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Read the following statements:
A. When small temperature difference between a liquid and its surrounding is doubled the rate of loss of heat of the liquid becomes twice.
B. Two bodies P and Q having equal surface areas are maintained at temperature $$10°C$$ and $$20°C$$. The thermal radiation emitted in a given time by P and Q are in the ratio $$1:1.15$$.
C. A Carnot Engine working between $$100 \text{ K}$$ and $$400 \text{ K}$$ has an efficiency of $$75\%$$.
D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below:
We need to check which statements (A, B, C, D) are correct.
Statement A: When the temperature difference is doubled, rate of heat loss becomes twice.
By Newton's law of cooling, the rate of heat loss is proportional to the temperature difference between the liquid and surroundings:
$$\frac{dQ}{dt} \propto \Delta T$$
If $$\Delta T$$ is doubled, the rate of heat loss doubles. Statement A is correct.
Statement B: Two bodies P and Q at 10°C and 20°C have thermal radiation ratio 1:1.15.
By Stefan's law, power radiated $$\propto T^4$$ (in Kelvin).
$$T_P = 10 + 273 = 283 \text{ K}$$, $$T_Q = 20 + 273 = 293 \text{ K}$$
$$\frac{E_P}{E_Q} = \left(\frac{283}{293}\right)^4 = \left(\frac{283}{293}\right)^4$$
$$\frac{283}{293} \approx 0.9659$$
$$(0.9659)^4 \approx 0.871$$
So $$E_P : E_Q \approx 0.871 : 1 \approx 1 : 1.15$$. Statement B is correct.
Statement C: Carnot engine between 100 K and 400 K has efficiency 75%.
Carnot efficiency: $$\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 - \frac{100}{400} = 1 - 0.25 = 0.75 = 75\%$$
Statement C is correct.
Statement D: When temperature difference is quadrupled, rate of heat loss becomes twice.
By Newton's law of cooling, rate $$\propto \Delta T$$. If $$\Delta T$$ is quadrupled, rate becomes 4 times (not twice).
Statement D is incorrect.
Statements A, B, and C are correct.
The correct answer is Option A: A, B, C only.
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