Q146 / 160 · TS EAMCET 2013 · Physics
A short bar magnet having magnetic moment $4 \mathrm{Am}^2$, placed in a vibrating magnetometer, vibrates with a time period of $8 \mathrm{~s}$. Another short bar magnet having a magnetic moment $8 \mathrm{Am}^2$ vibrates with a time period of $6 \mathrm{~s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$, the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field.)
- A$9 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$
- Correct answerB$8 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$
- C$5.33 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$
- D$12.2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$
Solution
We know that the time period of a vibrating bar magnet
$T=2 \pi \sqrt{\frac{I}{M B_H}}$
Given, $T_1=8 \mathrm{~s}, l_1=I, M=4 \mathrm{Am}^2$
$8=2 \pi \sqrt{\frac{1}{4 \times B_H}}$
Given, $T_2=6 \mathrm{~s}, I_2=9 \times 10^{-2} \mathrm{~kg} \mathrm{- \textrm {m } ^ { 2 }}, M=8 \mathrm{Am}^2$
$6=2 \pi \sqrt{\frac{9 \times 10^{-2}}{8 \times B_H}}$
Dividing Eqs. (i) and (ii)
$\frac{8}{6}=\sqrt{\frac{2 l}{9 \times 10^{-2}}}$
Squaring both sides and solving, we have
$I=8 \times 10^{-2} \mathrm{~kg} \mathrm{- \textrm {m } ^ { 2 }}$
$T=2 \pi \sqrt{\frac{I}{M B_H}}$
Given, $T_1=8 \mathrm{~s}, l_1=I, M=4 \mathrm{Am}^2$
$8=2 \pi \sqrt{\frac{1}{4 \times B_H}}$
Given, $T_2=6 \mathrm{~s}, I_2=9 \times 10^{-2} \mathrm{~kg} \mathrm{- \textrm {m } ^ { 2 }}, M=8 \mathrm{Am}^2$
$6=2 \pi \sqrt{\frac{9 \times 10^{-2}}{8 \times B_H}}$
Dividing Eqs. (i) and (ii)
$\frac{8}{6}=\sqrt{\frac{2 l}{9 \times 10^{-2}}}$
Squaring both sides and solving, we have
$I=8 \times 10^{-2} \mathrm{~kg} \mathrm{- \textrm {m } ^ { 2 }}$
Hence, the correct answer is option: 'B'.