Question

In the shown figure, two blocks one of mass $5\text{kg}$ and the other of mass $2\text{kg}$ are connected by light, inextensible strings and pulleys are light and frictionless. Choose the correct statement(s).

• A. The acceleration of $5\text{kg}$ mass is $\frac{5\text{g}}{11}{\text{ms}}^{-2}$
• B. The acceleration of $2\text{kg}$ mass is $\frac{5\text{g}}{11}{\text{ms}}^{-2}$
• C. Tension in the string is $\frac{12\text{g}}{11}\text{N}$
• D. Tension in the string is $\frac{10\text{g}}{11}\text{N}$

Answer 1

Answer: (B), (C)

Tension in the string is $\frac{12\text{g}}{11}\text{N}$

Total number of strings connected to $5\text{kg}$ mass and its pulley = $5$

So total tension force applied on $5\text{kg}$ mass $=5T$

Tension will be same throughout $\because$ same string.

The force equation for the $5\text{kg}$ mass:
$5T-5g=5a_1\dots \dots \left(1\right)$

The force equation for $2\text{kg}$ mass
$2g-T=2a_2\dots \dots \left(2\right)$

Equation from string constant is
$5a_1=a_2\dots \dots \left(3\right)$

Substituting in equation ($1$)

$\Rightarrow 5T-5g=a_2$

Multiplying equation ($2$) with $5$.

$\Rightarrow -5T+10g=10a_2\dots \left(4\right)$

Adding equation ($1$) and equation ($4$)

$\Rightarrow 5g=11a_2$

$\Rightarrow a_2=\frac{5g}{11}{\text{m/s}}^2$; $a_1=\frac{g}{11}{\text{m/s}}^2$

Solving equation ($2$) to get $T$:
$T=2g-2a_2$

$=2g-\frac{10}{11}g$

$\Rightarrow T=\frac{12g}{11}\text{N}$

Hence, options (b) and (c) are the correct answers.