wave_mechanics critical_thinking Question 49
Question: 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
[MH CET 2004]
Options:
A) 0.0007 kg wt
B) 0.0021 kg wt
C) 0.036 kg wt
D) 0.0029 kg wt
Show Answer
Answer:
Correct Answer: C
Solution:
Frequency of vibration of string is given by $ n=\frac{p}{2l}\sqrt{\frac{T}{m}} $
Therefore $ p\sqrt{T}= $ constant
Therefore $ \frac{p _{1}}{p _{2}}=\sqrt{\frac{T _{2}}{T _{1}}} $ Hence$ \frac{4}{6}=\sqrt{\frac{T _{2}}{(50+15)gm\text{-}force}} $
Therefore $ T _{2}=28.8gm\text{-}f $ Hence weight removed from the pan $ =T _{1}-T _{2}=65-28.8=3.62 $ gm-force = 0.036 kg-f.
BETA
