Important Questions

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 5 Laws of Motion Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 5 Laws of Motion

very short answer type questions

Question 1.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
According to the Newton’s second law of motion, F = ma, for given acceleration a, if m is large, F should be more i. e., greater force will be required to put a larger mass in motion.

Question 2.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Solution:
When S ∝ t, so acceleration = 0. Therefore, no external force is acting on the body.

Question 3.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 4.
If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?
Answer:
No change in speed, but change in direction is possible. Forces acting on a body in circular motion is an example.

Question 5.
An impulse is applied to a moving object with a force at an angle of 20° w.r.t. velocity vector, what is the angle between the impulse vector and change in momentum vector?
Answer:
Impulse and change in momentum are along the same direction. Therefore, angle between these two vectors is zero degree.

Question 6.
A body is moving in a circular path such that its speed always remains constant. Should there be a force acting on the body?
Answer:
When a body is moving along a circular path, speed always remains constant and a centripetal force is acting on the body.

Question 7.
The mountain road is generally made winding upwards rather than going straight up. Why?
Solution:
When we go up a mountain, the opposing force of friction
F = μR = μ mg cosθ.
where θ is angle of slope with horizontal. To avoid skidding, F should be large.
∴ cosθ should be large and hence, θ must be small.
Therefore, mountain roads are generally made winding upwards. The road straight up would have large slope.

Short answer type questions

Question 1.
A body of mass 500 g tied to a string of length 1 m is revolved in the vertical circle with a constant speed. Find the minimum speed at which there will not be any slack on the string. Take g = 10ms^-2
Solution:
The tension T in the string will provide the necessary centripetäl force
\(\frac{m v^{2}}{r}\) i.e., T = \(\frac{m v^{2}}{r}\)
Here, m = 500g = \(\frac{1}{2}\)kg; r = 1m
T = \(\frac{1}{2}\)υ²N ……………. (i)
There will not be slack 1f T ≥ weight of the body
i.e., T ≥ mg or \(\frac{1}{2}\)υ² ≥\(\frac{1}{2}\) × 10
υ² ≥ 10 or υ ≥ \(\sqrt{10}\) ms^-1
So the minimum speed = \(\sqrt{10}\) ms^-1 = 3.162 ms^-1

Question 2.
A light, inextensible string as shown in figure connects two blocks of mass M₁ and M₂. A force F as shown acts upon M₁. Find acceleration of the system and tension in string.

Solution:
Here as the string is inextensible, acceleration of two blocks will be same. Also, string is massless so tension throughout the string will be same. Contact force will be normal force only. Let acceleration of each block is a, tension in string is T and contact force between M₁ and surface is N₁ and contact force between M₂ and surface is N₂
Applying Newton’s second law for the blocks;
For M₁, F – T = M₁ a ……………. (i)
M₁ g – N₁ = 0 …………….. (ii)
For M₂, T = M₂ ……………… (iii)
M₂g – N = 0 ……………… (iv)
Solving equations (i) and (iii), we get
a = \(\frac{F}{M_{1}+M_{2}}\)
and T = \(\frac{M_{2} F}{M_{1}+M_{2}}\)

Question 3.
A block of mass m is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is μ and the acceleration due to gravity is g, calculate the minimum force required to be applied by finger to hold the block against the wall? (NCERT Exemplar)
Solution:
Given, mass of the block = m
Coefficient of friction between the block and the wall = μ
Let a force F be applied on the block to hold the block against the wall.
The normal reaction of mass be N and force of friction acting upward be f.
In equilibrium, vertical and horizontal forces should be balanced separately.
f = mg …………….. (i)
∴ and F = N …………… (ii)

But force of friction (f) = μN
= μF [using eq. (ii) ] ………….. (iii)
From eqs. (i) and (iii), we get
μF = mg
or F = \(\frac{m g}{\mu}\)

Question 4.
A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in the weight of the cage when the bird (i) starts flying in the cage with a constant velocity, (ii) flies upwards with acceleration and (iii) flies downwards with acceleration?
Solution:
In a closed glass cage, air inside is bound with the cage. Therefore,
(i) there would be no change in weight of the cage if the bird flies with a constant velocity.
(ii) the cage becomes heavier, when bird flies upwards with an acceleration.
(iii) the cage appears lighter, when bird flies downwards with an acceleration.

Question 5.
When walking on ice, one should take short steps rather than long steps. Why?
Solution:
Let R represent the reaction offered by the ground. The vertical component R cosθ will balance the weight of the person and the horizontal component R sinθ will help the person to walk forward.

Now, normal reaction = R cosθ
Friction force = R sinθ
Coefficient of friction, μ = \(\frac{R \sin \theta}{R \cos \theta}\) = tanθ
In a long step, θ is more. So tanθ is more. But μ has a fixed value. So, there is danger of slipping in a long step.

Question 6.
A body of mass m is suspended by two strings making angles α and β with the horizontal as shown in fig. Calculate the tensions in the two strings.

Solution:
Considering components of tensions T₁ and T₂ along the horizontal and vertical directions,
We have
-T₁cosα + T₂cosβ = 0
or T₁cosα = T₂cosβ …………… (i)
and T₁ sinα + T₂ sinβ = mg
From eq. (i) T₂ = \(\frac{T_{1} \cos \alpha}{\cos \beta}\) and substituting it in eq. (ii), we get

Question 7.
State the law of conservation of momentum. Establish the same for a ‘n’ body system.
Solution:
When no external force acts on a system the momentum will remain conserved. Consider a system of a n bodies of masses m₁ ,m₂ ,m₃ , ………… ,m_n. If p₁ , p₂ , P₃ , ………. ,P_n are the momentum associated then the rate of change of momentum with the system,
\(\frac{d p}{d t}=\frac{d p_{1}}{d t}+\frac{d p_{2}}{d t}+\frac{d p_{3}}{d t}\) + ………. + \(\frac{d p_{n}}{d t}=\frac{d}{d t}\) = (p+₁ +p₂ +p₃+ ………. +p_n )
If no external force acts, \(\frac{d p}{d t}\) = 0
∴ p = constant, i.e., P₁ + p₂ + P₃ +………… +P_n = constant.

Question 8.
A block slides down from top of a smooth inclined plane of elevation θ fixed in an elevator going up with an acceleration a0. The base of incline has length L. Find the time taken by the block to reach the bottom.

Solution:
The free body force diagram is shown. The forces are
(i) N normal to the plane (ii) mg acting vertically down (iii) ma₀ (pseudo-force).

If a is the acceleration of the body with respect to incline, taking components of forces parallel to the incline mg sinθ + ma0 sinθ = ma
a = (g + a₀)sinθ
This is the acceleration with respect to elevator.
The distance travelled is \(\frac{L}{\cos \theta}\) If t is the time for reaching the bottom of
incline, using equation of motion, s = ut + \(\frac{1}{2}\)at2, we get
\(\frac{L}{\cos \theta}\) = 0 + \(\frac{1}{2}\)(g + a₀)sinθ.t²
t = [latex]\frac{2 L}{\left(g+a_{0}\right) \sin \theta \cos \theta}[/latex]^1/2

Long answer type questions

Question 1.
Figure shows (x – t), (y – t) diagram of a particle moving in 2-dimensions.

If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle. (NCERT Exemplar)
Given, mass of the particle (m) = 500 g = 0.5 kg
x – t graph of the particle is a straight line.
Hence, particle is moving with a uniform velocity along x-axis, i. e., its acceleration along x-axis is zero and hence, force acting along x-axis is zero.
y – t graph of particle is a parabola. Therefore, particle is in accelerated motion along y – axis.
At t = 0, u_y = 0
Along y – axis, at t = 2s, y = 4m
Using equation of motion, y = u_yt + \(\frac{1}{2}\) a_yt²
4 = 0 × 2 + \(\frac{1}{2}\) × a_y × (2)²
or a_y = 2 m/s²
∴ Force acting along y – axis (f_y) = ma_y = 0.5 × 2 = 1.0 N (along y – axis)

Question 2.
When a body slides down from rest along a smooth inclined plane making an angle of 45° with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane. (NCERT Exemplar)
Solution:
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane, a = g sinθ
Here, θ = 45°
a = gsin45°= \(\frac{g}{\sqrt{2}}\)
Let the travelled distance be s.
Using the equation of motion, s = ut + \(\frac{1}{2}\) at² ,
We get
s = 0 .t + \(\frac{1}{2} \frac{g}{\sqrt{2}}\)T²
or s = \(\frac{g T^{2}}{2 \sqrt{2}}\) ………… (i)

On rough inclined plane
Acceleration of the body,
a = g (sinθ – μ cosθ)
= g (sin 45° – μ cos 45°)
= \(\frac{g(1-\mu)}{\sqrt{2}}\) [as sin 45°= cos 45° = \(\frac{1}{\sqrt{2}}\)]
Again using equation of motion,

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 7 System of Particles and Rotational Motion Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 7 System of Particles and Rotational Motion

very short answer type questions

Question 1.
(n – 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector \(\vec{a}\) with respect to the centre of the polygon. Find the position vector of centre of mass. (NCERT Exemplar)
Solution
Suppose, \(\vec{b}\) be the position vector of centre of mass of regular n-polygon. As (n – 1) equal point masses each of mass m are placed at (n – 1) vertices of regular polygon, therefore
\(\frac{(n-1) m b+m a}{(n-1+1) m}\) = 0
⇒ (n – 1)mb + ma = 0
⇒ b = \(\frac{-a}{(n-1)}\)

Question 2.
If net torque on a rigid body is zero, does it linear momentum necessary remain conserved?
Answer:
The linear momentum remain conserved if the net force on the system is zero.

Question 3.
When is a body lying in a gravitation field in stable equilibrium?
Answer:
A body in a gravitation field will be in stable equilibrium, if the vertical line through its centre of gravity passes through the base of the body.

Question 4.
Is centre of mass and centre of gravity body always coincide?
Ans.
No, if the body is large such that g varies from one point to another, then centre of gravity is offset from centre of mass.
But for small bodies, centre of mass and centre of gravity lies at their geometrical centres.

Question 5.
Why is moment of inertia also called rotational inertia?
Answer:
The moment of inertia gives a measure of inertia in rotational motion. So, it is also called rotational inertia.

Question 6.
In a flywheel, most of the mass is concentrated at the rim. Explain why?
Answer:
Concentration of mass at the rim increases the moment of inertia and thereby brings uniform motion.

Question 7.
Does the radius of gyration depend upon the speed of rotation of the body?
Answer:
No, it depends only on the distribution of mass of the body.

Question 8.
Can the mass of body be taken to be concentrated at its centre of mass for the purpose of calculating its rotational inertia?
Answer:
No, the moment of inertia greatly depends on the distribution of mass about the axis of rotation.

Short answer type questions

Question 1.
Does angular momentum of a body in translatory motion is zero?
Solution:
Angular momentum of a body is measured with respect to certain origin.

So, a body in translatory motion can have angular momentum.
It will be zero, if origin lies on the line of motion of particle.

Question 2.
Figure shows momentum versus time graph for a particle moving along x – axis. In which region, force on the particle is large. Why?

Solution:
Net force is given by F = \(\frac{d p}{d t}\)
Also, rate of change of momentum = slope of graph.
As from graph, slope AB = slope CD
And slope (BC) = slope (DE) = 0
So, force acting on the particle is equal in regions AB and CD and in regions BC and DE (which is zero).

Question 3.
Two cylindrical hollow drums of radii R and 2J2, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively.
Their axes, fixed are parallel and in a horizontal plane separated by (3R + δ). They are now brought in contact (δ → 0).
(i) Show the frictional forces just after contact.
(ii) Identify forces and torques external to the system just after contact.
(iii) What would be the ratio of final angular velocities when friction ceases? (NCERT Exemplar)
Solution:

(ii) F’ = F = F” where F and F” are external forces through support.
F_net = 0
External torque = F x 3 R, anti-clockwise.

(iii) Let ω₁ and ω₂ be final angular velocities (anti-clockwise and clockwise respectively).
Finally, there will be no friction.
Hence, Rω₁ = 2Rω₂ ⇒ \(\frac{\omega_{1}}{\omega_{2}}\) = 2

Question 4.
Angular momentum of a system is conserved if its M.I. is changed. Is its rotational K. E. also conserved?
Solution:
Kinetic energy of rotation = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\) (Iω)ω = \(\frac{1}{2}\)Lω

L = Iω is constant, if moment of inertia (I) of the system changes. It means as I changes, then ω also changes.
Hence K.E. of rotation also changes with the change in I. In other words, rotational K.E. is not conserved.

Question 5.
How much fraction of the kinetic energy of rolling is purely
(i) translational, (ii) rotational.
Solution:

Question 6.
Listening to the discussion on causes of pollution and due to which temperature on earth is rising, increase in temperature leads to melting of polar ice, Meenu realised that if each one of us contributed to create pollution free environment, then even small efforts can lead to big results. So, she decided to lead the step and instead of going to school by her car, she joined school bus and also asked her father to go to office using car pool.
(i) What do you think is mainly responsible for global warming?
(ii) If the ice on polar caps of the earth melts due to pollution, how will it affect the duration of the day?
Explain.
(iii) What values does Meenu show?
Answer:
(i) Pollution created by the people of world is the main cause of global warming.
(ii) Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis of rotation spreads out, therefore moment of inertia, I increases.
As no external torque acts,
∴ L = I = Iω = (\(\frac{2 \pi}{T}\)) = Constant
With increase of I, T will increase i.e., length of day will also increase,
(iii) Meenu is considerate towards environment and her thought of leading the steps to reduce pollution is commendable.

Question 7.
Explain how a cat is able to land on its feet after a fall taking the advantage of principle of conservation of angular momentum?
Answer:
When a cat falls to ground from a height, it stretches its body alongwith the tail so that its moment of inertia becomes high. Since, la is to remain constant, the value of angular speed a decreases and therefore the cat is able to’ land on the ground gently.

Question 8.
A uniform disc of radius R is resting on a table on its rim. The coefficient of friction between disc and table is μ (figure). Now, the disc is pulled with a force \(\overrightarrow{\boldsymbol{F}}\) as shown in the figure. What is the maximum value of \(\overrightarrow{\boldsymbol{F}}\) for which the disc rolls without slipping? (NCERT Exemplar)

Solution:
Let the acceleration of the centre of mass of disc be a, then
Ma = F – f
The angular acceleration of the disc is a = a/R (if there is no sliding).
Then, (\(\frac{1}{2}\)MR²)α = Rf
⇒ Ma = 2f
Thus, f =F/3. Since, there is no sliding.
⇒ f ≤ μ mg ⇒ F ≤ 3μ Mg

Question 9.
Two equal and opposite forces act on a rigid body. Under what condition will the body (i) rotate (ii) not rotate?
Answer:
(i) Two equal and opposite forces acting on a rigid body such that their lines of action do not coincide, constitute a couple. This couple produces the turning effect on the body. Hence, the rigid body will rotate.

(ii) If two equal and opposite forces act in such a way that their lines of action coincide, then these forces cancel out the effect of each other. Hence, the body will not rotate.

Long answer type questions

Question 1.
Find position of centre of mass of a semicircular disc of radius r. (NCERT Exemplar)
Solution:
As semicircular disc is symmetrical about its one of diameter, we take axes as shown. So, now we only have to calculate Y_CM (As X_CM is zero by symmetry and choice of origin).

Now, for a small element OAB, as element is small and it can be treated as a triangle so,
Area of sector OAB = \(\frac{1}{2}\) x r x rdθ
Height of triangle = r
Base of triangle = AB = rdθ
So, its mass dm = \(\frac{1}{2}\)r² dθ.ρ [∵ ρ = \(\frac{\text { mass }}{\text { area }}\)]
As centre of mass of a triangle is at a distance of \(\frac{2}{3}\) from its vertex (at centroid, intersection of medians). So, y = \(\frac{2}{3}\)rsinθ (location of CM of small sector AOB).


So, CM of disc is at a distance of \(\frac{4 r}{3 \pi}\)from its centre on its axis of symmetry.

Question 2.
Obtain an expression for linear acceleration of a cylinder rolling down an inclined plane and hence find the condition for the cylinder to roll down the inclined plane without slipping.
Solution:
When a cylinder rolls down on an inclind plane, then forces involved are (i) Weight mg (ii) Normal reaction N (iii) Friction f
From free body diagam,

From free body diagram,
N – mg cos θ = 0
or N = mg cosθ
Also, if a = acceleration of centre of mass down the plane, then
F_net = ma = mgsin θ – f …………… (i)
As friction produces torque necessary for rotation,
τ = Iα = f R

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 8 Gravitation Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 8 Gravitation

Very Short Answer Type Questions

Question 1.
By which law is the Kepler’s law of areas identical?
Answer:
The law of conservation of angular momentum.

Question 2.
Draw areal velocity versus time graph for mars. (NCERT Exemplar)
Answer:
Areal velocity of planet revolving around the Sun is constant with time (Kepler’s second law).

Question 3.
At what factor between the two particles gravitational force does not depend?
Answer:
Gravitational force does not depend upon the medium between the two particles.

Question 4.
Two particles of masses m₁ and m₂ attract each other gravitationally and are set in motion under the influence of the gravitational force? Will the centre of mass move?
Answer:
Since the gravitational force is an internal force, therefore the centre of mass would not move.

Question 5.
Work done in moving a particle round a closed path under the action of gravitation force is zero. Why?
Answer:
Gravitational force is a conservative force which means that work done by it, is independent of path followed.

Question 6.
What would happen if the force of gravity were to disappear suddenly?
Answer:
The universe would collapse. We would be thrown away because of the centrifugal force. Eating, drinking and in fact all activities would become impossible.

Question 7.
Why a body weighs more at poles and less at equator?
Answer:
The value of g is more at poles than at the equator. Therefore, a body weighs more at poles than at equator.

Question 8.
Give a method for the determination of the mass of the moon.
Solution:
Soli By making use of the relation, gm = \(\frac{G M_{m}}{R_{m}^{2}} \)

Short Answer Type Questions

Question 1.
A planet moving along an elliptical orbit is closest to the Sim at a distance r₁ and farthest away at a distance of r₂.
If v₁ and v₂ are the linear velocities at these points respectively, then find the ratio \(\frac{v_{1}}{v_{2}}\).
Solution:
From the law of conservation of angular momentum
mr₁v₁ = mr₂v₂
⇒ r₁v₁ = r₂v₂ or
\(\frac{v_{1}}{v_{2}}=\frac{r_{2}}{r_{1}}\)

Question 2.
A mass M is broken into two parts, m and (M – m). How is m related to M so that the gravitational force between two parts is maximum?
Solution:
Let =m,m₂ =M – m
F = G\(\frac{m(M-m)}{r^{2}}=\frac{G}{r^{2}}\left(M m-m^{2}\right)\)
Differentiating w.r:t. m, \(\frac{d F}{d m}=\frac{G}{r^{2}}(M-2 m)\)
For F to be maximum, \(\frac{d F}{d m}\) = 0

m₁ = m₂ = M/2

Question 3.
Two stationary particles of masses M₁ and M₂ are a distance d apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is the distance of this particle from M₁?
Solution:
The force on m towards Mi is F =G \(\frac{M_{1} m}{r^{2}}\)
The force on m towards Mi is F = G \(\frac{M_{2} m}{(d-r)^{2}} \)

Equating two forces, we have

So, distance of an particle from m is . r = d
r = d \(\left(\frac{\sqrt{M_{1}}}{\sqrt{M_{1}}+\sqrt{M_{2}}}\right)\).

Question 4.
Aspherical planet has mass M_p and clinometer D_p. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to whom?
Solution:
Force is given by
F = \(-\frac{G M m}{R^{2}}=\frac{G M_{p} m}{\left(D_{P} / 2\right)^{2}}=\frac{4 G M_{P} m}{D_{P}^{2}}\)
\(\frac{F}{m}=\frac{4 G M_{P}}{D_{P}^{2}}\)

Question 5.
What is the gravitational potential energy of a body at height h from the Earth surface?
Solution:
Gravitational potential energy, i. e.,
U_h = \(-\frac{G M m}{R+h}=-\frac{g R^{2} m}{R+h}\)
[where g = \(\frac{G M}{R^{2}}\) ]
= – \(\frac{g R^{2} m}{R\left(1+\frac{h}{R}\right)}=-\frac{m g R}{1+\frac{h}{R}}\)
.
Question 6.
An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from Earth.
Determine
(i) the height of satellite above Earth’s surface.
(ii) if the satellite is suddenly stopped, find the speed with
which the satellite will hit the Earth’s surface after falling down.
Solution:
Escape velocity = \(\sqrt{2 g R}\), where g is acceleration due to gravity on surface of Earth and R the radius of Earth.
Orbital velocity = \(\frac{1}{2} v_{e}=\frac{1}{2} \sqrt{2 g R}=\sqrt{\frac{g R}{2}} \) …………………. (i)

(i) If h is the height of satellite above Earth

h=R
(ii) If the satellite is stopped in orbit, the kinetic energy is zero and its
potential energy is – \(\frac{G M m}{2 R}\)
Total energy =-\(\frac{G M m}{2 R}\)

Let v be its velocity when it reaches the Earth.
Hence the kinetic energy = \(\frac{1}{2} m v^{2}\)
Potential energy = – \(\frac{G M m}{2 R}\)

Question 7.
Why do different planets have different escape velocities?
Solution:
Escape velocity, v = \(\sqrt{2 g R}=\sqrt{\frac{2 G M}{R}}\)
Thus escape velocity of a planet depends upon (i) its mass (M) and
(ii) its size (R).
As different planets have different masses and sizes, so they have different escape velocities.

Question 8.
Under what circumstances would your weight become zero?
Answer:
The weight will become zero under the following circumstances
(i) during free fall
(ii) at the centre of the Earth
(iii) in an artificial satellite
(iv) at a point where gravitational pull of Earth is equal to the gravitational pull of the Moon.

Long Answer Type Questions

Question 1.
A mass m is placed at P, a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r? (NCERT Exemplar)

Solution:
Consider a small element of the ring of mass dM, gravitational force between dM and m, distance x apart in figure i.e.,
dF = \(\frac{G(d m) m}{x^{2}}\)

dF can be resolved into two rectangular components.
(i) dF cos θ along PO and
(ii) dF sinθ perpendicular to PO (given figure)
The total force (F) between the ring and mass (m) can be obtained by integrating the effects of all the elements forming the ring, whereas all the components perpendicular to PO cancel out i.e., ∫dFsinθ=O, the component along PO add together to give F i.e.,

Question 2.
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication.
(i) Calculate height of such a satellite.
(ii) Find out the minimum number of satellites that are needed to cover entire Earth so that at least one of satellite is visible from any point on the equator.
[M = 6 x 10 ²⁴ kg, R = 6400 km, T = 24 h, G = 6.67 x 10^-11SI (NCERT Exemplar)
Solution:
(i) As, according co Kepler’s third law, we get
T² = \(\frac{4 \pi^{2} r^{3}}{G M}\)
⇒ r = \( \left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}\)

As we known =R +h
h=r-R
h=4.23 x 10⁷ m – 6.4 x 10⁶ m
h = 3.59 x 10⁷ m

(ii) In ΔOES,cos θ = \(\frac{O A}{O S}=\frac{R}{R+h}\)
= \(\frac{1}{\left(1+\frac{h}{R}\right)}\)
= \(\frac{1}{(1+5.609)}\)
=0.1513
(as,\(\frac{h}{R}=\frac{3.59 \times 10^{7} \mathrm{~m}}{6.4 \times 10^{6} \mathrm{~m}}\) = 5.609)
where, θ ≈ 81° or 2θ = 162°
Number of satellites required to cover entire the Earth.
= \(\frac{360^{\circ}}{162^{\circ}}=2.2\) ≈ 3.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 21 Neural Control and Coordination Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 21 Neural Control and Coordination

Very Short Answer Type Questions

Question 1.
Which part of the neuron is considered as afferent process?
Answer:
Dendrites conduct nerve impulses towards the cell body and are called afferent processes (receiving processes).

Question 2.
Give the name of the chemicals, which are released at the synaptic junction. [NCERT Exemplar]
Answer:
Neurotransmitters.

Question 3.
Name the small protein tubular structures between the two neurons.
Answer:
Gap junctions.

Question 4.
Which is the largest and most complex of all, the parts of the human brain?
Answer:
Cerebrum.

Question 5.
What is the role of afferent nerve fibers in the neural system?
Answer:
It transmits impulse (sensory) from tissues/organs to the CNS and form, the sensory or afferent pathway.

Question 6.
How does the efferent fibers work?
Answer:
The efferent nerve fibers transmit motor impulses from CNS to the concerned tissues/organs and form the motor or efferent pathway.

Question 7.
The autonomic neural system is considered as involuntary neural system. Why?
Answer:
This system transmits impulses from the CNS to the involuntary organs and smooth muscles of the body. So, it is also called involuntary neural system.

Question 8.
How does the somatic neural system works?
Answer:
The somatic neural system controls the movements of the body by acting on the skeletal muscles (i.e., relays voluntary impulses from the CNS to skeletal muscles).

Question 9.
Give the name of the covering that maintains the shape of the eyeball.
Answer:
Sclera (outermost layer).

Question 10.
Which is the bluish (pigmented) layer present beneath the sclera?
Answer:
Choroid.

Question 11.
Which part of our body helps us in maintaining the body balance?
Answer:
Ears.

Question 12.
Which of the photoreceptors is responsible for twilight vision?
Answer:
Rods.

Short Answer Type Questions

Question 1.
Give a brief description of the neural system.
Answer:
The neural system is composed of specialized cells called neurons. It detects stimuli and transmits neural signals. The neural system of complex animals is composed of two parts, viz. central neural system and peripheral neural system. The brain and nerve cord comprise the central neural system and other nerves comprise the peripheral neural system.

Question 2.
Explain parasympathetic neural system.
Answer:
The parasympathetic neural system is part of autonomic neural system. This system has some sort of inhibitory effect. The inhibitory effect minimises the over-functioning of certain functions. Functions, like salivating, digestion, are under control of parasympathetic neural system.

Question 3.
Give a description of the structure of neuron.
Answer:
A neuron is a microscopic structure composed of three major parts, namely, cell body, dendrites and axon.
Cell Body: The cell body contains cytoplasm with typical cell organelles and certain granular bodies called Nissl’s granules.

Dendrites: Short fibers which branch repeatedly, and project out of the cell body also contain Nissl’s granules and are called dendrites. These fibers transmit impulses towards the cell body.

Axon: The axon is a long fibre, the distal end of which is branched. Each branch terminates as a bulb-like structure called synaptic knob which possess synaptic vesicles containing chemicals called neurotransmitters. The axons transmit nerve impulses away from the cell body to a synapse or to a neuromuscular junction.

Question 4.
Describe reflex action.
Answer:
The entire process of response to a peripheral neural stimulation, that occurs involuntarily, i.e., without conscious effort or thought and requires the involvement of a part of the central neural system is called a reflex action.
The reflex pathway comprises at least one afferent neuron (receptor) and one efferent (effector or excitor) neuron appropriately arranged in a series.

The afferent neuron receives signals from a sensory organ and transmits the impulse via a dorsal nerve root into the CNS (at the level of spinal cord). The efferent neuron then carries signals from CNS to the effector. The stimulus and response thus forms a reflex arc.

Question 5.
What do you understand by olfactory receptors?
Answer:
The nose contains mucus-coated receptors which are specialized for receiving the sense of smell and are called olfactory receptors. These are made up of olfactory epithelium which consists of three kinds of cells. The neurons of the olfactory epithelium extend from the outside environment directly into a pair of broad bean-sized organs called olfactory bulb. Olfactory bulbs are extensions of the brain’s limbic system.

Long Answer Type Questions

Question 1.
(a) Give an account of spinal nerves in man.
(b) What biological functions are served by the skeletal system?
Answer:
(a) There are 31 pairs of spinal nerve in man. From each segment of the spinal cord, there arises two spinal nerves. Each spinal nerve is a mixed nerve, containing both sensory’ and motor nerve fibres. It runs between the spinal cord and peripheral tissue. The two roots, i. e., motor or ventral and sensory or dorsal connect the spinal nerve to the spinal cord.

The DORSAL ROOT carries sensory or afferent fibre and has dorsal root ganglion at its middle. The VENTRAL, ROOT contains motor or efferent nerve fibers. The dorsal root fibres bring impulses from the peripheral tissue and give rise to sensations like touch, temperature, and pain. The ventral nerve root fibres pass impulses to muscles and glands in the peripheral tissues. The spinal nerve has been named according to their relation with the vertebral column.

These are

• Eight pairs of cervical,
• 12 pairs of thoracic,
• 5 pairs of lumbar,
• 5 pairs of sacral and
• a pair of coccygeal or caudal.

(b)

• The skeletal system forms the rigid structural framework of the body and supports the weight of the body along with its limbs.
• It affords protection to the internal organs against mechanical injury by forming cage-like compartments, e.g., skull.
• It serves as a storage depot for calcium and phosphate, which are released for a number of functions of the body.
• It participates in movement and locomotion.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 22 Chemical Coordination and Integration Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 22 Chemical Coordination and Integration

very Short Answer Type Questions

Question 1.
There are many endocrine glands in human body. Name the gland, which is absent in male and the one absent in female. [NCERT Exemplar]
Answer:
The glands, which are absent in male are ovaries and which are absent in female are testes.

Question 2.
Which of the two adrenocortical layers, zona glomerulosa and zona reticularis lies outside enveloping the other?
[NCERT Exemplar]
Answer:
Zona glonierulosa (outer layer) envelopes zona reticularis (inner layer) from the outside.

Question 3.
Name the only hormone secreted by pars intermedia of pituitary gland. [NCERT Exemplar]
Answer:
Melanocyte Stimulating Hormone (MSH).

Question 4.
Mention the name of the largest and the smallest endocrine gland found in man.
Answer:
Thyroid gland is the largest endocrine gland and pituitary gland is the smallest endocrine gland.

Question 5.
A patient complains of constant thirst, excessive passing of urine and low the level blood pressure. When the doctor checked the patient’s blood glucose and blood insulin level, the level were normal or slightly low. The doctor diagnosed the condition as diabetes insipidus. But he decide to measure one more hormone in patient’s blood. Which hormone does the doctor intend to measure? [NCERT Exemplar]
Answer:
Glucagon.

Question 6.
The outermost layer of adrenal cortex is responsible for secretion of which hormone. Identify?
Answer:
Mineralocorticoids.

Question 7.
Identify the neurohormone that has its functioning in inhibiting the secretion of growth hormone from anterior lobe of pituitary.
Answer:
Somatostatin inhibits the secretion of growth hormone from anterior lobe of pituitary gland.

Question 8.
State the reason for the occurrence of diabetes insipidus in a individual.
Answer:
Deficiency in the secretion of vasopressin (ADH) leads to the disorder known as diabetes insipidus.

Question 9.
Define the term erythropoiesis. Also name the hormone that stimulates it. [NCERT Exemplar]
Answer:
Erythropoiesis is the process of formation of RBCs. The juxtaglomerular cells of kidney produce a peptide hormone called erythropoietin which stimulates it.

Question 10.
What do you understand by the term ANF?
Answer:
Atrial wall of human heart secretes a peptide hormone called atrial natriuretic factor which decreases blood pressure by dilation of the blood vessels.

Question 11.
Mention the name given to the hormones produced by some non-endocrine tissues.
Answer:
Hormones produced by some non-endocrine tissues are called growth factors.

Question 12.
Which two hormones are steroids chemically?
Answer:
Cortisol and testosterone are chemically steroid in nature.

Short Answer Type Questions

Question 1.
Explain the function of melanin.
Answer:
Melanin controls the circadian variations of the body. During 24 hours different organ system of our body works at different pace. During sleep certain body functions slow down. All of this is known as circadian rhythm. Additionally, melanin influences metabolism, pigmentation, menstruation and defence capability.

Question 2.
How does parathyroid hormone influences calcium uptake in the body?
Answer:
Parathyroid hormone (PTH) increases the Ca²⁺ levels in the blood. PTH acts on bones and stimulates the process of bone reabsorption (dissolution/demineralization). PTH also stimulates reabsorption of Ca2+ by the renal tubules and increases Ca²⁺ absorption from the digested food. It is, thus, clear that PTH is a hypercalcemic hormone, i.e., it increases the blood Ca²⁺ levels. Along with TCT, it plays a significant role in calcium balance in the body.

Question 3.
How do fight or flight hormones prepare our body to fight emergency?
Answer:
Adrenaline and noradrenaline are rapidly secreted in response to stress of any kind and during emergency situations and are called emergency hormones or hormones of fight or flight. These hormones increase alertness, pupillary dilation, piloerection (raising of hairs), sweating etc. Both the hormones increase the heartbeat, the strength of heart contraction, and the rate of respiration. Finally, the body is ready to counter the emergency situations.

Question 4.
What are secondary sexual characters?
Answer:
Characters which do not play direct role in sexual reproduction but are basically means of sexual differentiation are called secondary sexual characters. For example, facial hair and deep voice in males and thin voice in females are secondary sexual characters.

Question 5.
What is acromegaly?
Answer:
Excess secretion of growth hormone in adults, especially in middle age can result in severe disfigurement (especially of the face). This is called acromegaly. This can lead to serious complications and even death; if unchecked. The disease is hard to diagnose in the early stages and is frequently missed for many years, until changes in external features become noticeable.

Long Answer Type Questions

Question 1.
Hypothalamus is a super master endocrine gland. Elaborate. [NCERT Exemplar]
Answer:
Hypothalamus regulates a wide spectrum of body functions. It contains several groups of neurosecretory cells called nuclei, which produce hormones. These hormones regulate the synthesis and secretion of pituitary hormones. However, the hormones produced by hypothalamus are of two types, the releasing hormones (which stimulate secretion of pituitary hormones) and the inhibiting hormones (which inhibit secretions of pituitary hormones).
The hormones reach the pituitary gland through a portal circulatory system and regulate the functions of the anterior pituitary. The posterior pituitary is under the direct regulation of hypothalamus. The oxytocin and vasopressin are the two hormones synthesized by hypothalamus that are transported to posterior pituitary.

Question 2.
A sample of urine was diagnosed to contain high content of glucose and ketone bodies. Based on this observation, answer the following: (NCERT Exemplar)
(i) Which endocrine gland and hormone is related to this condition? %
(ii) Name the cells on which this hormone acts.
(iii) What is the condition called and how can it be rectified?
Answer:
(i) Pancreas gland and insulin hormone is related to this condition.
(ii) The (3-cells of islets of Langerhans of pancreas.
(iii) Prolonged hyperglycemia leads to a complex disorder, called diabetes mellitus, which is associated with loss of glucose through urine and formation of harmful compounds known as ketone bodies. Diabetic patients are successfully treated with insulin therapy.

Question 3.
(i) Give a diagrammatic representation of the mechanism of protein hormone (e. g., FSH) action.
(ii) Illustrate the differences between the mechanism of action of, a protein and a steroid hormone. [NCERT Exemplar]
Answer:

(ii) Differences between mechanism of action of a protein and a steroid hormone

Protein Hormone	Steroid Hormone
Protein hormones interact with membrane-bound receptors.	They interact with intracellular receptors.
They generate second messengers (cyclic AMP, IP3, Ca2+, etc.)	They regulate gene expression or chromosome function by the interaction of the hormone-receptor complex with the genome.
The second messengers regulate. cellular metabolism.	Cumulative biochemical action of hormone-receptor complex results in physiological and developmental effects.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 6 Work, Energy and Power Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 6 Work, Energy and Powere

Very short answer type questions

Question 1.
Under what condition is the work done by a force inspite of displacement being taking place?
Answer:
Work done by a force is zero inspite of displacement being taking place, if displacement is in a direction perpendicular to that of force applied.

Question 2.
Can acceleration be produced without doing any work? Give example.
Answer:
Yes, for uniform circular motion, no work done but a centripetal acceleration is present.

Question 3.
Does the amount of work done depend upon the fact that how fast is a load raised or moved in the direction of force?
Answer:
The amount of work does not depend upon the fact that how fast is a load raised or moved in the direction of force.

Question 4.
A body is moving along a circular path. How much work is done by the centripetal force?
Answer:
For a body moving along a circular path, the centripetal force acts along the radius while the displacement is tangential, i. e., θ = 90 °, therefore,
W = Fscos90° = 0.

Question 5.
What is the source of kinetic energy of the bulelt coming out of a rifle?
Answer:
The source of kinetic energy of bullet is the potential energy of the compressed spring in the loaded rifle.

Question 6.
A spring is cut into two equal halves. How is the spring constant of each half affected?
Answer:
Spring constant of each half becomes twice the spring constant of the original spring.

Question 7.
Is collision between two particles possible even without any physical contact between them?
Answer:
Yes, in atomic and subatomic particles collision without any physical contact between the colliding particles is taking place e. g., Rutherford’s alpha particles scattering.

Question 8.
Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case? (NCERT Exemplar)
Answer:
When the elevator is descending, then electric power is required to prevent it from falling freely under gravity.
Also, as the weight inside the elevator increases, its speed of descending – increases, therefore, there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with large velocity.

Short answer type questions

Question 1.
A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. Coefficient of friction between block and surface is μ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.

Solution:
The forces acting on the block are shown in figure. As the block moves with uniform velocity the forces add up to zero.
∴ Fcosθ = μN ………….. (i)
Fsinθ + N = Mg ……………. (ii)
Eliminating N from equations (i) and (ii)
F cosθ = μ(Mg – F sinθ)
F = [Latex]\frac{\mu M g}{\cos \theta+\mu \sin \theta}[/Latex]
Work done by this force during a displacement d
W = F. d cosθ = [Latex]\frac{\mu M g d \cos \theta}{\cos \theta+\mu \sin \theta}[/Latex]

Question 2.
Two springs have force constants K₂ and K₂ (K₁ > K₂ )• On which spring is more work done when they are stretched by the same force?
Solution:

As x₁ < x₂
∴ W₁ < W₂ or W₂ > W₁

Question 3.
A particle is moving in a circular path of radius r with constant speed. Due to change in the direction of motion of the particle continuously, the velocity of the particle is changing. But the kinetic energy of the particle remains the same. Explain why ?
Solution:
Kinetic energy is given by
E = \(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) m(\(\vec{v} \cdot \vec{v}\))
Since \(\vec{v} \cdot \vec{v}\) – υ², a scalar quantity, so it is the speed which is taken into account while calculating the kinetic energy of the particle. As the speed is constant, so kinetic energy of the particle will also remain constant.

Question 4.
Can a body have energy without momentum? If yes, then explain how they are related with each other?
Solution:
Yes, when p = 0,
Then, K = \(\frac{p^{2}}{2 m}\) = 0
But E = K + U = U (potential energy), which may or may not be zero.

Question 5.
Two bodies A and B having masses m_A and m_B respectively have equal kinetic energies. If p_A and p_B are their respective momenta, then prove that the ratio of momenta is equal to the square root of ratio of respective masses.
Solution:
Let υ_A and υ_B be the velocities of A and B respectively.
Since their kinetic energies are equal,

Question 6.
Two ball bearings of mass m each, moving in opposite directions with equal speed υ, collide head on with each other. Predict the outcome of the collision, assuming it to be perfectly elastic.
Solution:
Here, m₁ = m₂ = m
u₁ = υ,u₂ = -υ
Velocities of two balls after perfectly elastic collision between them are

After collision, the two ball bearings will move with same speeds, but their direction of motion will be reversed.

Question 7.
An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 kmh^-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of
energy of the wagon is lost due to friction, calculate the spring constant. (NCERT Exemplar)
Solution:
Given, mass of the system (m) = 50,000 kg
Speed of the system (υ) = 36 km/h
= \(\frac{36 \times 1000}{60 \times 60}\) = 10 m/s
Compression of the spring (x) = 1.0 m
KE of the system = \(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) × 50000 × (10)²
= 25000 × 100 J = 2.5 × 10⁶J
Since, 90% of KE qf the system is lost due to friction, therefore, energy transferred to shock absorber, is given by
ΔE = \(\frac{1}{2}\)kx² = 10% of total KE of the system
= \(\frac{10}{100}\) × 2.5 × 10⁶ J or k = \(\frac{2 \times 2.5 \times 10^{6}}{10 \times(1)^{2}}\)
= 5.0 × 10⁶ N/m

Question 8.
An adult weighting 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him is jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging. (NCERT Exemplar)
Solution:
Given, weight of the adult (w) = mg = 600 N
Height of each step = h = 0.25m
Length of each step = 1 m
Total distance travelled = 6 km = 6000 m
∴ Total number of steps = \(\frac{6000}{1}\) = 6000
Total energy utilised in jogging = n × mgh
= 6000 × 600 × 0.25J = 9 × 10⁵ J
Since, 10% of intake energy is utilised in jogging.
∴ Total intake energy = 10 × 9 × 10⁵J = 9 × 10⁶J

Long answer type questions

Question 1.
A body of mass 0.3 kg is taken up an inclined plane length 10 m and height 5 m, and then allowed to slide down the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the
(i) work done by gravitational force over the round trip?
(ii) work done by the applied force over the upward journey?
(iii) work done by the frictional force over the round trip?
(iv) kinetic energy of the body at the end of trip? (Take g = 10 ms^-2)

Solution:
Upward journey

Let us calculate work done by different forces over upward joume Work done by gravitational force
Wi = (mg sinθ)s cos 180°
W ₁= 0.3 × 10 sin30° × 10 (-1)
W₁ =-15J
Work donp by force of friction
W₂ = (μ mg cosθ)s cos180°
W₂ = 0.15 × 0.3 × 10 cos30° × 10 (-1)
W₂ =-3.879 J
Work done by external force
W₃ = F_ext × s × cos0°
W₃ = [mg sinθ + μ mg cosθ] × 10 × 1
W₃ = 18.897 J

Downward journey

mg sin30°> μ mg cos30°
Work done by the gravitational force
W₄ = mg sin 30° × scos0°
W₄ = 0.3 × 10 × \(\frac{1}{2}\) × 10 = +15J
Work done by the frictional force
W₅ = μmg cos30° × s cos180°
= 0.15 × 0.3 × \(\frac{10 \sqrt{3}}{2}\) × 10 × (-1) = – 3.897 J
(i) Work done by gravitational force over the round trip
= W₁ + W₄ = 0J
(ii) Work done by applied force over upward journey
= W₃ = 18.897J
(iii) Work done by frictional force over the round trip
W₂ + W₅ = – 3.897 + (-3.897) = – 7.794 J
(iv) Kinetic energy of the body at the end of the trip
W₄ + W₅ = 11.103 J

Question 2.
Prove that when a particle suffers an oblique elastic collision with another particle of equal mass anil initially at rest, the two particles would move in mutually perpendicular directions after collisions.
Solution:
Let a particle A of mass m and having velocity u collides with particle B of equal mass but at rest. Let the collision be oblique elastic collision and after collision the particles A and B move with velocities υ₁ and υ₂ respectively inclined at an angle 0 from each other.

Applying principle of conservation of linear momentum, we get
mu = mυ₁ +mυ₂ or u = υ₁+ υ₂
or u² = (υ₁ + υ₂) – (υ₁ + υ₂)
= υ₁² + υ₂² + 2υ₁υ₂cos0 ………….. (i)
Again as total KE before collision = Total KE after collision
∴ \(\frac{1}{2}\) mu² = \(\frac{1}{2}\)mυ₁² + \(\frac{1}{2}\)mυ₂²
⇒ u² = υ₁² + υ₂² ……………. (ii)
Comparing eqs. (i) and (ii), we get 2υ₁υ₂ cosθ = 0
As in an oblique collision both υ₁ and υ₂ are finite, hence cos0 = 0
⇒ θ = cos^-1(0) = \(\frac{\pi}{2}\)
Thus, particles A and B are moving in mutually perpendicular directions after the collision.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom

Very short answer type questions

Question 1.
What are coelomates? Give two examples.
Answer:
Animals possessing coelom are called coelomates, e.g., Annelids, molluscs, arthropods, etc.

Question 2.
Body cavity is the cavity present between body wall and gut wall. In some animals the body cavity is not lined by mesoderm. Such animals are called: [NCERT Exemplar]
Answer:
Pseudocoelomates.

Question 3.
In some animal groups, the body is found to be divided into compartments with atleast some organs/organ repeated. Name this characteristic feature. [NCERT Exemplar]
Answer:
The segmentation that simultaneously divides body both externally and internally is called metamerism.

Question 4.
Name the group which lack digestive tract.
Answer:
Porifera.

Question 5.
Name an animal having canal system and spicules. [NCERT Exemplar]
Answer:
Scypha (sycon)/Leucosolenia.

Question 6.
Give an example of animal which exhibit alternation of generation. [NCERT Exemplar]
Answer:
Obelia, shows alternation of asexual and sexual phases.

Question 7.
Name the animal which exhibits the phenomenon of bioluminescence. Mention the phylum to which it belongs.
[NCERT Exemplar]
Answer:
Pleurobrachia, phylum – Ctenophora.

Question 8.
How is mesoglea different in ctenophores that in cnidarians?
Answer:
The mesoglea in ctenophores also contains amoebocytes, elastic fibres and muscle cells.

Question 9.
What is the role of radula in Mollusca? [NCERT Exemplar]
Answer:
It is the rasping organ that help in feeding.

Question 10.
Identify the animal in which adults exhibit radial symmetry and larvae exhibit bilateral symmetry. [NCERT Exemplar]
Answer:
In Echinodermata, the adults are radially symmetrical and larvae are bilaterally symmetrical.

Question 11.
Why are urochordates called tunicates?
Answer:
This is because the soft body of urochordates is surrounded by a thick test or tunic, often transparent or translucent.

Question 12.
In which fishes cartilaginous skeleton is present?
Answer:
Chondrichthyes.

Question 13.
Which amphibians show branchial (gills) respiration?
Answer:
Young ones of most of the amphibians which are aquatic show branchial respiration.

Question 14.
What is the importance of pneumatic bones and air sacs in Aves? [NCERT Exemplar]
Answer:

• Pneumatic bones are light but strong, the feature which helps in flight,
• Air sacs increase the efficiency of respiration and provide buoyancy to the animal.

Short answer type questions

Question 1.
Define tissue level of organization in the Kingdom Animalia. Give some examples of organisms showing tissue level of organization.
Answer:
In tissue level of organization there are specified groups of cells to carry out specific functions. This is a start of division of labour in the Animal Kingdom. Examples: Aurelia and Hydra.

Question 2.
What are diploblastic and triploblastic organisation?
Answer:
Animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm, are called diploblastic animals, e.g., coelenterates. An undifferentiated layer, mesoglea, is present in between the ectoderm and the endoderm. Those animals in which the developing embryo has a third germinal layer, mesoderm, in between the ectoderm and endoderm, are called triploblasdc animals (platyhelminthes to chordates).

Question 3.
Write two characters on the basis of which you can say that Coelenterata is more evolved than Porifera.
Answer:
(a) Porifera shows cellular level of organization, while Coelenterata shows tissue level of organisation.
(b) All members of Porifera are sessile, i.e., they are attached to the substratum, while some members of Coelenterata are motile, showing further improvement.

Question 4.
On the basis of which characters you can say that Aschelminthes are more advanced compared to Platyhelminthes?
Answer:
(a) Platyhelminthes are acoelomate,while Aschelminthes are pseudocoelomates. This indicates development of mesoderm.
(b)In Platyhelminthes both sexes are on the same animal, while in Aschelminthes there is segregation of sexes. This shows another point of evolution.

Question 5.
Give a description of the water vascular system in Echinodermata..
Answer:
Water Vascular System in Echinodermata: The water vascular system is a hydraulic system used by echinoderms, such as starfish and sea urchins, for locomotion, food and waste transportation, and respiration. The system is composed of canals connecting numerous tube feet. Echinoderms move by alternately contracting muscles that force water into the tube feet, causing them to extend and push against the ground, then relaxing to allow the feet to retract.

Long answer type questions

Question 1.
Differentiate between polyp and medusa. [NCERT Exemplar]
Answer:
Differences between polyp and medusa are as follows :

Polyp	Medusa
1. It is a fixed zooid.	It is free swimming.
2. It is asexual.	It is sexual.
3. It is cylindrical in outline.	Medusa is umbrella-shaped.
4. Tentacles found at upper end of manubrium.	Tentacles occur along the margin of umbrella.
5. Mouth is circular and terminal over upright manubrium.	Mouth is four sided, lies at the lower end of hanging manubrium.
6. Velum and sense organs are absent.	Medusa have a-circular velum and eight sense organs or statocysts.

Question 2.
Differentiate between Chordates and Non-Chordates.
Answer:
Differences between chordates and non-chordates are given below:

Chordates	Non-Chordates
•» Notocord is present at stages in some stages of development.	Notochord is absent.
•» Central nervous system is dorsal, hollow, single and non-ganglionated.	Central nervous system is ventral, solid, double and ganglionated.
•» Gill slits present on lateral side of pharynx in sum stages of throughout life.	Gill slits are absent.
•» Tail is present in some stages and throughout life.	Tail generally absent.
•» Heart is ventral.	Heart is dorsal.
•» Haemoglobin is present in RBCs.	It is present in plasma.

Question 3.
Which features make mammals as most successful and dominant animals?
Answer:
Features which make mammals as dominant and successful animals are as follows:

• The presence of an insulating and protective hairy exoskeleton
• They are warm blooded so have high rate of metabolism.
• They are viviparous animals and show placentation and intrauterine development which increases the chances for survival of young ones.
• They show high degree of parental care.
• They have more developed hearing efficiency due to the presence of pinna, three ear-ossicles and coiled cochlea in the ear.
• They are able to speak through language.
• They have good power of learning due to the presence of more developed brain.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom

Very short answer type questions

Question 1.
Name the types of classification of plants.
Answer:
Artificial, natural and phylogenetic.

Question 2.
Which system indicates evolutionary as well as genetic relationships among organisms?
Answer:
Phylogenetic system of classification.

Question 3.
What is cytotaxonomy?
Answer:
Cytotaxonomy is a method of classification. It is based on cytological structure and their relatedness.

Question 4.
Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
[NCERT Exemplar]
Answer:
Phaeophyceae.

Question 5.
Holdfast, stipe and frond constitutes the plant body in case of
[NCERT Exemplar]
(a) Rhodophyceae
(b) Chlorophyceae
(c) Phaeophyceae
(d) All of these
Answer:
The lamina of Phaeophyceae members large sized body with differentiation of holdfast, stipe and lamina.

Question 6.
The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorptions. What is the equivalent of the roots in the less developed lower plants? [NCERT Exemplar]
Answer:
Rhizoids.

Question 7.
Most algal genera show haplontic life cycle. Name an alga which is (i) Haplodiplontic (ii) Diplontic [NCERT Exemplar]
Answer:
Haplodiplontic: Ulva, Dictyota.
Diplontic: Fucus, Cladophora, Glomerata.

Question 8.
How are mosses considered ecologically important?
Answer:
Mosses, along with lichens are the first organism to colonise rocks, hence are ecologically important.

Question 9.
A prothallus is
(a) a structure in pteridophyte formed before the thallus develops
(b) a sporophytic free-living structure formed in pteridophyte
(c) a gametophytic free-living structure formed in pteridophytes
(d) a primitive structure formed after fertilisation in pteridophyte
[NCERT Exemplar]
Answer:
Gametophyte is a free-living small thalloid structure called prothallus. In most ferns, the prothallus is green and autotrophic.

Question 10.
Where are seeds located in gymnosperm?
Answer:
Seeds lie naked or exposed on the surface of megasporophyll.

Question 11.
The embryo sac of an angiosperm is made up of:
(i) 8 cells
(ii) 7 cells and 8 nuclei
(iii) 8 nuclei
(iv) 7 cells and 7 nuclei [NCERT Exemplar]
Answer:
(ii) Embryo sac of angiosperm develops up to 8 nucleate state prior to fertilisation. There is a three celled egg apparatus, three antipodal cells and two polar nuclei.

Question 12.
What is alternation of generations?
Answer:
Alernation of generations is regular switch over from gamete bearing haploid gametophyte to haploid spore producing diploid sporophyte.

Short answer type questions

Question 1.
What are the main differences among Chlorophyceae, Phaeophyceae and Rhodophyceae?
Answer:

Question 2.
What is the general structure of bryophytes?
Answer:
Structure of Bryophytes: It is thallus-like and prostrate or erect, and attached to the substratum by unicellular or multicellular rhizoids. They lack true roots, stem or leaves. They may possess root-like, or stem-like structures.

Question 3.
What is the general structure of pteridophytes?
Answer:

• The main plant body is a sporophyte which is differentiated into true root, stem and leaves. These organs possess well-differentiated vascular tissues.
• The leaves in pteridophyta are small (microphylls) as in Selaginella or large (macrophylls) as in ferns.
• The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases sporophylls may form distinct compact structures called strobili or cones (Selaginella, Equisetum).

Question 4.
Write short notes on:
(a) Importance of carbon fixation by algae.
(b) Importance of Gymnosperms
(c) Importance of Angiosperms
(d) Medicinal use of algae
Answer:
(a) About 50% of carbon fixation is done by algae. This enables majority of sea organisms to get the required food.
(b) Gymnosperms are mainly used as decorative plants. Certain paints are
prepared from Gymnosperm plants.
(c) Angiosperms are the major providers of food-grains to the mankind.
(d) Spirullina is made by algae and is used as a nutritional supplement.

Long answer type questions

Question 1.
Algae are known to reproduce asexually by a variety of spores under different environmental condition. Name these spores and the conditions under which they are produced. [NCERT Exemplar]
Answer:
Zoospores: Flagellate spores, under favourable conditions.
Aplanospores: Non-flagellate, thin-walled spores under approaching unfavourable conditions.
Hypnospores: Thick-walled, resting spores in unfavourable conditions.
Akinetes: Thin-walled and thick-walled spores formed from whole cells in unfavourable conditions. .
Autospores: Spores which look exactly like parent cell formed under favourable conditions.

Question 2.
Write about habit and habitat of algae.
Answer:
Habit and Habitat of Algae: Algae are chlorophyll-bearing, simple, thalloid, autotrophic and largely aquatic (both fresh water and marine) organisms. They occur in a variety of other habitats : moist stones, soils and wood. Some of them also occur in association with fungi (lichen) and animals (e.g., on sloth bear).

Question 3.
What are the differences betweenpinus and cycas?
Answer:

Pinus	Cycas
1. Roots are micorrhizal.	1. Roots are not micorrhizal.
2. Stems are branched.	2. Stems are unbranched.
3. Male and female strobili are on same tree.	3. They are on different trees.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 20 Locomotion and Movement Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 20 Locomotion and Movement

Very short answer type questions

Question 1.
Give the name of the cells/tissues in human body which:
(i) exhibit amoeboid movement.
(ii) exhibit ciliary movement. [NCERT Exemplar]
Answer:
(i) Macrophages,
(ii) Ciliated, epithelium of trachea.

Question 2.
Which property of muscles is used effectively in muscular movement?
Answer:
Contractile property of muscles.

Question 3.
Give the name of the oxygen-carrying pigment present in skeletal muscle.
Answer:
Myoglobin or muscle hemoglobin.

Question 4.
Label the different components of actin filament in the diagram given below: [NCERT Exemplar]
Answer:

Question 5.
What causes muscle fatigue?
Answer:
Accumulation of lactic acid.

Question 6.
The three tiny bones present in middle ear are called ear ossicles. Write them in correct sequence beginning from eardrum. [NCERT Exemplar]
Answer:
Malleus, incus and stapes.

Question 7.
What is the difference between the matrix of bone and cartilage? [NCERT Exemplar]
Answer:
The matrix of bone is hard due to calcium salts, whereas, the cartilage has slightly pliable matrix due to chondroitin salts.

Question 8.
How many total bones are there in human body? Name the largest and strongest bone.
Answer:
Human body contains 206 bones. Femur is the largest and strongest bone of human body.

Question 9.
Give the name of the cavity in the girdle to which head of femur articulates.
Answer:
Acetabulum.

Question 10.
Give the name of the funny bone.
Answer:
Olecranon process on top of the ulna is called the funny bone.

Question 11.
Give the location of ball and socket joint in a human body. [NCERT Exemplar]
Answer:
Shoulder joint (between pectoral girdle and head of humerus).

Question 12.
What substance is responsible for lubricating the freely movable joint at the shoulder? ,
Answer:
Synovial fluid.

Short answer type questions

Question 1.
Explain anaerobic breakdown of glycogen in muscles and its effect.
Answer:
Anaerobic Breakdown of Glycogen: The reaction time of the fibres can vary in different muscles. Repeated activation of the muscles can lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen in them, causing fatigue.

Question 2.
Describe the structure of the human skull.
Answer:

• The skull is composed of two sets of bones-cranial and facial, that totals to 22 bones.
• Cranial bones are 8 in number. They form the hard protective outer covering, cranium for the brain.
• The facial region is made up of 14 skeletal elements which form thefront part of the skull.
• A single U-shaped bone called hyoid is present at the base of the buccal cavity and it is also included in the skull.
• Each middle ear contains three tiny bones-Malleus, Incus and Stapes, collectively called Ear Ossicles.
• The skull region articulates with the superior region of the vertebral column with the help of two occipital condyles (dicondylic skull).

Question 3.
Explain the structure of the vertebral column of human.
Answer:
Vertebral Column: Our vertebral column is formed by 26 serially arranged units called vertebrae and is dorsally placed. It extends from the base of the skull and constitutes the main framework of the trunk. Each vertebra has a central hollow portion (neural canal) through which the spinal cord passes.
First vertebra is the atlas and it articulates with the occipital condyles.

The vertebral column is differentiated into following regions starting from the skull:

1. cervical (7),
2. thoracic (12),
3. lumbar (5),
4. sacral (1-fused) and
5. coccygeal (1-fused) regions

The number of cervical vertebrae are seven in almost all mammals including human beings. The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back. Sternum is a flat bone on the ventral midline of thorax.

Question 4.
Describe the structure of the rib cage of human.
Answer:
Rib Cage: There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic.
First seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. These are called verte brochondral (false) ribs. Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are, therefore, called floating ribs. Thoracic vertebrae, ribs and sternum together form the rib cage.

Question 5.
Give a description of the appendicular skeleton in human.
Answer:
Appendicular Skeleton: The bones of the limbs alongwith their girdles constitute the appendicular skeleton. Each limb is made of 30 bones.

Bones of Limbs
Fore’ Limb	Hind Limb
Humerus,	Femur,
Radius,	Tibia,
Ulna	Fibula,
Carpals (8)	Tarsals (7)
Metacarpals (5)	Metatarsals (5)
Phalanges (14)	Phalanges (14)
	Patella

Question 6.
Write a short note on disorders of muscular and skeletal systems.
Answer:
Disorders of Muscular System

• Myasthenia gravis: It is an auto-immune disorder, affecting the neuromuscular junction leading to progressive weakening and paralysis of skeletal muscles.
• Muscular dystrophy: It is a genetic disorder resulting in progressive degeneration of skeletal muscles.
• Tetany: It refers to the continued state of contraction or wild contractions of muscles due to low Ca++ in body fluids.

Disorders of Skeletal System:

• Arthritis: Inflammation of joints.
• Osteoporosis: Age-related disorder characterized by decreased bone mass and increased chances of fractures. Decreased levels of estrogen is a common cause.
• Gout: Inflammation of joints due to accumulation of uric acid crystals.

Long answer type questions

Question 1.
Give answer for the following:
(i) Female pelvis is larger and has a broader front than male pelvis. Why?
(ii) Name the different curves of vertebral column.
(iii) What is a sesamoid bone? Name it.
(iv) Which bones have become modified to form ear ossicles?
Answer:
(i) Female pelvis is larger and has a broader front. This is an adaptation for childbirth.
(ii) Vertebral column forms four curves, i.e., cervical, thoracic, lumbar, and sacral located in the neck, thorax, abdomen, and pelvis respectively.
(iii) A bone embedded within a tendon is called a sesamoid bone, e.g., Patella which covers the knee ventrally.
(iv) Articular bone of lower jaw modifies to be malleus. Quadrate bone of upper jaw becomes incus and hyomandibular gets modified to become stapes.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification

Very short answer type questions

Question 1.
Name the organisms living in salty areas. [NCERT Exemplar]
Answer:
Halophiles live in habitats having high salinity and high light intensity.

Question 2.
Name the kingdom under which cyanobacteria have been classified.
Answer:
Cyanobacteria are prokaryotic organisms belong to kingdom – Monera.

Question 3.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement? [NCERT Exemplar]
Answer:
Some cyanobacteria live in mutually beneficial relationship with plants, obtaining food and shelter and fixing nitrogen for the plant. They also reduce soil alkalinity and improve soil texture.

Question 4.
Are chemosynthetic bacteria autotrophic or heterotrophic? [IMCERT Exemplar]
Answer:
Chemosynthetic bacteria are- autotrophs because they are able to synthesize their food from inorganic raw material with the help of energy obtained from chemical reaction.

Question 5.
Fusion of two gametes which are dissimilar in size is termed as. [NCERT Exemplar]
Answer:
Anisogamy, it occurs in Ceratium a dinoflagellate.

Question 6.
Why are cysts formed in protistans?
Answer:
Cysts formation helps to over come unfavourable condition.

Question 7.
What kind of nutrition occurs in a parasite?
Answer:
Parasites have phagotrophic and absorptive type of nutrition.

Question 8.
An association between roots of higher plants and fungi is called.
Answer:
Mycorrhiza is a symbiotic association between a fungus and the root of a plant. Mycorrhiza does not penetrate deep but remains in the superficial layers of the soil.

Question 9.
How the saprotrophic Basidiomycetes are able to decompose plant matter?
Answer:
The saprotrophic Basidiomycetes can decompose plant matter because they have enzymes for metabolising both cellulose and lignin.

Question 10.
Which are the most advanced group of fungi?
Answer:
Basidiomycetes.

Question 11.
What is capsid and how it is useful for viruses?
Answer:
Capsid is a proteinaceous covering around the virus. It protects the nucleoid from damage from physical and chemical agents.

Question 12.
Which enzyme is present in bacteriophages?
Answer:
Lysozyme is present in the region that comes in contact with host cell.

Short answer type questions

Question 1.
Write five beneficial usage or effects of bacteria.
Answer:
(a) Curdling of milk
(b) Lactobacillus is an important commensal in the gut flora of humans.
(c) Penicillin antibiotics are prepared by bacteria.
(d) Bacteria is a good decomposer, so it assists in completing the energy cycle.
(e) Rhizobium bacteria helps in nitrogen fixation.

Question 2.
What is the role of methanogens?
Answer:
Methanogens are a type of bacteria which live in the gut of ruminating animals. They assist those animals in digestion and the byproduct of that digestive process is methane. More number of livestock population results in increased methane level in the environment leading to global warming. So, indirectly methanogens can be responsible for global warming.

Question 3.
Cyanobacteria plays a major role in our ecology. Discuss.
Answer:
Cyanobacteria, also known as ‘blue-green algae’ help in carbon fixation in a major way on the ocean surface. They are helpful in nitrogen fixation in paddy fields leading to a better harvest. About 80% of photosynthesis on ocean surface is done by cyanobacteria. So, it can be said that they play a major role in our ecology.

Question 4.
Give a diagrammatic representation of classification of Protista.
Answer:

Long answer type questions

Question 1.
Answer the following:
(i) A poisonous mushroom having white spores
(ii) Edible Basidiomycetes
(iii) Used in brewing industry
(iv) It is searched by trained pets
(v) Root like, cards like hyphal masses having a distinct growing point.
(vi) Non-motile meiospores develop in Basidiomycetes
(vii) Compact groups of hyphal produced to overcome unfavourable conditions.
Answer:
(i) toadstools
(ii) mushrooms and young pufballs
(iii) brewer’s yeast Saccharomyces cerevisiae
(iv) Truffles (tuber like underground fungus)
(v) Rhizomorphs perennate during periods of scarcity of food and water.
(vi) Basidiospores (develop exogenously)
(vii) Sclerotia under favourable conditions each one forms a new mycelium.