Two hot bodies B₁ and B₂ have temperature 100° C and 80° C respectively at t = 0. The temperature of the surroundings is 40° C. The ratio of the respective rates of cooling R₁ and R₂ of these two bodies at t = 0 will be (a) R₁: R₂ = 3: 2 (b) R₁: R₂ = 5: 4 (c) R₁: R₂ = 2: 3 (d) R₁: R₂ = 4: 5

Answer: (a). Newton's law of cooling: rate ∝ (T - T₀). (R₁)/(R₂) = (100 - 40)/(80 - 40) = 60/40 So the answer is (a) R₁: R₂ = 3: 2.

Two hot bodies B₁ and B₂ have temperature 100° C and 80° C… | MHT CET 2026 (13 May, Shift 2)

Q1MHT CET 2026 · 13 MAY · SHIFT 2Medium

Newton's Law of Cooling

Two hot bodies $B_1$ and $B_2$ have temperature $100^\circ$ C and $80^\circ$ C respectively at $t = 0$ . The temperature of the surroundings is $40^\circ$ C. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t = 0$ will be

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Answer: (a)

Newton's law of cooling: rate $\propto (T - T_0)$ .
$\frac{R_1}{R_2} = \frac{100 - 40}{80 - 40} = \frac{60}{40}$
So the answer is (a) $R_1 : R_2 = 3 : 2$ .

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Newton's Law of Cooling question from MHT CET 2026 (13 May, Shift 2)

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