Question

Two hot bodies \(B_{1}\) and \(B_{2}\) have temperature \(100^{\circ }\text{C}\) and \(80^{\circ }\text{C}\) respectively at \(t=0\). The temperature of the surroundings is \(40^{\circ }\text{C}\). The ratio of the respective rates of cooling \(R_{1}\) and \(R_{2}\) of these two bodies at \(t=0\) will be

• A. \(R_{1}:R_{2}=3:2\)
• B. \(R_{1}:R_{2}=5:4\)
• C. \(R_{1}:R_{2}=2:3\)
• D. \(R_{1}:R_{2}=4:5\)

Hint:

By Newton's law of cooling, rate of cooling is proportional to the temperature excess over the surroundings.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Newton's law of cooling says the rate of cooling is proportional to the temperature difference with the surroundings.

Step 2: Temperature Excess
For \(B_1\): \(100-40=60^{\circ}\text{C}\). For \(B_2\): \(80-40=40^{\circ}\text{C}\).

Step 3: Ratio
\(\dfrac{R_1}{R_2}=\dfrac{60}{40}=\dfrac{3}{2}\).

Final Answer:
The ratio of the cooling rates is 3 to 2. \[ \boxed{\text{(A) }R_{1}:R_{2}=3:2} \]