Question

Two blocks of masses $m$ and $2m$ are placed on triangular wedge by means of a massless inextensible string over a frictionless fixed pulley. Surface of contact between the masses and wedge is frictionless and wedge makes ${45}^{\circ }$ with horizontal on both sides as shown in figure. The tensions in the string $\left(T\right)$ is

• A. $\frac{4mg}{3\sqrt{2}}\text{N}$
• B. $\frac{2mg}{3\sqrt{2}}\text{N}$
• C. $\frac{mg}{3\sqrt{2}}\text{N}$
• D. $\frac{3mg}{\sqrt{2}}\text{N}$

Answer 1

The correct option is A $\frac{4mg}{3\sqrt{2}}\text{N}$

By the FBD of mass $m$:

$R_1=mg\cos {45}^{\circ }$
$T-mg\sin 45=ma...\left(i\right)$

By the FBD of mass $3m$:

$R_2=2mg\cos {45}^{\circ }$
$2mg\sin {45}^{\circ }-T=2ma...\left(ii\right)$

Adding both the equation $\left(i\right)$ and $\left(ii\right)$
We get, $\frac{2mg}{\sqrt{2}}-\frac{mg}{\sqrt{2}}=3ma$

$\Rightarrow a=\frac{g}{3\sqrt{2}}{\text{m/s}}^2$

Substituting the value of ${}^{\prime }{a}^{\prime }$ in either of the equation $\left(i\right)$ and $\left(ii\right)$

$T-mg\sin 45=ma$

$T=m\times \frac{g}{3\sqrt{2}}+\frac{mg}{\sqrt{2}}$

$\Rightarrow T=\frac{4mg}{3\sqrt{2}}\text{N}$

Hence option A is the correct answer.