$a)\; v=\sqrt {\large\frac{gm}{k}}\\ b)\;v=g \sqrt {\large\frac{m}{R}} \\ c)\;v=\sqrt {\large\frac{m}{Rg}}\\ d)\;v=\sqrt {\large\frac{mk}{g}}$

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At equillibrium position when net forcea acting on the block is zero.

$kx_0=mg$

$x_0=\large\frac{mg}{k}$

Let v be speed of block in this position.

Applying conservation of mechanical energy

$mgx_0=\large\frac{1}{2}$$mv_0^2+\large\frac{1}{2}$$kx_0^2$

$v=\sqrt {2gx_0-\large\frac{k}{m} \normalsize x_0^2}$

Substituting $x_0=\large\frac{mg}{k}$

We get $v=g \sqrt {\large\frac{m}{k}}$

Hence b is the correct answer.

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