Question

A heater boils 1 kg of water in lime $t_1$. Another heater boils it in lime $t_2$. If both are connected in series, the combination will boils same water in time

• A. $\frac{t_1+t_2}{2}$
• B. $(t_1+t_2)$
• C. $\frac{t_1{t}_2}{t_1+t_2}$
• D. $\frac{t_1{t}_2}{t_1-t_2}$

Answer 1

The correct option is B $(t_1+t_2)$

We know
$P=\frac{Q}{t}$ ......................................................(1)
here, both heaters are connected in series and the combination will boils same water in time $t_1$ and $t_2$ with powers $P_1=\frac{Q}{t_1}$ and $P_2=\frac{Q}{t_2}$

Hence,
$\frac{1}{P^{\prime }}=\frac{1}{P_1}+\frac{1}{P_2}$
$\frac{1}{P^{\prime }}=\frac{1}{\left(\frac{Q}{t_1}\right)}+\frac{1}{\left(\frac{Q}{t_2}\right)}$
$\frac{1}{P^{\prime }}=\frac{t_2}{Q}+\frac{t_2}{Q}$

$\Rightarrow P^{\prime }=\frac{Q}{t_1+t_2}$

Comparing with (1), we get,
$t=t_1+t_2$