Test: Laws of Motion (July 10) - NEET MCQ

# Test: Laws of Motion (July 10) - NEET MCQ

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## 15 Questions MCQ Test Daily Test for NEET Preparation - Test: Laws of Motion (July 10)

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Test: Laws of Motion (July 10) - Question 1

### 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 [1988]

Detailed Solution for Test: Laws of Motion (July 10) - Question 1

In presence of friction a = (g sinθ – μg cosθ)

∴   Time taken to slide down the plane

In absence of friction

Test: Laws of Motion (July 10) - Question 2

### A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s.The velocity of heaviest fragment in m/s will be

Detailed Solution for Test: Laws of Motion (July 10) - Question 2

Masses of the pieces are 1, 1, 3 kg. Hence

(1 x 21)2 + (1 x 21)2 = (3 x V)2

That is,

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Test: Laws of Motion (July 10) - Question 3

### A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 ms–1 , the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is [1990]

Detailed Solution for Test: Laws of Motion (July 10) - Question 3

Thrust

Test: Laws of Motion (July 10) - Question 4

Physical independence of force is a consequence of[1991]

Detailed Solution for Test: Laws of Motion (July 10) - Question 4

Newton’s first law of motion is related to physical independence of force.

Test: Laws of Motion (July 10) - Question 5

A particle of mass M is moving in a horizontal circle of radius R with uniform speed V. When it moves from one point to a diametrically opposite point, its [1992]

Detailed Solution for Test: Laws of Motion (July 10) - Question 5

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore, change in momentum is MV – (– MV) = 2MV

Test: Laws of Motion (July 10) - Question 6

A block has been placed on an inclined plane with the slope angle θ, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to [1993]

Detailed Solution for Test: Laws of Motion (July 10) - Question 6

When the block slides down the plane with a constant speed, then the inclination of the plane is equal to angle of repose (θ).
Coeff. of friction = tan of the angle of repose = tanθ.

Test: Laws of Motion (July 10) - Question 7

A satellite in a force free space sweeps stationary interplanetary dust at a rate (dM/dt) = αv. The acceleration of satellite is [1994]

Detailed Solution for Test: Laws of Motion (July 10) - Question 7

Thrust on the satellite,

Acceleration

Test: Laws of Motion (July 10) - Question 8

What will be the maximum speed of a car on a road turn of radius 30 m if the coefficient of friction between the tyres and the road is 0.4 (Take g = 9.8 m/s2) [1995]

Detailed Solution for Test: Laws of Motion (July 10) - Question 8

r = 30 m and μ = 0.4.

Test: Laws of Motion (July 10) - Question 9

A ball of mass 150 g, moving with an acceleration 20 m/s2, is hit by a force, which acts on it for 0.1 sec. The impulsive force is [1996]

Detailed Solution for Test: Laws of Motion (July 10) - Question 9

Mass = 150 gm =

Force = Mass × acceleration

Impulsive force

Test: Laws of Motion (July 10) - Question 10

A bullet is fired from a gun. The force on the bullet is given by F = 600 – 2 × 105 t where, F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel.What is the average impulse imparted to the bullet?[1998]

Detailed Solution for Test: Laws of Motion (July 10) - Question 10

Given F = 600 – (2 x 105t ) The force is zero at time t, given by
0 = 600 – 2 x 105t

= 600 x 3 x 10 –3 – 105 (3 x10 –3)2
= 1.8 – 0.9 = 0.9Ns

Test: Laws of Motion (July 10) - Question 11

Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light frictionless pulley as shown here. What is the acceleration of the masses when they are left free? [2000]       (where g is acceleration due to gravity)

Detailed Solution for Test: Laws of Motion (July 10) - Question 11

Let T be the tension in the string.

Test: Laws of Motion (July 10) - Question 12

A person slides freely down a frictionless inclined plane while his bag falls down vertically from the same height. The final speeds of the man (VM) and the bag (VB) should be such that [2000]

Detailed Solution for Test: Laws of Motion (July 10) - Question 12

As there is only gravitational field which works.
We know it is conservative field and depends only on the end points.  So, VM = VB

Test: Laws of Motion (July 10) - Question 13

If a cricketer catches a ball of mass 150 gm moving with a velocity of 20 m/s, then he experiences a force of (Time taken to complete the catch is 0.1 sec.)[2001]

Detailed Solution for Test: Laws of Motion (July 10) - Question 13

Net force experienced =

Test: Laws of Motion (July 10) - Question 14

A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction  μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]

Detailed Solution for Test: Laws of Motion (July 10) - Question 14

Test: Laws of Motion (July 10) - Question 15

A man weighing 80 kg, stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5m/s2. What would be the reading on the scale ? (g = 10 m/s2) [2003]

Detailed Solution for Test: Laws of Motion (July 10) - Question 15

Reading of the scale = Apparent wt. of the man = m(g + a) = 80 (10 + 5) = 1200 N

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