When two short magnets having magnetic moments in the ratio $27 : 8$ are placed on opposite sides of deflection magnetometer, there is no deflection in the D.M. The distance of the stronger magnet from the centre is $15 c m$ . The distance of the weaker magnet from the center is :

15 c m

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7.5 c m

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10 c m

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\frac{10}{3} c m

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Solution

The correct option is C $10 c m$
By null deflection method,
$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

$\Rightarrow \frac{27}{8} = \frac{15^{3}}{r_{2}^{3}}$

$\Rightarrow r_{2} = \sqrt{\frac{15^{3} . 2^{3}}{3^{3}}}$
$= \frac{15 \times 2}{3}$
$= 5 \times 2$
$= 10 c m .$

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When two short magnets having magnetic moments in the ratio 27:8 are placed on opposite sides of deflection magnetometer, there is no deflection in the D.M. The distance of the stronger magnet from the centre is 15 cm. The distance of the weaker magnet from the center is :

The correct option is C 10 cmBy null deflection method, ⇒M1M2=r31r32⇒278=153r32⇒r2=√153.2333 =15×23 =5×2 =10cm.

When two short magnets having magnetic moments in the ratio $27 : 8$ are placed on opposite sides of deflection magnetometer, there is no deflection in the D.M. The distance of the stronger magnet from the centre is $15 c m$ . The distance of the weaker magnet from the center is :

The correct option is C $10 c m$
By null deflection method,
$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

$\Rightarrow \frac{27}{8} = \frac{15^{3}}{r_{2}^{3}}$

$\Rightarrow r_{2} = \sqrt{\frac{15^{3} . 2^{3}}{3^{3}}}$
$= \frac{15 \times 2}{3}$
$= 5 \times 2$
$= 10 c m .$