Question

A machine which is 75% efficient, uses 12J of energy in lifting 1 Kg mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the bottom of its fall is:

• A. $\sqrt{12}m/s$
  
• B. $\sqrt{18}m/s$
  
• C. $\sqrt{24}m/s$
  
• D. $\sqrt{32}m/s$

Answer 1

The correct option is B $\sqrt{18}m/s$


$\eta =\frac{E_0}{E_{in}}{E}_0=\frac{75}{100}{E}_{in}=\frac{3}{4}\times 12=9Joule$
This $E_0$ get converted into K.E. during fall by energy conservation
$E_0=\frac{1}{2}m{v}^{2}9=\frac{1}{2}\times 1v^{2}v=\sqrt{18}m/s$