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Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 7 System of Particles and Rotational Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

PSEB 11th Class Physics Guide System of Particles and Rotational Motion Textbook Questions and Answers

Question 1.
Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Answer:
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
No, The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc. lies outside the body.

Question 2.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1Å = 10^-10 m). Find the approximate location of the C.M. of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Solution:
The given situation can be shown as:

Distance between H and Cl atoms = 1.27 Å
Mass of H atom = m
Mass of Cl atom = 35.5 m
Let the centre of mass of the system lie at a distance x Å from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x) Å.
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
\(\frac{m(1.27-x)+35.5 m x}{m+35.5 m}\) = 0
m (1.27 – x) + 35.5mx = 0
1.27 – x = -35.5x
x = \(\frac{-1.27}{(35.5-1)}\) = -0.037 Å
[the negative sign indicates that the centre of mass lies at the left of the molecule, -ve sign negligible.]
Hence, the centre of mass of the HC1 molecule lies 0.037Å from the Cl atom.
Hence, the centre of mass of the HC1 molecule lies
(1.27 – x) = 1.27 – 0.037 = 1.24 Å from the H atom.

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Solution:
The child is running arbitrarily on a trolley moving with velocity υ. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy-trolley system, the boy’s motion will produce no change in the velocity of the centre of mass’of the trolley.

Question 4.
Show that the area of the triangle contained between the vectors [Latex]\vec{a}[/Latex] and [Latex]\vec{b}[/Latex] is one half of the magnitude of \(\vec{a} \times \vec{b}\).
Consider two vecters \(\overrightarrow{O K}=|\vec{a}|\) = and \(\overrightarrow{O M}=|\vec{b}|\) inclined at an angle θ as, shown in the following figure.

In Δ OMN , we can write the relation:
sinθ = \(\frac{M N}{O M}=\frac{M N}{|\vec{b}|}\)
MN = \(|\vec{b}|\) sinθ
\(|\vec{a} \times \vec{a}|=|\vec{a}||\vec{b}|\) sinθ
= OK.MN x \(\frac{2}{2}\)
= 2 x Area of Δ OMK
∴ Area of Δ𝜏 OMK = \(\frac{1}{2}\) \(|\vec{a} \times \vec{b}|\)

Question 5.
Show that \(\vec{a} \cdot(\vec{b} \times \vec{c})\) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution:
A parallelepiped with origin O and sides \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) is shown in the following figure.

Volume of the given parallelepiped = abc
\(\overrightarrow{O C}=\vec{a}\)
\(\overrightarrow{O B}=\vec{b}\)
\(\overrightarrow{O C}=\vec{c}\)
Let n̂ be a unit vector perpendicular to both \(\) and \(\). Hence, n̂ and \(\)
have the same direction.
∴ \(\vec{b} \times \vec{c}\) = bc sin n̂
= bc sinθ n̂ bcsin90° n̂ = bc n̂
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = a.(bc n̂)
= abc cosθ n̂
= abc cos0°= abc
= Volume of the parallelepiped

Question 6.
Find the components along the x, y, z axes of the angular momentum \(\) of a particle, whose position vector is \(\) with components x, y, z and momentum is \(\) with components p_x, p_y and p_z. Show
that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution:


The particle moves in the x – y plane. Hence, the z – component of the position vector and linear momentum vector becomes zero, i. e.,
z = P_z = 0
Thus, equation (i) reduces to

Therefore, when the particle is confined to move in the x – y plane, the direction of angular momentum is along the z – direction.

Question 7.
Two particles, each of mass m and speed υ, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution:
Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular momentum of the system about point P:
\(\vec{L}\)_P = mυ × 0 + mυ × d
= mυd …………. (i)
Angular momentum of the system about point Q:
\(\vec{L}\)_Q = mυ × d + mυ × 0 = mυd …………… (ii)
Consider a point R, which is at a distance y from point Q, i. e.,
QR = y
∴ PR = d – y
Angular momentum of the system about point R:
\(\vec{L}\)_R = mυ × (d – y) + mυ × y = mυd – mυy + mυy
= mυd ……………. (iii)
Comparing equations (i), (ii), and (iii), we get
\(\vec{L}\)_P = \(\vec{L}\)_Q = \(\vec{L}\)_R ……….. (iv)
We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

Question 8.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure given below. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Solution:
The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m
T₁ and T₂ are the tensions produced in the left and right strings respectively.
At translational equilibrium, we have:
T₁ sin 36.9° = T₂ sin 53.1°
\(\frac{T_{1}}{T_{2}}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}=\frac{0.800}{0.600}=\frac{4}{3}\)
⇒ T₁ = \(\frac{4}{3}\) T₂ ……………… (i)
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
T₁ cos 36.9° × d = T₂ cos 53.1° (2 – d)
T₂ × 0.800 d = T₂ 0.600 (2 – d)
\(\frac{4}{3}\) × T₂ × 0.800 d = T₂ [0.600 × 2 – 0.600 d] [from eq. (i)]
1.067 d+ 0.6 d = 1.2
∴ d = \(\frac{1.2}{1.67}\) = 0.72 m
Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.

Question 9.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solutio:
Mass of the car, m = 1800 kg
Distance between the front and back axles, d = 1.8 m
Distance between the C.G. (centre of gravity) and the back axle
= 1.05 m
The various forces acting on the car are shown in the following figure.

R_f and R_b are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
R_f + R_b = mg
= 1800 × 9.8 = 17640 N ………….. (i)
For rotational equilibrium, on taking the torque about the C.G.,
we have
R_f (1.05) = R_b (1.8 – 1.05)
R_f × 1.05 = R_b × 0.75
\(\frac{R_{f}}{R_{b}}=\frac{0.75}{1.05}=\frac{5}{7}\)
\(\frac{R_{b}}{R_{f}}=\frac{7}{5}\)
R_b = 1.4 R_f …………… (ii)
Solving equations (i) and (ii), we get
1.4 R_f + R_f =17640
⇒ 2.4 R_f = 17640
⇒ R_f = \(\frac{17640}{2.4}\)= 7350N
∴ Rb = 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel = \(\frac{7350}{2}\) = 3675 N and
The force exerted on each back wheel = \(\frac{10290}{2}\)= 5145 N

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/15, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR² /4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
(a) The moment of inertia (M.I.) of a sphere about its diameter = \(\frac{2}{5}\)MR²

M.I.= \(\frac{2}{5}\)MR²
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The M.I. about a tangent of the sphere = \(\frac{2}{5}\) MR² + MR² = \(\frac{7}{5}\) MR²

(b) The moment of inertia of a disc about its diameter = \(\frac{1}{4}\) MR²
According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes – concurrent with perpendicular axis and lying in the plane of the body.
The M.I. of the disc about its centre = \(\frac{1}{4}\) MR² + \(\frac{1}{4}\) MR² = \(\frac{1}{4}\) MR²
The situation is shown in the given figure.

Applying the theorem of parallel axes,
The moment of inertia about an axis normal to the disc and passing through a point on its edge = \(\frac{1}{2}\)MR² + \(\frac{1}{2}\)MR² = \(\frac{3}{2}\) MR²

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Solution:
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,
I₁ = mr²
The moment of inertia of the solid sphere about an axis passing through its centre, I₂ = \(\frac{2}{5}\) mr²
We have the relation:
τ = Iα
where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ₁ = I₁α₁
For the solid sphere, τ₂ = I₂τ₂
As an equal torque is applied to both the bodies, τ₁ = τ₂
∴ \(\frac{\alpha_{2}}{\alpha_{1}}=\frac{I_{1}}{I_{2}}=\frac{m r^{2}}{\frac{2}{5} m r^{2}}=\frac{5}{2}\)
⇒ α₂ = \(\frac{5}{2}\)α₁
⇒ α₂ > α₁ …………… (i)
Now, using the relation:
ω = ω₀ + αt
where,
ω₀ = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω and t, we have:
ω ∝ α ……………. (ii)
From equations (i) and (ii), we can write:
ω₂ > ω₁
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution:
Mass of the cylinder, m = 20 kg
Angular speed, ω = 100 rad s^-1
Radius of the cylinder, r = 0.25 m
The moment of inertia of the solid cylinder:
I = \(\frac{m r^{2}}{2}\) = \(\frac{1}{2}\) × 20 × (0.25)²
= 0.625 kg-m²
∴ Kinetic energy = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) × 0.625 × (100)² = 3125 J
∴ Angular momentum, L = Iω = 0.625 × 100 = 62.5 J-s

Question 13.
(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution:
(a) Initial angular velocity,ω₁ = 40 rev/min
Let, Final angular velocity = ω₂
The moment of inertia of the child with.stretched hands = I₁
The moment of inertia of the child with folded hands = I₂
The two moments of inertia are related as:
I₂ = \(\frac{2}{5}\)I₁
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I₂ω₂ = I₁ω₁
ω₂ = \(\frac{I_{1}}{I_{2}}\) ω₁
= \(\frac{I_{1}}{\frac{2}{5} I_{1}}\) × 40 = \(\frac{5}{2}\) × 40
= 100 rev/min

∴ E_F = 2.5 E_I
The increase in the rotational kinetic energy is attributed to the internal energy of the child.

Question 14.
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Solution:
Mass of the hollow cylinder, m = 3 kg
Radius of the hollow cylinder, r = 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis,
I = mr² = 3 × (0.4)² = 0.48 kg-m²
Torque, τ = F × r = 30 × 0.4 =12 N-m
For angular acceleration α, torque is also given by the relation
τ = Iα
α = \(\frac{\tau}{I}=\frac{12}{0.48}\)= 25 rad s^-2
Linear acceleration = rα = 0.4 × 25 = 10 ms^-2

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N-m. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient.
Solution:
Angular speed of the rotor, ω = 200 rad / s
Torque required, τ = 180 N-m
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 10³ W = 36 kW
Hence, the power required by the engine is 36 kW.

Question 16.
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
Let, Mass per unit area of the original disc = σ
Radius of the original disc = R
∴ Mass of the original disc, M = πR² σ
The disc with the cut portion is shown in the following figure:

Radius of the smaller disc = \(\frac{R}{2}\)
Mass of the smaller disc, M’= π(\frac{R}{2})² σ = \(\frac{1}{4}\)πR² σ = \(\frac{M}{4}\)

Let O and O'[]be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O’.
It is given that:
00′ = \(\frac{R}{2}\)
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and
– M = (= \(\frac{M}{4}\)) concentrated at O’
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
x = \(\frac{m_{1} r_{1}+m_{2} r_{2}}{m_{1}+m_{2}}\)
For the given system, we can write:

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)
Hence, the centre of gravity is located at the distance of R/6 from the original centre of the body and opposite to the centre of the cut portion.

Question 17.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to he balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
Let W and W’ be the respective weights of the metre stick and the coin.

The mass of the metre stick is concentrated at its mid-point, i. e., at the 50 cm mark.
Let, mass of the metre stick = m’
Given, mass of each coin, m = 5 g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
10 × g (45 -12) – m’ g (50 – 45) = 0
⇒ 10 × 33 = m’ × 5
∴ m’ = \(\frac{10 \times 33}{5}\) = 66 g
Hence, the mass of the metre stick is 66 g.

Question 18.
A solid sphere rolls down two different inclined planes of the same heights hut different angles of inclination, (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
Solution:
(a) Yes (b) Yes (c) on the smaller inclination
(a) Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane = υ
At the top of the plane, the total energy of the sphere
= Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy = –\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\)Iω²
Using the law of conservation of energy, we can write:
\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\)Iω² = mgh ……………. (i)
For a solid sphere, the moment of inertia about its centre, I = \(\frac{2}{5}\) mr²
Hence equation (i) becomes,
\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) (\(\frac{2}{5}\)mr²)ω² = mgh
\(\frac{1}{2}\)υ² + \(\frac{1}{5}\)r²ω² = gh
But we have the relation, υ = rω
\(\frac{1}{2}\)υ² + \(\frac{1}{5}\)υ² = gh
∴ υ²(\(\frac{7}{10}\)) = gh
υ = \(\sqrt{\frac{10}{7} g h}\)
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

(b), (c) Consider two inclined planes with inclinations θ₁ and θ₂, related as
θ₁ < θ₂
The acceleration produced in the sphere when it rolls down the plane inclined at θ₁,
θ₁ = g sinθ₁
The various forces acting on the sphere are shown in the following figure.

R₁ is the normal reaction to the sphere.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ₂,
a₂ = gsinθ₂
The various forces acting on the sphere are shown in the following figure.

R₂ is the normal reaction to the sphere.
θ₂ > 0₁; sinθ₂ > sinθ₁ ……….(i)
∴ a₂ > a₁ ……………. (ii)
Initial velocity, u = 0
Final velocity, υ = Constant
Using the first equation of motion, we can obtain the time of roll as,
υ = u + at
t ∝ \(\frac{1}{a}\)

From equations (ii) and (iii), we get:
t₂ < t₁
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to he done to stop it?
Solution:
Radius of the hoop, r = 2 m
Mass of the hoop, m = 100 kg
Velocity of the hoop, υ = 20 cm/s = 0.2 m/s
Total kinetic energy of the hoop = Translational KE + Rotational KE
E_T = \(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) Iω²
Moment of inertia of the hoop about its centre, I = mr²
E_T = \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)(mr² )ω²
But we have the relation, υ = rω
E_T = \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)mr² ω²
= \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)mυ² = mυ²
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
□Required work to be done,
W = mυ² =100 × (0.2)² = 4 J

Question 20.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia of 1.94 × 10^-46 kg-m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Solution:
Mass of an oxygen molecule, m = 5.30 × 10^-26 kg
Moment of inertia, I =1.94 × 10^-46 kg-m²
Velocity of the oxygen molecule, υ = 500 m/s
The separation between the two atoms of the oxygen molecule = 2 r
Mass of each oxygen atom = \(\frac{m}{2}\)
Hence, moment of inertia I, is calculated as

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the , cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Solution:
(a) A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, υ = 5 m/s
Angle of inclination, θ = 30°
Height reached by the cylinder = h
Energy of the cylinder at point A
KE_rot = KE_trans
\(\frac{1}{2}\) Iω² = \(\frac{1}{2}\)mυ²
Energy of the cylinder at point B = mgh
Using the law of conservation of energy, we can write
\(\frac{1}{2}\) Iω² + \(\frac{1}{2}\)mυ² =mgh

Hence, the cylinder will travel 3.82 m up the inclined plane.

(b) For radius of gyration K, the velocity of the cylinder at the instance
when it rolls back to the bottom is given by the relation:


Therefore, the total time taken by the cylinder to return to the bottom is 2 × 0.764 = 1.53 s.

Question 22.
As shown in figure given below, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s²)
(Hint: Consider the equilibrium of each side of the ladder separately.)

Solution:
The given situation can be shown as:

where, N_B = Force exerted on the ladder by the floor point B
N_C = Force exerted on the ladder by the floor point C
T = Tension in the rope
BA = CA =1.6 m
DE = 0.5 m
BF =1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
Δ ABI and Δ AIC are similar
∴ BI = IC
Hence, I is the mid-point of BC.
In Δ ABC, DE || BC
∴ BC = 2 × DE = 1m
and AF = BA – BF= 1.6 – 1.2 = 0.4 m …………… (i)
D is the mid-point of AB.
Hence, we can write:
AD = \(\frac{1}{2}\) × BA = 0.8 m ………….. (ii)
Using equations (i) and (ii), we get
DF = AD – AF = 0.8 – 0.4 = 0.4 m
Hence, F is the mid-point of AD.
FG || DH and F is the mid-point of AD. Hence, G will also be the mid-point ofAH.
Δ AFG and Δ ADH are similar
∴ \(\frac{F G}{D H}=\frac{A F}{A D}\)
\(\frac{F G}{D H}=\frac{0.4}{0.8}=\frac{1}{2}\)
FG = \(\frac{1}{2}\)DH
= \(\frac{1}{2}\) × 0.25 = 0.125 m , [∵ DH = \(\frac{1}{2}\)DE]
In Δ ADH,
AH = \(\sqrt{A D^{2}-D H^{2}}\)
= \(\sqrt{(0.8)^{2}-(0.25)^{2}}\) = 0.76 m
For translational equilibrium of the ladder, the upward force should be equal to the downward force.
N_B + N_C = mg = 40 × 9.8 = 392 …………….. (iii)
For rotational equilibrium of the ladder, the net moment about A is
-N_B × BI + mg × FG + N_C × CI + T × AG – T × AG = 0
-N_B × 0.5 + 40 × 9.8 × 0.125 + N_C × (0.5) = 0
(N_B – N_C) × 0.5 = 49
N_B – N_C = 98 …………. (iv)
Adding equations (iii) and (iv), we get:
N_B = 245 N
N_C = 147N
For rotational equilibrium of the side AB, consider the moment about A
-N_B × BI + mg × FG + T × AG = 0
-245 × 0.5 + 40 × 9.8 × 0.125 + T × 0.76 = 0
0.76 T = 122.5 – 49 = 73.5
T = \(\frac{73.5}{0.76}\)= 96.7N

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man ‘ then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20cm. The moment *of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg-m² .
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Solution:
(a) Moment of inertia of the man-platform system
= 7.6 kg-m²
Moment of inertia when the man stretches his hands to a distance of 90 cm
= 2 × mr² = 2 × 5 × (0.9)²
= 8.1 kg-m²
Initial moment of inertia of the system, I_i = 7.6 + 8.1 = 15.7 kg-m²
Initial angular speed, ω_i = 30 rev/min
Initial angular momentum, L_i = I_iω_i = 15.7 × 30 …………….. (i)
Moment of inertia when the man folds his hands to a distance of 20 cm
= 2 × mr² = 2 × 5(0.2)² = 0.4 kg-m²
Final moment of inertia, I_f = 7.6 + 0.4 = 8 kg-m²
Let, final angular speed = ω_f
Final angular momentum, L_f = I_fω_f = 8ω_f ………….. (ii)
From the conservation of angular momentum, we have
I_iω_i = I_fω_f
∴ ω_f = \(\frac{15.7 \times 30}{8}\)= 58.88 rev/min
Hence, new angular speed is 58.88 revolutions per minute.

(b) No, kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML² /3.)
Solution:
Mass of the bullet, m = 10 g = 10 x 10^-3 kg
Velocity of the bullet, υ = 500 m/s
Thickness of the door, L = 1 m
Radius of the door, r = \(\frac{1}{2}\) m
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door,
α = mυr
= (10 × 10^-3) × (500) × \(\frac{1}{2}\) = 2.5kg-m²s^-1 ………………. (i)
Moment of inertia of the door,
I = \(\frac{1}{2}\)ML² = \(\frac{1}{3}\) × 12 × (1)² = 4 kg-m²
But α = Iω
∴ ω = \(\) = 0.625 rad s^-1

Question 25.
Two discs of moments of inertia I\ and /2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident, (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies ‘of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂.
Solution:
(a) Moment of inertia of disc I = I₁
Angular speed of disc I = ω₁
Moment of inertia of disc II = I₂
Angular speed of disc II = ω₂
Angular momentum of disc I, L₁ = I₁ω₁
Angular momentum of disc II, L₂ = I₂ω₂
Total initial angular momentum, L _i = I₁ω₁ + I₂ω₂
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the systme of two discs, I = I₁ + I₂
Let ω be the angular speed of the system.
Total final angular momentum, L_f = (I₁ + I₂)ω
Using the law of conservation of angular momentum, we have
L_i = L_f
I₁₁ + I₂₂ (I₁ + I₂)ω
< ω = \(\frac{I_{1} \omega_{1}+\bar{I}_{2} \omega_{2}}{I_{1}+I_{2}}\)

(b) Kinetic energy of disc I, E₁ = \(\frac{1}{2}\)I₁ω₁²
Kinetic energy of disc II, E₂ = \(\frac{1}{2}\)I₂ω₂²
Total initial kinetic energy, E_i = E₁ + E₂ = \(\frac{1}{2}\) (I₁ω₁² + I₂ω₂²)
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, I = I₁ + I₂
Angular speed of the system = ω
Final kinetic energy E_f.

All the quantities on RHS are positive
∴ E_i – E_f > 0
E_i > E_f
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.

Question 26.
(a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x – y plane from an axis through the origin perpendicular to the plane is x² + y² )
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin Z m_ir_i = 0).
Solution:
(a) The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m, in the x – y plane at , (x, y) is shown in the following figure.

Moment of inertia about x – axis, I_x = mx²
Moment of inertia about y – axis, I_y = my²
Moment of inertia about z – axis, I_z = \(\left(\sqrt{x^{2}+y^{2}}\right)^{2}\)
I_x + I_y = mx² + my²
= m(x² + y²)
= m\(\left(\sqrt{x^{2}+y^{2}}\right)^{2}\)
I_x + I_y = I_z
Hence, the theorem is proved.

(b) The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel . axes.

Suppose a rigid body is made up of n particles, having masses m₁, m₂, m₃, …… , m_n, at perpendicular distances r₁, r₂, r₃, ………….. , m_n respectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O,

I_IRS = \(\sum_{i=1}^{n}\) m_ir_i

The perpendicular distance of mass mi, from the axis QP = a + r_i
Hence, the moment of inertia about axis QP,
I_QP = \(\sum_{i=1}^{n}\)_i(a + r_i)²

Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,
2 \(\sum_{i=1}^{n}\) m_iar_i = 0
∴ m_ir_i = 0
a ≠ 0
Σm_ir_i = 0
Also, \(\sum_{i=1}^{n}\) m_i = M; M = Total mass of the rigid body
∴ I_QP = I_RS + Ma²
Hence, the theorem is proved.

Question 27.
Prove the result that the velocity v of translation of a rolling hody (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by
υ² = \(\frac{2 g h}{\left(1+k^{2} / R^{2}\right)}\)
using dynamical consideration (i. e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Solution:
A body rolling on an inclined plane of height h, is shown in the following figure :

m = Mass of the body
R = Radius of the body
k = Radius of gyration of the body
υ = Translational velocity of the body
h = Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E_T = mgh
Total energy at the bottom of the plane,
E_b = KE_rot + KE_trains
= \(\frac{1}{2}\)Iω² + \(\frac{1}{2}\)mυ²

Hence, the given result is proved.

Question 28.
A disc rotating about its axis with angular speed (ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure? Will the disc roll in the direction indicated?

Solution:
Angular speed of the disc = ω₀
Radius of the disc = R
Using the relation for linear velocity, υ = ω₀R
For point A:υ_A = Rω₀; in the direction tangential to the right
For point B:υ_B = Rω₀; in the direction tangential to the left
For point C:υ_C = (\(\frac{R}{2}\))ω₀; in the direction same as that of vA.
The directions of motion of points A, B and C on the disc are shown in the following figure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.

Question 29.
Explain why friction is necessary to make the disc in figure given in question 28 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
Solution:
A torque is required to roll the given disc. As per the definition of torque,
the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 πrad s^-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is μ_k = 0.2
Solution:
Radii of the ring and the disc, r- = 10 cm = 0.1 m
Initial angular speed, ω₀ = 10 πrad s-^-1
Coefficient of kinetic friction, μ_k = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma.
μ_kmg = ma
where,
a = Acceleration produced in the objects
m = Mass
∴ a = μ_kg …………… (i)
As per the first equation of motion, the final velocity of the objects can be obtained as
υ = u + at
= 0 + μ_kgt
= μ_kgt ……………. (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ = -Iα
where, α = Angular acceleration
μ_kmgr = -Iα
∴ α = \(\frac{-\mu_{k} m g r}{I}\) ……………… (iii)
Using the first equation of rotational motion to obtain the final angular speed,

Question 31.
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μ_g, = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Solution:
Given, mass of the cylinder, m =10 kg
Radius of the cylinder, r = 15cm = 0.15m
Coefficient of static friction, μ_s = 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = \(\frac{1}{2}\)mr²
The various forces acting on the cylinder are shown in the following figure:

= \(\frac{2}{3}\) × 9.8 × 0.5 = 3.27 m/s²

(a) Using Newton’s second law of motion, we can write net force as
f_netnet = ma
mg sin30° – f = ma
f = mgsin30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
= 49 – 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:
μ = \(\frac{1}{3}\)tanθ
tanθ = 3μ = 3 × 0.25
∴ θ = tan^-1 (0.75) = 36.87°

Question 32.
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Solution:
(a) False
Reason: Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

(b) True
Reason: Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False
Reason: When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True
Reason: When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True
Reason: The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

Question 33.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(a) Show \(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}=m_{i} \vec{V}\)
where, \(\overrightarrow{p_{i}}\) is the momentum of the i^th particle (of mass m_i) and \(\overrightarrow{p_{i}^{\prime}}=\vec{m}_{i} \vec{v}_{i}^{\prime}\). Note \(\overrightarrow{\boldsymbol{v}_{\boldsymbol{i}}^{\prime}}\) is the velocity of the i^th particle relative to the centre of mass.
Also, prove using the definition of the centre of mass \(\sum_{i} \overrightarrow{p_{i}^{\prime}}\) = 0

(b) Show K = K’ + \(\frac{1}{2}\) MV²
where, K is the total kinetic energy of the system of particles, K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV² /2 is the kinetic energy of the translation of the system as a whole (i. e., of the centre of mass motion of the system). The result has been used in Sec. 7.14.

(c) Show \(\overrightarrow{\boldsymbol{L}}^{\prime}=\overrightarrow{\boldsymbol{L}}^{\prime}+\overrightarrow{\boldsymbol{R}} \times \boldsymbol{M} \overrightarrow{\boldsymbol{V}}\)
where, \(\overrightarrow{\boldsymbol{L}}^{\prime}=\Sigma \overrightarrow{\boldsymbol{r}_{\boldsymbol{i}}^{\prime}} \times \overrightarrow{\boldsymbol{p}_{\boldsymbol{i}}^{\prime}}\) is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember \(\overrightarrow{\boldsymbol{r}_{i}^{\prime}}=\overrightarrow{\boldsymbol{r}_{i}}-\overrightarrow{\boldsymbol{R}}\) rest of the notation is the standard notation used in the chapter. Note \(\overrightarrow{\boldsymbol{L}}\) and \(\boldsymbol{M} \overrightarrow{\boldsymbol{R}} \times \overrightarrow{\boldsymbol{V}}\) can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show = \(\frac{d \vec{L}^{\prime}}{d t}=\sum_{i} \overrightarrow{r_{i}^{\prime}} \times \frac{d}{d t}\left(\overrightarrow{p_{i}^{\prime}}\right)\)
Further, show that
\(\frac{d \vec{L}^{\prime}}{d t}\) = τ’_ext
where, τ’_ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:
(a) Take a system of i moving particles.
Mass of the i^th particle = m_i
Velocity of the t^th particle = υ_i
Hence, momentum of the i^th particle, \(\overrightarrow{p_{i}}\) = m_iυ_i
Velocity of the centre of mass = V
The velocity of the i^th particle with respect to the centre of mass of the system is given as:
\(\overrightarrow{v_{i}^{\prime}}=\overrightarrow{v_{i}}-\vec{V}\) ……………. (i)
Multiplying m; throughout equation (i), we get
\(m_{i} \overrightarrow{v_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}}-m_{i} \vec{V}\)
\(\overrightarrow{p_{i}^{\prime}}=\overrightarrow{p_{i}}-m_{i} \vec{V}\)
where, \(\overrightarrow{p_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}^{\prime}}\) = Momentum of the i^th particle with respect to the centre of mass of the system
∴ \(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}+m_{i} \vec{V}\)
We have the relation: \(\overrightarrow{p_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}^{\prime}}\)
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get
\(\Sigma \overrightarrow{p_{i}^{\prime}}=\Sigma_{i} m_{i} \overrightarrow{v_{i}^{\prime}}=\Sigma_{i} m_{i} \frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\)
where, \(\overrightarrow{r_{i}^{\prime}}\) = Position vector of i^th particle with respect to the centre of mass
\(\overrightarrow{v_{i}^{\prime}}=\frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\)
As per the definition of the centre of mass, we have
\(\sum_{i} m_{i} \overrightarrow{r_{i}^{\prime}}\) = 0
\(\sum_{i} m_{i} \frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\) = 0
\(\sum_{i} \overrightarrow{p_{i}^{\prime}}\) = 0

(b) KE. of a system consists of two parts translational K.E. (K_t) and rotational K.E. (K’) i.e., K.E. of motion of C.M. (\(\frac{1}{2}\)mυ²) and K.E. of rotational motion about the C.M. of the system of particles (K’), thus total K.E. of the system is given by
K = \(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\)mυ² + K’
= K’ + \(\frac{1}{2}\)mυ²

(c) Position vector of the i th particle with respect to origin = \(\overrightarrow{r_{i}}\)
position vector of the i th particle with respect to the centre of mass = \(\overrightarrow{r_{i}^{\prime}}\)
Position vector of the centre of mass with respect to the origin = \(\vec{R}\)
It is given that:
\(\overrightarrow{r_{i}^{\prime}}=\overrightarrow{r_{i}}-\vec{R}\)
\(\overrightarrow{r_{i}}=\overrightarrow{r_{i}^{\prime}}+\vec{R}\)
We have from part (a),
\(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}+m_{i} \vec{V}\)
Taking the cross product of this relation by r_i, we get:

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 19 Excretory Products and their Elimination Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination

PSEB 11th Class Biology Guide Excretory Products and their Elimination Textbook Questions and Answers

Question 1.
Define Glomerular Filtration Rate (GFR).
Answer:
Glomerular Filtration Rate (GFR) is the amount of the filtrate formed by the kidneys per minute. GFR in a healthy individual is approximately 125 mt/mm, i.e., 180 L day.

Question 2.
Explain the autoregulatory mechanism of GFR.
Answer:
Regulation of GFR: The kidneys have built-in mechanisms for the regulation o glomerular filtration rate. One such efficient mechanism is carried out by the juxtaglomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.

Question 3.
Indicate whether the following statements are true or false:
(a) Micturition is carried out by a reflex.
(b) Al)H helps In water elimination, making the urine hypotonia.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.
Answer:
(a) True,
(b) False,
(C) True,
(d) True,
(e) True.

Question 4.
Give a brief account of the counter-current mechanism.
Answer:
Counter-current Mechanism

• The counter-current mechanism takes place in Henle’s loop and vasa recta.
• The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus, forms a counter-current.
• The flow of blood through the two limbs of vasa recta is also in a counter-current pattern.
• NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.
• NaCl is returned to the interstitium by the ascending portion of vasa recta.
• Similarly small amount of urea enters the thin segment of the ascending limb of Henle’ loop, which is transported back to the interstitium by the collecting tubule.
• This transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter-current mechanism,
• It helps to maintain a concentration gradient in the medullary interstitium, which facilitates an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).

Question 5.
Describe the foie of liver, lungs and skin in excretion.
Answer:
Liver, lungs and skin also play an important role in the process of excretion. Role of the Liver: Liver is the largest gland in vertebrates. It helps in the excretion of cholesterol, steroid hormones, vitamins, drugs, and other waste materials through bile. Urea is formed in the liver by the ornithine cycle. Ammonia-a toxic substance is quickly changed into urea in the liver and thence eliminated from the body. Liver also changes the decomposed haemoglobin pigment into bile pigments called bilirubin and biliverdin.

Role of the Lungs: Lungs help in the removing waste materials such as carbon dioxide from the body.
Role of the Skin: Skin has many glands which help in excreting waste products through pores. It has two types of glands-sweat and sebaceous glands.

Sweat glands are highly vascular and tubular glands that separate the waste products from the blood and excrete them in the form of sweat. Sweat excretes excess salt and water from the body.
Sebaceous glands are branched glands that secrete an oily secretion called sebum.

Question 6.
Explain micturition.
Answer:
Micturition: The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex. An adult human excretes on an average 1 to 1.5 L of urine per day. The urine formed is a light yellow coloured watery fluid which is slightly acidic (pH 6.0) and has a characteristic odour.

Question 7.
Match the items of Column-I with those of Column-II:

Column-I	Column-II
(a) Ammonotelism	(i) Birds
(b) Bowman’s capsule	(ii) Water reabsorption
(c) Micturition	(iii) Bony fish
(d) Uricotelism	(iv) Urinary bladder
(e) ADH	(v) Renal tubule

Answer:

Column-I	Column-II
(a) Ammonotelism	(iii) Bony fish
(b) Bowman’s capsule	(v) Renal tubule
(c) Micturition	(iv) Urinary bladder
(d) Uricotelism	(i) Birds
(e) ADH	(ii) Water reabsorption

Question 8.
What is meant by the term osmoregulation?
Answer:
It is the phenomenon of regulation of change in the concentration of body fluids according to the concentration of external environment. Le., most marine invertebrates, some freshwater invertebrates, hagfish (a vertebrate).

Question 9.
Terrestrial animals are generally either republic or uricotelic, not ammonotelic, why?
Answer:
Terrestrial adaptation requires the production of less toxic nitrogenous wastes like urea and uric acid for the conservation of water. Aquatic animals excret atnmonia which requires large amounts of water to dissolve. This huge quantity of water is easily available to such animals from surroundings.

However, for terrestrial animals, such a huge quantity of water is not available, hence, they modify their original water product NH3 to comparatively less toxic products like urea and uric acid which requires less amount of water for their excretion.

Question 10.
What is the significance of juxtaglomerular apparatus (JGA) in Sidney function?
Answer:
The juxtaglomerular apparatus (JGA) plays an important role in monitoring and regulation of kidney functioning by hormonal feedback, mechanism, involving the hypothalamus and to a certain extent, the heart.

Question 11.
Name the following:
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Answer:
(a) Flarworms,
(b) Columns of Bertini,
(c) Vasa recta

Question 12.
Fill in the gaps:
(a) Ascending limb of Henle’s loop is ………………………… to water whereas the descending limb is …………………….. to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ……………………………. .
(c) Dialysis fluid contains all the constituents as in plasma except …………………………… .
(d) A healthy adult human excretes (on an average) ………………………… gm of urea/day.
Answer:
(a) impermeable, permeable;
(b) vasopressin or ADH.;
(c) the nitrogenous wastes;
(d) 25-30.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 4 Animal Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 4 Animal Kingdom

PSEB 11th Class Biology Guide Animal Kingdom Textbook Questions and Answers

Question 1.
What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
Answer:
The major difficulties in the classification of animals are on the following lines :

• Some show cellular level of organisation, some have tissue level and even some have organ system level of organisation.
• Regarding symmetry, some are radially symmetrical, while some have bilateral symmetry.
• Some have open circulatory system, while others have closed type.
• Regarding digestion, some animals have extracellular digestion, while others have intracellular digestion.
• In case of body cavity, some have true coelom while others are pseudocoelomates.
• Regarding reproduction, some have only asexual reproduction, while others reproduce both by sexual and asexual means.
  So, these are difficulties that zoologists face in the classification of animals.

Question 2.
If you are given a specimen, what are the steps that you would follow to classify it?
Answer:
If I am given an animal specimen, then I will classify it on the basis of fundamental features which are common to all animal types inspite of the presence of some major differences in the structure and form of animals. The features taken into consideration during classification of animal are as follows:

• The type of arrangement of cells.
• Body symmetry.
• Nature of coelom.
• Pattern of digestive system.
• Type of circulatory system.
• Type of methods of reproduction.

Question 3.
How useful is the study of the nature of body cavity and coelom in the classification of animals?
Answer:
Presence or absence of a cavity between the body wall and the gut wall is very important in classification. The body cavity which is lined by, mesoderm, is called coelom.

• Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates.
• In some animals, the body cavity is not mesoderm, instead the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., Aschelminthes.
• The animals in which the body cavity is absent are known as acoelomates, e.g. , Platyhelminthes.

Question 4.
Distinguish between intracellular and extracellular digestion?
Answer:
Differences between intracellular and extracellular digestion are as follows:
(i) Intracellular Digestion: It occurs inside the living cells with the help of lysosomal enzymes. Food particle is taken in through endocytosis. It forms a phagosome which fuses with a lysosome. The digested material pass into the cytoplasm. The undigested matter is throw out by exocytosis. It occurs in Amoeba, Paramecium, etc.,

(ii) Extracellular Digestion: In case of coelentrates digestion occurs in gastrovascular cavity. This cavity has gland cells which secret digestive enzymes over the food. The partially digested fragmented food particles are ingested by nutritive cells. It occurs in Hydra, Aurelia, etc.

Question 5.
What is the difference between direct and indirect development?
Answer:
In direct development the embryo directly develops into an adult, while in indirect development there is an intermediate larval stage. Certain members of arthropods show larval stage of development.

Question 6.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
Hooks and suckers are present in the parasitic forms. They are parasitic flatworms commonly called flukes. The body is unsegmented leaf like, which is covered by a thick living tegument. There is no epidermis. The mouth is anterior and is armed with suckers for attachment in the host. Life history includes larval stage and involves more than one hosts.
Examples : Fasciola (the liver fluke), Schistosoma (the blood fluke.)

Question 7.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods constitute the largest group of the animal kingdom. It is estimated that the Arthropoda population of the world is approximately a billion (10¹⁸) individuals, in terms of species diversity, number of individuals and geographical distribution. It is the most successful phylum on the Earth that have ever existed. Arthropods are equipped with jointed appendages, which are variously adapted for walking, swimming, feeding, sensory reception and defence. The appendages of abdomen are associated with locomotion, reproduction and in some cases with defence as well.

The appendages of head are related to defence, whereas those of thorax are mainly associated with locomotion. These features are responsible for its large diversity.

Question 8.
Water vascular system is the characteristic of which group of the following?
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
Echinodermata have the water vascular system.

Question 9.
‘All vertebrates are chordates but all chordates are not vertebrates’.’Justify the statement.
Answer:
Notochord is a characteristic feature of all chordates. The members of sub-phylum – Vertebrata possess notochord during the embryonic stage. But in adults the notochord is replaced by a cartilaginous or bony vertebral column. Whereas in member of other Sub-phyla of Chordata the notochord remain as such. The urochordate and cephalochordates retain the notochord during their entire life cycle. Thus, the absence of notochord in adult vertebrates suggest that all vertebrates are chordates but all chordates are not vertebrates.

Question 10.
How important is the presence of air bladder in Pisces?
Answer:
In fishes, air bladder regulates buoyancy and helps in floating in water. If it is absent, animals need to swim constantly to avoid sinking.

Question 11.
What are the modifications that are observed in birds that help them fly?
Answer:
Flight adaptations in birds are as follows :

• Boat-shaped body helps to propel through the air easily.
• Feathery covering of body to reduce the friction of air.
• Holding the twigs automatically by hindlimbs.
• Extremely powerful muscles that enables the wings to work during flight.
• Bones are light, hollow and provide more space for muscle attachment. Presence of pneumatic bones which reduce the weight of body and help in flight.
• The first four thoracic vertebrae are fused to form a furculum for walking of the wings.
• Lungs are solid and elastic and have associated air sacs.
• The power of accomodation of eyes is well developed due to the presence of comb-like structure pecten.
• A single left ovary and oviduct to reduce the body weight.

Question 12.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
The number of eggs or young ones produced by an oviparous or viviparous mother cannot be equal. An oviparous mother gives rise to more number of eggs as some of them die during hatching and as they have to pass through a large number of developmental stages before becoming an adult. On the other hand, a viviparous mother gives rise to fewer number of young ones because there are less chances of their death. Moreover, they did not have to pass through any larval stage.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Annelida
(c) Aschelminthes
(d) Arthropoda
Answer:
Segmentation in the body is first observed in Annelida. This phenomenon is known as metamerism.

Question 14.
Match the following:

A. Operculum	1. Ctenophora
B. Parapodia	2. Mollusca
C. Scales	3. Porifera
D. Comb plates	4. Reptilia
E. Radula	5. Annelida
F. Hairs	6. Cyclostomata and Chondrichthyes
G. Choanocytes	7. Mammalia
H. Gill slits	8. Osteichthyes

Answer:

A. Operculum	8. Osteichthyes
B. Parapodia	5. Annelida
C. Scales	4. Reptilia
D. Comb plates	1. Ctenophora
E. Radula	2. Mollusca
F. Hairs	7. Mammalia
G. Choanocytes	3. Porifera
H. Gill slits	6. Cyclostomata and Chondrichthyes

Question 15.
Prepare a list of some animals that are found parasitic on human heings.
Answer:
A list of parasitic animals on human beings:

Parasite	In Part of Human Body
Leishmania donovani	Blood
Trichomonas vaginalis	Vagina of human female
Plasmodium vivax	Blood
Taenia solium	Intestine
Ascaris lumbricoides	Small intestine
Wuchereria bancrofti	Lymphatic and muscular system
Loa loa	Eyes
Fasciola hepatica	Liver and bile ducts
Entamoeba histolytica	Intestine
Trypanosoma gambiense	Blood

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 3 Plant Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 3 Plant Kingdom

PSEB 11th Class Biology Guide Plant Kingdom Textbook Questions and Answers

Question 1.
What is the basis of classification of algae?
Answer:
Basis of classification of algae are as follows:

• Kinds of pigments.
• Nature of reserve food.
• Kinds, number and points of insertion of flagella of motile cells.
• Presence or absence of organised nucleus in the cell.

Question 2.
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Answer:
Reduction division in the life cycle of a liverwort, a moss, a fern and a gymnosperm take place during the production of spores from spore mother cells. In case of an angiosperm, the reduction division occurs during pollen grain formation from anthers and during production of embryo sac from ovule.

Question 3.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Answer:
Three groups of plants that bear archegonia are bryophytes, pteridophytes and gymnosperms.
Life Cycle of a Pteridophyte: The life cycle of a pteridophyte consists of two morphologically distinct phases:
1. The gametophytic phase
2. The sporophytic phase.
These two phases come one after another in the life cycle of a pteridophyte. This phenomenon is called alternation of generation. The gametophyte is haploid with single set of chromosomes. It produces male sex organs antheridia and female sex organs archegonia.

• The antheridia may be embedded or projecting type. Each antheridium has single layered sterile jacket enclosing a mass of androcytes.
• The androcytes are flask-shaped, sessile or shortly stalked and differentiated into globular venter and tubular neck.
• The archegonium contains large egg, which is non-mo tile.
• The antherozoids after liberation from antheridium, reaches up to the archegonium fuses with the egg and forms a diploid structure known as zygotes.
• The diploid zygote is the first cell of sporophytic generation. It is retained inside the archegonium and forms the embryo.
• The embryo grows and develop to form sporophyte which is differentiated into roots, stem and leaves.
• At maturity the plant bears sporangia, which encloses spore mother cells.
• Each spore mother cell gives rise to four haploid spores which are usually arranged in tetrads.
• The sporophytic generation ends with the production of spores.
• Each spore is the first cell of gametophytic generation. It germinates to produce gametophyte and completes its life cycle.

Question 4.
Mention the ploidy of the following:
Protonemal cell of a moss, primary endosperm nucleus in dicot, leaf cell of a moss, prothallus cell of a fern, gemma cell in Marchantia, meristem cell of monocot, ovum of a liverwort and zygote of a fern.
Answer:
Protonemal cell of a moss – haploid
Primary endosperm nucleus in dicot – triploid
Leaf cell of a moss – haploid
Prothallus cell of a fern – haploid
Gemma cell in Marchantia – haploid
Meristem cell of monocot – diploid
Ovum of a liverwort – haploid
Zygote of a fern – diploid

Question 5.
Write a note on economic importance of algae and gymnosperms.
Answer:
Economic Importance of Algae

• Red algae provides food, fodder and commercial products. Porphyra tenera is rich in protein, carbohydrates and vitamin-A, B, E and C.
• Corallina has vermifuge properties.
• Agar-agar a gelatin substance used as solidifying agent in culture media is obtained from Gelidium and Gracilaria algae. Funori is a glue used as adhesive and in sizing textiles, papers, etc. Chondrus is most widely used in sea weed in Europe.
• Mucilage extracted from Chondrus is used in sampoos, shoe polish and creams.
• Carrageenin is a sulphated polysaccharide obtained from cell wall of Chondrus crispus and Gigartina and is used in confectionary, bakery, jelly, creams, etc.

Economic Importance of Gymnosperms

• Gymnosperms hold soil particles and thus check soil erosion.
• Many gymnosperms are grown in gardens as ornamental plants, i.e., Cycas, Thiya, Araucaria, Taxus, Agathis, Maiden hair tree, etc.
• Sago is a kind of starch obtained from cortex and pith of stem and seeds of Cycas. Roasted seeds of Pinus geradiana (chilgoza) are used as dry fruit.
• Paper pulp is obtained from wood of Picea (spruce), Gnetum, Pinus (pine) and Larix (larck).
• The wood of Juniperus virginiana (red cedar) is used to make pencils, holders and cigar boxes. Wood of Taxus is heaviest amongst soft woods and is used for making bows for archery.
• Dry leaves of Cycas are used to make baskets and brooms. Needles of Pinus in making fibre board. Electric and telephone poles are made of stem of conifers.
• Essential oils are obtained from Juniperus, Tsugo, Picea, Abies, Cedrus, etc. Resins are obtained from many species of Pinus.

Question 6.
Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Answer:
Both gymnosperms and angiosperms bear seeds, but they are yet classified separately. Because, in case of gymnosperms the seeds are naked, i.e., the seeds are not produced inside the fruit but in case of angiosperms the seeds are enclosed inside the fruit.

Question 7.
What is heterospory? Briefly comment on its significance. Give two examples.
Answer:
Heterospory is the phenomenon in which a plant produces two types of spores, namely microspores and megaspores.
Heterospory is significant in the following ways:

• Microspores give rise to male gametophyte and megaspores give rise to female gametophyte.
• Female gametophyte is retained on the parent plant. The development of zygote takes place within the female gametophyte.
• This leads to formation of seeds.
  Examples: All gymnosperms and all angiosperms, Pinus, Gnetum, neem, peepal, etc.

Question 8.
Explain briefly the following terms with suitable examples:
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Diplontic
(v) Sporophyll
(vi) Isogamy.
Answer:
(i) Protonema: It is the juvenile stage of a moss. It results from the germinating meiospore. When fully grown, it consists of a slender green, branching system of filaments called the protonema.

(ii) Antheridium: The male sex organ of bryophyte and pteridophyte is known as antheridium. It has a single-layered sterile jacket enclosing in a large number of androcytes. The androcytes’ metamorphose into flagellated motile antherozoids.

(iii) Archegonium: The female sex organ of bryophytes, which is multicellular and differentiated into neck and venter. The neck consists of neck canal cells and venter contains the venter canal cells and egg.

(iv) Diplontic: A kind of life cycle in which the sporophyte is the dominant, photosynthetic, independent phase of the plant and alternate with haploid gametophytic phase is known as diplontic life cycle.

(v) Sporophyll: The sporangium bearing structure in case of Selaginella is known as sporophyll.

(vi) Isogamy: It is the process of fusion between two similar gametes, i.e., Chlamydomonas.

Question 9.
Differentiate between the following:
(i) red algae and brown algae,
(ii) liverworts and moss,
(iii) homosporous and heterosporous pteridophytes.
(iv) syngamy and triple fusion.
Answer:
(i) Differences between Red Algae and Brown Algae

Red Algae	Brown Algae
1. It belongs to the It belongs to the class-Rhodophyceae	It belongs to the It belongs to the class Phaeophyceae.
2. It is red in colour due to the presence of pigments chlorophyll-a, c and phycoerythrin. Example: Stylolema, Rhodela.	It is brown in colour due to the presence Example: Sargassum, Microcystis.

(ii) Differences between Liverworts and Moss

Liverwort	Moss
1. These are the member of class-Hepaticopsida of bryophyta.	These belongs to class-Bryop’sida of bryophyta.
2. Thallus is dorsoventrally flattened and lobed liver like	Thallus is leafy and radially symmetrical.
3. Rhizoids are unicellular.	Rhizoids are multicellular
4. Elaters are present in capsule to assist dispersal of spores.	Elaters are absent, but peristome teeth are present in the capsule to assist dispersal of spores.

(iii) Differences between Homosporous and Heterosporous Pteridophytes

Homosporous Pteridophyte	Heterosporous Pteridophyte
Pteridophytes, which produce only one kind of spores. Example: Lycopodium	These produce two kinds of spores, i.e., large megaspore and smaller microspore. Example: Selaginella

(iv) Differences between Syngamy and Triple Fusion

Syngamy	Triple Fusion
It is the act of fusion of one male gamete with the egg cell to form zygote.	The act of fusion of second male gamete with secondary nucleus to form triploid enddsperm is called triple endosperm is called triple fusion.

Question 10.
How would you distinguish monocots from dicots?
Answer:

Dicotyledons (Dicots)	Monocotyledons (Monocots)
•» Tap root system	Fibrous root system
•» Two cotyledons	One cotyledon
•» Reticulate Venation	Parallel venation
•» Tetramerous or Pentamerous flowers	Trimerous flowers

Question 11.
Match the followings (column I with column II)

Column I	Column II
(a) Chlamydomonas	(i) Moss
(b) Cycas	(ii) Pteridophyte
(c) Selaginella	(iii) Algae
(d) Sphagnum	(iv) Gymnosperm

Answer:

Column I	Column II
(a) Chlamydomonas	(iii) Algae
(b) Cycas	(iv) Gymnosperm
(c) Selaginella	(ii) Pteridophyte
(d) Sphagnum	(i) Moss

Question 12.
Describe the important characteristics of gymnosperms.
Answer:
Characteristics of gymnosperms are as follows :

• Naked-seeded plants, i.e., their ovules are exposed and not enclosed in ovaries. Hence, the seeds are naked without fruits.
• Tap root system is present. They show symbiotic as speciation with fungi
  to form mycorrhizae or with N₂-fixing cyanobacteria to form colloidal roots as in Cycas.
• Leaves are large and needle-shaped.
• Vascular tissues are well developed.
• Gymnosperms are heterosporous.
• Pollination by wind and deposited in ovules.
• Fertilisation occurs in archegonia.
• Retention of female gametophyte inside the ovule and the ovules on the sporophytic plant for complete development is responsible for the development of seed habit.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 20 Locomotion and Movement Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

PSEB 11th Class Biology Guide Locomotion and Movement Textbook Questions and Answers

Question 1.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:

Question 2.
Define sliding-filament theory of muscle contraction.
Answer:
The sliding-filament, theory states that the contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Answer:
Mechanism of Muscle Contraction:

• The mechanism of muscle contraction is explained by the sliding filament theory.
• This theory states that contraction of a muscle fibre is due to the sliding of the thin (actin) filaments over the thick (myosin) filaments.
• Muscle contraction is initiated by a neural signal from the central nervous system through a motor neuron.
• When the neural signal reaches the neuromuscular junction, it releases a neurotransmitter, i.e., acetylcholine, which generates an action potential in the sarcolemma.
• This spreads through the muscle fibre and causes the release of Ca⁺⁺ ions from the sarcoplasmic reticulum into the sarcoplasm.
• The Ca⁺⁺ ions bind to the subunit of troponin and brings about conformational changes; this removes the masking of the active site for myosin.
• The myosin head binds to the active site on actin to form a cross-bridge; this utilises energy from the hydrolysis of ATP.
• This pulls the actin filaments towards the centre of A-band.
• As a result, the Z-lines limiting the sarcomere are pulled closer together, causing a shortening of the sarcomere or contraction of muscle.
• Thus, during muscle contraction, the length of A band remains unchanged, while that of I-band decreases.
• The myosin goes back to its relaxed state.
• A new ATP binds and the cross-bridge is broken and the actin filaments slide out of A-band.
• The cycle of cross bridge-formation and cross bridge breakage continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic reticulum which leads to the masking of the active site on F-actin.

Question 4.
Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.
Answer:
(a) True
(b) False, H-zone represents thick filaments
(c) True
(d) False, there are 12 pairs of ribs in man.
(e) True

Question 5.
Write the difference between:
(a) Actin and Myosin
(b) Red and White Muscles
(c) Pectoral and Pelvic Girdle
Answer:
(a) Differences between Actin and Myosin Filament

Actin Filaments	Myosin Filaments
1. These are found in I-band.	These are found in A-band.
2. These are thin.	These are thick.
3. Cross bridges (heads) are absent.	Cross bridges (heads) are present.
4. It is a globular protein with low molecular weight.	It is a heavy molecular weight polymerised protein.

(b) Differences between Red and White Muscles

Red Muscles	White Muscles
1. In some muscles, myoglobin content is high, which gives a reddish colour to them, such muscles are called red muscles.	Some muscles possess very less quantity of myoglobin, so they appear whitish called as white muscles.
2. These contain plenty of mitochondria.	These have less number of mitochondria but amount of sarcoplasmic reticulum is high.
3. These are called aerobic muscles.	They depend on anaerobic process of energy.

(c) Differences between Pectoral and Pelvic Girdle

Pectoral Girdle	Pelvic Girdle
1. It helps in the articulation of upper limbs.	It helps in the articulation of lower limbs.
2. It is situated in the pectoral region of the body.	It is situated in the pelvic region of the body.
3. Each half of pectoral girdle is formed of a clavicle and a scapula.	Pelvic girdle consists of two coxal bones.
4. Scapula is a large triangular flat bone and clavicle is a long slender bone.	Each coxal bone is formed of three bones, ilium, ischium and pubis.
5. An expanded process, acromion from scapula forms a depression called glenoid cavity, which articulates with the head of humerus to form shoulder joint.	Ilium, ischium and pubis fuse at a point to form a cavity called acetabulum to which the thigh bone articulates.

Question 6.
Match Column-I with Column-II

Column-I	Column-II
(a) Smooth muscle	(i) Myoglobin
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(iii) Sutures
(d) Skull	(iv) Involuntary

Answer:

Column-I	Column-II
(a) Smooth muscle	(iv) Involuntary
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(i) Myoglobin
(d) Skull	(iii) Sutures

Question 7.
What are the different types of movements exhibited by the cells of human body?
Answer:
Cells of the human body exhibit three main types of movements-amoeboid, ciliary and muscular.
(i) Amoeboid Movement: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

(ii) Ciliary Movement: Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

(iii) Muscular Movement: Movement of our limbs, jaws, tongue, etc., require muscular movement. Locomotion requires a perfect coordinated activity of muscular, skeletal and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Answer:

Skeletal Muscle	Cardiac Muscle
1. The, cells of skeletal muscles are unbranched.	1. The cells of cardiac muscles are branched.
2. Intercalated disks are absent.	2. The cells are joined with one another by intercalated disks that help in coordination or synchronization of the heartbeat.
3. Alternate light and dark bands are present.	3. Faint bands are present.
4. They are voluntary muscles.	4. They are involuntary muscles.
5. They contract rapidly and get fatigued in a short span of time.	5. They contract rapidly but do not get fatigued easily.
6. They are present in body parts such as the legs, tongue, hands, etc.	6. These muscles are present in the heart and control the contraction and relaxation of the heart.

Question 9.
Name the type of joint between the following:
(i) Atlas/Axis
(ii) Carpal/Metacarpal of thumb
(iii) Between phalanges
(iv) Femur/Acetabulum
(v) Between cranial bones
(vi) Between pubic bones in the pelvic girdle.
Answer:
(i) Pivot joint
(ii) Saddle joint
(iii) Hinge joint
(iv) Ball and socket joint
(v) Fibrous joint
(vi) Cartilagenous joint

Question 10.
Fill in the blank spaces.
(a) All mammals (except a few) have ………………………………. cervical vertebra.
(b) The number of phalanges in each limb of human is ……………………………………
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ………………………. and ………………..
(d) In a muscle fibre Ca²⁺ is stored in ………………………..
(e) ………………….. and ……………………………….. pairs of ribs are called floating ribs.
(f) The human cranium is made of …………………………. bones.
Answer:
(a) seven
(b) fourteen.
(c) troponin and tropomyosin
(d) sarcoplasm
(e) 11 th; 12th
(f) eight