Question 21

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 $$\boxed{53}$$.

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests