A compass needle oscillates 20 times per minu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $

Vedclass · Accepted Answer

$2 \sqrt{2}: 3 \sqrt{3}$

Vedclass · Answer

(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos 	heta$ is the horizontal component of the Earth's magnetic field,$\mu$ is the magnetic moment,and $I$ is the moment of inertia.Since the same needle is used,$\mu$ and $I$ are constant. Thus,$f \propto \sqrt{B \cos 	heta}$.Given $f_1 = 20$ oscillations/min at $	heta_1 = 45^{\circ}$ and $f_2 = 30$ oscillations/min at $	heta_2 = 30^{\circ}$.Taking the ratio: $\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}} \Rightarrow \frac{20}{30} = \sqrt{\frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}}$.Squaring both sides: $\frac{4}{9} = \frac{B_1}{B_2} 	imes \frac{1}{\sqrt{2}} 	imes \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} 	imes \frac{\sqrt{2}}{\sqrt{3}}$.Therefore,$\frac{B_1}{B_2} = \frac{4}{9} 	imes \frac{\sqrt{3}}{\sqrt{2}} = \frac{2 	imes 2 	imes \sqrt{3}}{3 	imes 3 	imes \sqrt{2}} = \frac{2 \sqrt{2} 	imes \sqrt{2} 	imes \sqrt{3}}{3 \sqrt{3} 	imes \sqrt{3} 	imes \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}}$.Thus,$B_1: B_2 = 2 \sqrt{2}: 3 \sqrt{3}$.

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

Similar Questions

If a brass bar is placed on a vibrating magnet, then its time period

$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $19\%$. By doing this,the periodic time of the magnetometer will

$A$ tangent galvanometer has $80$ turns of wire. The internal and external diameters of the coil are $19 \, cm$ and $21 \, cm$ respectively. The reduction factor of the galvanometer at a place where $H = 0.32 \, \text{oersted}$ will be $(1 \, \text{oersted} = 80 \, A/m)$.

With a standard rectangular bar magnet of length $l$, breadth $b$ $(b \ll l)$ and magnetic moment $M$, the time period of the magnet in a vibration magnetometer is $4 \ s$. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is

Vedclass Test Series

Exam Paper Generator

Online Exam Module