The work done in joules in increasing the extension of a spring of stiffness $10\,N/cm$ from $4\,cm$ to $6$$cm$ is:
A) $1$
B) $10$
C) $50$
D) $100$

Hint: According to that the electric force is proportional to the difference in potential energies between the starting position and the final position by moving a charge into an electric field. We know the two positions here so that we can quantify the work performed by spring rigidity.

Formula used:
Work done by spring is $\dfrac{1}{2}K\left( {{x_2}^2 - {x_1}^2} \right)$.
where, $K$ is the stiffness of the spring
${x_1}$ is initial position and ${x_2}$ is final position

Complete step by step solution:
Given by,
Spring stiffness $K = \dfrac{{10N}}{{cm}}$
Initial position ${x_1} = 4\,cm$
Final position ${x_2} = 6\,cm$
According to the work done by spring formula,
When a force that is applied to an object moves that object, work is done.
$\Rightarrow$ $\dfrac{1}{2}K\left( {{x_2}^2 - {x_1}^2} \right)$
The spring stiffness can be written as,$K = \dfrac{{10N}}{{cm}} = 1000N/m$
We convert the centimeter to meter,
Also,
Initial and final position can be written as,
${x_1} = 4\,cm = 0.04\,m$
${x_2} = 6\,cm = 0.06\,m$
Now,
Substituting the given value in above formula,
We get,
$\Rightarrow$ $K = \dfrac{1}{2}1000\left( {{{0.06}^2} - {{0.04}^2}} \right)$
We know that the work performed is equal to the spring's shift in potential energy.
On simplifying, Here,
$\Rightarrow$ $1.0\,J$
$1$ joule is defined as the work performed in displacing a body in the direction of force through \[1{\text{ }}m\] under the influence of \[1{\text{ }}N\] force.
It can be written as $1\,J$.

Hence, option A is the correct answer.

Note: When a force of one newton is applied over a distance of one meter, one joule of work is performed on an object. A watt-second is a derived energy unit equal to the joule. The power equal to the power of one watt sustained for one second is the watt-second.