1
JEE Main 2026 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm . The moment of inertia of this pair of spheres about the tangent passing through the point of contact is $\_\_\_\_$ $\mathrm{kg} \cdot \mathrm{m}^2$.

A

0.18

B

0.72

C

0.36

D

0.63

2
JEE Main 2026 (Online) 21st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is ________ (assume no slipping of string on pulley).

JEE Main 2026 (Online) 21st January Evening Shift Physics - Rotational Motion Question 26 English

A
$$ \dfrac{(M-m) g}{\left[\left(\frac{8}{3}\right) M+m\right]} $$
B

$ \dfrac{(M - m)g}{2M + m} $

C

$ \dfrac{(M - m)g}{M + m} $

D
$$ \dfrac{(M-m) g}{\left[\left(\frac{13}{6}\right) M+m\right]} $$
3
JEE Main 2026 (Online) 21st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A uniform rod of mass $m$ and length $l$ suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is $\_\_\_\_$ . $(g$ acceleration due to gravity)

JEE Main 2026 (Online) 21st January Morning Shift Physics - Rotational Motion Question 19 English
A

$\mathrm{mg} / \mathrm{s}$

B

$m g / 4$

C

$m g$

D

$m g / 2$

4
JEE Main 2025 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

A

$ \frac{\lambda L^3}{8 \pi^2} $

B

$ \frac{\lambda L^3}{16 \pi^2} $

C

$ \frac{\lambda L^3}{4 \pi^2} $

D

$ \frac{\lambda L^3}{12} $

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