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1

AIEEE 2006

MCQ (Single Correct Answer)
A player caught a cricket ball of mass $$150$$ $$g$$ moving at a rate of $$20$$ $$m/s.$$ If the catching process is completed in $$0.1s,$$ the force of the blow exerted by the ball on the hand of the player is equal to
A
$$150$$ $$N$$
B
$$3$$ $$N$$
C
$$30$$ $$N$$
D
$$300$$ $$N$$

Explanation

We know, Force$$ \times $$ time = Impulse = Change in momentum

$$\therefore$$ $$F \times t = m\left( {v - u} \right)$$

$$ \Rightarrow $$ $$F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$$
2

AIEEE 2005

MCQ (Single Correct Answer)
A block is kept on a frictionless inclined surface with angle of inclination $$'\,\alpha \,'.$$ The incline is given an acceleration $$a$$ to keep the block stationary. Then $$a$$ is equal to
A
$$g$$ $$cosec$$ $$\alpha $$
B
$$g/tan$$ $$\alpha $$
C
$$g$$ $$tan$$ $$\alpha $$
D
$$g$$

Explanation

Acceleration of the block is to the right. Pseudo force acting on the block to the left.


From diagram we can say,

m$$a$$cos$$\alpha $$ = m$$g$$sin$$\alpha $$

$$ \Rightarrow a = g\tan \alpha $$
3

AIEEE 2005

MCQ (Single Correct Answer)
Consider a car moving on a straight road with a speed of $$100$$ $$m/s$$. The distance at which car can be stopped is $$\left[ {{\mu _k} = 0.5} \right]$$
A
$$1000$$ $$m$$
B
$$800$$ $$m$$
C
$$400$$ $$m$$
D
$$100$$ $$m$$

Explanation

Acceleration due to friction = $$\left( { - {\mu _k}g} \right)$$

We know, $${v^2} = {u^2} + 2as$$

$$ \Rightarrow $$ $${0^2} = {u^2} + 2\left( { - {\mu _k}g} \right)s$$

$$ \Rightarrow $$ $$2 { {\mu _k}g}s$$ = $${u^2}$$

$$ \Rightarrow s = {{{{100}^2}} \over {2 \times 0.5 \times 10}}$$

$$ \Rightarrow s = 1000\,m$$
4

AIEEE 2005

MCQ (Single Correct Answer)
A particle of mass 0.3 kg subjected to a force $$F=-kx$$ with $$k=15$$ $$N/m$$. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
A
$$15\,\,\,\,m/{s^2}$$
B
$$3\,\,\,m/{s^2}$$
C
$$10\,\,\,m/{s^2}$$
D
$$5\,\,\,m/{s^2}$$

Explanation



Given F = - kx

$$\Rightarrow$$ F = - 15$$ \times {{20} \over {100}}$$ = - 3 N

F = m.a = 3 N

$$\Rightarrow$$ a = $${3 \over m}$$ = $${3 \over {0.3}}$$ = 10 m/s2

Questions Asked from Laws of Motion

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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