Answer

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**Hint:**Use the condition for the no deflection between the magnets, frame the relation between the length and the distance of the magnet from the compass needle. Substitute the known values to find the value of the distance of the second magnet from the compass needle.

**Useful formula:**

The formula of the no deflection in the position of the $\tan \,A$ from the compass needle is

$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$

Where ${M_1}$ is the length of the first magnet, ${M_2}$ is the length of the first magnet, ${d_1}$ is the distance of the first magnet from the compass needle and ${d_2}$ is the distance of the second magnet from the compass needle.

**Complete step by step solution:**

It is given that the

The shorter magnet is placed at a distance from $\tan \,A$ , ${d_1} = 20\,cm$

The longer magnet is twice the length of the shorter magnet.

Using the formula of the no deflection,

$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$

By cancelling the similar terms on both sides,

$\dfrac{{{M_1}}}{{{d_1}^3}} = \dfrac{{{M_2}}}{{{d_2}^3}}$

By rearranging the terms in the above step, we get

$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{{d_1}^3}}{{{d_2}^3}}$

Substituting the known values in the above equation,

$\dfrac{1}{2} = \dfrac{{{{20}^3}}}{{{d_2}^3}}$

By simplifying the above values,

${d_{{2^{}}}}^3 = 4000$

Hence the value of the distance of the second magnet from the compass needle is obtained as follows.

${d_2}^3 = 20 \times {\left( 2 \right)^{1/3}}\,cm$

**Thus the option (B) is correct.**

**Note:**In the question, it is given that the Two short magnets have equal pole strengths but one is twice as long as the other. Hence in the above calculation, the value of the length of the first magnet is taken as $1$ and that of the second magnet is $2\left( 1 \right) = 2$.

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