A) \[\sqrt{12}\] m/s
B) \[\sqrt{18}\] m/s
C) \[\sqrt{24}\]m/s
D) \[\sqrt{32}\] m/s
Correct Answer: B
Solution :
Potential energy of the mass at a height above the surface is given by \[=\frac{75}{100}\times 12=9J\] ...(1) Now K.E. of the mass at the end of fall \[K.E.=\frac{1}{2}m{{\upsilon }^{2}}\] ...(2) Applying law of conservation of energy \[\frac{1}{2}m{{\upsilon }^{2}}=9\] \[\upsilon =\sqrt{\frac{2\times 9}{m}}=\sqrt{\frac{18}{1}}=\sqrt{18}\,m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
