MHT CET · Physics · Mathematics in Physics
A body of mass \(2 \mathrm{~kg}\) is acted upon by two forces each of magnitude \(1 \mathrm{~N}\) and inclined at \(60^{\circ}\) with each other. The acceleration of the body in \(\frac{\mathrm{m}}{\mathrm{s}}\) is \(\left[\cos 60^{\circ}=0.5\right]\)
- A \(\sqrt{0.35}\)
- B \(\sqrt{0.65}\)
- C \(\sqrt{0.75}\)
- D \(\sqrt{0.20}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{0.75}\)
Step-by-step Solution
Detailed explanation
Let us apply the parallelogram law of vectors to find the resultant of the forces.
\(\mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \times \mathrm{F}_{1} \mathrm{~F}_{2} \cos \theta}\)
Given: \(\mathrm{F}_{1}=1, \mathrm{~F}_{2}=1\) and \(\theta=60^{\circ}\)
So, \(\mathrm{F}_{\text {net }}=\sqrt{(1)^{2}+(1)^{2}+2 \times 1 \times 1 \times \cos 60^{0}}=\) \(\sqrt{1+1+2 \times 1 / 2}=\sqrt{3} \mathrm{~N}\)
Now
\(\mathrm{F}=\mathrm{ma}\)
or, \(a=\frac{F}{m}=\frac{\sqrt{3}}{2} m / s^{2}\)
Thus, the net acceleration of the body is \(\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrm{s}^{2}\)
\(\mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \times \mathrm{F}_{1} \mathrm{~F}_{2} \cos \theta}\)
Given: \(\mathrm{F}_{1}=1, \mathrm{~F}_{2}=1\) and \(\theta=60^{\circ}\)
So, \(\mathrm{F}_{\text {net }}=\sqrt{(1)^{2}+(1)^{2}+2 \times 1 \times 1 \times \cos 60^{0}}=\) \(\sqrt{1+1+2 \times 1 / 2}=\sqrt{3} \mathrm{~N}\)
Now
\(\mathrm{F}=\mathrm{ma}\)
or, \(a=\frac{F}{m}=\frac{\sqrt{3}}{2} m / s^{2}\)
Thus, the net acceleration of the body is \(\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrm{s}^{2}\)
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