JEE Main Rotational Motion PYQ

16. (JEE Main 2023 (Online) 25th January Morning Shift )

$I_{CM}$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $I_{AB}$ is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance $\frac{2}{3}$ R from center. Where R is the radius of the disc. The ratio of $I_{AB}$ and $I_{CM}$ is $x : 9$ . The value of $x$ is _____________.

JEE Main 2023 (Online) 25th January Morning Shift Physics - Rotational Motion Question 16 English

Correct answer is (17)

JEE Main 2023 (Online) 25th January Morning Shift Physics - Rotational Motion Question 16 English Explanation

I_{A B} = I_{cm} + M \times {(\frac{2}{3} R)}^{2}

= \frac{1}{2} M R^{2} + \frac{4}{9} M R^{2}

\begin{aligned} = \frac{(9 + 8) M R^{2}}{18} = (\frac{17}{18}) M R^{2} \end{aligned}

\frac{I_{A B}}{I_{cm}} = \frac{17 / 18}{1 / 2} = (\frac{17}{9})

Value of

x = 17

17. (JEE Main 2023 (Online) 24th January Evening Shift )

A uniform solid cylinder with radius R and length L has moment of inertia I $_{1}$ , about the axis of the cylinder. A concentric solid cylinder of radius $R^{'} = \frac{R}{2}$ and length $L^{'} = \frac{L}{2}$ is carved out of the original cylinder. If I $_{2}$ is the moment of inertia of the carved out portion of the cylinder then $\frac{I_{1}}{I_{2}} =$ __________.

(Both I $_{1}$ and I $_{2}$ are about the axis of the cylinder)

Correct answer is (32)

$I_{1} = \frac{(ρ π R^{2} L) R^{2}}{2}$ ( $ρ$ : density of cylinder)

$\begin{aligned} I_{2} = \frac{[ρ π {(\frac{R}{2})}^{2} \frac{L}{2}] {(\frac{R}{2})}^{2}}{2} \\ ∴ \frac{I_{1}}{I_{2}} = \frac{32}{1} \end{aligned}$

18. (JEE Main 2023 (Online) 24th January Morning Shift )

Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be $\sqrt{x}$ cm. The value of $x$ is ________.

JEE Main 2023 (Online) 24th January Morning Shift Physics - Rotational Motion Question 14 English

Correct answer is (110)

$\begin{aligned} I_{cm} = \frac{2}{5} {MR}^{2} \\ I_{PQ} = I_{cm} + {md}^{2} \\ I_{PQ} = \frac{2}{5} {mR}^{2} + m (10 cm)^{2} \end{aligned}$

For radius of gyration $I_{PQ} = {mk}^{2}$

$\begin{aligned} k^{2} = \frac{2}{5} R^{2} + (10 cm)^{2} \\ = \frac{2}{5} (5)^{2} + 100 \\ = 10 + 100 = 110 \\ k = \sqrt{110} cm \\ x = 110 \end{aligned}$

19. (JEE Main 2022 (Online) 28th June Morning Shift)

Match List-I with List-II

	List-I		List-II
(A)	Moment of inertia of solid sphere of radius R about any tangent.	(I)	$\frac{5}{3} M R^{2}$
(B)	Moment of inertia of hollow sphere of radius (R) about any tangent.	(II)	$\frac{7}{5} M R^{2}$
(C)	Moment of inertia of circular ring of radius (R) about its diameter.	(III)	$\frac{1}{4} M R^{2}$
(D)	Moment of inertia of circular disc of radius (R) about any diameter.	(IV)	$\frac{1}{2} M R^{2}$

Choose the correct answer from the options given below :

A. A - II, B - I, C - IV, D - III

B. A - I, B - II, C - IV, D - III

C. A - II, B - I, C - III, D - IV

D. A - I, B - II, C - III, D - IV

Correct Answer is Option (A)

(A) Moment of inertia of solid sphere of radius R about a tangent $= \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$

$\Rightarrow$ A $-$ (II)

(B) Moment of inertia of hollow sphere of radius R about a tangent $= \frac{2}{3} M R^{2} + M R^{2} = \frac{5}{3} M R^{2}$

$\Rightarrow$ B $-$ (I)

(C) Moment of inertia of circular ring of radius (R) about its diameter = $\frac{(M R^{2})}{2}$

$\Rightarrow$ C $-$ (IV)

(D) Moment of inertia of circular disc of radius (R) about any diameter

$= \frac{M R^{2} / 2}{2} = \frac{M R^{2}}{4}$

$\Rightarrow$ D $-$ (III)

20. (JEE Main 2022 (Online) 28th July Morning Shift )

Four identical discs each of mass ' $M$ ' and diameter ' $a$ ' are arranged in a small plane as shown in figure. If the moment of inertia of the system about ${OO}^{'}$ is $\frac{x}{4} {Ma}^{2}$ . Then, the value of $x$ will be ____________.

JEE Main 2022 (Online) 28th July Morning Shift Physics - Rotational Motion Question 27 English

Correct answer is (3)

$I = 2 \times (\frac{M {(\frac{a}{2})}^{2}}{4}) + 2 \times (\frac{M {(\frac{a}{2})}^{2}}{4} + M {(\frac{a}{2})}^{2})$

$= \frac{M a^{2}}{8} + \frac{5 M a^{2}}{8}$

$= \frac{6 M a^{2}}{8} = \frac{3}{4} M a^{2}$