A) 3 m/s
B) 5 m/s
C) 9 m/s
D) 14 m/s
Correct Answer: D
Solution :
Given:\[m=0.25\text{ }kg,\text{ }l=1.96\text{ }m\]and \[t=25N\] Centrifugal force (tension) in the string is given by \[F=\frac{m{{\upsilon }^{2}}}{r}\] \[\Rightarrow \] \[{{\upsilon }^{2}}=\frac{Fr}{m}\] \[\Rightarrow \] \[{{\upsilon }^{2}}=\frac{25\times 1.96}{0.25}=196\] \[\Rightarrow \] \[{{\upsilon }^{2}}=\sqrt{196}\] \[=14\text{ }m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
