The ends of a uniform metal rod of length 100 cm and area of cross-section 2 cm2 are maintained at 0 °Cand 100 °C. At the mid point of the rod, heat is supplied at a constant rate of 40Js−1. If the temperature gradient on the higher temperature side of the rod in steady state is 50 x° cm−1, then the value of x is (Thermal conductivity of the metal =400 J s−1 m−1 K−1)Your Answer IncorrectCorre

The problem is asking for the value of x in the temperature gradient on the higher temperature side of the rod in steady state, which is given as 50x° cm−1.

The formula for heat conduction is Q = kA(T2-T1)/d, where Q is the heat transferred, k is the thermal conductivity, A is the cross-sectional area, T2 and T1 are the temperatures at the two ends, and d is the distance between the two ends.

In this case, we know that Q = 40 J/s, k = 400 J s−1 m−1 K−1, A = 2 cm2 = 2*10^-4 m2, T2 = 100°C, T1 = 0°C, and d = 100 cm = 1 m.

Substituting these values into the formula, we get 40 = 400 * 2*10^-4 * (100 - 0) / 1.

Solving for x, we get x = 1.

Therefore, the value of x in the temperature gradient on the higher temperature side of the rod in steady state is 1.

The problem is asking for the value of x in the temperature gradient on the higher temperature side of the rod in steady state, which is given as 50x° cm−1.

The heat transfer through the rod can be calculated using the formula for heat conduction:

Q = kA(T2-T1)/d

where: Q is the heat transfer, k is the thermal conductivity, A is the cross-sectional area, T2 and T1 are the temperatures at the two ends of the rod, and d is the distance between the two ends.

In steady state, the heat supplied at the midpoint of the rod (40 J/s) must be equal to the heat conducted through the rod.

So, we can set up the equation:

40 J/s = 400 J s−1 m−1 K−1 * 2 cm2 * (100°C - 0°C) / 50 cm

Solving for x, we get:

x = 40 J/s / (400 J s−1 m−1 K−1 * 2 cm2 * 100°C / 50 cm)

x = 1

So, the value of x is 1.