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

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} $

2
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Which of the following are correct expression for torque acting on a body?

A. $\vec{\tau}=\vec{r} \times \vec{L}$

B. $\vec{\tau}=\frac{d}{d t}(\vec{r} \times \vec{p})$

C. $\vec{\tau}=\vec{r} \times \frac{d \vec{p}}{d t}$

D. $\vec{\tau}=I \vec{\alpha}$

E. $\vec{\tau}=\vec{r} \times \vec{F}$

( $\vec{r}=$ position vector; $\vec{p}=$ linear momentum; $\vec{L}=$ angular momentum; $\vec{\alpha}=$ angular acceleration; $I=$ moment of inertia; $\vec{F}=$ force; $t=$ time)

Choose the correct answer from the options given below:

A
A, B, D and E Only
B
C and D Only
C
B, C, D and E Only
D
B, D and E Only
3
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass ' $m$ ' having position vector as $\vec{r}=a(\hat{i} \cos \omega t+\hat{j} \sin \omega t)$. The direction of force is

A
Opposite to the direction of $\vec{L}$
B
Opposite to the direction of $\vec{L} \times \vec{P}$
C
Opposite to the direction of $\vec{r}$
D
Opposite to the direction of $\vec{P}$
4
JEE Main 2025 (Online) 3rd April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Rotational Motion Question 5 English

A

$$ 2.5 \mathrm{~m} / \mathrm{s}^2 $$

B

$$ 3.5 \mathrm{~m} / \mathrm{s}^2 $$

C

$$ 0.25 \mathrm{~m} / \mathrm{s}^2 $$

D

$$ 0.35 \mathrm{~m} / \mathrm{s}^2 $$

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