How much time will the heater take to increase the temperature of $100g$ water by ${50^ \circ }C$, if the resistance of the heating coil is $484\Omega $ and supply voltage is $220V$.
A. $210S$
B. $42S$
C. $200S$
D. $50S$

Hint: In the given data is, time taken by a heater to increase the temperature, as it has a heating coil and supply voltage is given and also the temperature of the water also given. For this calculation we are using joules law.

Complete step by step answer:
Given data is, heating coil and supply voltage from this we are writing them as,
Heating coil in resistance, $I = \dfrac{{220}}{{484}}$
Now, total heat generated in resistance of heating coil = Heat taken by water
In the above derivative heat taken by water values are an increase of temperature and water that means, $100g$ and ${50^ \circ }C$. Total heat generated in resistance of heating coil is assumed as joules law,

Joule's law: Heat produced by electric current and flowing through resistance at certain time which is proportional to it $\left( {{I^2}Rt} \right)$
Now we are applying all the values,
Therefore, ${I^2}Rt = 100 \times 50$
Already we have I value whereas R is nothing but the resistance of heating coil,
${\left( {\dfrac{{220}}{{484}}} \right)^2} \times 484 \times t = 100 \times 50 \\
\Rightarrow t = \dfrac{{100 \times 50 \times 484 \times 484}}{{220 \times 220 \times 484}} \\
\therefore t = 50\sec $
We have found the time taken by the heater to heat the water is $50\sec $.

Hence, the correct option is D.

Note: we have discussed the time taken by the heater to heat water at some temperature having their resistance in the heating coil and some supply voltage. For this kind of derivation we have used the joule's law, from that formula only we have proved the time taken by the heater is $50\sec $.