Question

A massless string of length $l$ passes over a frictionless pulley with horizontal axis. Two monkeys hang from the ends of the string at the same distance $l/2$ from the pulley, the monkeys start climbing upwards simultaneously. First monkey climbs with speed $v$ relative to the string and the second with speed of $2v$. Both monkeys have got same masses. The time taken by the first and second monkeys in reaching the pulley is respectively:

• A. $\left(\frac{l}{v}\right),\left(\frac{l}{2v}\right)$
• B. $\sqrt{\frac{2l}{v}},\sqrt{\frac{l}{v}}$
• C. ${\left(\frac{l}{2v}\right)}^{1/2},{\left(\frac{l}{v}\right)}^{1/2}$
• D. $\left(\frac{l}{3v}\right),\left(\frac{l}{3v}\right)$

Answer 1

Answer: (D)

$\left(\frac{l}{3v}\right),\left(\frac{l}{3v}\right)$

As the pulley is frictionless and masses of monkeys are equal so it takes same amount of time for both monkeys to reach the pulley.
Total relative velocity is $v_t=v+2v=3v$
Total distance covered ,$S=l/2+l/2=l$
The time taken by each monkey is, $t=\frac{S}{v_t}=\frac{l}{3v}$