A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{\frac{5}{4}}$ s. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in second is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

When two identical bar magnets are held perpendicular to each other
$\begin{array}{l} M_{1} = \sqrt{M^{2} + M^{2}} = M \sqrt{2}, I_{l} = I \\ and T_{1} = 2^{5 / 4} s, T_{2} = ? \end{array}$
andM2 M = (as one magnet is removed)
$\begin{array}{l} I_{2} = I_{1} / 2 \\ \frac{T_{2}}{T_{1}} = \sqrt{\frac{I_{2}}{M_{2}} \frac{M_{1}}{I_{1}}} = \sqrt{\frac{1}{2} . \frac{M \sqrt{2}}{M}} = {(\frac{1}{\sqrt{2}})}^{1 / 2} \\ = \frac{1}{2^{1 / 4}} \\ T_{2} = T_{1} \times \frac{1}{2^{1 / 4}} = 2^{5 / 4} \times \frac{1}{2^{1 / 4}} = 2 s \end{array}$