What is the dimension of the coefficient of thermal conductivity in the equation dQ/dt = KA(dT/dx)?

The question "What is the dimension of the coefficient of thermal conductivity in the equation dQ/dt = KA(dT/dx)?" relates to dimensional analysis in physics.

The given equation is Fourier's law of heat conduction:
dQ/dt = KA(dT/dx)

Where:
- dQ/dt = rate of heat transfer (energy per time)
- K = coefficient of thermal conductivity
- A = area
- dT/dx = temperature gradient

Rearranging for K:
K = (dQ/dt × dx) / (A × dT)

Now, substitute the dimensional formula:
- Heat (Q) has dimension of energy = [ML²T⁻²]
So, dQ/dt = [ML²T⁻³].
- dx = [L]
- Area (A) = [L²]
- Temperature difference (dT) = [Θ]

Now,
K = [ML²T⁻³] × [L] / ([L²] × [Θ]) = [MLT⁻³Θ⁻¹]

So, the dimension of coefficient of thermal conductivity is [MLT⁻³Θ⁻¹].

This tells us that thermal conductivity depends directly on mass and length, inversely on cubic time, and inversely on temperature. Students should note this as an important derivation in Class 11 physics and competitive exams like JEE/NEET.