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 \[=\frac{75}{100}\times 12=9\,J\] ...(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 \[\frac{1}{2}m{{\upsilon }^{2}}=9\] \[{{\upsilon }^{2}}=\sqrt{\frac{2\times 9}{m}}=\sqrt{\frac{18}{1}}=\sqrt{18}m/s\]
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