Question

# A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed âˆ’â†’vrel (relative to rope). Find velocity of balloon relative to the ground ?

A
V=mMvrel
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B
V=mMvrel
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C
V=mm+Mvrel
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D
V=+mm+Mvrel
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Solution

## The correct option is C →V=−mm+M→vrelGiven that initially the system is at rest, i.e., →VCM=0 So →VCM=constant=0 i.e., m→v+M→Vm+M=0 or m→v+M→V=0[as(m+M)=finite)] i.e., M→V=−m→v (1) Furthermore, here it is given that; →vrel=→v−→V (2) Putting the value of →v from Eq. (2) in Eq. (1), we get; M→V=−m(→vrel+→V) or →V=−m→vrel(m+M) (3)

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