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1.Units, Dimensions and Measurement
hard
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
A$\,\eta \cdot \left( {\frac{{S\Delta \theta }}{h}} \right)\left( {\frac{1}{{\rho g}}} \right)$
B$\,\left( {\frac{{S\Delta \theta }}{{\eta h}}} \right)\left( {\frac{1}{{\rho g}}} \right)$
C$\,\frac{{S\Delta \theta }}{{\eta h}}$
D$\eta \,\frac{{S\Delta \theta }}{h}$
(JEE MAIN-2015)
Solution
$\begin{array}{l}
Let\,\frac{Q}{A} = {\eta ^a}{\left( {\frac{{S\Delta \theta }}{h}} \right)^b}{\left( {\frac{1}{{\rho g}}} \right)^c}\\
{\rm{Using}}\,{\rm{dimensional}}\,method\\
M{T^{ – 3}} = {\left[ {M{L^{ – 1}}{T^{ – 1}}} \right]^a}{\left[ {L{T^{ – 2}}} \right]^b}{\left[ {{M^{ – 1}}{L^2}{T^2}} \right]^c}\\
or,\,M{T^{ – 3}} = \left[ {{M^{a – c}}{L^{ – a + b + 2c}}{T^{ – a – 2b + 2c}}} \right]\\
Equating\,powers\,and\,solving\\
we\,get,a = 1,b = 1,c = 0\\
\therefore \frac{Q}{A} = \eta \frac{{S\Delta \theta }}{h}
\end{array}$
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}$
Standard 11
Physics
