JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass \(m\) is suspended by two strings making angles \(\theta_1\) and \(\theta_2\) with the horizontal ceiling with tensions \(T_1\) and \(T_2\) simultaneously. \(T_1\) and \(T_2\) are related by \(T_1=\sqrt{3} T_2\). the angles \(\theta_1\) and \(\theta_2\) are
- A \(\theta_1=30^{\circ} \theta_2=60^{\circ}\) with \(\mathrm{T}_2=\frac{3 \mathrm{mg}}{4}\)
- B \(\theta_1=60^{\circ} \theta_2=30^{\circ}\) with \(\mathrm{T}_2=\frac{\mathrm{mg}}{2}\)
- C \(\theta_1=45^{\circ} \theta_2=45^{\circ}\) with \(\mathrm{T}_2=\frac{3 \mathrm{mg}}{4}\)
- D \(\theta_1=30^{\circ} \theta_2=60^{\circ}\) with \(\mathrm{T}_2=\frac{4 \mathrm{mg}}{5}\)
Answer & Solution
Correct Answer
(B) \(\theta_1=60^{\circ} \theta_2=30^{\circ}\) with \(\mathrm{T}_2=\frac{\mathrm{mg}}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{T}_1 \sin \theta_1+\mathrm{T}_2 \sin \theta_2=\mathrm{mg} \& \mathrm{~T}_1=\sqrt{3} \mathrm{~T}_2 \\ & \Rightarrow \mathrm{~T}_2\left[\sqrt{3} \sin \theta_1+\sin \theta_2\right]=\mathrm{mg} \\ & \text { for } \theta_1=60^{\circ} \& \theta_2=30^{\circ}…
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