A) \[\sqrt{\frac{4T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}\]
B) \[\sqrt{\frac{2T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}\]
C) \[\sqrt{\frac{T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}\]
D) \[\sqrt{\frac{6T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}\]
Correct Answer: D
Solution :
According to question \[\frac{4}{3}\pi {{a}^{3}}n=\frac{4}{3}\pi {{b}^{3}}\] \[n={{\left( \frac{b}{a} \right)}^{3}}\] \[W=T.4\pi [n{{a}^{2}}-{{n}^{2}}]\] \[W=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}.\frac{4}{3}\pi {{b}^{3}}p{{v}^{2}}\] \[\therefore \] \[\frac{1}{2}\times \frac{4}{3}\pi {{b}^{3p{{v}^{2}}=}}T.4\pi [n{{a}^{2}}-{{b}^{2}}]\] or \[v=\sqrt{\frac{6T}{\rho }\left( \frac{n{{a}^{2}}}{{{b}^{3}}}-\frac{{{b}^{2}}}{{{b}^{3}}} \right)}\] \[\therefore \] \[v=\sqrt{\frac{6T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}\]
You need to login to perform this action.
You will be redirected in
3 sec
