A) 1.5
B) 1.1
C) 0.75
D) 0.55
Correct Answer: B
Solution :
\[T.l=v\rho g\] or \[T.l=\pi {{r}^{2}}l\rho g\] or \[T=\pi {{r}^{2}}\rho g\] \[\therefore \] \[70\times {{10}^{-3}}=3.14\times {{r}^{2}}\times 8\times {{10}^{-3}}\times 9.8\] or \[{{r}^{2}}=\frac{70\times {{10}^{-3}}}{3.14\times 8\times {{10}^{-3}}\times 9.8}=0.28\] \[\therefore \] Minimum diameter of the wire =2r \[=2\times 0.53\] \[\approx 1.1\,m\]You need to login to perform this action.
You will be redirected in
3 sec