Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $$\frac{x}{20}$$ kg m$$^2$$, where the value of $$x$$ is
Correct Answer: 53
Two spheres: each mass 2kg, radius 0.5m, separation 1.5m (center to center), so each is 0.75m from middle.
$$I=2\left[\frac{2}{5}mr^2+md^2\right]=2\left[\frac{2}{5}(2)(0.25)+(2)(0.5625)\right]=2[0.2+1.125]=2(1.325)=2.65=\frac{53}{20}$$ kg m².
So $$x=53$$.
The answer is $$\boxed{53}$$.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation