A) 30%
B) 20%
C) 69%
D) \[\sqrt{69}%\]
Correct Answer: A
Solution :
\[f=\frac{1}{2l}\sqrt{T/\mu }\] T= tension \[\mu =\] mass/length \[l=\] length. __ \[{{f}_{1}}=\frac{1}{2{{l}_{1}}}\sqrt{\frac{{{T}_{1}}}{\mu }}\] \[{{T}_{1}}=T+0.69T=1.69T\] \[{{l}_{1}}=?,f={{f}_{1}}\] \[\frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{2{{l}_{1}}}\sqrt{\frac{1.69T}{\mu }}\] \[\frac{1}{l}=\frac{1.3}{{{l}_{1}}}\] \[{{l}_{1}}=1.3l\] Increase in length \[=1.3l-l=0.3l\] \[=30%\]
You need to login to perform this action.
You will be redirected in
3 sec
