NCERT Exemplar: Laws of Motion - Physics Class 11 - NEET

MULTIPLE CHOICE QUESTIONS 

Q.1. 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 speed.
(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.
Ans. (c)
Solution.
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.

Q.2. A metre scale is moving with uniform 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.
Ans. (b)
Solution.
As the meter, the scale is moving with uniform velocity.
∴ No change in its velocity i.e., acceleration of it zero by Newton’s second law
∴ Hence net or resultant force must act on body zero.

So torque must be zero.
Hence, for uniform motion force and torque, both must be zero.So, It verifies the option (b).

Q.3. A cricket ball of mass 150 g has an initial velocity  and a final velocity  after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s¹)
(a) zero
(b)
(c)
(d)
Ans. (c)
Solution.
m = 150g = 0.15kg

Change in momentum momentum-Initial momentum



Hence, this verifies the option(c).

Q.4. In the previous problem (5.3), the magnitude of the momentum transferred during the hit is
(a) Zero
(b) 0.75 kg m s^–1 
(c) 1.5 kg m s^–1 
(d) 14 kg m s^–1
Ans. (c)
Solution.
By previous solution,


= 1.5 kg ms^-1
So, this verifies the option (c).

Q.5. 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.
Ans. (d)
Solution.
(i) By Newton’s second law 

As  in law of conservation of momentum is zero.
i.e.,  = 0
 = 0
⇒
(ii) By Newton’s third law action force is equal to reaction force in magnitude but in opposite direction.
∴  ( = 0)


So proves the law of conservation of momentum and verifies the option (d).

Q.6. 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.
Ans. (c)
Solution.
As we know rate of change of momentum is force So in this case direction of force acting on player will be the same as the direction of change in momentum. Now to understand this in better let’s look the image below

shows player’s northward and westward directions respectively. Now shows south direction sois towards south-west which means force on player will act on this direction.
Therefore, option (c) is correct.

Q.7. A body of mass 2kg travels according to the law x(t) = pt + qt² + rt³ where p = 3ms⁻¹ , q = 4ms⁻² = and r = 5ms⁻³.
The force acting on the body at t = 2 seconds is
(a) 136 N
(b) 134 N
(c) 158 N
(d) 68 N
Ans. (a)
Solution.
We have given in question x(t) = pt + qt² + rt³ so to find the force we will differentiate above position equation two times i.e.

We have x(t) = pt + qt² + rt³ where p = 3m^s-1, q = 4ms^-2, r = 5ms^-3
So x(t) = 3t + 4t² + 5t³
Now first derivative of above equation




Therefore, option (a) is correct.

Q.8. A body with mass 5 kg is acted upon by a force If its initial velocity at t = 0 is  the time at which it will just have a velocity along the y-axis is
(a) Never
(b) 10 s
(c) 2 s
(d) 15 s
Ans. (b)
Solution.
We have to find the time at which the final velocity along the y axis which means x-component will be zero. We have now X-component = v_x + a_xt = 0

Therefore, option (b) is correct.

Q.9. A car of mass m starts from rest and acquires a velocity along east  in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is
(a) mv/2  eastward and is exerted by the car engine.
(b) mv/2 eastward and is due to the friction on the tyres exerted by the road.
(c) more than mv/2 eastward exerted due to the engine and overcomes the friction of the road.
(d) mv/2 exerted by the engine.
Ans.
Solution.
Car start from rest mean initial velocity u = 0, also and t = 2s mass is given as m.
As we know
Now force exerted on the car i.e. here this force is due to engine of car so whatever this force is actually internal force in another words due to friction force the car moves in eastward direction.

Q.10. The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin4p t for 0 < t <(1/4) s (A > o), and x = 0 for t >(1/4) s. Which of the following statements is true?
(a) The force at t = (1/8) s on the particle is –16π² A m.
(b) The particle is acted upon by on impulse of magnitude 4π² A m at t = 0 s and t = (1/4) s.
(c) The particle is not acted upon by any force.
(d) The particle is not acted upon by a constant force.
(e) There is no impulse acting on the particle.
Ans. (a, b, d)
Solution.
We have given mass = m, x(t) = 0 for t < 0


Now let’s look for time interval
x(t) = A sin 4πt
Also we know,
1^st derivative of x(t) w.r.t. time will give velocity

2^nd derivative of x(t) w.r.t. time will give velocity

Then,

Here, this force is obviously time dependent so force is not constant hence option (d) is correct.
Now let’s look to option (a) i.e. At


So from this we can say option (a) also correct.
Now for option (b) i.e. between t = 0 s and t = (1/4) s, as we know impulse is change in momentum or  so at t = (1/4) s impulse will be
 here we have to keep in mind that F(t) varies from t=0 to maximum 1/8 s, so F (1/4) will be replaced by F (1/8) also from above we have found


so option(b) is also correct.
Option (c) is absolutely wrong as we have above found that force is acting.
Option (e) is also wrong because we have already calculated the impulse.

Q.11. In Fig, 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.
Ans. (a, b, d, e)
Solution.
We have given  Let acceleration in body A and B is ‘a’.
Body A will move along with body B by force F till the force of friction between surface of A and B is larger or equal to zero.
Now taking system A+B then acceleration will be



So force on A



If F_AB is equal or smaller than f₂ then body A will move along with body B.


N = Reaction force by B on A

[N₂ = Normal reaction on B along with A by surface]

From (i)
 so adding this with equation (ii) we can say F = 0.45mg ...(iii)
F = 0.45mg Newton is the maximum force on B. so that A and B can move together. So option (e) is correct.
Both bodies can move together if F is less than or equal to 0.45mg Newton.
So options (a) and (b) are also correct and rejects the option (c) as 0.5mg>0.45mg.
For option (d): Minimum force which can move A and B together



Given force in option (d) 0.1mg Newton <0.25mg Newton. So body A and B will not move i.e.
Bodies A and B will remain in rest hence option (d) is also correct.

Q.12. Mass m₁ 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 Fig. The co-efficient of friction between m₁ and the sloping surface is µ.
(a) If m₂ > m₁ sin θ, the body will move up the plane.
(b) If m₂ > m₁ > + (sinθ + µ cos θ), the body will move up the plane.
(c) If m₂ < m₁ + (sinθ + µ cos θ), the body will move up the plane.
(d) If m₂ < m₁ (sinθ - µ cos θ), the body will move down the plane.
Ans. (b, d)
Solution.
Let’s consider a case in which normal reaction i.e.
N = m₁g cosθ from figure also we know friction

Now from figure taking whole as a system then

when m₁ will up and m₂ will down.
Putting f in this equation


 from this option (a) is totally wrong while option (b) is correct.
Now if body m₁ moves down and m₂ moves up then, direction of friction force (f) becomes upward (opp. to motion).



Hence option (d) is correct but option (c) is wrong.

Q.13. In Fig, a body A of mass m slides on plane inclined at angle θ1 to the horizontal and µ1 is the coefficent of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle θ2 to the horizontal. Which of the following statements are true?(a) A will never move up the plane.
(b) A will just start moving up the plane when

(c) For A to move up the plane, θ₂ must always be greater than θ₁.
(d) B will always slide down with constant speed.
Ans. (b, c)
Solution.

Condition	Free body diagrams
Equation	Tension and acceleration
T - m1g sin α = m1a
m2a = m2g sin β - T

In this problem first we have to decide the direction of motion. Let block A moves up the plane friction force on A will be downward (along the plane) as shown.



Q.14. Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of 5m 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?
(a) The impulse imparted to each ball is 0.25 kg m s^–1 and the force on each ball is 250 N.
(b) The impulse imparted to each ball is 0.25 kg m s^–1 and the force exerted on each ball is 25 × 10^–5 N.
(c) The impulse imparted to each ball is 0.5 Ns.
(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.
Ans. (c, d)
Solution.

According to the problem, balls are identical.

Initial velocity (u) = u₁ = u₂ = 5 m/s
Final velocity (v) = v₁ = v₂ = -5 m/s
Time duration of collision = 10^-3s
Change in linear momentum = m(v - u)



Impulse and force are opposite in directions.

Q.15. A body of mass 10kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is
(a) 1 m s^–2 at an angle of tan^-1 (4/3) w.r.t. 6N force.
(b) 0.2 m s^–2 at an angle of tan^-1 (4/3) w.r.t. 6N force.
(c) 1 m s^–2 at an angle of tan^-1 (3/4) w.r.t. 8N force.
(d) 0.2 m s^–2 at an angle of tan^-1 (3/4) w.r.t. 8N force.
Ans. (a, c)
Solution.
Recall the concept of resultant of two vectors, when they are perpendicular

As they are perpendicular, cos 90° = 0
So resultant will be
As shown in the diagram
According to the problem, mass = m = 10 kg
F₁ = 6 N, F₂ = 8 N


Let θ₁ be the angle between F and F₁.

Let θ₂ be angle between F and F₂.

VERY SHORT ANSWER TYPE QUESTIONS

Q.1. A girl riding a bicycle along a straight road with a speed of 5 m s^–1 throws a stone of mass 0.5 kg which has a speed of 15 m s^–1 with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?
Ans. Let m₁ and m₂ be the mass of girl with bicycle and stone respectively i.e. m₁ = 50kg & m₂ = 0.5kg
Also we have given u₁ = 5m/s, u₂ = 5m/s and v₂ = 15 m/s and we have to find v₁
As we know by law of conservation of momentum initial momentum (girl, bicycle, stone) = final momentum (cycle and girl) + stone





Therefore, the speed of cycle and girl decreased by


Q.2. A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m s^–2, what would be the reading of the weighing scale? (g = 10 m s^–2)
Ans. Let the acceleration be ‘a’ when the lift is descending, then the apparent weight decreases on weighing scale
So W' = R
= mg - ma
= m(g - a)
Apparent weight due to reaction force by the lift on weighing scale will be



Reading of weighing scale = R/g = 50/10 = 5 kg

Q.3. The position time graph of a body of mass 2 kg is as given in Fig. What is the impulse on the body at t = 0 s and t = 4 s.Ans. As we know impulse is change in momentum which can be written in terms of force

We have given mass of body m=2 kg, initial velocity (v1) at t=0 is zero. Now in the interval t ≥ 0 to t ≤ 4 x-t graph is straight line that mean velocity of body will remain constant.

And when t ≤ 4 the slope of graph is zero so velocity will also be zero i.e. v₃ = 0
Now  or Impulse=change in momentum
Impulse at t = 0

Impulse at t = 4

So from above observation impulse at t = 0 increased by  + 1.50 kg ms^-1 and at t = 4 it is decreased by -150 kg ms^-1.

Q.4. A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?
Ans. When a person applies breaks suddenly, the lower part of person slows rapidly with the car, but the upper part of driver continue to move with same speed in the same direction due to the inertia of motion and his head car hit with steering.

Q.5. The velocity of a body of mass 2 kg as a function of t is given by Find the momentum and the force acting on it, at time t = 2s.
Ans. we have mass m = 2kg and

So momentum at 2 sec p (2) = mv (2)





Hence momentum and force at 2 sec will be  and  respectively.

Q.6. A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Ans. Let f (frictional force) at y-axis and F at x-axis
Now when a small force F₁ is applied on a heavier box, this box does not move, at this state, force of friction f₁ is equal to F₁. Increasing force on box does not move till F = f_s (the maximum static frictional or limiting force).
After force f_s, the frictional force decrease i.e. less kinetic force F_k < f_s is applied on body and it starts to move with less friction
In diagram A is equivalent to limiting frictional force and a B is equivalent to kinetic frictional force.

Q.7. Why are porcelain objects wrapped in paper or straw before packing for transportation?
Ans. Porcelain (or glass) objects are brittle in nature and can crack even small jerk on it. During transportation sudden jerks or even fall takes place.
When objects are packed in paper or straw etc. the objects takes more time to stop or change velocity during jerks (due to breaks, or uneven road) so acceleration (v - u)/t decreased. So the force on objects will be smaller and objects become safer.

Q.8. Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?
Ans. The effect of force F = ma. i.e., if the mass is constant for a system to decrease force, the ‘a’ should be decreased a = (v - u)/t initial and final velocity of falling body on a surface are u and zero. so it cannot be changed. If time during hitting is increased, the acceleration decreased and force will decrease.
On cemented hard floor the time to stop after fall on it is very-very small. But when she/he falls on soft ground of garden she/he sinks in ground and takes more time to stop hence smaller force or pain acts on her/him.

Q.9. A woman throws an object of mass 500 g with a speed of 25 m s¹.
(a) What is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
Ans. (a) Mass of object m = 500 g = 0.5 kg
u = 0, v = 25 m/s

Hence, the  or force is opposite to the initial velocity of ball.

Q.10. Why are mountain roads generally made winding upwards rather than going straight up?
Ans. On an inclined plane force of friction on a body going upward is f_s = μN cos θ where θ is angle of inclination of a plane with horizontal if θ is small, the force of friction is high and there is a less chance of skidding. The road straight up would have a larger angle and smaller would be the value of friction, hence more are the chances of skidding. 

SHORT ANSWER TYPE QUESTIONS

Q.11. A mass of 2kg is suspended with thread AB (Fig). Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward directon so as to apply force on AB. Which of the threads will break and why?

Ans. Thread AB will break. Force on CD is equal to the force (f) applied at D downward, but the force on thread AB is equal to the force F along with force due to mass 2 kg downward. so the force on AB is 2kg more than applied force at D.Hence the thread AB will break up.

Q.12. In the above given problem if the lower thread is pulled with a jerk, what happens?
Ans. Thread CD will break up if CD is pulled with the jerk because pull on thread CD is not transmitted from CD to AB by mass 2 kg due to its inertia of rest.

Q.13. Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in Fig. Calculate T₁ and T₂ when whole system is going upwards with acceleration = 2 m s² (use g = 9.8 m s^–2).

Ans. As the whole system is going up with acceleration = a= 2ms^-2

Tension in a string is equal and opposite in all parts of a string.
Forces on mass m₁


Forces on mass m₂


Q.14. Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° (Fig.). A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.
Ans. Main concept used: On balanced condition i.e., no motion then no frictional force or

f = 0
Solution.
During equilibrium of A or B
mg sin30° = F


For B is at rest W = F = 50N.

Q.15. 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 the finger to hold the block against the wall ?
Ans. Let F force is applied by the finger on a body of mass M to hold rest against the wall. Under the balanced condition

F = N
And f = Mg
⇒ μF = Mg
or  is the minimum force to hold the block against the wall at rest.

Q.16. A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 500m. It falls on the ground at a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 m s^–2)
Ans. Given, the height of the cliff is,
Height of the cliff, h=500 m
Mass of the ball, m= 1 kg
Mass of the gun, M=100 kg
Range of the ball, R= 400 m
Acceleration due to gravity, g= 10 m s^-2
Let the final velocity of the ball be u and the of the gun be v.
Now we know, time of flight is


And range is given by:
R = ut
So, u=400/10 = 40 m s^-2
Therefore, the velocity is:



The recoil velocity of the gun is 0.4 m s^-1.

Q.17. 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.
Ans. From graph (a)

From figure (b) y = t²





F = 1 N toward Y-axis.

Q.18. A person in an elevator accelerating upwards with an acceleration of 2 m s^–2, tosses a coin vertically upwards with a speed of 20 m s¹. After how much time will the coin fall back into his hand? (g = 10 m s^–2)
Ans. Here, the initial velocity of the coin is, u = 20 m s¹
And the acceleration of the elevator is, a = 2 m s^-2
Here, let’s think this problem from the elevator reference frame. For the person inside the elevator going up with acceleration 2 m s^-2 will experience a net acceleration of (g+a) which is 12 m s^-2.
And as the coin return back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.
So we can use the laws of motion to solve the problem.

Here, a' = a + g = 12 m s^-2


So the time of flight of the coin is 3.33 seconds.

LONG ANSWER TYPE QUESTIONS

Q.1. There are three forces F₁, F₂ and F₃ acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.
(a) Show that the forces are coplanar.
(b) Show that the torque acting on the body about any point due to these three forces is zero.
Ans. (a) We know force is a vector quantity. From the theory of planes, we know that two intersecting lines form a plane.
Let these lines be F₁ and F₂. This means that the vector sum, F₁ + F₂ is also on this plane.
As the body is moving with uniform speed, the acceleration is zero. So the net of all forces is zero.
F₁ + F₂ + F₃ = 0
So, F₃ = - (F₁ + F₂)
This means even F₃ lies on the same plane as F₁ and F₂.
Thus the forces are coplanar.
(b) As the sum of all forces are zero, the net torque in any direction is zero. For example, torque about “O” is
Τ = OP x (F₁ + F₂ + F₃)
Τ = 0 as (F₁ + F₂ + F₃) = 0
Hence, torque acting about any point is also zero.

Q.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 co-efficient of friction between the body and the rough plane
Ans. General case

Smooth case

Rough case
Acceleration a = g sinθ − µg cosθ



Q.3. Figure shows (v_x,t),and(v_y,t) diagrams for a body of unit mass. Find the force as a function of time.Ans. Here, as the velocity shows different behavior at different times, we have to divide time into intervals to specify velocity.


Now, we can differentiate velocity to calculate Forces(F).


Hence, the total force is



Q.4. A racing car travels on a track (without banking) ABCDEFA (Fig.). ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The co-effecient of friction on the road is µ = 0.1. The maximum speed of the car is 50 m s^–1. Find the minimum time for completing one round.
Ans. To find the total time, we find the time of each segments.
Segment: ABC
Here, the frictional force is
f = μmg
And the centripetal force is

Therefore,


Segment:DEF
Again, the centripetal force is equal to friction.


Now, the time taken is distance/speed.
For ABC, the distance is

And for DEF we have

Therefore,
 = 86.3
Hence the time taken is 86.3 seconds.

Q.5. The displacement vector of a particle of mass m is given by
(a) Show that the trajectory is an ellipse.
(b) Show that Fr = −mω²r.
Ans. (a) Here, the x and y coordinates are
x = A cos ωt
y = B sin ωt
The equation of an ellipse is:

Substituting x and y, we find that LHS = RHS.
So the trajectory is an ellipse.
(b) To find force, we need to find acceleration.
This can be found by differentiating r two times with respect to t.

And


We know,
F = ma
⇒ F = -mrω²

Q.6. A cricket bowler releases the ball in two different ways
(a) Giving it only horizontal velocity, and
(b) Giving it horizontal velocity and a small downward velocity.
The speed v_s at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.
Ans.


For (b) also
 is the total energy of the ball when it hits the ground
So the speed would be the same for both (a) and (b).

Q.7. There are four forces acting at a point P produced by strings as shown in Fig. which is at rest. Find the forces F₁ and F₂ 
Ans. As the particle is rest or a = 0. So resultant force due to all forces will be zero.
∴ Net components along X and Y-axis will be zero.
Resolving all forces along X-axis

Resolving all forces along Y-axis



Q.8. A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is µ. Let the mass of the box be m.
(a) At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to α > θ ?
(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?
Ans. (a) As the box just start to slide down the plane then μ = tan θ (by angle of repose)

(b) If angle α > θ, the angle of inclination of the plane with horizontal it will slide down (f upward) as θ is the angle of repose. So net force downward

(c) To keep the box either stationary or just move it up with uniform velocity (1 =0) upward f downward)


(d) The force applied f.to move the box upward with acceleration a,


Q.9. A helicopter of mass 2000kg rises with a vertical acceleration of 15 m s^–2. The total mass of the crew and passengers is 500 kg.
Give the magnitude and direction of the (g = 10 m s^–2)
(a) Force on the floor of the helicopter by the crew and passengers.
(b) Action of the rotor of the helicopter on the surrounding air.
(c) Force on the helicopter due to the surrounding air.
Ans. (a) From the diagram, the force on the floor(F) is given by:
F - mg = ma
Where m is the mass of the crew and a is the upwards acceleration.
(b) Here, we have to consider all the weight if the system. The force due to rotor F_r is:


(c) This force is just the reaction force of Fr. So the force due to surrounding air is 6.25 x 10⁴ N.