Physics · Rotational Dynamics

JEE Main 2024 — 31 January, Shift 2 — Question 55

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 $\frac{x}{20} \mathrm{~kg} \mathrm{~m}^{2}$ , where the value of x is $\_\_\_\_\_\_\_$

Answer: 53

Numerical answer — enter this value.

Step-by-step solution

$I=\left(\frac{2}{5} m R^{2}+m d^{2}\right) \times 2$

$\mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^{2}+2 \times\left(\frac{3}{4}\right)^{2}\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^{2}$

$X=53$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia

A body of mass ' m ' is projected with a speed ' u ' making an angle of 45^circ with the ground. The angular momentum of the body about the…

Answer: 53

Step-by-step solution

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