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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A goalkeeper in a game of football pulls his hands backwards while holding the ball shot at the goal. This enables the goalkeeper toA. exert larger force of ballB. reduce the force exerted by the ball on handsC. increae the rate of change of momentumD. decrease the rate of change of momentum |
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Answer» Correct Answer - B::D A goalkeeper pulls his hands backwards while holding the ball shot at the goal to decrease the rate of change of momentum of ball and hence reduse the force exerted by the ball on the hands.both option b and d |
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| 52. |
Though the force of action and reaction are equal and opposite, yet the acceleration produced by them may be of different magnitudes. Why? |
| Answer» As `a= F//m`, therefore when `F` is same, but `m` is different, `a` will be different. | |
| 53. |
There are three solids made up of aluminum, steel and wood, of the same shape and same volume. Which of them would have highest inertia? |
| Answer» Out of three solids of same shape and same volume, mass of steel solid is maximum as density of steel is maximim. As mass is a measure of inertia, therefore, steel solid would have highest inertia. | |
| 54. |
Name the various effects of force. |
| Answer» Force may (i) move a stationary body, (ii) stop a moving body, (iii) change the direction of motion of the body, (iv)change the shape and size of the body. | |
| 55. |
An automobile vehcile has a mass of `1500 kg` What must be the force between the vehcile and road if the vehcile is to be stopped with a negative acceleration of `1.7 m//s^(2)` ? |
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Answer» Here, `m=1500 kg, a = -1.7 m//s^(-2), F = ?` clearly, `F= ma = 1500 (-1.7) N = -2550N` Negative sign indicates that force is opposing the motion of the vehcile. |
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| 56. |
What is the momentum of an object of mass `m` , moving with a velocity`v` ?A. `(mv)^(2)`B. `mv^(2)`C. `1//2 mv^(2)`D. `mv` |
| Answer» (d) Momentum = mass `xx` velocity `= mxxv= mv` | |
| 57. |
Are action-reaction forces simultaneous? |
| Answer» Yes, action-reaction forces acts simultaneous. | |
| 58. |
How are action-reaction forces related in magnitude and direction? |
| Answer» Action and reaction forces are equal in magnitude, but opposite in direction. | |
| 59. |
Calculate momentum of a toy car of mass `300g` moving with a speed of `18km//h`. |
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Answer» Here, `m=300g=0.3kg, v=18km//h=(18xx1000m)/(60xx60s)=5m//s` `p=mv=0.3xx5=1.5kgm//s`. |
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| 60. |
The speed time graph of a body of mass `300g` is shown in (fugure). Calculate the force actiong on the body. |
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Answer» Here, `m=300g= 0.3 kg` From the graph, acceleration of body `a=(dv)/(dt)= ((50-0))/((3-0))= 5/3m//s^(2)` `F= ma= 0.3xx5/3= 0.5N`. |
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| 61. |
It is a common experience that to initiate motion in a body at rest, force required is much greater than the force required to maintain the motion. Read the above passage and answer the following question: (i) Do you agree with the statement? (ii) How do you justify the statement? (iii) What values of day to day life do you learn from this? |
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Answer» (i) Yes, the statement is true (ii) To initiate the motion in a body at rest, inertia of rest has to be overcome before the actual motion starts. The opposing force of friction is the limiting friction, which is larger than the dynamic friction that comes into play when the actual motion starts. (iii) In a day to day life, we find that to initiate any activity, more intense efforts are needed to plug all the loop holes. Once the activity has actually begun, routine efforts are often enough to maintain it. |
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| 62. |
Do action reaction forces act on the same body? |
| Answer» No, action-reaction forces act always on different bodies. | |
| 63. |
Which of the following statements is not true in respect of forces of action and reaction?A. They are always equalB. they are always oppositeC. They can cancel each otherD. none of these |
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Answer» Correct Answer - C The force of action and reaction never cancel eachother. |
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| 64. |
What mass of a body can attain an acceleration of `5 ms^(-2)` under a force of `250N`?A. `5kg`B. `250`kgC. `50kg`D. `10kg` |
| Answer» From `F=ma, m=F/a=(250)/5= 50kg` | |
| 65. |
What is one Newton force? |
| Answer» One newton force is that much force which produces an acceleration of `1m//s^(2)` in a body of mass `1kg`. | |
| 66. |
The velocity time graph of a ball of mass `20g` moving along a straight line on a long table is given in (figure) How much force does the table exert on the ball to bring it to rest? |
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Answer» Here, `m=20g = 20xx10^(-3)kg, F=?` From the graph, (figure) initial speed, `u=20 cm//s = 0.2 m//s`, final speed, `v=0` time taken, `t=10s` As, `F=ma = (m(v-u))/t= (20xx10^(-3)(0-0.2))/10= -4xx10^(4)N` |
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| 67. |
A body moving along a straight line is brought to rest in `2 sec` by a force `F_(1)` and in `3 sec`. By a force `F_(2)` the ratio `F_(1)//F_(2)` isA. `2:3`B. `1:1`C. `3:2`D. `9:4` |
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Answer» Correct Answer - C As force `=("change in momentum")/("time taken") :. (F_(1))/(F_(2))=(t_(2))/(t_(1))=3/2`. |
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| 68. |
A body of mass `2kg` moving with a velocity of `10m//s` is brought to rest in `5 sec`. Calculate the stopping force applied. |
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Answer» `F=m a = (m(v-u))/t=(2(0-10))/5= -4N` Negative sign indicates opposite direction of force. |
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| 69. |
A spring balance A is suspended vertically from a rigid support. Another spring balance B is suspended from the hook of A. Now, a weight of `1kg` is suspended from hook of `B` The readings of balances A and B would beA. `1/2kg` eachB. `2kg` eachC. `1kg` eachD. none of the above. |
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Answer» Correct Answer - C As force of action= force of reacation, the balances A and B will read `1kg` each |
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| 70. |
Why is it difficult to walk on a slippery road? |
| Answer» This is because on a slippery road, both the force of action and reaction reduce. | |
| 71. |
Walking becomes difficult, When the ground is covered with snow or sand this is becauseA. ground become slipperyB. our foot can exert much smaller force in the from of backward actionC. either (a) and (b)D. neither (a) or (b). |
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Answer» Correct Answer - B When ground is covered with snow or sand, our foot can exert much smaller force in the form of backward action. |
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| 72. |
A hunter has a machine gun that can fire `50g` bullet with a velocity of `150m//s`. A `60kg` tiger springs at him with a velocity of `10m//s`. How many bullets must the hunter fire per second into the tiger in order to stop him in the track? |
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Answer» Let `n` bullets be fired per second. To stop the tiger in his track, rate of change of momentum of bullets= rate of change of momentum of tiger `n(50xx10^(-3))xx150=60xx10` `n=(60xx10)/(50xx10^(-3)xx150)=80`. |
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| 73. |
The momentum of a person weighing `60kg` walking with a velocity of `2m//s` isA. `60 N-s`B. `300N-s`C. `120 kg m//s`D. none of these. |
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Answer» Correct Answer - C `p=mv = 60xx2= 120 kg m//s`. |
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| 74. |
A force of `2N` gives a mass `m_(1)` an acceleration of `5m//s^(2)` and a mass `m_(2)` an acceleration of `7m//s^(2)`. What acceleration would be produced if both the masses are tied together? |
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Answer» From `m_(1)= F/(a_(1))=2/5kg , m_(2)=F/(a_(2))=2/7kg` Total mass, `m=m_(1)+m_(2)= 2/5+2/7= (14+10)/(35)=(24)/(35)kg` Acceleration, `a=F/m= 2/(24//35)= (35)/(12)= 2.9m//s^(2)` |
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| 75. |
A force of `5N` gives a mass `m_(1)`, an acceleration of `m//s^(2)`, and a mass `m_(2)`, an acceleration of `24 m//s^(2)`. What acceleration would it give if both the masses are tied together? |
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Answer» Here, force `F=5N` For masses `m_(1)`, accelaration, `a_(1)=8 m//s^(2)`. For masses `m_(2)`, accelaration, `a_(2) = 24 m//s^(2)` For combined mass `(m_(1)+m_(2))`, acceleration, `a=?` From `F = m_(1)a_(1)=m_(2)a_(2), m_(1)= F/(a_(1))= 5/8 kg` and `m_(2)= F/(a_(2))= 5/24 kg` Now, `(m_(1)+m_(2))= 5/8+5/24 = (15+5)/24=20/24 = 5/6 kg`, Thus `a = F/(m_(2)+m_(2))= 5/(5//6) = 6 m//s^(2)` |
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| 76. |
A force of `5N` changes the velocity of a body from `10 ms^(-1)` to `20 ms^(-1)` in `5 sec`. How much force is required to bring about the same change in `2 sec`? |
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Answer» From `F = (dp)/(dt)`, we have `F_(1)=(dp)/(dt_(1))` and `F_(2)=(dp)/(dt_(2))` `:. (F_(2))/F_(1)=(dt_(1))/(dt_(2))=5/2 F_(2)=5/2F_(1)=5/2xx5=12.5N` |
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| 77. |
Is force a scalar or Vector? |
| Answer» Force is vector quantity. | |
| 78. |
What is the SI unit of force? |
| Answer» Answer:- Newton`(N)` | |
| 79. |
When a branch of a tree is shaken, some of the fruits may fall down. Why? |
| Answer» The fruits fall down due to inertia of rest. | |
| 80. |
What is the SI unit of displacement ? |
| Answer» Answer : - `kg ms^(-1)`. | |
| 81. |
What force would be needed to produce an acceleration of `4m//s^(2)` on a ball of mass 6`kg` |
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Answer» Here, acceleration, `a = 4m//s^(2)`, mass, m =6 kg, Force, F=? Clearly, `F =ma = 6xx4=24N` |
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| 82. |
(a)When a galloping horse stops suddenly, what happens to the riders and why? (b)Why do some of the leaves get deatched from a tree when we shake its branch vigorously? |
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Answer» (a)The rider tends to fall forward due to inertia of motion of upper part of his body. (b)This is due to inertia of rest of the leaves. |
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| 83. |
Why does a person in a bus tends to fall forward when it stops suddenly? |
| Answer» This is due to inertia of motion of upper part of his body. | |
| 84. |
A ball set roilling on the ground stops after sometime. This is due toA. unbalanced force of frictionB. unbalance force of air resistanceC. both (a) and (b)D. neither (a) or (b). |
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Answer» Correct Answer - C A rolling ball stops after some time due to unbalanced forces of friction and air resistance. |
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| 85. |
Two friends on roller-skates are standing `5m` apart facing eachother. One of them throws a ball of `2kg` towards the other, who catches it. How will this activity affect the position of the two? Explain your answer. |
| Answer» In the process, sepration between the two friends on roller skates will increase. To start with. Both, are at rest. When one throws the ball, he aquires a momentum (equal to momentum of the ball) in opposite direction and moves in opposite direction. The other friend who catches the ball acquires the momentum of the ball and moves away from his friends in the direction of motion of the ball. | |
| 86. |
Which would requires a greater force: accelerating a `2kg` mass at `5m//s^(2)` or a `4kg` mass at `2m//s^(2)`? |
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Answer» `F_(1)=m_(1)a_(1)= 2xx5 = 10N, F_(2) = m_(2)a_(2)= 4xx2=8N` Clearly, `F_(1)gtF_(2)`, i.e., greater force would be required in accelerating a `2kg` mass at `5m//s^(2)`. |
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| 87. |
A bullet of mass `20g` is fired horizontly with a velocity of `150ms^(-1)` from a pistol of maass `2kg`. What is the recoil velocity of the pistol? |
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Answer» Here, `m_(1)= 20g =20/(1000)kg= 1/50kg, v_(1)= 150ms^(-1), m_(2)= 2 kg, v_(2)=?` As `m_(1)v_(1)+m_(2)v_(2)=0 v_(2) = -(m_(1)v_(1))/(m_(2))= -1/50xx(150)/2= -1.5m//s` |
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| 88. |
The masses of two objects are `2kg` and `20kg`. Which has greater inertia? |
| Answer» The object of mass `20kg` has greater inertia. | |
| 89. |
Rocket works on the principle of coservation ofA. massB. energyC. momentumD. velocity |
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Answer» Correct Answer - C A rocket works on the principle of conservation of linear momentum. |
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| 90. |
An object of mass `2kg` is sliding with a constant velocity of `4ms^(-1)` on a frictionless horizontal table. The force required to keep the object moving with the same velocity isA. 32NB. 0 NC. 2ND. 8N |
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Answer» Correct Answer - B As velocity is to be kept the same, acceleration = 0. Force, F=ma= zero. |
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| 91. |
While extinguishing fire, the fireman has hold the hose pipe very strongly. This is because as water rushes out at a great speedA. the hose pipe tends to move backward due to strong force of reactionB. the hose pipe exerts huge pressure in the forward directionC. either (a) and (b)D. neither (a) or (b). |
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Answer» Correct Answer - A The hose pipe has be held strongly as it tends to move backward due to strong force of reaction. |
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| 92. |
When reasultant of all forces acting on a body is zero, the forces are calledA. unbalanced forcesB. balanced forceC. neither (a) nor (b)D. either (a) or (b) |
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Answer» Correct Answer - B When Resultant `=0`, forces are called balanced forces. |
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| 93. |
A ball of mass `0.5kg` moving with a speed of `10m//s` rebounds after striking normally a perfectly elastic wall. Find the change in momentum of the ball . If contact time with the wall is `0.1s`, what force is exerted by wall? |
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Answer» Here, `m =0.5 kg, u=m//s v=10m//s, dt=0.1s, F=?` Change in momentum of ball, `dp=m(v-u)=0.5(-10-10)= -10kg m//s` `F=(dp)/(dt)=(-10)/(0.1)= -100N` Negative sign indicates that force is in a direction opposite to the initial direction of motion of the ball. |
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| 94. |
A person strikes a nail with a hammer of mass `500g` moving with a velocity of `10m//s`. The hammer comes to rest in `0.01 s` after striking the nail. Calculate (i) force exerted by the hammer on the nail (ii) distance moved by the nail into the plank. |
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Answer» Here, `m=500g= 0.5 kg, u=10m//s, v-0, t=0.01 s`. From `v= u+at` `0=10+axx0.01, a= (-10)/(0.01)= - 1000m//s^(2)` Force exerted by hammer `F= ma= 0.5xx(-1000)= -500N` From `v^(2)-u^(2)= 2as` `s=(v^(2)-u^(2))/(2a)= (0-10^(2))/(2(-1000))= 0,05m` |
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| 95. |
A stone of mass `500g` is thrown with a velocity of `20ms^(-1)` across the frozen surface of a lake. It comes to rest after travelling a distance of `10*1km`. Calculate force of friction between the stone and frozen surface of lake. |
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Answer» Here, `m=500g=0.5kg, u= 20m//s, v=0, s=0.1km= 100m, F=?` From `v^(2)-u^(2)= 2as` `0-20^(2)= 2a(100)` `a=(-400)/(200)= -2m//s^(2)` From `F=ma=(0.5(-2)= -N` Negative sign shows that force of friction is in a direction opposite to the direction of motion. |
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| 96. |
Name any two applications of the law of conservation of liner momentum. |
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Answer» (i) Flight of jet planes and rockets (ii)Recoiling of an gun on firing. |
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| 97. |
Akhtar, Kiran and Rahul were riding in a motorcar that was a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen . Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. |
| Answer» The suggestion made by Kiran that insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong. The suggestion made by Akhtar that motor car exerted a large force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motorcar, both experience the same force as action and reaction are always equal and opposite. Further, change in their momenta are also the same. Only the sign of changes in momenta are opposite,i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same. | |
| 98. |
Two persons manage to push a motorcar of mass`1200kg` at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of `0.2m//s^(2)`. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort). |
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Answer» Here, mass of motorcar, `m= 1200kg` Let each person exert a push `F` on the motorcar. Total push of two person `= F+F=2F` As this push gives a uniform velocity to the motorcar along a level road, it must be a measure of the force of friction`(f)` between the motorcar and the road,i.e., `f=2F` As `spropt^(3)`, `v=(ds)/(dt)prop3t^(2)` Further, `a=(dv)/(dt)prop6t, i.e.,a propt`. When three person push, total force applied `=F+F+F=3F` Force that produces acceleration `(a=0.2m//s^(2))`, ie., `ma=3F-f=3F-2F=F` or `F=ma=1200xx0.2= 240N` |
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| 99. |
When a car rounds a curve suddenly, the person inside is pushed outwards.Why? |
| Answer» It happens on account of inertia of direction. While the car turns, the person sitting inside tries to maintain his original direction of motion, due to inertia of direction. | |
| 100. |
A girl of mass `40kg` jumps with a horizontal velocity of `5m//s` onto a stationary cart with frictionless wheels. The mass of the cart is `3kg`. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction. |
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Answer» Here, `m_(1)= 40kg, u_(1)=5m//s` , `m_(2)= 3kg, u_(2)=0` If `v` is velocity of the girl on the cart, then applying principle of conservation of linear momentum, we get `(m_(1)+m_(2))v = m_(1)u_(1)+m_(2)u_(2)` or `(40+3)v = 40xx5+0= 200` or `v=(200)/(43)= 4.65m//s` |
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