In Melde's experiment, the string vibrates in 4 loops when a 50 gram weight is placed in the pan of weight 15 gram . To make the string to vibrates in 6 loops the weight that has to be removed from the pan is

Options

A. 0.0007 kg wt
B. 0.0021 kg wt
C. 0.036 kg wt
D. 0.0029 kg wt

Answer

C

Explanation

Frequency of vibration of string in 'P' loops is given as
`n = (p)/(2l)sqrt((T)/(m))`
From Melde's law
`psqrt(T)` = constant
implies `(p_(1))/(p_(2)) = sqrt((T_(2))/(T_(1)))`
Hence `(4)/(6) = sqrt((T_(2))/((50 + 15)gm - foce) )implies = 28.8 gm - f`
Hence weight removed from the pan = `T_(1) - T_(2) = 65 - 28.8 = 36.8` gm force cong 0.36 kg -f.