MCQ
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 $30^o$ 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 ............ $m/s^2$ (neglect friction and take $g = 10\ m / s^2$ )
  • A
    $2$
  • B
    $4$
  • $6$
  • D
    $8$

Answer

Correct option: C.
$6$
c
Let $T$ be the tension in the string.

The upward force exerted on the clamp $=T \sin 30^{\circ}=\frac{T}{2}$

$\text { Given: } \frac{T}{2}=40 N \text { or } T=80 N$        $...(i)$

If $a$ is the acceleration of monkey in upward direction,

$a=\frac{T-m g}{m}=\frac{80-5 \times 10}{5}=6 m / s^{2}$

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