Temperature gradient in a rod of $0.5\ m$ long is $80{}^\circ C/m$. If the temperature of hotter end of the rod is $30{}^\circ C$, then the temperature of the cooler end is
(a) $0{}^\circ C$
(b) $-10{}^\circ C$
(c) $10{}^\circ C$
(d) $40{}^\circ C$

Hint: We know that temperature gradient is given by $k=\dfrac{\Delta T}{x}$, which means the difference of temperatures at two ends divided by the length or distance x. Using this formula, we will find the value of temperature of the cooler ends.

Formula used: $k=\dfrac{\Delta T}{x}$

Complete step-by-step answer:
In the question it is given that the temperature gradient in a rod of $0.5\ m$ long is $80{}^\circ C/m$. If the temperature of the hotter end of the rod is $30{}^\circ C$ and we are asked to find the temperature at the hotter end. So, based on this we will draw a figure for our simplicity.

Now, as shown in the figure, let us assume that ${{T}_{1}}$ is the temperature of rod at hotter end and ${{T}_{2}}$ is the temperature of rod at colder end. Now, the length of rod is 0.5 m as given in the question, so this values can be given mathematically as, ${{T}_{1}}=30{}^\circ C$, $L=0.5\ m$ value of temperature gradient is $k=80{}^\circ C/m$.
Now, we know that temperature gradient can be given by the formula,
$k=\dfrac{\Delta T}{x}$
So, on substituting the values we will get,
$k=\dfrac{{{T}_{1}}-{{T}_{2}}}{x}\Rightarrow 80=\dfrac{30-{{T}_{2}}}{0.5}$
$\Rightarrow 80\times 0.5=30-{{T}_{2}}$
$\Rightarrow {{T}_{2}}=-10{}^\circ C$
Thus, we can say that the temperature at the cooler end of the rod is $-10{}^\circ C$.
Hence, option (b) is correct.

Note: Student should not get confuse between Heat rate Q and Temperature gradient k as the formula of Q is $\dfrac{dQ}{dt}=\dfrac{bA}{x}\left( {{T}_{1}}-{{T}_{2}} \right)$ and formula of k is $k=\dfrac{\Delta T}{x}$, both are different terms though temperature gradient is a part of heat rate Q. So, student must take care and read the question properly and work accordingly.