MHT CET · Physics · Mathematics in 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 \(0.9 \mathrm{~m} / \mathrm{s}^{2}\)
- D \(0.5 \mathrm{~m} / \mathrm{s}^{2}\)
Answer & Solution
Correct Answer
(D) \(0.5 \mathrm{~m} / \mathrm{s}^{2}\)
Step-by-step Solution
Detailed explanation
\(F_{n e t}=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta}\)
\(=\sqrt{1+1+2 \cos 60}\)
\(=\sqrt{3}\)
Now,
\(a=\mathrm{F} / \mathrm{m}=\sqrt{3} / 2 \sqrt 3=0.5 \mathrm{~m} / \mathrm{s}^{2}\)
\(=\sqrt{1+1+2 \cos 60}\)
\(=\sqrt{3}\)
Now,
\(a=\mathrm{F} / \mathrm{m}=\sqrt{3} / 2 \sqrt 3=0.5 \mathrm{~m} / \mathrm{s}^{2}\)
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