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1

AIEEE 2003

MCQ (Single Correct Answer)
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $$49$$ $$N,$$ when the lift is stationary. If the lift moves downward with an acceleration of $$5 m/{s^2}$$, the reading of the spring balance will be
A
$$24$$ $$N$$
B
$$74$$ $$N$$
C
$$15$$ $$N$$
D
$$49$$ $$N$$

Explanation


When lift is stationary then,

T1 = $$mg$$ = 49 N

$$m$$ = 5

For the bag accelerating down

$$mg-T=ma$$

$$\therefore$$ $$T=m(g-a)$$

$$ = 5\left( {9.8 - 5} \right) = 24\,N$$
2

AIEEE 2002

MCQ (Single Correct Answer)
One end of a mass-less rope, which passes over a mass-less and friction-less pulley $$P$$ is tied to a hook $$C$$ while the other end is free. Maximum tension that the rope can bear is $$360$$ $$N.$$ With what value of maximum safe acceleration (in $$m{s^{ - 2}}$$) can a man of $$60$$ $$kg$$ climb on the rope?
A
$$16$$
B
$$6$$
C
$$4$$
D
$$8$$

Explanation

Assuming acceleration $$a$$ of the man is downwards. So the equation will be

$$mg - T = ma$$

$$\therefore$$ $$a = g - {T \over m} = 10 - {{360} \over {60}} = 4$$ $$ m/{s^2}$$

So the maximum acceleration of man is 4 m/s2 downwards.
3

AIEEE 2002

MCQ (Single Correct Answer)
The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is
A
$$60$$
B
$$30$$
C
$$15$$
D
$$25$$

Explanation

For no skidding along curved track,

The maximum velocity possible $${v_{\max }} = \sqrt {\mu rg} $$

Here $$\mu = 0.6,\,r = 150m,\,g = 9.8$$

$$\therefore$$ $${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$$
4

AIEEE 2002

MCQ (Single Correct Answer)
Three identical blocks of masses $$m=2$$ $$kg$$ are drawn by a force $$F=10.2$$ $$N$$ with an acceleration of $$0.6$$ $$m{s^{ - 2}}$$ on a frictionless surface, then what is the tension (in $$N$$) in the string between the blocks $$B$$ and $$C$$?
A
$$9.2$$
B
$$3.4$$
C
$$4$$
D
$$7.8$$

Explanation

Assume tension between block A and B is T1 and block B and C is T2. Acceleration a = 0.6 m/s2.

For block A :

F - T1 = ma

For block B :

T1 - T2 = ma

By adding both those equation we get,

F - T2 = 2ma

T2 = F - 2ma

T2 = 10.2 - 2$$ \times $$2$$ \times $$0.6 = 10.2 - 2.4 = 7.8 N

Questions Asked from Laws of Motion

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