Q91 / 125 · BITSAT 2011 · Physics
Three blocks of masses $\mathrm{m}_{1}, \mathrm{~m}_{2}$ and $\mathrm{m}_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $\mathrm{T}_{3}$ $=40 \mathrm{~N}$. If $m_{1}=10 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}$ and $\mathrm{m}_{3}=4 \mathrm{~kg},$ the
tension $\mathrm{T}_{2}$ will be
- A$20 \mathrm{~N}$
- B$40 \mathrm{~N}$
- C$10 \mathrm{~N}$
- Correct answerD$32 \mathrm{~N}$
Solution
For equilibrium of all 3 masses,
$$
a=\frac{\mathrm{T}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}
$$
For equilibrium of $\mathrm{m}_{1} \& \mathrm{~m}_{2}$ $\mathrm{T}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \cdot \mathrm{a}$ or, $\mathrm{T}_{2}=\frac{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{T}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}$
Given $m_{1}=10 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}, \mathrm{~m}_{3}=4 \mathrm{~kg}$
$\mathrm{T}_{3}=40 \mathrm{~N}$
$\therefore \mathrm{T}_{2}=\frac{(10+6) \cdot 40}{10+6+4}=32 \mathrm{~N}$
Hence, the correct answer is option: 'D'.