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1

AIEEE 2003

MCQ (Single Correct Answer)
A horizontal force of $$10$$ $$N$$ is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is $$0.2$$. The weight of the block is
A
$$20N$$
B
$$50N$$
C
$$100N$$
D
$$2N$$

Explanation

Assume weight of the block = $$W$$ and friction force = $$f$$ = $$\mu N$$

For the block to remain stationary with the wall, weight of the block should balanced by the force of friction.

$$\therefore$$ $$f=W$$

$$ \Rightarrow $$ $$\mu N = W$$

$$ \Rightarrow $$ $$W$$ = 0.2 $$ \times $$ 10 = 2 N
2

AIEEE 2003

MCQ (Single Correct Answer)
Three forces start acting simultaneously on a particle moving with velocity, $$\overrightarrow v \,\,.$$ These forces are represented in magnitude and direction by the three sides of a triangle $$ABC$$. The particle will now move with velocity
A
less than $$\overrightarrow v \,$$
B
greater than $$\overrightarrow v \,$$
C
$$\left| v \right|$$ in the direction of the largest force $$BC$$
D
$$\overrightarrow v \,\,,$$ remaining unchanged

Explanation

As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order. So the particle will be in equilibrium under the three forces.

Therefore the resultant force is zero.

We know, $${\overrightarrow F_{net}} = m\overrightarrow a .$$

$$\therefore$$ 0 = $$ m\overrightarrow a .$$

$$ \Rightarrow \overrightarrow a = 0$$

Therefore acceleration will be zero.

Hence the particle velocity remains unchanged.
3

AIEEE 2003

MCQ (Single Correct Answer)
A rocket with a lift-off mass $$3.5 \times {10^4}\,\,kg$$ is blasted upwards with an initial acceleration of $$10m/{s^2}.$$ Then the initial thrust of the blast is
A
$$3.5 \times {10^5}N$$
B
$$7.0 \times {10^5}N$$
C
$$14.0 \times {10^5}N$$
D
$$1.75 \times {10^5}N$$

Explanation

Here, thrust force is responsible to accelerate the rocket,

So initial thrust of the blast

= (Lift-off mass) × acceleration

= (3.5 × 104) × (10)

= 3.5 × 105 N
4

AIEEE 2003

MCQ (Single Correct Answer)
A block of mass $$M$$ is pulled along a horizontal frictionless surface by a rope of mass $$m.$$ If a force $$P$$ is applied at the free end of the rope, the force exerted by the rope on the block is
A
$${{Pm} \over {M + m}}$$
B
$${{Pm} \over {M - m}}$$
C
$$P$$
D
$${{PM} \over {M + m}}$$

Explanation

Taking the rope and the block as a system
Acceleration of block($$a$$) = $${{Force\,applied} \over {total\,mass}}$$

$$\therefore$$ $$a = {P \over {m + M}}$$

Force on block(T) = Mass of block $$ \times $$ $$a$$

$$T=Ma$$

$$\therefore$$ $$T = {{MP} \over {m + M}}$$

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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AIEEE 2010 (1)
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AIEEE 2006 (1)
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AIEEE 2003 (7)
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AIEEE 2002 (7)
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