At two places A and B using vibration | vedclass

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Question

(a)$T = 2\pi \sqrt {\frac{I}{{MH}}} \Rightarrow T \propto \frac{1}{{\sqrt H }}$ $==>$ $\frac{{{T_A}}}{{{T_B}}} = \sqrt {\frac{{{H_B}}}{{{H_A}}}} $

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(a)$T = 2\pi \sqrt {\frac{I}{{MH}}} \Rightarrow T \propto \frac{1}{{\sqrt H }}$ $==>$ $\frac{{{T_A}}}{{{T_B}}} = \sqrt {\frac{{{H_B}}}{{{H_A}}}} $
$==>$ $\frac{{{H_A}}}{{{H_B}}} = $${\left( {\frac{{{T_B}}}{{{T_A}}}} \right)^2} = {\left( {\frac{3}{2}} \right)^2} = \frac{9}{4}$.

At two places $ A $ and $B$ using vibration magnetometer, a magnet vibrates in a horizontal plane and its respective periodic time are $2$ $sec$ and $3$ $sec$ and at these places the earth's horizontal components are $H_A$ and $H_B$ respectively. Then the ratio between $H_A$ and $H_B$ will be

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