A) 0.0007 kg wt
B) 0.0021 kg wt
C) 0.036 kg wt
D) 0.0029 kg wt
Correct Answer: C
Solution :
Frequency of vibration of string is given by \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\]Þ \[p\sqrt{T}=\]constant Þ \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] Hence\[\frac{4}{6}=\sqrt{\frac{{{T}_{2}}}{(50+15)gm\text{-}force}}\]Þ \[{{T}_{2}}=28.8\,gm\text{-}f\] Hence weight removed from the pan \[={{T}_{1}}-{{T}_{2}}=65-28.8=3.62\]gm-force = 0.036 kg-f.
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