Category: Class 11 Physics

NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-4-laws-motion/

NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion

Multiple Choice Questions
Single Correct Answer Type

Q1. A ball is travelling with uniform translatory motion. This means that
(a) it is at rest.
(b) the path can be a straight line or circular and the ball travels with uniform
(c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.
(d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.
Sol: (c) When a body moves in such a way that the linear distance covered by each particle of the body is same during the motion, then the motion is said to be translatory or translation motion.
Translatory motion can be, again of two types viz., curvilinear (shown in fig. (a)) or rectilinear (shown in fig. (b)), accordingly as the paths of every constituent particles are similarly curved or straight line paths. Here it is important that the body does not change its orientation. Here we can also define it further in uniform and non-uniform translatory motion. Here figure
(b) is uniformly translatory motion.

Q2. A metre scale is moving with uiiiform velocity. This implies
(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.
(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.
(c) the total force acting on it need not be zero but the torque on it is zero.
(d) neither the force nor the torque need to be zero.
Sol: (b)
Key concept: To solve these types of problem we have to apply Newton’s second law of motion!
Newton’s Second Law of Motion
According to this law: The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the force applied.

takes place always in the direction of the force applied.
We know that F = df/ dt
According to the question that the meter scale is moving with uniform velocity, hence, change in momentum will be zero, i.e. dp = 0
This implies momentum will remains same. So, Force = F= 0.
So, we can say that all parts of the meter scale is moving with uniform velocity because total force is zero and if there is any torque acting on the body this means that the body will be in rotational motion which means that the direction of velocity will be changing continuously. So, the torque acting about centre of mass of the scale is also zero.

Important point: Change in velocity = final velocity – initial velocity

Q5. Conservation of momentum in a collision between particles can be understood from
(a) Conservation of energy
(b) Newton’s first law only
(c) Newton’s second law only
(d) Both Newton’s second and third law
Sol: (d)
Key concept: If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.

This equation shows that in absence of external force for a closed system the linear momentum of individual particles may change but their sum remains unchanged with time.
Conservation of linear momentum is equivalent to Newton’s third law of motion.
For a system of two particles in absence of external force by law of conservation of linear momentum.

i.e., for every action there is equal and opposite reaction which is Newton’s third law of motion.
In case of collision between particles equal and opposite forces will act on individual particles by Newton’s third law.
Hence total force on the system will be zero.

Important point: We should not confuse with system and individual particles. As total force on the system of both particles is zero but force acts on individual particles.
Law of conservation of linear momentum is independent of frame of reference though linear momentum depends on frame of reference.

Q6. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is
(a) frictional force along westward (b) muscle force along southward (c) frictional force along south-west (d) muscle force along south-west
Sol: (c)
Key concept: According to Newton’s second law of motion only external forces can change linear momentum of the system. The internal forces cannot change linear momentum of system under consideration. If we take hockey player as a system, the external force which can change the direction of motion of the player is the force must be friction between the ground and shoes of player. The muscle force is the internal force, this cannot change the linear momentum of the player. According to Newton’s Second Law, The rate of change of linear momentum of a body is equal to the external force applied on the body or F = dp/dt .So, the external force must be in the direction of change in momentum.

According to the problem, mass = 2 kg

Position of the particle is given here as a function of time, x(t) =pt + qt² + rt³ By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

v = dx/dt = p + 2 qt + 3 rt²

If we again differentiate this equation w.r.t. time, we will get acceleration of the particle as a function of time.

a  = dv/dt = 0+ 2q + 6rt

At t = 2 s; a = 2q + 6x2xr
= 2 q+ 12r
= 2×4+12×5
= 8 + 60 = 68 m/s
Force =F=ma
= 2×68= 136 N






Let us assume the eastward
direction as x-axis.
A car is able to move towards due to friction acting between its tyres and the road.
The force of friction of the road on the tyre acts in the forward direction and is equal but in the opposite direction to the force of friction of the tyre on the road.
Mass of the car = m
As car starts from rest, its initial velocity u = 0 Velocity acquired along east = vi
Time interval (in which car acquired that velocity) t = 2 s.
As acceleration is uniform, so by applying kinematic equation (v = u + at), we get

More Than One Correct Answer Type

Q10. The motion of a particle of mass m is given by x – 0 for t < 0 s, x(t) = A sin 4πt for 0 < t< (1/4) s (A > 0),
and x = 0 for / > (1/4) s.

Which of the following statements is true?

• The force at t = (1/8) s on the particle is -16π²A-m.
• The particle is acted upon by an impulse of magnitude 4/rA-m at t = 0 s and t = (1/4) s.
• The particle is not acted upon by any force.
• The particle is not acted upon by a constant force.
• There is no impulse acting on the particle.

Sol:

Q11. In figure the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m/2 and of B is m. Which of the following statements are true?

(a) The bodies will move together if F = 0.25 mg.
(b) The body A will slip with respect to B if F = 0.5 mg.
(c) The bodies will move together if F = 0.5 mg.
(d) The bodies will be at rest if F = 0.1 mg.
(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

Sol: (a, b, d, e)

Q12. Mass m, moves on a slope making an angle θ with the horizontal and is attached to mass m₂ by a string passing over a frictionless pulley as shown in figure. The coefficient of friction between m_u and the sloping surface is µ. Which of the following statements are true?

Sol: (b, d)
Key concept: When a mass mx is placed on a rough inclined plane: Another mass m2 hung from the string connected by ftictionless pulley, the tension IT) produced in string will try to start the motion of mass w,.
At limiting condition,

Simplified situation is shown in the diagram.
Let m1 moves up the plane. Different forces involved are shown in the diagram.

Q13. In figure a body A of mass m slides on a plane inclined at angle θ₁ to the horizontal and µ is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body 5, also of mass m, sliding on a frictionless plane inclined at an angle θ₂ to the horizontal. Which of the following statements are true?

Q14. Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m s^-1 each, collide and rebound with the same speed. If the collision lasts for 10 ^-3 s, which of the following statements are true?
The impulse imparted to each ball is 0.25 kg-ms ¹ and the force on each ball is 250 N.
(a) The impulse imparted to each ball is 0.25 kg-ms ¹ and the force exerted on each ball is 25 x 10 ⁵
(b) The impulse imparted to each ball is 0.5 N-s.
(c) The impulse and the force on each ball are equal in magnitude and opposite in directions.

Q15. Abody of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is

Sol: (a, c) Recall the concept of resultant of two vectors, when they are perpendicular

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We hope the NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-3-motion-plane/

NCERT Exemplar Problems Class 11 Physics Chapter 3 Motion in a Plane

Multiple Choice Questions
Single Correct Answer Type

Motion In A Plane Class 11 Numericals With Solutions 

Sol: (b)
Key concept: Scalar Product of Two Vectors:
(1) Definition : The scalar product (or dot product) of two vectors is defined as the product of the magnitude of two vectors with cosine of angle between them.





Q2. Which one of the following statements is true?
(a) A scalar quantity is the one that is conserved in a process.
(b) A scalar quantity is the one that can never take negative values.
(c) A scalar quantity is the one that does not vary from one point to another in space.
(d) A scalar quantity has the same value for observers with different orientation of the axes.
Sol: (d) A scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.
For example, a car is traveling along +x axis, it travels 10 m. If the same car is moving with the same speed for the same time interval along -x axis, then the distance meter of car shows the same travelled distance. The path length is same in both the cases.

Q3. Figure shows the orientation of two vectors u and v in the TY-plane.

which of the following is correct?
(a) a and p are positive while b and q are negative
(b) a, p and b are positive while q is negative
(c) a, q and b are positive while p is negative
(d) a, b, p and q are all positive

Here it is worthy to note once a vector is resolved into its components, the components themselves can be used to specify the vector as:

In such type of problems we have to resolve the rectangular components according to the diagram.

Q4. The component of a vector r along X-axis will have maximum value if
(a) r is along positive Y-axis
(b) r is along positive X-axis
(c) r makes an angle of 45° with the X-axis
(d) r is along negative Y-axis

Q5. The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(a) 60 m (b) 71m (c) 100 m (d) 141m

Q6. Consider the quantities pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are
(a) Impulse, pressure and area
(b) Impulse and area
(c) Area and gravitational potential
(d) Impulse and pressure
Sol: (b) We know that impulse J = F. ∆t = ∆p, where F is force, At is time duration and Ap is change in momentum. As ∆p is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector direction of area vector is perpendicular to its plane.
Q7. In a two dimensional motion, instantaneous speed ==vo== is a positive constant. Then, which of the following are necessarily true?
(a) The average velocity is not zero at any time
(b) Average acceleration must always vanish
(c) Displacements in equal time intervals are equal
(d) Equal path lengths are traversed in equal intervals

Sol :(d) Speed (Instantaneous Speed): The magnitude of the velocity at any instant of time is known as Instantaneous Speed or simply speed at that instant of time. It is denoted by v.
Quantitatively: Speed = distance/ time
Mathematically, it is the time rate at which distance is being travelled by the particle.
• Speed is a scalar quantity. It can never be negative (as shown by speedometer of our vehicle).
• Instantaneous speed is the speed of a particle at a particular instant of time.
Hence, Total distance travelled = Path length = (speed) x time taken Important point: We should be very carefttl with the fact that speed is related with total distance covered not with displacement.

Q8. In a two-dimensional motion, instantaneous speed v₀ is a positive constant. Then, which of the following are necessarily true?
‘(a) The acceleration of the particle is zero.
(b) The acceleration of the particle is bounded.
(c) The acceleration of the particle is necessarily in the plane of motion.
(d) The particle must be undergoing a uniform circular motion.
Sol. (c) This motion is two dimensional and given that instantaneous speed v₀  is positive constant. Acceleration is defined as the rate of change of velocity (instantaneous speed), hence it will also be in the plane of motion.

Sol : (c) These types of problems can be solved by hit and trial method by picking up options one by one






More Than One Correct Answer type

Q11. Two particles are projected in air with speed v₀ at angles θ₁ and θ ₂ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices.

(a) Angle of projection: θ₁ > θ₂
(b) Time of flight: T₁ > T₂
(c) Horizontal range: R_x> R₂                  
(d) Total energy: U₁> U₂

Q12. A particle slides down a frictionless parabolic (y = x²) track (A – B – C)
starting from rest at point A (figure). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) height at P = height at A
(c) total energy at P = total energy at A
(d) time of travel from AtoB


Solution : (c)

Key concept: In such type of problems, we have to observe the nature of track that if there is a friction or not, as friction is not present in this track, total energy of the particle will remain constant throughout the journey.
According to the problem, the path traversed by the particle on a frictionless track is parabolic, is given by the equation y = x², thus total energy (KE + PE) will be same throughout the journey.
Hence, total energy at A = total energy at P
At B the particle is having only KE but at P some KE is converted to PE.
So, (KE)_B  > (KE)_P
Total energy at A = PE = Total energy at B = KE = Total energy at P = PE + KE
The potential energy at A is converted to KE and PE at P, hence (PE)P < (PE)A
Hence, (Height) P < (Height) A
As, Height of P < Height of A
Hence, path length AB > path length BP
Hence, time of travel from A to B ≠ Time of travel from B to P.

Q14. For a particle performing uniform circular motion, choose the correct
statement(s) from the following.
(a) Magnitude of particle velocity (speed) remains constant.
(b) Particle velocity remains directed perpendicular to radius vector.
(c) Direction of acceleration keeps changing as particle moves.
(d) Angular momentum is constant in magnitude but direction keeps changing.

Sol: (a, b, c) While a particle is in uniform circular motion. Then the following statements are true.

(i) speed will be always constant throughout.
(ii) velocity will be always tangential in the direction of motion at a particular point.
(iii) the centripetal acceleration a = v²/r and its direction will always towards centre of the circular trajectory.
(iv) angular momentum (mvr) is constant in magnitude and direction. And its direction is perpendicular to the plane containing r and v.
Important point: In uniform circular motion, magnitude of linear velocity and centripetal acceleration is constant but direction changes continuously.

Very Short Answer Type Questions

Q16 .A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown in figure. If he maintains constant speed of 10 ms”¹, what is his acceleration at point R in magnitude and direction?

Ans: According to the problem the path of the cyclist is O-P- R-Q-O.

The cyclist is in uniform circular motion and it is given that linear velocity = 10 m/s, R = 1 km = 1000 m. As we know whenever an object is performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.So cyclist experiences a centripetal force (acceleration) at point R towards centre.

Q17. A particle is projected in air at some angle to the horizontal, moves along parabola as shown in figure where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.

Sol: In projectile motion horizontal component of velocity will always be constant and acceleration is always vertically downward and is equal to g. Direction of velocity will always be tangential to the curve in the direction of motion.
As shown in the diagram in which a particle is projected at an angle 0.


Q18. A ball is thrown from a roof top at an angle
of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) greatest speed
(b) smallest speed
(c) greatest acceleration Explain.

Ans: In this problem total mechanical energy of the ball is conserved. As the ball is projected from point O, and covering the path OABC.
At point A it has both kinetic and potential energy.
But at point C it have only kinetic energy, (keeping the ground as reference where PE is zero.)
(a) At point B, it will gain the same speed u and after that speed increases and will be maximum just before reaching C.
(b) During upward journey from OtoA speed decreases and smallest speed attained by it is at the highest point, i.e., at point A.
(c) Acceleration is always constant throughout the journey and is vertically downward equal to g.

Q19. A football is kicked into the air vertically upwards. What is its (a) acceleration and (b) velocity at the highest point?

Sol: (a) The situation is shown in the diagram below in which a football is kicked into the air vertically upwards. Acceleration of the football will always be vertical downward and is called acceleration due to gravity (g).
(b) When the football reaches the highest point it is momentarily at rest and at that moment its velocity will be zero as it is continuously retarded by acceleration due to gravity (g).

Key concept: Collinear vectors: When the vectors under consideration can share the same support or have a common support then the considered vectors are collinear.
Coplanar vectors: Three (or more) vectors are called coplanar vector if they lie in the same plane. Two (free) vectors are always coplanar.

Q21. A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.
Sol: With respect to the observer standing on the footpath ball is thrown with velocity u at an angle θ with the horizontal, hence it seems as a projectile. So path of the ball will be parabolic. The horizontal speed of the ball is same as that of the car, therefore, ball as well car travels equal horizontal ult distance. Due to its vertical speed, the ball follows a parabolic path.



Important point: We must be very clear that we are working with respect to ground. When we observe with respect to the car, motion will be along vertical direction only.

Q22. A boy throws a ball in air at 60° to the horizontal along a road with a speed of 10 m/s (36 km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of 18 km/h. Give explanation to support your diagram.
Sol: The situation is shown in the below diagram.

According to the problem the boy standing on ground throws the ball at an angle of 60° with horizontal at a speed of 10 m/s.

Speed of the car = 18 km/h = 5 m/s
As horizontal speed of ball and car is same, hence relative velocity of ball w.r.t car in the horizontal direction will be zero.
Only vertical motion of the ball will be observed by the boy in the car, as shown in above diagram.

Q23. In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.
Sol:

When we are dealing with projectile motion generally we neglect the air resistance. But if air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.
Due to air resistance, particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.
When we are neglecting air resistance path was symmetric parabola (OAC). When air resistance is considered path is asymmetric parabola (OAB).

Short Answer Type Questions
Q24. A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target ?


2. When an object is dropped/released by any moving vehicle. Then initial velocity of the object is same as the moving vehicle.

When the bomb is dropped from Plane the plane which is moving horizontally. So, the bomb will have same initial velocity as that of plane along horizontal direction.
The situation is shown in the diagram below. Let a fighter plane, when it be plane at position P, drops a bomb to hit a target T.
Let the target is seen at an angle θ with horizontal.

Important point: Angle is with respect to target. As seen by observer in the plane, motion of the bomb will be vertically downward below the plane. .

Q25. (a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of the earth due to the earth rotation (period 1 day). What is the acceleration of object on the surface of the earth (at equator) towards its centre? What is it at latitude 9? How does these accelerations compare with g = 9.8 m/s²? (b) Earth also moves in circular orbit around the sun once every year with an orbital radius of 1.5 x 10¹¹ What is the acceleration of the earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s²?

Q26. Given below in Column I are the relations between vectors a, b and c and in Column II are the orientations of a, b and c in the AY-plane. Match the relation in Column I to correct orientations in Column II.


Sol: We apply triangulr law of addition
Triangular law of vector addition: Two vectors are considered as two sides of a triangle taken in the same order. The third side or completing side of the triangle is the resultant taken in the opposite order.
or
We can say that vectors are arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition.

As shown in the diagram below in which vectors A and B are corrected by head and tail. Resultant vector C = A + B
(a) from (iv), it is clear that c = a + b
(b) from (iii), c + b = a => a- c = b
(c) from (i), b = a + c => b-a = c
(d) from (ii), -c = a + b => a + b + c = 0

Long Answer Type Questions

Q29. A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s, so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g= 10 m/s².
Sol.According to the problem, speed of packets =125 m/s , height of the hill = 500 m, distance between the cannon and the foot of the hill, d = 800 m

To cross the hill in shortest time, then the vertical component of the velocity should be minimum so that it just crosses the height of hill.

Distance through which canon has to be moved = 800 – 750 = 50 m Speed with which canon can move = 2 m/s

Important point: We should not confuse with the positive direction of motion. May be vertically upward direction or vertically downward direction is taken as positive according to convenience. And this problem can also be solved by taking motion from point P to T. From point P in the diagram projection at speed v0 at an angle θ below horizontal with height h and horizontal range ∆x. Then this will be analyzed from the alongside diagram

Q31. A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (figure).
(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
(b) Time of flight.
(c) β at which range will be maximum

Q32. A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.

Q33. A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?
Sol: V_rgis the velocity of rain appears to the girl.
We must draw all vectors in the reference frame of ground-based observer.

Q34. A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s (figure).
(a) If swimmer, starts swimming due north, what will be his resultant velocity (magnitude and direction)?
(b) If he wants to start from point A on south bank –
and reach opposite point B on north bank,
(i) which direction should he swim?
(ii) what will be his resultant speed?
(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

Q32. A cricket fielder can throw the Cricket ball with a speed v₀. If he throws the ball while running with speed u at an angle θ to the horizontal, find
(a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find θ at which he should throw the ball that would maximize the horizontal range as found in (c).
(e) how does θ for maximum range change if u > v₀, u = v₀, u < v₀?
(f) how does θ in (e) compare with that for u = 0 (i.e., 45°)?
Sol:. The observer on ground (spectator) observes that the x-component of ball is more because of the speed of fielder. As shown in the adjacent diagram,
So, initial velocity in x-direction

Q37. A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m x 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Solu:

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We hope the NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-2-motion-straight-line/

NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line

Multiple Choice Questions
Single Correct Answer Type
Motion In A Straight Line Numericals NCERT Class 11

Q1. Among the four graphs shown in the figure, there is only one graph for which average velocity over the time interval (0, 7) can vanish for a suitably chosen T. Which one is it?

Key concept: Average velocity : It is defined as the ratio of displacement to time taken by the body.
Displacement /Time taken
According to this problem, we need to identify the graph which is having same displacement for two timings. When there are two timings for same displacement, the corresponding velocities should be in opposite directions.
As shown in graph (b), the first slope is decreasing that means particle is going in one direction and its velocity decreases, becomes zero at highest point of curve and then increasing in backward direction. Hence the particle
return to its initial position. So, for one value of displacement there are two different points of time and we know that slope of x, x-t graph gives us the average velocity. Hence, for one time, slope is positive then average velocity is A also positive and for other time slope is negative then average velocity is also negative.

As there are opposite velocities in the interval 0 to T, hence average velocity can vanish in (b).
This can be seen in the figure given alongside.
As shown in the graph, OA = BT (same displacement) for two different points of time.

Important points:
Various position-time graphs and their interpretation

1. Graph: Line parallel to time axis

Interpretation: It represents that the particle is at rest.

2. Graph: Line perpen¬dicular to time axis

Interpretation: It represents that particle is changing its position but time does not change, it means the particle possesses infinite velocity.
This situation is practically not possible.

3. Graph: Line with constant slope

Interpretation: It represents uniform velocity of the particle.

4. Graph: Parabola bending towards position axis

Interpretation: It represents increasing velocity of the particle. It means the particle possesses acceleration.
Hence slope of position-time graph goes on increasing.

5. Graph: Parabola bending towards time axis

Interpretation: It represents decreasing velocity of particle. It means the particle possesses retardation.
Hence slope of position-time graph goes on decreasing.

6. Graph: Line with negative slope

Interpretation: It represents that the particle returns towards the point of reference (negative displacement) with uniform velocity.

Problems On Motion In A Straight Line NCERT Class 11 

2. A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
(a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0
(c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a < 0
Sol:(a)
Key concept: The time rate of change of velocity of an object is called acceleration of the object.
It is a vector quantity. Its direction is same as that of change in velocity (Not of the velocity).
In the table: Possible ways of velocity change

When only direction of velocity changes	When only magnitude of velocity changes	When both magnitude and direction of velocity change
Acceleration perpendicular to velocity	Acceleration parallel or anti­parallel to velocity	Acceleration has two components—one is perpendicular to velocity and another parallel or anti­parallel to velocity
E.g.: Uniform circular motion	E.g.: Motion under gravity	E.g: Projectile motion

Here we will take upward direction positive. As. the lift is coming in downward direction, the displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion displacement will be negative. When the lift reaches 4th floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature. Thus, x < 0; a > 0. Asdisplacementisinnegativedirection, velocity will also be negative, i.e. v < 0.
The motion of lift will be shown like this.

Motion In A Straight Line Problems NCERT Class 11

Q3. In one dimensional motion, instantaneous speed v satisfies 0 < v < v₀
(a )The displacement in time T must always take non-negative values.
(b) The displacement x in time T satisfies -v₍₎T < x < v₀
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.
Sol: (b) .
Key concept: Instantaneous speed: It is the speed of a particle at a particular instant of time. When we say “speed”, it usually means instantaneous speed. The instantaneous speed is average speed for infinitesimally small time interval (i.e., ∆ –> 0).

As instantaneous speed is less than maximum speed. Then either the velocity is increasing or it is decreasing. For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is v₀, magnitude of maximum velocity in opposite direction is also v₀.
Maximum displacement in one direction = v₀T Maximum displacement in opposite directions = -v₀T Hence,-v₀T<x< v₀T.
Important point: We should not confuse with direction of velocities, i.e., in one direction it is taken as positive and in another direction it is taken as negative.

Motion In Straight Line Numericals NCERT Class 11 

Q4. A vehicle travels half the distance L with speed V₁ and the other half with speed v₂, then its average speed is

Let the vehicle travels from A to B. Distances, velocities and time taken are shown. To calculate average speed we will calculate total distance covered and will divide by time interval in which it covers that total distance.

Motion In A Straight Line Class 11 Numericals NCERT

Q5. The displacement of a particle is given by x = (t- 2)² where x is in metres and t in seconds. The distance covered by the particle in first 4 seconds is
(a) 4 m
(b) 8 m                    
(c) 12 m                  
(d) 16 m
Sol: (b)
Key concept: Instantaneous velocity : Instantaneous velocity is defined as the rate of change of position vector of particles with time at a certain instant of time.

i.e., if x  is given as a function of time, second time derivative of displacement gives acceleration.
In such type of problems we have to analyze whether the motion is accelerating or retarding. When acceleration is parallel to velocity, velocity of particle increases with time, i.e. motion is accelerated. And when acceleration is anti-parallel to velocity, velocity of particle decreases with time, i.e. motion is retarded. During retarding journey, particle will stop in between.
According to the problem, displacement of the particle is given as a function of time.
x = (t-2)²
By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
v = dx/ dt = d/dt ( t-2)² = 2(t – 2) m/s
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.

Class 11 Motion In A Straight Line Important Questions

Q6. At a metro station, a girl walks up a stationary escalator in time t₁ If she remains stationary on the escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be
(a) (t_l+t₂)/2 (b) t
(b) t₁t₂ /(t₂ – t₁)
(c) t₁t₂ /(t₂ +t₁)                                     
(d) t₁ —t₂
Sol: (c)

Key concept: Net velocity when object is moving on the moving frame in One Dimension:
We will define this concept by taking an example.
River-Man problem in one dimension:
Velocity of river water current is u and velocity of man in still water is v, i.e. man can swim in water with velocity v.

Problem-Solving Tips for Relative Velocity

• If the velocity is mentioned without specifying the frame, assume it is with respect to the ground.
• In many cases, a body travels on water or in air. Depending on the context you will have to figure out whether the velocity is with respect to the water/air or with respect to the ground.
• In some situations you have to presume the velocities. For example, if the problem says that a man can walk at a maximum of 8 kmlf¹ and if it asks you to find the velocity on a train, then you have to assume that the velocity of the man with respect to the surface he is on (in this case the tram is 8 kmh^-1). Similarly the velocity of a bullet is always measured with respect to the gun. If the gun is mounted on a truck, the bullet will have a different velocity.

If particle is moving with constant velocity towards right (+x-axis):

Equation to be used: x = x_Q + vt. Graph will be a straight line.

Let the particle be at some point P initially at time t – 0 which is at a distance of x₀ from origin. Since the particle is moving towards right so its distance from origin goes on increasing. Hence position-time graph for a particle moving with constant velocity towards right will be a straight line inclined to time axis making an acute angle α.
Recall that tan α is slope of position-time graph which is equal to velocity of the particle.

For uniform motion velocity is constant, hence slope will be positive. Hence quantity A is displacement.
If particle is moving with a constant positive acceleration:
Equation to be used: v = u + at
As the time passes velocity goes on increasing. Hence velocity-time graph for a particle moving with constant positive acceleration is a straight line inclined to time axis making an acute angle a. Here tan a is the slope of velocity-time graph (Figure).

For uniformly accelerated motion, slope will be positive and A will represent velocity.

Q9. A graph of x versus t is shown in figure.

Choose the correct alternatives given below.
(a) The particle was released from rest at t =0
(b) At B, the acceleration a >0
(c) At C, the velocity and the acceleration vanish.
(d) Average velocity for the motion between Aand D is positive.
(e) The speed at D exceeds that at E

Sol: (a, c, e)
Key concept: We know that velocity v = dx/dt and slope of x-t graph gives ‘
us velocity. This implies slope = dx/dt for the graph.
As per the diagram, at point A the graph is parallel to time axis, hence dx
v = dx/dt = 0. As the starting point is A, hence we can say that the particle is starting from rest. Thus option (a) is correct.
At C, the graph changes slope, hence velocity also changes. As graph at C is almost parallel to time axis, hence we can say that velocity vanishes. Hence option (c) is correct.
As direction of acceleration changes, hence we can say that it may be zero in between.
From the graph it is clear that | slope at D| > | slope at E |
Hence, speed at D will be more than at E. Hence option (e) is correct.
Important point: Here, negative slope does not mean less value. It represents change in direction of velocity.

Q9. For the one-dimensional motion, described by x = t – sin t.
(a) x(t) > 0 for all t > 0 (b) v(t) > 0 for all r > 0
(c) a(t) > 0 for all t > 0 (d) v(t) lies between 0 and 2

Sol: (a, d) Position of the particle is given as a function of time i.e. x = t – sint By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

Important points:
(i) When sinusoidal function is involved in an expression we should be careful about sine and cosine functions.
(ii) We should be very careful when calculating maximum and minimum value of velocity because it is in inverse relation with cost in the given expression.

Q10. A spring with one end attached to a mass and the other to a rigid support is stretched and released.
(a) Magnitude of acceleration, when just released is maximum.
(b) Magnitude of acceleration, when at equilibrium position, is maximum.
(c) Speed is maximum when mass is at equilibrium position.
(d) Magnitude of displacement is always maximum whenever speed is minimum
Sol:(a, c)

Q11. A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to the walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,
(a) the direction of motion of the ball changes every 10 seconds.
(b) speed of ball changes every 10 seconds.
(c) average speed of ball over any 20 second interval is fixed.
(d) the acceleration of ball is the same as from the train.
Sol: (b, c, d) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains (10 m/s) speed of ball is less (1 m/s). (b, c, d) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains (10 m/s) speed of ball is less (1 m/s).
The speed of the ball before collision with side of train is 10 + 1 = 11 m/s Speed after collision with side of train =10-1=9 m/s As speed is changing after travelling 10 m and speed is 1 m/s, hence time duration of the changing speed is 10 s.
Since, the collision of the ball is perfectly elastic there is no dissipation of energy, hence total momentum and kinetic energy are conserved.
Since, the train is moving with a constant velocity, hence it will act as an inertial frame of reference as that of Earth and acceleration will be same in both frames.
Remember: We should not confuse with non-inertial and inertial frame of reference. A frame of reference that is not accelerating will be inertial.

Very Short Answer Type Questions

Ans. (a) (iii); (b) (ii); (c) (iv);(d) -(i)
Sol: Let us pick graphs one by one.

In graph (a),
There is a point (B) on the curve for which displacement is zero. So curve, (a) matches with (iii).

In graph (b),
In this graph, x is positive (> 0) throughout and at point B the highest point of curve the slope of curve is zero. It means at
this point v = dx/dt = 0 . Also at point C the dt
curvature changes, it means at this point the acceleration of the particle should be zero or a = 0, So curve (b) matches with (ii).

In graph (c),
In this graph the slope is always negative, hence velocity will be negative or v < 0. Also x-t graph opens up, it represents positive acceleration. So curve (c) matches with (iv).

In graph (d),
In this graph the slope is always positive, hence velocity will be positive or v > 0. Also x-t graph opens down, it represents negative acceleration. So curve (d) matches with (i).

13. A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time (Take acceleration in the backward direction as positive).
Sol: Impulsive Force is generated by the bat: If we ignore the effect of gravity just by analyzing the motion of ball in horizontal direction only, then ball moving uniformly will return back with the same speed when a bat hits it.
Acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.

Q14. Give examples of a one-dimensional motion where
(a) the particle moving along positive x-direction comes to rest periodically and moves forward.
(b) the particle moving along positive x-direction comes to rest periodically and moves backward.
Sol: The equation which contains sine and cosine functions is periodic in nature.
(a) The particle will be moving along positive x-direction only if t > sin t We have displacement as a function of time, x(t) = t – sin t By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

At t = 2π, x = 0, v = 1 (positive) and a = 0
Hence the particle moving along positive x-direction comes to rest periodically and moves backward.
As displacement and velocity is involving sin t and cos t, hence these equations represent periodic nature.

Q15. Give example of a motion where x > 0, v < 0, a > 0 at a particular instant.
Sol: Let the motion is represented by

Q16. An object falling through a fluid is observed to have acceleration given by a = g – bvwhere g= gravitational acceleration and b is constant After a long time of release, it is observed to fall with constant speed. What must be the value of the constant speed?
Sol: Key concept: If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity.
According to the problem, acceleration of object is given by the relation
a=g-bv
When speed becomes constant acceleration a = dv/dt = 0 (uniform motion).
where, g = gravitational acceleration

Clearly, from above equation as speed increases acceleration will decrease. At a certain speed say v₀, acceleration will be zero and speed will remain constant. Hence, a = g- bv₀ = 0 => v₀ = g/b

Short Answer Type Questions

Q17. A ball is dropped and its displacement versus time graph is as shown (Displacement x from ground and all quantities are positive upwards).
(a) Plot qualitatively velocity versus time graph.
(b) Plot qualitatively acceleration versus time graph.

Q19. A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 km/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird?
Sol:
Concept of relative velocity (for 1-D): If two objects are moving along the same straight line and we are observing the motion from the frame of one object. Then for the relative velocity, it will be subtracted for velocities in same direction and added for velocities in opposite directions. (Remember: add or subtract them with proper sign conventions).

Q20. A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is at a lower height than the first. If his speed is 9 m/s, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building? (Take g = 10 m/s²)
Sol: Key concept: Horizontal Projectile:
When a body is projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time.

Time of flight: If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then

Horizontal range: Let R be the horizontal distance travelled by the body

We will apply kinematic one by one along downward and along horizontal. We first consider motion along horizontal and there is no horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time.
According the problem, horizontal speed of the man (u _x) = 9 m/s Horizontal distance between the two buildings = 10 m
Height difference between the two buildings = 9 m and g =10 m/s²

Horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.

Q21. A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed of 40 m/s. Calculate the relative speed of the balls as a function of time.
Sol: In motion under gravity, if the ball is released or dropped that means its initial velocity is zero. In this problem as ball is dropped, so its initial velocity will be taken as zero. We will apply kinematic equations.

Important point: Sign Convention:
Any vector quantity directed upward will be taken as positive and directed downward will be taken as negative. According to this sign convention:

(i) Displacement will be taken as positive if final position lies above initial position and negative if final position lies below initial position.
(ii) Velocity(initial or final) will be taken as positive if it is upward and negative if it is downward.
(iii) Acceleration a is always taken to be -g.
In equations of motions we replace a by -g (minus sign, because acceleration is always directed downward)
Q22. The velocity-displacement graph of a particle is shown in figure.
(a) Write the relation between v and x.
(b) Obtain the relation between acceleration and displacement and plot it.

Long Answer Type Questions

Q23. It is a common observation that rain clouds can be at about a kilometer altitude above the ground.

• If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h (g = 10 m/s²).
• A typical rain drop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits the ground.
• Estimate the time required to flatten the drop.
• Rate of change of momentum’s force. Estimate how much force such a drop would exert on you.
• Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm. (Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through)

Sol: Key concept: This problem can be solved by kinematic equations of
motion and Newton’s second law that F_ext = dp/dt will be used, where dp is change in momentum over time dt.
(a) According to the problem (h) =1 km = 1000 m and we know that the initial velocity of the ball is zero. And displacement covered by rain drop in downward direction, so we will taking h as negative. (We are neglecting the air resistance.)

(c) Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground

Q24. A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway, the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s.

Q25. A monkey climbs up a slippery pole for 3 and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 – t); 0 < t < 3 and v(t) = -(t – 3)(6 – t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20 m.

(a) At what time is its velocity maximum?
(b) At what time is its average velocity maximum?
(c) At what time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top? Sol. We have to calculate time corresponding to maximum velocity. So we first need to find the maximum velocity in this problem. To calculate maximum dv velocity we will use dv/dt=0

Q26. A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Sol: We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take difference of displacements.

Important note: We should be very careful when we are applying the equation of rectilinear motion. These equations are applicable only in case of constant acceleration.

Some important observations for motion under gravity:

• The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity, i.e., t = √(2h/g) and v = √2gh.
• In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.

Time of descent (t₁) = time of ascent (t₂) = u/g
Total time of flight T=t_x + t₂ = .2u / g

• In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
  As well as the magnitude of velocity at any point on the path is same whether the body is moving in upwards or downward direction.

• A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than the time of descent t₂^>t₁

where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will act vertically downward.
For downward ‘motion a and g will act in opposite direction because a always act in direction opposite to motion and g always act vertically downward.

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NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-1-units-measurements/

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

Single Correct Answer Type

Q1. The number of significant figures in 0.06900 is
(a) 5 (b) 4 (c) 2 (d) 3
Sol: (b)
Key concept: Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the number of significant figures in a given measured quantity.
1. All non-zero digits are significant.
2. A zero becomes significant figure if it appears between two non¬zero digits.
3. Leading zeros or the zeros placed to the left of the number are never significant.
4. Trailing zeros or the zeros placed to the right of the number are significant.
5. In exponential notation, the numerical portion gives the number of significant figures.
Leading zeros or the zeros placed to the left of the number are never

Hence, number of significant figures are four.

Q2. The sum of the numbers 436.32, 227.2 and 0.301 inappropriate significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Sol: (b) The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

The final result should, therefore, be rounded off to one decimal place, i.e. 664.

Q3. The mass and volume of a body are 4.237 g and 2.5 cm³, respectively. The density of the material of the body in correct significant figures is
(a) 1. 6048 g cm^-3
(b) 1.69 g cm^-3
(c) 1.7 g cm ³                                           
(d) 1.695 g cm^-3

Sol: (c) The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures

After rounding off the number, we get density =1.7

Q4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give
(a) 2.75 and 2.74
(b) 2.74 and 2.73
(c) 2.75 and 2.73
(d) 2.74 and 2.74
Sol: (d)
Key concept: While rounding off measurements, we use the following rules by convention:
1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.
2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.
3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one.
4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd.
Units and Measurements
Let us round off 2.745 to 3 significant figures.
Here the digit to be dropped is 5, then preceding digit is left unchanged, if it is even.
Hence on rounding off 2.745, it would be 2.74.
Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

Q5. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, re­spectively. The area of the sheet in appropriate significant figures and error is
(a) 164 ±3 cm²
(b) 163.62 ± 2.6 cm²
(c) 163.6 ±2.6 cm²                                  
(d) 163.62 ±3 cm²
Sol: (a)

Q6. Which of the following pairs of physical quantities does not have same dimensional formula?
(a) Work and torque
(b) Angular momentum and Planck’s constant
(c) Tension and surface tension
(d) Impulse and linear momentum

Q7. Measure of two quantities along with the precision of respective measuring instrument is
A = 2.5 ms^-1 ± 0.5 ms^-1, B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m
(b)  (0.25       ± 0.5) m
(c) (0.25 ± 0.05) m
(d)  (0.25    ± 0.135) m

Q8. You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for √AB  as
(a) 1.4 m± 0.4 m
(b) 1.41 m± 0.15 m
(c) 1.4 m + 0.3 m
(d) 1.4 m± 0.2 m

Q9. Which of the following measurements is most precise?
(a) 5.00 mm
(b) 5.00 cm
(c) 5.00 m
(d) 5.00 km
Sol:(a)
Key concept: Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision.
Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct upto two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

Q10. The mean length of an object is 5 cm. Which of the following measurements is most accurate?
(a) 4.9 cm
(b) 4.805 cm
(c) 5.25 cm
(d) 5.4 cm
Sol: (a)
Key concept: Accuracy describes the nearness of a measurement to the standard or true value, i.e. a highly accurate measuring device will provide measurements very close to the standard, true or known values.
Example: In target shooting, a high score indicates the nearness to the bull’s eye and is a measure of the shooter’s accuracy.

Q11. Young’s modulus of steel is 1.9 x 10¹¹ N/m². When expressed in CGS units of dyne/cm², it will be equal to (1 N = 10⁵ dyne, 1 m² = 10⁴ cm²)                     .
(a) 1.9 xlO¹⁰                                            
(b) 1.9×10¹²
(c) 1.9 xlO¹²                                           
(d) 1.9 xlO¹³

Q12. If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula

More Than One Correct Answer Type
Q13. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

Hence, (c) is not the correct option.
=> LHS ≠ RHS.
So, option (b) is also not correct.

Q14. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
(a) (P-Q)/R          
(b) PQ-R
(c) PQ/R                                  
(d) (PR-Q²)/R
(e)(R + Q)/P
Sol: (a, e)
Key concept: Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation.
According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful.
So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q² may have same dimensions as those of R.
Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Q15. Photon is a quantum of radiation with energy E = hv, where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of
(a) Linear impulse
(b) Angular impulse
(c) Linear momentum                          
(d) Angular momentum

Q16. If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?
(a) Mass of electron (m_e)             
(b) Universal gravitational constant (G)
(c) Charge of electron (e)             
(d) Mass of proton (m_p)

Q17. Which of the following ratios express pressure?
(a) Force/Area                                       
(b) Energy/Volume
(c) Energy/Area                                     
(d) Force/Volume

Sol: (a, b) Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

Q18. Which of the following are not a unit of time?
(a). Second
(b) Parsec
(c) Year
(d) Lightyear
Sol: (b, d) Parsec and light year are those practical units which are used to measure large distances. For example, the distance between sun and earth or other celestial bodies. So they are the units of length not time. Here, second and year represent time.
Important point: 1 light year (distance that light travels in 1 year with speed = 3 x 10⁸ m/s.) = 9.46 x 10¹¹ m And 1 par see = 3.08 x 10¹⁶ m

Very Short Answer Type Questions

Q19. Why do we have different units for the same physical quantity?
Sol: Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required.
For example:
1.Mass ranges from 10^-30 kg (for an electron) to 10⁵³ kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc.
2.The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

Q20. The radius of atom is of the order of 1 A and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Q21. Name the device used for measuring the mass of atoms and molecules.
Sol: A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q22. Express unified atomic mass unit in kg.
Sol: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Q24. Why length, mass and time are chosen as base quantities in mechanics?
Sol: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because
(i) Length, mass and time cannot be derived from one another, that is these quantities are independent.
(ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Short Answer Type Questions
25. (a) The earth-moon distance is about 60 earth radius. What will be the . diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of (1/2)° diameter from the earth. What must be the relative size compared to the earth?
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Q26. Which of the following time measuring devices is most precise?
(a) A wallclock                                        
(b) A stop watch
(c) A digital watch                                  
(d) An atomic clock
Given reason for your answer.
Sol: Option (d) is correct because a clock can measure time correctly up to one second. A stop watch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as precision of this clock is about 1 s in 10¹³ s.

Q27. The distance of a galaxy is of the order of 10²⁵ Calculate the order of magnitude of time taken by light to reach us from the galaxy.
Sol: According to the problem, distance of the galaxy = 10²⁵m.
Speed of light = 3 x 10⁸ m/s
Hence, time taken by light to reach us from galaxy is

Q28. The Vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

Q29. During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.
Sol: Key point: In geometry, a solid angle (symbol: Ω or w) is the two­dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

Q30. If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

Q31. Give an example of
(a) a physical quantity which has a unit but no dimensions
(b) a physical quantity which has neither unit nor dimensions
(c) a constant which has a unit
(d) a constant which has no unit

Q32. Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

Q33. Calculate the solid angle subtended by the periphery of an area of 1 cm² at a point situated symmetrically at a distance of 5 cm from the area.

Important point: Please keep in mind that solid angle is for 3-D figure like sphere, cone etc and plane angle is for plane objects or 2-D figures like circle, arc etc.

Q34. The displacement of a progressive wave is represented by y =A sin(wt– kx), where x is distance and / is time. Write the dimensional formula of (i) w and (ii) k 
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal.
According to the problem

Q35. Time for 20 oscillations of a pendulum is measured as t₁ =39.6 s; t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement?
Sol: According to the problem, time for 20 oscillations of a pendulum,
t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s
It is quite obvious from these observations that the least count of the watch is 0.1 s. As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument= 0.1 s
Precision in 20 oscillations = 0.1

Long Answer Type Questions

Q36. A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s). How much will 5J measure in this new system?
Sol: For solving this problem, dimensions of physical quantity will remain same whatever be the system of units of its measurement.

Q40. If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal,
We know that, dimensions of

Q41. An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional

Q42. In an experiment to estimate ‘the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions.

• Why do we dissolve oleic acid in alcohol?
• What is the role of lycopodium powder?
• What would be the volume of oleic acid in each mL of solution prepared?
• How will you calculate the volume of n drops of this solution of oleic
• What will be the volume of oleic acid in one drop of this solution?

Sol: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.

(b)Lycopodium powder spreads on the entire surface of water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.

(c)Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution =1/20 mL. Further, as this 1 mL is diluted to 20 mL by adding alcohol. In each mL of solution prepared, volume of oleic acid = 1/20 mL x 1/20  = 1/400 mL

(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.

(e) As 1 mL of solution contains n number of drops, then the volume of oleic acid in one drop will be = 1/(400)n mL

Q43. (a) How many astronomical units (AU) make 1 parsec?
(b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what size it will appear when seen through the same telescope.

Q44. Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E= mc², where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV = 1.6 x 10^-13 J; the masses are measured in unified atomic mass unit (u) where, 1 u = 67 x 10^-27 kg.
(a) Show that the energy equivalent of 1 u is 931.5 MeV.
(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
Sol: (a) We can apply Einstein’s mass-energy relation in this problem, E = mc², to calculate the energy equivalent of the given mass.
Here

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