Question

A uniform rod of length $2m$ and mass $5kg$ is lying on a horizontal surface. The work done in raising one end of the rod with the other end in contact with the surface until the rod makes an angle ${30^ \circ }$ with the horizontal is,
$\left( A \right){\text{ 25J}}$
$\left( B \right){\text{ 50J}}$
$\left( C \right){\text{ 25}}\sqrt 3 {\text{J}}$
$\left( D \right){\text{ 50}}\sqrt 3 {\text{J}}$

Answer 1

Here in this question we have to find the work done and as we know that the work is done is calculated by taking out the difference between the initial and final energy. So by using this we will get to the solution for obtaining the work done.

Formula used:
$\text{Work done = Final potential energy - Initial potential energy}$

Complete step by step answer
Here in this question, we can see that the initial potential energy is zero and the final potential will be equal to,
$\Rightarrow Final\,P.E = mg \times \dfrac{{L\sin {{30}^ \circ }}}{2}$
So on solving the above equation, we will get the equation as
$\Rightarrow Final\,P.E = \dfrac{{MgL}}{4}$
Now on substituting the values, we get
$\Rightarrow Final\,P.E = \dfrac{{5 \times 10 \times 2}}{4}$
So on solving the above equation, we will get the equation as
$\Rightarrow Final\,P.E = 25J$
Therefore, by using the formula of work done, it will be calculated as
$\Rightarrow Work\,done = 25J - 0$
And on solving it, it will be equal to
$\Rightarrow Work\,done = 25J$
Therefore, the work done in raising one end of the rod with the other end in contact with the surface until the rod makes an angle ${30^ \circ }$ with the horizontal is $25J$ .
Hence, the option $\left( A \right)$ is correct.

Note:
In a simple line if we need to understand the work then it will be as when there is a change in a system causing energy transfer, the transfer of energy into or out of the system is called the energy in the transfer. When a system has energy being relocated into or out of it, it can be of two procedures, either in the form of heat or in the form of work.