The time period of a vibration magnetometer i | vedclass

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

Question

(A) The time period of a vibration magnetometer is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is th

Vedclass · Accepted Answer

$3T_0$

Vedclass · Answer

(A) The time period of a vibration magnetometer is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.Given that the new magnet has a moment of inertia $I' = 3I$ and a magnetic moment $M' = M/3$.The new time period $T'$ is given by: $T' = 2\pi \sqrt{\frac{I'}{M'B_H}}$.Substituting the values: $T' = 2\pi \sqrt{\frac{3I}{(M/3)B_H}} = 2\pi \sqrt{9 \cdot \frac{I}{MB_H}} = 3 \cdot 2\pi \sqrt{\frac{I}{MB_H}}$.Therefore,$T' = 3T_0$.

The time period of a vibration magnetometer is $T_0$. Its magnet is replaced by another magnet whose moment of inertia is $3$ times and magnetic moment is $1/3$ of the initial magnet. The new time period will be:

Similar Questions

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$,then $(p, q)$ is equal to

$ACTH$ is secreted by

Which of the following statements is incorrect?

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $\left(\frac{\lambda_2}{\lambda_1}\right)$ is:

How many of the following plants produce a composite fruit that develops from an inflorescence? Walnut,Poppy,Radish,Fig,Pineapple,Apple,Tomato,Mulberry.

Vedclass Test Series

Exam Paper Generator

Online Exam Module