JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2015
- Hard
- MCQ
A beaker contains a fluid of density \(\rho \, kg / m^3\), specific heat \(S\, J / kg\,^oC\) and viscosity \(\eta \). The beaker is filled upto height \(h\). To estimate the rate of heat transfer per unit area \((Q / A)\) by convection when beaker is put on a hot plate, a student proposes that it should depend on \(\eta \,,\,\left( {\frac{{S\Delta \theta }}{h}} \right)\) and \(\left( {\frac{1}{{\rho g}}} \right)\) when \(\Delta \theta \) (in \(^oC\)) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for \((Q / A)\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\eta \,\frac{{S\Delta \theta }}{h}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Let\,\frac{Q}{A} = {\eta ^a}{\left( {\frac{{S\Delta \theta }}{h}} \right)^b}{\left( {\frac{1}{{\rho g}}} \right)^c}\\
{\rm{Using}}\,{\rm{dimensional}}\,method\\
M{T^{ - 3}} = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ {{M^{ - 1}}{L^2}{T^2}} \right]^c}\\
or,\,M{T^{ - 3}} = \left[ {{M^{a - c}}{L^{ - a + b + 2c}}{T^{ - a - 2b + 2c}}} \right]\\
Equating\,powers\,and\,solving\\
we\,get,a = 1,b = 1,c = 0\\
\therefore \frac{Q}{A} = \eta \frac{{S\Delta \theta }}{h}
\end{array}\)
Let\,\frac{Q}{A} = {\eta ^a}{\left( {\frac{{S\Delta \theta }}{h}} \right)^b}{\left( {\frac{1}{{\rho g}}} \right)^c}\\
{\rm{Using}}\,{\rm{dimensional}}\,method\\
M{T^{ - 3}} = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ {{M^{ - 1}}{L^2}{T^2}} \right]^c}\\
or,\,M{T^{ - 3}} = \left[ {{M^{a - c}}{L^{ - a + b + 2c}}{T^{ - a - 2b + 2c}}} \right]\\
Equating\,powers\,and\,solving\\
we\,get,a = 1,b = 1,c = 0\\
\therefore \frac{Q}{A} = \eta \frac{{S\Delta \theta }}{h}
\end{array}\)
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