MHT CET · Physics · Magnetic Properties of Matter
A torque of \(1 \cdot 732 \times 10^{-5} \mathrm{Nm}\) is required to hold a magnet at \(90^{\circ}\) with the horizontal component of earth's magnetic field. The torque required to hold it at \(60^{\circ}\) will be
\(\left[\sin \frac{\pi}{2}=1, \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right][\sqrt{3}\) \(=1 \cdot 732]\)
- A \(1 \cdot 5 \times 10^{-5} \mathrm{Nm}\)
- B \(1 \times 10^{-5} \mathrm{Nm}\)
- C \(1 \cdot 732 \times 10^{-5} \mathrm{Nm}\)
- D \(0 \cdot 5 \times 10^{-5} \mathrm{Nm}\)
Answer & Solution
Correct Answer
(A) \(1 \cdot 5 \times 10^{-5} \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}\tau_{1}=1.732 \times 10^{-5} \mathrm{Nm} & \theta=90^{\circ} \\ \tau_{2}=? & \theta=60^{\circ}\end{array}\)
\(\tau_{1}=\mathrm{BM} \sin \theta=\mathrm{BM} \sin 90^{\circ}=\mathrm{BM}\)
\(\tau_{2}=\mathrm{BM} \sin 60^{\circ}=\mathrm{BM} \frac{\sqrt{3}}{2}=1.732 \times 10^{-5} \times \frac{1.732}{2}\)
\(=1.4999 \times 10^{-5}=1.5 \times 10^{-5} \mathrm{Nm}\)
\(\tau_{1}=\mathrm{BM} \sin \theta=\mathrm{BM} \sin 90^{\circ}=\mathrm{BM}\)
\(\tau_{2}=\mathrm{BM} \sin 60^{\circ}=\mathrm{BM} \frac{\sqrt{3}}{2}=1.732 \times 10^{-5} \times \frac{1.732}{2}\)
\(=1.4999 \times 10^{-5}=1.5 \times 10^{-5} \mathrm{Nm}\)
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