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Standard VIII

Physics

Thrust & Pressure

Two blocks of...

Question

Two blocks of masses $m$ and $M$ = $2 m$ are connected by means of a metal wire of cross sectional area A, passing over a frictionless fixed pulley as shown in figure. The system is then released.
The stress produced in the wire is :

\frac{m g}{A}

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\frac{2 m g}{3 A}

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\frac{3 m g}{4 A}

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\frac{4 m g}{3 A}

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Solution

The correct option is D $\frac{4 m g}{3 A}$
Let tension in string $= T$ , acceleration of the system $= a$
For $M \to$ $2 m g$ - $T = 2 m a$
For $m \to$ $T$ - $m g = m a$
Solving both equations $\Rightarrow$ $T = \frac{4 m g}{3}$
Stress produced in wire $= \frac{T}{A} = \frac{4 m g}{3 A}$
Option D

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Two blocks of masses m and M = 2m are connected by means of a metal wire of cross sectional area A, passing over a frictionless fixed pulley as shown in figure. The system is then released.The stress produced in the wire is :

The correct option is D 4mg3ALet tension in string =T, acceleration of the system =aFor M→ 2mg - T=2maFor m→ T - mg=ma Solving both equations ⇒ T=4mg3 Stress produced in wire =TA=4mg3AOption D

Two blocks of masses $m$ and $M$ = $2 m$ are connected by means of a metal wire of cross sectional area A, passing over a frictionless fixed pulley as shown in figure. The system is then released.
The stress produced in the wire is :

The correct option is D $\frac{4 m g}{3 A}$
Let tension in string $= T$ , acceleration of the system $= a$
For $M \to$ $2 m g$ - $T = 2 m a$
For $m \to$ $T$ - $m g = m a$
Solving both equations $\Rightarrow$ $T = \frac{4 m g}{3}$
Stress produced in wire $= \frac{T}{A} = \frac{4 m g}{3 A}$
Option D