JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 4)
The temperature of a body in air falls from $40^{\circ} \mathrm{C}$ to $24^{\circ} \mathrm{C}$ in 4 minutes. The temperature of the air is $16^{\circ} \mathrm{C}$. The temperature of the body in the next 4 minutes will be :
$\frac{28}{3}{ }^{\circ} \mathrm{C}$
$\frac{56}{3}{ }^{\circ} \mathrm{C}$
$\frac{42}{3}{ }^{\circ} \mathrm{C}$
$\frac{14}{3}{ }^{\circ} \mathrm{C}$
Explanation
$$\begin{aligned}
& \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right] \\
& \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\
& \frac{40-24}{4}=\mathrm{K}[32-16] \\
& \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\
& \text { Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\
& 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\
& \frac{3 \mathrm{~T}}{2}=28 \\
& \mathrm{~T}=\frac{56}{3} \mathrm{C}
\end{aligned}$$
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