A short magnet placed end on to a compass needle with its center at $0.15 m$ from the needle deflects through $45^{o}$ . The distance that the magnet should be moved away so that the deflection decreases by $15^{o}$ is
(take ( $\sqrt{3})^{1 / 3} = 1.2)$

0.18 m

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0.2 m

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0.24 m

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0.03 m

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Solution

The correct option is A $0.18 m$
From comparison of magnetic moments.
By null deflection method:
$\Rightarrow \frac{tan θ_{1}}{tan θ_{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$
$\Rightarrow \frac{tan 45 °}{tan 60 °} = \frac{{(0.15)}^{3}}{r_{2}^{3}}$ $[θ_{1} = 45 °, r_{1} = 0.15, θ_{2} 45 ° - (- 15) ° = 60]$
$\Rightarrow r_{2}^{3} = (0.15)^{3} \frac{tan 60}{tan 45}$
$\Rightarrow r^{2} = (0.15) (\sqrt{3})^{1 / 3}$
$\Rightarrow r^{2} = (0.15) \times (1.2)$
$= 0.18 m .$
Hence, the answer is $0.18 m .$

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A short magnet placed end on to a compass needle with its center at 0.15m from the needle deflects through 45o. The distance that the magnet should be moved away so that the deflection decreases by 15o is (take (√3)1/3=1.2)

The correct option is A 0.18mFrom comparison of magnetic moments.By null deflection method:⇒tanθ1tanθ2=r31r32⇒tan45°tan60°=(0.15)3r32 [θ1=45°,r1=0.15,θ245°−(−15)°=60]⇒r32=(0.15)3tan60tan45⇒r2=(0.15)(√3)1/3⇒r2=(0.15)×(1.2) =0.18m.Hence, the answer is 0.18m.

A short magnet placed end on to a compass needle with its center at $0.15 m$ from the needle deflects through $45^{o}$ . The distance that the magnet should be moved away so that the deflection decreases by $15^{o}$ is
(take ( $\sqrt{3})^{1 / 3} = 1.2)$