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

Concept: In dimensional analysis, we express every physical quantity in terms of the fundamental dimensions \(M\), \(L\), \(T\), and \(\Theta\). The correct option must ultimately reduce to the dimension of time \([T]\).

Step 1: Write the dimensions of heat capacity \(C\). Heat capacity is defined as: \[ C=\frac{\text{Heat}}{\text{Temperature}} \] Since heat (energy) has dimensions: \[ [ML^{2}T^{-2}] \] Therefore, \[ [C]=[ML^{2}T^{-2}\Theta^{-1}] \]

Step 2: Write the dimensions of thermal conductivity \(K\). Using Fourier's law: \[ \frac{Q}{t}=KA\frac{\Delta T}{L} \] Therefore, \[ [K] = \frac{\left(ML^{2}T^{-3}\right)L} {L^{2}\Theta} \] \[ [K] = [MLT^{-3}\Theta^{-1}] \]

Step 3: Determine dimensions of Option (A). \[ \frac{C}{Kr} \] Dimensions: \[ \frac{ML^{2}T^{-2}\Theta^{-1}} { (MLT^{-3}\Theta^{-1})(L) } \] \[ = \frac{ML^{2}T^{-2}\Theta^{-1}} {ML^{2}T^{-3}\Theta^{-1}} \] \[ =T \] Hence Option (A) has dimensions of time.

Step 4: Verify remaining options. For Option (B): \[ \frac{mC}{Kr} = \frac{M\cdot ML^{2}T^{-2}\Theta^{-1}} {ML^{2}T^{-3}\Theta^{-1}} \] \[ =MT \] Not time. For Option (C): \[ \frac{mC}{K} = \frac{M\cdot ML^{2}T^{-2}\Theta^{-1}} {MLT^{-3}\Theta^{-1}} \] \[ =MLT \] Not time. For Option (D): \[ \frac{Cr}{K} = \frac{ML^{2}T^{-2}\Theta^{-1}\cdot L} {MLT^{-3}\Theta^{-1}} \] \[ =L^{2}T \] Not time.

Conclusion: Only Option (A) reduces exactly to the dimension of time. \[ \boxed{\frac{C}{Kr}} \] Therefore, option (A) is correct.