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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2024 · 31 January · Evening Shift · Physics · Rotational Motion

Two identical spheres each of mass 2kg and radius 50cm are fixed at...

Two identical spheres each of mass 2kg2kg and radius 50cm50cm are fixed at the ends of a light rod so that the separation between the centers is 150cm150cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x20kgm2\dfrac{x}{20}kg{m}^{2}, where the value of xx isTwo identical spheres each of mass 2kg and radius 50cm are fixed at the ends of a light rod so that the separation

Correct answer

0

Step-by-step solution

Two identical spheres each of mass 2kg and radius 50cm are fixed at the ends of a light rod so that the separation

Using parallel axis theorem, we can write total moment of inertia as, I=(25mR2+md2)×2I = \left(\dfrac{2}{5}m{R}^{2} + m{d}^{2}\right) \times 2

⇒I=2(25×2×(12)2+2×(34)2)=5320kgm2⇒ I = 2\left(\dfrac{2}{5} \times 2 \times {\left(\dfrac{1}{2}\right)}^{2} + 2 \times {\left(\dfrac{3}{4}\right)}^{2}\right) = \dfrac{53}{20}kg{m}^{2}

Therefore, x=53x = 53.

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