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Physics Question on Magnetism and matter

A magnet makes 40 oscillation per minute at a place having magnetic intensity of 0.1×10−50.1 \times 10^{-5} tesla. At another place it takes 2.5 sec to complete one oscillation. The value of earth's horizontal field at that place is

A

0.76×10−60.76 \times 10^{-6} tesla

B

0.18×10−60.18 \times 10^{-6} tesla

C

0.09×10−60.09 \times 10^{-6} tesla

D

0.36×10−60.36 \times 10^{-6} tesla

Answer

0.36×10−60.36 \times 10^{-6} tesla

Explanation

Solution

Time period of vibration of a magnet is T=2πIMHT = 2 \pi \sqrt{\frac{I}{MH}} For the same magnet, I and M are constant where M = magnetic moment, I = moment of inertia of magnet. ⇒ T∝1H\Rightarrow \, T \propto \frac{1}{\sqrt{H}} First case T1=6040=32=1.5sec⁡.T_{1} = \frac{60}{40} = \frac{3}{2}= 1.5 \sec. H1=0.1×10−5T,T2=2.5sec⁡,H2=? H_{1} = 0.1 \times10^{-5} T , T_{2} = 2.5 \sec, H_{2} = ? ⇒1.52.5=H2H1\Rightarrow \frac{1.5}{2.5} = \sqrt{\frac{H_{2}}{H_{1}}} ⇒1525=H210−6\Rightarrow \frac{15}{25} = \sqrt{\frac{H_{2}}{10^{-6}}} ⇒H2=(35)2×10−6 \Rightarrow H_{2} = \left(\frac{3}{5}\right)^{2} \times10^{-6} ⇒H2=925×10−6=0.36×10−6T\Rightarrow H_{2} = \frac{9}{25} \times10^{-6} = 0.36 \times10^{-6} T