The ratio of magnetic moments of two short magnets which give null deflection in tan B position at $12$ cm and $18$ cm from the centre of a deflection magnetometer is?

2 : 3

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8 : 27

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27 : 8

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4 : 9

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Solution

The correct option is A $8 : 27$
According to question,
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\Rightarrow \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3} = {(\frac{12}{18})}^{3}$
$∴ \frac{M_{1}}{M_{2}} = \frac{8}{27}$ .

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The ratio of magnetic moments of two short magnets which give null deflection in tan B position at 12cm and 18cm from the centre of a deflection magnetometer is?

The correct option is A 8:27According to question,M1d31=M2d32⇒M1M2=(d1d2)3=(1218)3∴M1M2=827.

The ratio of magnetic moments of two short magnets which give null deflection in tan B position at $12$ cm and $18$ cm from the centre of a deflection magnetometer is?

The correct option is A $8 : 27$
According to question,
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\Rightarrow \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3} = {(\frac{12}{18})}^{3}$
$∴ \frac{M_{1}}{M_{2}} = \frac{8}{27}$ .