A light string passing over a smooth light pulley connects two blocks of masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ (where $\mathrm{m}_{2}>\mathrm{m}_{1}$). If the acceleration of the system is $\frac{\mathrm{g}}{\sqrt{2}}$, then the ratio of the masses $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ is :

(A) $\frac{\sqrt{2}-1}{\sqrt{2}+1}$

(B) $\frac{1+\sqrt{5}}{\sqrt{5}-1}$

(C) $\frac{1+\sqrt{5}}{\sqrt{2}-1}$

(D) $\frac{\sqrt{3}+1}{\sqrt{2}-1}$

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Laws of Motion

A light string passing over a smooth light pulley connects two blocks of masses

m_{1}

and

m_{2}

(where

m_{2} > m_{1}

). If the acceleration of the system is

\frac{g}{2}

, then the ratio of the masses

\frac{m _{1}}{m _{2}}

is :

(A) $\frac{2 - 1}{2 + 1}$ ✓ Correct
(B) $\frac{1 + 5}{5 - 1}$
(C) $\frac{1 + 5}{2 - 1}$
(D) $\frac{3 + 1}{2 - 1}$

Answer: (A)

Solution

For the Atwood machine,

a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{g}{2}

.
So

\frac{m _{2} - m _{1}}{m _{2} + m _{1}} = \frac{1}{2}

, giving

2 (m_{2} - m_{1}) = m_{1} + m_{2}

.
Thus

m_{2} (2 - 1) = m_{1} (2 + 1)

, and

\frac{m _{1}}{m _{2}} = \frac{2 - 1}{2 + 1} .

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option

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