Question

A wire of length $50cm$ and cross sectional area of $1sq.mm$ is extended by $1mm$. The required work will be $\left( {Y = 2 \times {{10}^{10}}N{m^2}} \right)$

(A) $6 \times {10^{ - 2}}J$
(B) $4 \times {10^{ - 2}}J$
(C) $2 \times {10^{ - 2}}J$
(D) $1 \times {10^{ - 2}}J$

Answer 1

Hint: First start with finding the force in terms of Young’s modulus, area and length of wire. All these values are already given in the question, put it in the formula of force and after finding force use the formula of work done in terms of force and displaced distance and get the required answer.



Complete answer:
Let start with the given information from the question:
Area, $A = 1m{m^2} = 1 \times {10^{ - 6}}{m^2}$
Length of the wire, $l = 50cm = 0.5m$
Change in length, $\Delta l = 1mm = 1 \times {10^{ - 3}}m$
Young’s modulus, $Y = 2 \times {10^{10}}N/{m^2}$

Now we know force is given by:
$F = \dfrac{{YA\Delta l}}{l}$
Putting all the values in above question, we get;
$F = \dfrac{{2 \times {{10}^{10}} \times {{10}^{ - 6}} \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 1}}}} = 40N$

Now, work done is given by:
$W = F.\Delta x$
$W = 40 \times 1 \times {10^{ - 3}}$
$W = 4 \times {10^{ - 2}}J$

Hence the correct answer is Option(B).



Note: Be careful about all the units of all the variable quantities given in the question, change all the units in meters and then put all the values in the respective formula and finally you will get the required answer. Here all the values were already given in the question. If any of the values were missing then we do not get the answer.