Radiation and Newton's Law of Cooling | JEE Main previous year questions with Solutions

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Topic Question Set

Q 1 :
The temperature of a body in air falls from 40°C to 24°C in 4 minute. The temperature of the air is 16°C. The temperature of the body in the next 4 minutes will be [2025]

14/3°C
28/3°C
56/3°C
42/3°C

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2

3

4

(3)

$\frac{T_{2} - T_{1}}{t} = K [T_{avg} - T_{s}]$

$T_{1} = 24 ° C; T_{2} = 40 ° C; t = 4, T_{s} = 16 ° C$

$\frac{40 - 24}{4} = K [32 - 16]$

$K = \frac{4}{16} = \frac{1}{4}$

Now $\frac{24 - T}{4} = K [\frac{T + 24}{2} - 16]$

$24 - T = \frac{T - 8}{2}$

$\frac{3 T}{2} = 28 \Rightarrow T = \frac{56}{3} ° C^{}$

Q 2 :
A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is: [2025]

$\frac{13}{5} t$
$\frac{10}{13} t$
$\frac{13}{10} t$
$\frac{5}{13} t$

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2

3

4

(1)

By using average form of Newton's law of cooling

$\frac{90 - 80}{t} = c (\frac{90 + 80}{2} - 20)$ ... (i)

and $\frac{80 - 60}{t_{1}} = c (\frac{80 + 60}{2} - 20)$ ... (ii)

From (i) and (ii)

$t_{1} = \frac{13}{5} t$

Q 3 :
A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where x = ________.

(Given $σ = 6.0 \times 10^{- 8} {Wm}^{- 2} K^{- 4}, π = 3.14$ and assume that the emissivity of wire material is same at all wavelength.) [2025]

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5

L = 10 cm, d = 0.5 mm, T = 1727°C = 2000 K

Power, P = 94.2 W

$P = ε σ A T^{4}$

$94.2 = ε \times (16 \times 10^{- 8}) (π d L) {(2000)}^{4}$

$94.2 = ε \times (6 \times 10^{- 8}) (3.14) (0.5) (10^{- 3}) (10 \times 10^{- 2}) {(2000)}^{4}$

$ε = \frac{5}{8} \Rightarrow x = 5$

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