Question

Consider a metal cylinder of heat capacity \(C\), thermal conductivity \(K\), mass \(m\) and radius \(r\). The combination that will have the dimension of time is:

• A. \(\dfrac{C}{Kr}\)
• B. \(\dfrac{mC}{Kr}\)
• C. \(\dfrac{mC}{K}\)
• D. \(\dfrac{Cr}{K}\)

Hint:

In dimensional analysis questions, first write the dimensions of every physical quantity separately. Most mistakes occur when students try to substitute directly without writing dimensions explicitly.

Answer 1

The Correct Option is A

Step 1: Decide on the strategy.
We want the combination whose dimensions reduce to time $[T]$. Rather than testing all four options blindly, let us build the dimensions of $C$ and $K$, then form the ratio $C/K$ and see what extra length factor is needed.
Step 2: Dimensions of heat capacity $C$.
Heat capacity is heat per unit temperature, and heat is energy $[ML^2T^{-2}]$. Hence \[ [C] = [ML^2 T^{-2} \Theta^{-1}] \]
Step 3: Dimensions of thermal conductivity $K$.
From Fourier's law $\dfrac{Q}{t} = K A \dfrac{\Delta T}{L}$, solving for $K$ gives power times length over (area times temperature): \[ [K] = \frac{[ML^2 T^{-3}][L]}{[L^2][\Theta]} = [MLT^{-3}\Theta^{-1}] \]
Step 4: Form the bare ratio $C/K$.
\[ \frac{[C]}{[K]} = \frac{ML^2 T^{-2}\Theta^{-1}}{MLT^{-3}\Theta^{-1}} = L\,T \] So $C/K$ already carries the dimension of time, but it has one stray factor of length $L$ too many.
Step 5: Cancel the extra length with $r$.
Dividing by the radius $r$ (dimension $L$) removes exactly that surplus length: \[ \left[\frac{C}{Kr}\right] = \frac{L\,T}{L} = T \] which is pure time.
Step 6: Confirm the others fail.
Multiplying by mass $m$ injects an unwanted $[M]$ (ruling out $mC/Kr$ and $mC/K$), and $Cr/K = L^2 T$ keeps a leftover length (ruling out $Cr/K$). Only $C/Kr$ is clean, agreeing with option (A).
\[ \boxed{\frac{C}{Kr}} \]