Initially :
\(\frac{1}{4}\)=1−\(\frac{300}{T_H}\)
⇒ TH = 400 K
Finally, : Efficiency becomes \(\frac{1}{2}\)
∴\(\frac{1}{2}\)=1−\(\frac{300}{T_H}\)
∴TH=600 K
⇒ The temperature of the source increases by 200°C.
The correct option is (B) : Increases by 200°C.

Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to

From the second law of thermodynamics, two important results are derived where the conclusions are taken together to constitute Carnot’s theorem. It may be stated in the following forms.
