A) 2 : 7
B) 3 : 1
C) 1 : 2
D) 1 : 1
Correct Answer: B
Solution :
\[{{E}_{k}}=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}\] ?(i) \[\therefore \] Rotational energy \[{{E}_{Rotational\,}}=\frac{1}{2}I{{\omega }^{2}}\] ?(ii) Also we know that \[I=m{{K}^{2}}\] so \[{{K}^{2}}=\frac{{{R}^{2}}}{2}\] and \[v=R\omega \] Putting the value of \[I\] in equation (i) and (ii), we get \[\frac{{{E}_{k}}}{{{E}_{rotational}}}=\frac{\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\frac{m{{R}^{2}}}{2}\times \frac{{{\upsilon }^{2}}}{{{R}^{2}}}}{\frac{1}{2}\times \frac{m{{R}^{2}}}{2}\times \frac{{{\upsilon }^{2}}}{{{R}^{2}}}}\] \[=\frac{\frac{3}{4}m{{\upsilon }^{2}}}{\frac{1}{4}m{{\upsilon }^{2}}}=\frac{3}{1}\] hence ratio 3 : 1.
You need to login to perform this action.
You will be redirected in
3 sec
