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.
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) 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.