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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 $g = 10\, m/s^2$)
$2$
$4$
$6$
$8$
Solution
Let $T$ be the tension in the string.
The upward force exerted on the clamp
${=\mathrm{T} \sin 30^{\circ}=\frac{\mathrm{T}}{2}}$
Given ${\frac{\mathrm{T}}{2}=40 \mathrm{N} \text { or } \mathrm{T}=80 \mathrm{N}}$ $…(i)$
If a is the acceleration of monkey in upward direction,
$a=\frac{T-m g}{m}=\frac{80-5 \times 10}{5}=6 m / s^{2}$
