If the strength of the magnetic field is increased by 21%, the frequency of a magnetic needle oscillating in that field :

increased by 10%

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

decreases by 10%

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

increases by 11%

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

decreased by 21%

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

Open in App

Solution

The correct option is C increased by 10%
The time period of oscillation of a bar magnet freely suspended in air is given by:
$T_{1} = 2 π \sqrt{\frac{I}{m B_{1}}}$
$B_{2} = B + 0.21 B_{1}$
$B_{2} = 1.21 B_{1}$
$\frac{T_{2}}{T_{1}} = \sqrt{\frac{B_{1}}{B_{2}}}$
$\frac{T_{2}}{T_{1}} = \frac{1}{1.1}$
$\frac{f_{2}}{f_{1}} = 1.1$
$\frac{Δ f}{f} = 0.1$
So frequency increases by $10$ percent.

Suggest Corrections

Similar questions

A magnetic needle suspended by a silk thread is oscillating in the earth’s magnetic field. If the temperature of the needle is increased by $500^{o} C$ then :

Q. The fundamental frequency of a sonometer wire increases by 5 Hz if its tension is increased by 21%. Instead if the length of wire is decreased by 10%, then the fundamental frequency becomes

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. A magnetic needle suspended by a silk thread is vibrating in the earth's magnetic field if the temperature of the needle is increased by

700^{o}

C, then

Q. A magnetic needle suspended by a silk thread is vibrating in the earth's magnetic field. If temperature of the needle is increased to

1000^{0} C

, then

Join BYJU'S Learning Program

If the strength of the magnetic field is increased by 21%, the frequency of a magnetic needle oscillating in that field :

The correct option is C increased by 10%The time period of oscillation of a bar magnet freely suspended in air is given by:T1=2π√ImB1B2=B+0.21B1B2=1.21B1T2T1=√B1B2T2T1=11.1f2f1=1.1Δff=0.1So frequency increases by 10 percent.