Question

The work done in twisting a steel wire of length $25\,cm$ and radius $2\,mm$ through will be$45^\circ $$\left( {\eta = 8 \times {{10}^{10}}\,N\,{m^{ - 2}}} \right)$
A. $2.48\,joule$
B. $3.1\,joule$
C. $15.47\,joule$
D. $18.79\,joule$

Answer 1

Hint:We are given a steel wire of length $25\,cm$ and radius $2\,mm$. Also, we are given the coefficient of rigidity in the question. Therefore, it is clear that we will use the formula of work done which will be in the form of torsional rigidity. Therefore, at first, we will calculate the torsional rigidity and then the work done for twisting the steel wire.

Formula used:
The formula used for calculating the work done for twisting the wire is given below
$W = \dfrac{1}{2}C{\theta ^2}$
Here, $W$ is the work done, $C$ is the torsional rigidity and $\theta $ is the angular displacement.
Now, the formula of the shear modulus is given below
$C = \dfrac{{\pi \eta {r^4}}}{{2l}}$
Here, $C$ is the torsional rigidity, $\eta $ is the coefficient of rigidity, $r$ is the radius of the wire and $l$ is the length of the wire.

Complete step by step answer:
Consider a steel wire of length $25\,cm$ and radius $2\,mm$. Now, consider that we have twisted the wire through an angle of $45^\circ $.
Therefore, the length of the wire, $l = 25\,cm = 0.25\,m$
Also, the radius of wire, $r = 2\,mm = 2 \times {10^{ - 3}}m$
Also, the angle through which we have twisted the wire is given below
$\theta = 45 \times \dfrac{\pi }{{180}}$
$ \Rightarrow \,\theta = \dfrac{\pi }{4}$
Now, Now, the formula of the shear modulus used in the formula of work done is given below
$C = \dfrac{{\pi \eta {r^4}}}{{2l}}$
$ \Rightarrow \,C = \dfrac{{3.14 \times 8 \times {{10}^{10}} \times {{\left( {2 \times {{10}^{ - 3}}} \right)}^4}}}{{2 \times 0.25}}$
$ \Rightarrow \,C = \dfrac{{3.14 \times 8 \times {{10}^{10}} \times 16 \times {{10}^{ - 12}}}}{{0.5}}$
$ \Rightarrow \,C = 8.042\,Nm$
Now, the formula used for calculating the work done for twisting the wire is given below
$W = \dfrac{1}{2}C{\theta ^2}$
$ \Rightarrow \,W = \dfrac{1}{2} \times 8.042 \times {\left( {\dfrac{\pi }{4}} \right)^2}$
$ \Rightarrow \,W = 4.021 \times {\left( {\dfrac{{3.14}}{4}} \right)^2}$
$ \therefore \,W = 2.48\,Joule$
Therefore, the work done for twisting a wire of steel is $2.48\,Joule$.

Hence, option A is the correct option.

Note:While solving these types of questions, just remember to change the smaller units into bigger units. In this question, we have changed the units of radius and length of the wire into bigger units. We can also say that the coefficient of rigidity is given in $meter$, that is why, we have changed the units in $meters$.