JEE Advanced · Physics · 5. Laws of Motion

A block of mass $2 M$ is attached to a massless spring with spring-constant $k .$ This block is connected to two other blocks of masses $M$ and $2 M$ using two massless pulleys and strings. The accelerations of the blocks are $a_{1}, a_{2}$ and $a_{3}$ as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_{0} .$ Which of the following option(s) is/are correct?
[ $g$ is the acceleration due to gravity. Neglect friction]

A $x_{0} = \frac{4 M g}{k}$
B When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 g \sqrt{\frac{M}{5 k}}$
C $a_{2} - a_{1} = a_{1} - a_{3}$
D At an extension of $\frac{x_{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$

Verified Solution

Answer & Solution

Correct Answer

(C) $a_{2} - a_{1} = a_{1} - a_{3}$

Step-by-step Solution

Detailed explanation

Using string constraint

a_{1} = \frac{a_{2} + a_{3}}{2}

\Rightarrow 2 a_{1} = a_{2} + a_{3}

a_{1} - a_{3} = a_{2} - a_{1} \Rightarrow (C)

is correct.
for other options use

m

equivalent

Equivalent mass
Or Reduced mass

m_{e q} = \frac{2 m_{1} \times m_{2}}{m_{1} + m_{2}}

(for a two-mass system)

= \frac{2 (2 m) \times m}{2 m + m} = \frac{4 m}{3}

\Rightarrow T = (\frac{4 m}{3}) . g

∴ 2 T = 2 . \frac{4 m g}{3} = (\frac{8 m}{3}) g

\Rightarrow m_{e q} = \frac{8 m}{3}

Using mechanical energy conservation

\Rightarrow \frac{1}{2} k x_{0}^{2} = \frac{8 m g}{3} x_{0}

x_{0} = \frac{16 m g}{3 k} \Rightarrow (A)

is incorrect
Option (B) $\Rightarrow V_{\frac{x_0}{2}}=V_{\max }=\frac{x_0}{2} \omega=\frac{x_0}{2} \sqrt{\frac{k}{2 m+\frac{8 m}{3}}}=$ $\frac{x_0}{2} \sqrt{\frac{3 k}{14 m}}=g \sqrt{\frac{32}{21 k}}$
Option (D) $\Rightarrow a_{\frac{x_0}{4}}=\frac{x_0}{4} \omega^2=\frac{x_0}{4} \frac{3 k}{14 m}=\frac{3 k x_0}{42 m}=\frac{8 g}{21}$
For a simple harmonic motion

x_{0} \Rightarrow

maximum displacement from extreme to extreme

\Rightarrow x_{0} = 2 A \Rightarrow A = \frac{x_{0}}{2}

and

ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{3 k}{14 m}}

∴ v_{m a x} = A . ω

and

a = ω^{2} . x

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.

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From JEE Advanced

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = - B_{0} \hat{z}$

In which case will the particle move in a straight line with a constant velocity?

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