Question

A rod of length $1m$ and mass $0.5kg$ hinged at one end, is initially hanging vertical. The other end is now raised slowly until it makes an angle ${60}^o$ with the vertical. The required work is : (use $g=10m/s^2$)

• A. $\frac{5}{2}J$
• B. $\frac{5}{4}J$
• C. $\frac{17}{8}J$
• D. $\frac{5\sqrt{3}}{2}J$

Answer 1

The correct option is B $\frac{5}{4}J$

Given, $length=1mmass=0.5kg,angle={60}^{\circ },g=10m/s^2$

Since the rod has been raised to${60}^0$ So the centre of the gravity is the mid point is raised through height $\frac{1}{4}m$

Its weighgt is $mg=0.5\times 10=5N$

Thus the work-done is $=5\times \frac{1}{4}=\frac{5}{4}J$