Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Applications of Dimensional Analysis›Question 72

JEE Main 2015 (11 Apr, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2015 (11 Apr, Shift 1)

Q72JEE Main 2015 · 11 APR · SHIFT 1Difficult

Applications of Dimensional Analysis

A beaker contains a fluid of density ρ\rho kg / m3, specific heat S J/kgoC and viscosity η\eta. The beaker is filled up to height h. To estimate the rate of heat transfer per unit area ( Q˙/A\dot{Q}/A) by convection when beaker is put on a hot plate, a student proposes that it should depend on η⋅(SΔθh)\eta \cdot \left(\frac{S\Delta\theta}{h}\right) and (1ρg)\left(\frac{1}{\rho g}\right) when Δθ\Delta\theta (inoC) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for (Q˙/A\dot{Q}/A) is:

Show answer and solution

Answer: (a)

  1. [Q˙/A]=ML2T−3L2=MT−3[\dot{Q}/A] = \frac{ML^{2}T^{-3}}{L^{2}} = MT^{-3}
  2. [η]=ML−1T−1[\eta] = ML^{-1}T^{-1}, [SΔθ]=L2T−2[S\Delta\theta] = L^{2}T^{-2}, [h]=L[h] = L
  3. [ηSΔθh]=ML−1T−1⋅L2T−2⋅L−1=MT−3\left[\eta\frac{S\Delta\theta}{h}\right] = ML^{-1}T^{-1}\cdot L^{2}T^{-2}\cdot L^{-1} = MT^{-3}, which already matches; multiplying by 1ρg\frac{1}{\rho g} would spoil it.
  4. So the answer is (a) ηSΔθh\eta\frac{S\Delta\theta}{h}.
Report a mistake

More questions on Applications of Dimensional Analysis