JEE Main Rotational Motion PYQ

21. (JEE Main 2022 (Online) 26th July Evening Shift )

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ___________ $m$ .

Given, the length of the rod is $10 \sqrt{3} m$ .

Correct answer is (5)

$l = \frac{M L^{2}}{12} = M K^{2}$

$K = \frac{L}{\sqrt{12}} = \frac{10 \sqrt{3}}{\sqrt{12}} = 5 m$

22. (JEE Main 2022 (Online) 30th June Morning Shift )

Four particles with a mass of 1 kg, 2 kg, 3 kg and 4 kg are situated at the corners of a square with side 1 m (as shown in the figure). The moment of inertia of the system, about an axis passing through the point O and perpendicular to the plane of the square, is ______________ kg m².

JEE Main 2022 (Online) 30th June Morning Shift Physics - Rotational Motion Question 33 English

Correct answer is (5)

$I = (m_{A} + m_{B} + m_{C} + m_{D}) r^{2}$

$\begin{aligned} I_{net} = (1 + 2 + 3 + 4) \cdot {(\frac{a}{\sqrt{2}})}^{2}, where a = side of
square \\ = 10 \times \frac{1^{2}}{2} = 5 kg m^{2} \end{aligned}$

23. (JEE Main 2022 (Online) 29th June Evening Shift )

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I₁. The same rod is bent into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}}$ is $\frac{x π^{2}}{3}$ , then the value of x will be ____________.

Correct answer is (8)

$I_{1} = \frac{M L^{2}}{3}$ ..... (1)

For ring : $I_{2} = \frac{M R^{2}}{2}$

and $2 π R = L$

$\Rightarrow I_{2} = \frac{M}{2} (\frac{L^{2}}{4 π^{2}})$ ...... (2)

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{8 π^{2}}{3}$

$\Rightarrow x = 8$

24. (JEE Main 2022 (Online) 25th June Evening Shift )

Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:

I₁ = M.I. of solid sphere about its diameter

I₂ = M.I. of solid cylinder about its axis

I₃ = M.I. of solid circular disc about its diameter

I₄ = M.I. of thin circular ring about its diameter

If 2(I₂ + I₃) + I₄ = x . I₁, then the value of x will be __________.

Correct answer is (5)

$2 (\frac{1}{2} + \frac{1}{4}) \times M (2 R)^{2} + \frac{1}{2} M (2 R)^{2} = x \frac{2}{5} M (2 R)^{2}$

$\Rightarrow 1 + \frac{1}{2} + \frac{1}{2} = x \times \frac{2}{5}$

$\Rightarrow x = 5$

25. (JEE Main 2021 (Online) 31st August Evening Shift)

A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the end of light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?

A. 18.75 kgm²

B. 1.905 $\times$ 10⁵ kgm²

C. 19.05 kgm²

D. 1.875 $\times$ 10⁵ kgm²

Correct Answer is Option (C)

JEE Main 2021 (Online) 31st August Evening Shift Physics - Rotational Motion Question 48 English Explanation

M = 1.5 kg, r = 0.5 m, d = $\frac{5}{2}$ m

$I = 2 (\frac{2}{5} M r^{2} + M d^{2})$

= 19.05 kgm²