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Q.

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

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a

0.0007 kg wt

b

0.0021 kg wt

c

0.036 kg wt

d

0.0029 kg wt

answer is C.

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Detailed Solution

Frequency of vibration of string is given by
$n = \frac{p}{2 l} \sqrt{\frac{T}{m}} \Rightarrow p \sqrt{T} =$ constant $\Rightarrow \frac{p_{1}}{p_{2}} = \sqrt{\frac{T_{2}}{T_{1}}}$
Hence $\frac{4}{6} = \sqrt{\frac{T_{2}}{(50 + 15) gm . force}} \Rightarrow T_{2} = 28 .8 gm . f$
Hence weight removed from the pan
$= T_{1} - T_{2} = 65 - 28 .8 = 36.2$ gm-force = 0.036 kg-f.

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