Magnetic field of the earth is $H = 0.3 g$ . A magnet is vibrating $5$ oscillations per min then the dippreciation required in the magnetic field of the earth of increase time period upto $10$ Oscillations per minute is

2.25 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.9 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.12 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $0.12 g$
The time period of vibration of a magnet in magnetic field $H$ is:
$T = 2 π \sqrt{\frac{I}{M B}}$

Hence, $\frac{T_{2}}{T_{1}} = \sqrt{\frac{B_{1}}{B_{2}}}$
$⟹ \frac{1}{2} = \sqrt{\frac{B_{1}}{B_{2}}}$
$⟹ B_{2} = 4 B_{1} = 0.12 g$

Suggest Corrections

Similar questions

A magnetic needle has a frequency of 60 oscillations per minute in the earth horizontal field. When a field B supports the earths horizontal field its frequency increases by 30 oscillations per minute. Then the ratio of the field B to that of earth is :

Q. Magnetic field of earth is