Question

The length of an inclined plane is halved and the angle of inclination is changed from ${30}^o$ to ${60}^o$ . If the work done in pulling a load up the first inclined plane is ${}^{\prime }{w}^{\prime }$ , then the work done in pulling the same load up the second inclined plane is :

• A. $2w$
• B. $\frac{\sqrt{3}w}{2}$
• C. $\frac{w}{2}$
• D. $\frac{2w}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{\sqrt{3}w}{2}$

In the first case, the work done to pull the mass is given as,

$w=mgl\sin {30}^{\circ }$

$2w=mgl$

The work done in the second case is given as,

$w^{\prime }=mg\frac{l}{2}\sin {60}^{\circ }$

$w^{\prime }=w\sin {60}^{\circ }$

$w^{\prime }=\frac{\sqrt{3}w}{2}$

Thus, the work done in pulling the same load up the second inclined plane is $\frac{\sqrt{3}w}{2}$.