A) 5 N
B) 20 N
C) 40 N
D) 80 N
Correct Answer: C
Solution :
The fundamental frequency (n) of a wire of length I and tension T is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}},\] where m is mass per unit length of wire. Given, \[{{n}_{1}}=n,{{T}_{1}}=T,{{n}_{2}}=2n\] \[\therefore \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[{{T}_{2}}={{\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)}^{2}}{{T}_{1}}\] \[\therefore \] \[{{T}_{2}}={{(2)}^{2}}\times 10=40N\]
You need to login to perform this action.
You will be redirected in
3 sec
