Q154 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
A smooth pulley fixed to a frictionless axle carries two masses attached to a massless intensible string as shown in figure. If the tension in the string is 45 N, the mass M is
$[M < 6\text{kg}, \text{ take } g = 10 \text{ m/s}^2]$

$[M < 6\text{kg}, \text{ take } g = 10 \text{ m/s}^2]$

- A$5.6 \text{ kg}$
- B$5 \text{ kg}$
- C$4.2 \text{ kg}$
- Correct answerD$3.6 \text{ kg}$
Solution
Let the acceleration of the system be $a$. Since $M < 6\text{ kg}$, the $6\text{ kg}$ mass moves downwards and mass $M$ moves upwards.
For the $6\text{ kg}$ mass, the equation of motion is:
$6g - T = 6a$
Given $T = 45\text{ N}$ and $g = 10\text{ m/s}^2$:
$6(10) - 45 = 6a$
$60 - 45 = 6a$
$15 = 6a \Rightarrow a = 2.5\text{ m/s}^2$
For the mass $M$, the equation of motion is:
$T - Mg = Ma$
Substituting the values of $T$, $g$, and $a$:
$45 - M(10) = M(2.5)$
$45 = 10M + 2.5M$
$45 = 12.5M$
$M = \dfrac{45}{12.5} = 3.6\text{ kg}$
Answer: $3.6 \text{ kg}$
For the $6\text{ kg}$ mass, the equation of motion is:
$6g - T = 6a$
Given $T = 45\text{ N}$ and $g = 10\text{ m/s}^2$:
$6(10) - 45 = 6a$
$60 - 45 = 6a$
$15 = 6a \Rightarrow a = 2.5\text{ m/s}^2$
For the mass $M$, the equation of motion is:
$T - Mg = Ma$
Substituting the values of $T$, $g$, and $a$:
$45 - M(10) = M(2.5)$
$45 = 10M + 2.5M$
$45 = 12.5M$
$M = \dfrac{45}{12.5} = 3.6\text{ kg}$
Answer: $3.6 \text{ kg}$
Hence, the correct answer is option: 'D'.