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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=p2lTm⇒pT=constant ⇒p1p2=T2T1Hence 46=T2(50+15) gm.force⇒T2=28.8 gm.fHence weight removed from the pan =T1−T2=65−28.8=36.2 gm-force = 0.036 kg-f.
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