Question

A rubber cord has a cross-sectional area $1m{m^2}$ and total unstretched length $10cm$. It is stretched to $12cm$ and then released to project a mass of $80g$. The Young’s modulus for rubber is $5 \times {10^8}N/{m^2}$. Find the velocity of mass (in $m/s$)?
A) $5$
B) $3$
C) $7$
D) $2$

Answer 1

When a wire is stretched its potential energy decreases and kinetic energy increases. If we have both potential and kinetic energy then we can easily find the velocity of the motion.
If a mass of some gram or kilograms is added to the wire then the wire is stretched and longitudinal stress produced, which is equal to the applied force. So we can find the average potential energy which is stored in the wire.

Formula used: for finding force which is equal to the stress. so we can apply Young's modulus.
According to the hooke's law, Young's modulus of any wire is equal to the ratio of longitudinal stress to the longitudinal strain produced in the wire.
$Y = \dfrac{{Stress(longitudinal)}}{{Strain(longitudinal)}} \\\ \Rightarrow Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} \\\ \Rightarrow Y = \dfrac{{FL}}{{A\Delta L}} \\\$

Complete step by step solution: -
In this question, when a rubber cord is stretched to $12cm(12cm = 12 \times {10^{ - 2}}m)$, the original length of the wire is $10cm(10cm = 10 \times {10^{ - 2}}m)$, the change in length is $2cm(2cm = 2 \times {10^{ - 2}}m)$ and the cross section area of the wire is $1m{m^2}(1m{m^2} = 1 \times {10^{ - 6}}m)$, then by using the Young’s modulus formula, we have-
$Y = \dfrac{{FL}}{{A\Delta L}}$
Or we can write it in the terms of force-
$F = YA\left( {\dfrac{{\Delta L}}{L}} \right)$
Substituting the values in SI system, we get-

$$F = 5 \times {10^8} \times 1 \times {10^{ - 6}} \times \left( {\dfrac{{2 \times {{10}^{ - 2}}}}{{10 \times {{10}^{ - 2}}}}} \right) \\\ \Rightarrow F = 5 \times {10^2} \times \dfrac{1}{5} \\\ \Rightarrow F = 100N \\\$$

If the average potential energy when the body is released from the rubber cord is $\dfrac{{0 + U}}{2} = \dfrac{U}{2}$
Because when mass is released the potential energy becomes zero.
Potential energy of the wire is equal to the work done in stretching wire to the $\Delta L$(length difference).
When a mass of 20kg is projected by this rubber cord then the potential energy will be converted into kinetic energy. So,
Average potential energy$=$kinetic energy
$\Rightarrow \dfrac{1}{2} \times$ Work done$= \dfrac{1}{2}m{v^2}$
$\Rightarrow \dfrac{1}{2} \times F\Delta L = \dfrac{1}{2} \times m{v^2}$
Now, substituting the values in the above equation, we have-

$$\Rightarrow \dfrac{1}{2} \times 100 \times 2 \times {10^{ - 2}} = \dfrac{1}{2} \times 80 \times {10^{ - 3}} \times {v^2} \\\ \Rightarrow {v^2} = \dfrac{{200 \times {{10}^{ - 2}}}}{{80 \times {{10}^{ - 3}}}} \\\ \Rightarrow {v^2} = 2.5 \times 10 \\\ \Rightarrow {v^2} = 25 \\\ \Rightarrow v = 5m/s \\\$$

Hence, the velocity is projected mass is$5m/s$.

Therefore, option A is correct.

Note: - It should be remembered that the potential energy is considered as average potential energy otherwise the answer may be wrong. If there is any change in the dimension of wire, then the other changes occur respectively. But Young’s modulus remains constant. If the mass of the body is increased then the velocity will be decreased.