A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes an angle of $30^{\circ}$ with the ground. A monkey of mass $5 kg$ climbs up the rope. The clamp can tolerate a vertical force of $40 N$ only. The maximum acceleration in upward direction with which the monkey can climb safely is (Neglect friction and take $g = 10 {ms}^{- 2})$

2 {ms}^{- 2}

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4 {ms}^{- 2}

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6 {ms}^{- 2}

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8 {ms}^{- 2}

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Solution

The correct option is C $6 {ms}^{- 2}$
Let $T$ be the tension in the string. The upward force exerted on the clamp is $T sin 30^{\circ} = \frac{T}{2} .$

Given $\frac{T}{2} = 40 N$
$\Rightarrow T = 80 N . . . . . . . . . . . . . . (1)$
If $a$ is the acceleration of monkey in upward direction.
$a = \frac{T -mg}{m}$
$a = \frac{80 - (5) (10)}{5} = 6 {ms}^{- 2}$
Hence, the correct answer is (C).

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Q. A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes an angle of

30^{\circ}

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g = 10 {ms}^{- 2})

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The correct option is C 6 ms−2Let T be the tension in the string. The upward force exerted on the clamp is T sin 30∘=T2. Given T2=40 N ⇒T=80 N..............(1) If a is the acceleration of monkey in upward direction. a=T -mgm a=80−(5)(10)5=6 ms−2 Hence, the correct answer is (C).