JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the length of the water pipe, L = 1 m
The cross-sectional area of the water pipe, A=1cm2=10−4m2
The temperature of the ice = -10ºC
Current passing in the conductor, I = 0.5 A
Resistance of the conductor, R = 4 kΩ
The latent heat of fusion for ice, ,Li=3.33×105J∕kg
The density of the ice, d =1000‌kg∕m3
The specific heat of the ice, cp,ice=2×103J∕kg
Heat required to melt the ice at 10°C to 0°C
Q=mcpΔT+mLi
Q=dVcpΔT+dVLf
(∵V=A×L)
= 35300 J
According to the Joule’s law of heating,
H=I2Rt
⇒ 35300=(0.5)2(4000)(t)
t = 35.3 s
Thus, the minimum time required to melt the ice is 35.3 s.