Work done in increasing the extension of spring of spring constant $10N/cm$ from $4\;cm$ to $6\;cm$ is:
\[\begin{align}
  & A.1J \\
 & B.10J \\
 & C.50J \\
 & D.100J \\
\end{align}\]

Hint: We know that the spring constant is the minimum force which must be applied on the spring to disturb the equilibrium of the spring. This force then displaces the spring from its equilibrium position. Hooke's law gives the relationship between the displacements of the spring the external force applied.

Formula used:
$F=-kx$ and $dW=k\dfrac{x^{2}}{2}$

Complete step by step answer:
We know that when an external force $F$ is applied to a spring it produces a harmonic oscillation. The force applied on the spring produces a displacement $x$.
We also know from Hooke’s law that the magnitude of the force is directly proportional to the displacement of the spring. It is mathematically given as $F=-kx$ , $k$ is the spring constant. It is also the force applied on the spring to produce unit displacement. The negative sign indicates that the spring resists the applied force.
Then we can say that the work done by the spring is nothing but the change in the potential energy of the spring. Then for a small displacement \[dx\] the small work done \[dW\]is given as $dW=\int kxdx=k\dfrac{x^{2}}{2}$
Here it is given that $k=10N/cm=1000N/cm$, initial position of the spring is given as $x_{1}=0.04m$ and the final position of the spring is $x_{2}=0.06m$
Then the work done is given as $W=\dfrac {k}{2}(x_{2}^{2}-x_{1}^{2})=\dfrac{1000}{2}((0.06)^{2}-(0.04)^{2})=500\times (0.36-0.16)=500\times 0.20=1J$
Thus the required answer is \[A.1J\]

Note:
The force on the spring is either due to the compression or due to the extension of the spring. We can say that $k$ is a restoring force which tries to restore the spring to its equilibrium position. The spring constant is expressed in terms of $N/m$.