Physics - Laws Motion - Quiz-6

Assuming the coefficient of friction between the road and tyres of a car to be 0.5, the maximum speed with which the car can move round a curve of 40.0 m radius without slipping, if the road is unbanked, should be

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25 m/s
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19 m/s
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14 m/s
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11 m/s

Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g = 10 m/s²)

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30 m
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40 m
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72 m
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20 m

On the horizontal surface of a truck (μ = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at the rate of 5m/sec² then frictional force on the block will be

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5 N
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6 N
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5.88 N
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8 N

A vehicle of mass m is moving on a rough horizontal road with momentum p. If the coefficient of friction between the tyres and the road be μ, then the stopping distance is

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$\frac{p}{2 μmg}$
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$\frac{p^{2}}{2 μmg}$
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$\frac{p}{2 {μm}^{2} g}$
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$\frac{p^{2}}{2 {μm}^{2} g}$

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40 N is applied, the acceleration of the block will be (g = 10 m/s²)

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5.73 m/sec²
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8.0 m/sec²
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3.17 m/sec²
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10.0 m/ $s e c^{2}$

A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

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$(1 - \frac{1}{n^{2}})$
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$\frac{1}{1 - n^{2}}$
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$\sqrt{(1 - \frac{1}{n^{2}})}$
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$\sqrt{\frac{1}{1 - n^{2}}}$

Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

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0.33
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0.25
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0.75
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0.80

A force of 750 N is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle 30° with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the frictional force acting on the block is

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750 N
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500 N
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345 N
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250 N

A body takes time t to reach the bottom of an inclined plane of angle θ with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of friction of the rough surface is

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$\frac{3}{4} \tan θ$
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$\frac{2}{3} \tan θ$
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$\frac{1}{4} \tan θ$
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$\frac{1}{2} \tan θ$

A block is kept on an inclined plane of inclination θ of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is μ)

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$\sqrt{2 g l (μ \cos θ - \sin θ)}$
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$\sqrt{2 g l (\sin θ - μ \cos θ)}$
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$\sqrt{2 g l (\sin θ + μ \cos θ)}$
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$\sqrt{2 g l (\cos θ + μ \sin θ)}$

A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle θ to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is:

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$\frac{(P + Q \sin θ)}{(m g + Q \cos θ)}$
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$\frac{(P \cos θ + Q)}{(m g - Q \sin θ)}$
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$\frac{(P + Q \cos θ)}{(m g + Q \sin θ)}$
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$\frac{(P \sin θ - Q)}{(m g - Q \cos θ)}$

A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is

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2.5 N
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0.98 N
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4.9 N
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0.49 N

What is the maximum value of the force F such that the block shown in the arrangement, does not move

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20 N
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10 N
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12 N
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15 N

A body of mass m rests on horizontal surface. The coefficient of friction between the body and the surface is μ. If the mass is pulled by a force P as shown in the figure, the limiting friction between body and surface will be

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μmg
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$μ [m g + (\frac{P}{2})]$
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$μ [m g - (\frac{P}{2})]$
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$μ [m g - (\frac{\sqrt{3} P}{2})]$

A 40 kg slab rests on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force 100 N. If g = 9.8 m/s², the resulting acceleration of the slab will be

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0.98 m/s²
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1.47 m/s²
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1.52 m/s²
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6.1 m/s²

A body of weight 50 N placed on a horizontal surface is just moved by a force of 28.2 N. The frictional force and the normal reaction are

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10 N, 15 N
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20 N, 30 N
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2 N, 3 N
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5 N, 6 N

A rough vertical board has an acceleration ‘a’ so that a 2 kg block pressing against it does not fall. The coefficient of friction between the block and the board should be

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> g/a
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< g/a
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= g/a
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> a/g

It is easier to draw up a wooden block along an smooth inclined plane than to haul it vertically, principally because

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The friction is reduced
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The mass becomes smaller
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Only a part of the weight has to be overcome
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‘g’ becomes smaller

A lead ball strikes a wall and falls down, a tennis ball having the same mass and velocity strikes the wall and bounces back. Check the correct statement

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The momentum of the lead ball is greater than that of the tennis ball
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The lead ball suffers a greater change in momentum compared with the tennis ball
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The tennis ball suffers a greater change in momentum as compared with the lead ball
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Both suffer an equal change in momentum

A ball of weight 0.1 kg coming with speed 30 m/s strikes with a bat and returns in opposite direction with speed 40 m/s, then the impulse is (Taking final velocity as positive)

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$- 0 .1 \times (40) - 0 .1 \times (30)$
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$0 .1 \times (40) - 0 .1 \times (- 30)$
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$0 .1 \times (40) + 0 .1 \times (- 30)$
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$0 .1 \times (40) - 0 .1 \times (20)$

A ball of mass m falls vertically to the ground from a height h₁ and rebound to a height h₂. The change in momentum of the ball on striking the ground is

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$m g (h_{1} - h_{2})$
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$m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$
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$m \sqrt{2 g (h_{1} + h_{2})}$
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$m \sqrt{2 g} (h_{1} + h_{2})$

An elevator and its load have a total mass of 800 kg .Find the tension in the supporting cable when the elevator, moving downward at 10 m/s is brought to rest with constant acceleration in a distance of 25 m. ( Take g = 10m/s² ) :-

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6400 N
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8000 N
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9600 N
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Zero

Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is is suspended by a massless spring as shown in figure. the magnitudes of acceleration of A and B immediately after the string is cut, are respectively

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g, g/3
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g/3, g
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g, g
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g/3, g/3

A car is negociating a curved road of radius R. The road is banked at angle $θ$ . The coefficient of friction between the car and the road is $μ_{s}$ . The maximum safe velocity on this road is

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() $\sqrt{g R (\frac{\tan θ + μ s}{1 - μ_{s} \tan θ})}$
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() $\sqrt{\frac{g}{R} (\frac{\tan θ + μ s}{1 - μ_{s} \tan θ})}$
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() $\sqrt{\frac{g}{R^{2}} (\frac{\tan θ + μ s}{1 - μ_{s} \tan θ})}$
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() $\sqrt{g R^{2} (\frac{\tan θ + μ s}{1 - μ_{s} \tan θ})}$

The length of a spring is l₁ and l₂, when stretched with a force of 4 N and 5 N respectively. Its natural length is

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l₂ + l₁
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2(l₂-l₁)
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5l₁ - 4l₂
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5l₂ - 4l₁

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius $\frac{r}{2}$ and the lighter one in the radius r.The tangential speed of lighter stone is n times that of heavier stone when they experience the same centripetal forces. The value of n is:

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2
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3
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4
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1

One end of the string of length l is connected to a particle of mass m and the other end is connected to a small peg on a smooth horizontal table.If the particle moves in a circle with speed v, the net force on the particle (directed towards the centre) will be: (T represents the tension in the string)

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T
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$T + \frac{m v^{2}}{l}$
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T- $\frac{m v^{2}}{l}$
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Zero

A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is $k^{'}$ . If they are connected in parallel and force constant is $k^{''}, t h e n k^{'} : k^{''}$ is

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1:6
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1:9
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1:11
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6:11

Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

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2N
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6N
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8N
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18N

A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μ_k. When the block A is sliding on the table, the tension in the string is

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(m₂+μ_km₁)g/(m₁+m₂)
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(m₂-μ_km₁)g/(m₁+m₂)
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m₁m₂(1+μ_k)g/(m₁+m₂)
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m₁m₂(1-μ_k)g/(m₁+m₂)

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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