Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Test If the Following Equations Are Dimensionally Correct: V = π P R 4 T 8 η L - Physics

Sum

Test if the following equation is dimensionally correct:
\[V = \frac{\pi P r^4 t}{8 \eta l}\]

where v = frequency, P = pressure, η = coefficient of viscosity.

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Solution

\[V = \frac{\left( \pi P r^4 t \right)}{\left( 8 \eta l \right)}\]
Volume, [V] = [L3]
Pressure,
\[P = \frac{\left[ F \right]}{\left[ A \right]} = \left[ {ML}^{- 1} T^{- 2} \right]\]
[r]= [L] and [t] = [T]
Coefficient of viscosity,
\[\left[ \eta \right] = \frac{\left[ F \right]}{6\pi\left[ r \right]\left[ v \right]} = \frac{\left[ {MLT}^{- 2} \right]}{\left[ L \right]\left[ {LT}^{- 1} \right]} = \left[ {ML}^{- 1} T^{- 1} \right]\]
Now,
\[\frac{\pi\left[ P \right] \left[ r \right]^4 \left[ t \right]}{8\left[ \eta \right]\left[ l \right]} = \frac{\left[ {ML}^{- 1} T^{- 2} \right] \left[ L^4 \right] \left[ T \right]}{\left[ {ML}^{- 1} T^{- 1} \right] \left[ L \right]} = \left[ L^3 \right]\]
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 1 Introduction to Physics
Exercise | Q 18.3 | Page 10
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