At the point of Equilibrium for Spring of constant (K) and attached Mass (M) below it, the extension of spring $(x_{o})$ is given by:

\frac{M g}{K}

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\frac{M^{2} g}{K}

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\frac{M g^{2}}{K}

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\frac{M g}{K^{2}}

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Solution

The correct option is A $\frac{M g}{K}$
Suppose at equilibrium it extends by $x$
as spring is in equilibrium so net force should be zero.
so, $M g = k x$
$x = \frac{M g}{K}$

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At the point of Equilibrium for Spring of constant (K) and attached Mass (M) below it, the extension of spring (xo) is given by:

The correct option is A MgKSuppose at equilibrium it extends by xas spring is in equilibrium so net force should be zero.so, Mg=kxx=MgK

At the point of Equilibrium for Spring of constant (K) and attached Mass (M) below it, the extension of spring $(x_{o})$ is given by:

The correct option is A $\frac{M g}{K}$
Suppose at equilibrium it extends by $x$
as spring is in equilibrium so net force should be zero.
so, $M g = k x$
$x = \frac{M g}{K}$