Q137 / 150 · MHT CET 2020 (20 Oct Shift 2) · Physics
A mass $2 \sqrt{3} \mathrm{~kg}$ is acted upon by two forces which are inclined to each other at $60^{\circ}$ and each of magnitude $1 \mathrm{~N}$. The acceleration of that mass in SI system is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right]$
- A$0.7 \mathrm{~m} / \mathrm{s}^{2}$
- B$0.3 \mathrm{~m} / \mathrm{s}^{2}$
- C$0.9 \mathrm{~m} / \mathrm{s}^{2}$
- Correct answerD$0.5 \mathrm{~m} / \mathrm{s}^{2}$
Solution
$F_{n e t}=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta}$
$=\sqrt{1+1+2 \cos 60}$
$=\sqrt{3}$
Now,
$a=\mathrm{F} / \mathrm{m}=\sqrt{3} / 2 \sqrt 3=0.5 \mathrm{~m} / \mathrm{s}^{2}$
Hence, the correct answer is option: 'D'.