Question

A steel rod (density $ = 8000\,kg/{m^3}$) has a length of 2 m. It is bolted at both ends between immobile supports. Initially, there is no tension in the rod because the rod just fits between the supports. If the tension is 100x (in N) that develops when the rod loses 3600J of heat, then find the value of x. Take specific heat capacity of steel as $450J/kg\;^\circ C$, coefficient of linear expansion $ = {10^{ - 5}}/^\circ C$, Young's modulus $ = 2 \times {10^{11}}{\text{ }}N/{m^2}$.

Answer 1

Hint: Determine the change in length of the rod when it loses heat. Since the rod is bolted at both ends, the tension that develops will be due to the change in length of the rod.

Formula used:
$\Rightarrow Q = mc\Delta T$ where $Q$ is the loss in heat, $m$ is the mass of the rod, $\Delta T$ is the change in temperature of the rod
$\Rightarrow \Delta L = L\alpha \Delta T$ where $\Delta L$ is the change in length of the rod of length $L$, coefficient of linear expansion $\alpha = {10^{ - 5}}/^\circ C$ due to temperature change $\Delta T$
$\Rightarrow Y = \dfrac{{TL}}{{A\Delta L}}$ where $Y$ is the young’s modulus, $T$ is the tension in the rod, $A$ is the cross-sectional area of the rod

Complete step by step solution:
When the rod is initially bolted, there is no tension in the rod. After it loses energy, the rod will try to lower its length however since it is bolted, instead, tension will be created in the rod. First, let us determine the change in temperature of the rod due to the loss of heat.
To use the formula $Q = mc\Delta T$, we need to first find the mass of the rod. Since density=mass x volume, we can write $m{\text{ = }}\rho \times Al$ that gives us
$\Rightarrow m = (8000)(2A) $
$\Rightarrow 16000A $
We can then calculate the change in temperature from
$Q = mc\Delta T$
$3600 = (16000A)(450)\Delta T$
On dividing both sides by $16000 \times 450$, we get
$A\Delta T = 5 \times {10^{ - 4}} \Rightarrow \Delta T = \dfrac{{5 \times {{10}^{ - 4}}}}{A}$
Since we don’t know the cross-sectional area of the rod, we will keep the temperature in this form.
Now to calculate the change in length due to the change in temperature, we can use $\Delta L = L\alpha \Delta T$
which gives us
$\Rightarrow \Delta L = \dfrac{{2 \times {{10}^{ - 5}} \times 5 \times {{10}^{ - 4}}}}{A} $
$\Rightarrow \dfrac{{{{10}^{ - 8}}}}{A} $
Then to calculate the tension in the rod, we use the relation $Y = \dfrac{{TL}}{{A\Delta L}}$ as $\Rightarrow 2 \times {10^{11}} = \dfrac{{T(2)}}{{A\Delta L}}$
Substituting the value of $\Delta L$ we derived earlier, we get
$\Rightarrow 2 \times {10^{11}} = \dfrac{{T(2)(A)}}{{A \times {{10}^{ - 8}}}}$
Solving for the tension gives us
$\Rightarrow T = 1000\,N$.

Note:
While the lack of information about the cross-sectional formula in the question makes it seem incomplete, we saw that it gets cancelled out in the problem and we can solve the question without it which means that the tension developed in the rod is independent of the cross-sectional area of the rod.