
(a) False
Reason: Despite being a scalar quantity, energy is not conserved in inelastic collisions.

(b) False
Reason: Despite being a scalar quantity, temperature can take negative values.

(c) False
Reason: Total path length is a scalar quantity. Yet it has the dimension of length.

(d) False
Reason: A scalar quantity such as gravitational potential can vary from one point to another in space.

(e) True
Reason: The value of a scalar does not vary for observers with different orientations of axes.

Question 25.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:
The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ =30°
Time = 10 s
In Δ PRO:
tan15° = \(\frac{P R}{O R}\)
PR = OR tan 15°
= 3400 × tan15°
Δ PRO is similar to Δ RQO.
PR =RQ
Motion in a Plane 81
PQ = PR + RQ
= 2PR = 2 × 3400 tanl5°
= 6800 × 0.268 = 1822.4 m
speed of the aircraft = \(\frac{1822.4}{10}\) = 182.24 m/s

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical ‘ effects? Give examples in support of your answer.
Answer:
No; Yes; No
Generally speaking, a vector has no definite locations in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However, a position vector has a definite location in space.

A vector can vary with time. For example, the displacement vector of a particle moving with a certain velocity varies with time.

Two equal vectors located at different locations in space need not produce the same physical effect. For example, two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Answer:
No; No
A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.

Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

Question 28.
Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.
Answer:
No; Yes; No
One cannot associate a vector with the length of a wire bent into a loop. One can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area.
One cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.
Solution:
No
Range, R = 3 km
Angle of projection, θ = 30°
Acceleration due to gravity, g = 9.8 m/s²
Horizontal range for the projection velocity u0, is given by the relation :
R = \(\frac{u_{0}^{2} \sin 2 \theta}{g}\)
3 = \(\frac{u_{0}^{2}}{g}\) sin 60°
\(\frac{u_{0}^{2}}{g}\) = 2√3 ……………… (i)
The maximum range (R_max) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,
R_max = \(\frac{u_{0}^{2}}{g}\) = ………………. (ii)
On comparing equations (i) and (ii), we get:
R_max = 2√3 × 1.732 = 3.46 km
Hence, the bullet will not hit a target 5 km away.

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an dnti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 ms^-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10ms^-2)
Solution:
Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, υ = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The
situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = u_xt
Distance travelled by the plane = υt
The shell hits the plane. Hence, these two distances must be equal.
u_xt = υt
usinθ = υ
sinθ = \(\frac{v}{u}=\frac{200}{600}=\frac{1}{3}\) 0.33
θ = sin^-1 (0.33) = 19.5°
In order to avoid being hit by the shell, the pilot must fly the plane at
an altitude (H) higher than the maximum height achieved by the shell.

Question 31.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Solution:
Speed of the cyclist, υ = 27 km/h = 7.5 m/s
Radius of the circular turn , r = 80m
Centripetal acceleration is given as:
a = \(\frac{v^{2}}{r}\)
= \(\frac{(7.5)^{2}}{80}\) = 0.7 m/s²
The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the brakes and decelerates the speed of the bicycle by 0.5 m/s².
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between a_c and a_T is 90°, the resultant acceleration a is given by:
a = \(\sqrt{a_{c}^{2}+a_{T}^{2}}=\sqrt{(0.7)^{2}+(0.5)^{2}}=\sqrt{0.74}\) = 0.86 m/s²
and
tan θ = \(\frac{a_{c}}{a_{T}}\)
where θ is the angle of the resultant with the direction of velocity
tanθ = \(\frac{0.7}{0.5}\) = 1.4
θ = tan^-1 (1.4) = 54.46°
Hence, the net acceleration of the cyclist is 0.86 rn/s², 54.60 0 with the direction of velocity.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is
θ(t) = tan (\(\frac{v_{0 y}-g t}{v_{0 x}}\))

(b) Show that the projection angle θ₀ for a projectile launched from the origin is given by
θ₀ = tan^-1 (\(\frac{\mathbf{4} \boldsymbol{h}_{\boldsymbol{m}}}{\boldsymbol{R}}\))
where the symbols have their usual meaning.
Solution:
Let y Ox and y 0, respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.
Let y and y , respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = t.
Applying the first equation of motion along the vertical and horizontal directions, we get: