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Byju's Answer

Standard XII

Physics

Magnetic Field

A compass nee...

Question

A compass needle placed at a distance $r$ from a short magnet in a position shows a deflection of $60^{o}$ . If the distance is increased to $r {(3)}^{1 / 3}$ , then deflection of compass needle, in degrees, is :

30

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60 \times 3^{1 / 3}

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60 \times 3^{2 / 3}

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60 \times 3

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Solution

The correct option is D $30$
$\frac{tan θ_{2}}{tan θ_{1}} = \frac{d_{1}^{3}}{d_{2}^{3}} = \frac{r^{3}}{{[r {(3)}^{1 / 3}]}^{3}} = \frac{1}{3}$
$tan θ_{2} = \frac{1}{3} tan θ_{1} = \frac{tan 60}{3} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$
$∴ θ_{2} = 30^{o}$

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A compass needle placed at a distance r from a short magnet in a position shows a deflection of 60o. If the distance is increased to r(3)1/3, then deflection of compass needle, in degrees, is :

The correct option is D 30tanθ2tanθ1=d31d32=r3[r(3)1/3]3=13tanθ2=13tanθ1=tan603=√33=1√3∴θ2=30o

A compass needle placed at a distance $r$ from a short magnet in a position shows a deflection of $60^{o}$ . If the distance is increased to $r {(3)}^{1 / 3}$ , then deflection of compass needle, in degrees, is :

The correct option is D $30$
$\frac{tan θ_{2}}{tan θ_{1}} = \frac{d_{1}^{3}}{d_{2}^{3}} = \frac{r^{3}}{{[r {(3)}^{1 / 3}]}^{3}} = \frac{1}{3}$
$tan θ_{2} = \frac{1}{3} tan θ_{1} = \frac{tan 60}{3} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$
$∴ θ_{2} = 30^{o}$