Two heavy bodies of mass ‘m’ and ‘3m’ tied together with a light string of mass density $μ$ are dropped from a helicopter as shown. A constant air friction force ‘F’ acts on both of them. If the length of the string is $l .$ Determine the time taken by a pulse to go from top to bottom of the string : (acceleration due to gravity = g)

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$l \sqrt{\frac{μ}{F}}$

$l \sqrt{\frac{2 μ}{F}}$

$l \sqrt{\frac{4 μ}{F + 4 m g}}$

$l \sqrt{\frac{μ}{2 F + 4 m g}}$

answer is B.

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Detailed Solution

$3 m g - (T + F) = 3 m g T + m g - F = m a 3 m g - T - F = 3 T + 3 m g - 3 F$

$2 F = 4 T \Rightarrow T = \frac{F}{2} v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{F}{2 μ}} t = \frac{l}{v} = l \sqrt{\frac{2 μ}{F}}$

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