A conducting rod of length 1 ~m has area of cross-section 10^-3 ~m^2 . One end is immersed in boiling water (100^° C ) and the other end in Ice (0^° C ) . If coefficient of thermal conductivity of rod is 96 cal / s . m^° C and latent heat for ice is 8 × 10^4 cal / kg then the amount of ice which will melt in one minute is

Correct answer: (B) 7.2 × 10^-3 ~kg. Q=(KA(Δ θ) t)/(ℓ)=mL m=(KA(Δ θ) t)/(ℓ L)=frac96 × 10^-3 × 100 × 601 × 8 × 10^4 =7.2 × 10^-3 ~kg

Q116 / 150 · MHT CET 2021 (20 Sep Shift 1) · Physics

A conducting rod of length $1 \mathrm{~m}$ has area of cross-section $10^{-3} \mathrm{~m}^2$. One end is immersed in boiling water $\left(100^{\circ} \mathrm{C}\right)$ and the other end in Ice $\left(0^{\circ} \mathrm{C}\right)$. If coefficient of thermal conductivity of rod is $96 \mathrm{cal} / \mathrm{s} . \mathrm{m}^{\circ} \mathrm{C}$ and latent heat for ice is $8 \times 10^4 \mathrm{cal} / \mathrm{kg}$ then the amount of ice which will melt in one minute is

A
$5.4 \times 10^{-3} \mathrm{~kg}$
Correct answer
B
$7.2 \times 10^{-3} \mathrm{~kg}$
C
$1.8 \times 10^{-3} \mathrm{~kg}$
D
$3.6 \times 10^{-3} \mathrm{~kg}$

Solution

$\begin{aligned} & \mathrm{Q}=\frac{\mathrm{KA}(\Delta \theta) \mathrm{t}}{\ell}=\mathrm{mL} \\ & \mathrm{m}=\frac{\mathrm{KA}(\Delta \theta) \mathrm{t}}{\ell \mathrm{L}}=\frac{96 \times 10^{-3} \times 100 \times 60}{1 \times 8 \times 10^4} \\ & =7.2 \times 10^{-3} \mathrm{~kg}\end{aligned}$

Hence, the correct answer is option: 'B'.

Solution

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