JEE Advanced · Physics · 8. Rotational Motion

Paragraph :
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
Question :
When disc $B$ is brought in contact with disc $A$, they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is

A $\frac{2 I \omega}{3 t}$
B $\frac{9 I \omega}{2 t}$
C $\frac{91 \omega}{4 t}$
D $\frac{3 I \omega}{2 t}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{2 I \omega}{3 t}$

Step-by-step Solution

Detailed explanation

Let $\omega^{\prime}$ be the common velocity. Then from conservation of angular momentum, we have
$
\begin{aligned}
(i+2 I) \omega^{\prime} & =I(2 \omega)+2 I(\omega) \\
\omega^{\prime} & =\frac{4}{3} \omega
\end{aligned}
$
From the equation,
Angular impulse = change in angular momentum for any of the disc, we have
$
\begin{aligned}
\tau t & =I(2 \omega)-I\left(\frac{4}{3} \omega\right)=\frac{2 I \omega}{3} \\
\tau & =\frac{2 I \omega}{3 t}
\end{aligned}
$
$\therefore$ Option (a) is correct.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

One mole of a monatomic ideal gas undergoes the cyclic process $\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}$, as shown in the P-T diagram.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
[$\mathcal{R}$ is the gas constant.]

List-I	List-II
(P) Work done in the complete cyclic process	(1) $R T_0-4 R T_0 \ln 2$
(Q) Change in the internal energy of the gas in the process JK	(2) $0$
(R) Heat given to the gas in the process KL	(3) $3 K T_0$
(S) Change in the internal energy of the gas in the process MJ	(4) $-2 R T_0 \ln 2$
	$(5)-3 R T_0 \ln 2$

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From JEE Advanced

Let

α, β

and

γ

be real numbers. Consider the following system of linear equations

x + 2 y + z = 7

x + α z = 11

2 x - 3 y + β z = γ

Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
$(P)$	If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ then the system has	$(1)$	a unique solution
$(Q)$	If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq 28$ , then the system has	$(2)$	no solution
$(R)$	If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ \neq 28$ , then the system has	$(3)$	infinitely many solutions
$(S)$	If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then the system has	$(4)$	$x = 11, y = - 2$ and $z = 0$ as a solution
		$(5)$	$x = - 15, y = 4$ and $z = 0$ as a solution

The correct option is:

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