Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Test If the Following Equations Are Dimensionally Correct: H = 2 S C O S θ Prg , - Physics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Test if the following equation is dimensionally correct:
$h = \frac{2S cos\theta}{\text{ prg }},$
where h = height, S = surface tension, ρ = density, I = moment of interia.

Advertisement Remove all ads

#### Solution

$h = \frac{2S \cos \theta}{\text{ prg }}$
Height, [h] = [L]
Surface Tension,
$\left[ S \right] = \frac{\left[ F \right]}{\left[ L \right]} = \frac{\left[ {MLT}^{- 2} \right]}{\left[ L \right]} = \left[ {MT}^{- 2} \right]$
Density,
$\left[ \rho \right] = \frac{\left[ M \right]}{\left[ I \right]} = \left[ {ML}^{- 3} T^0 \right]$
Radius, [r] = [L], [g]= [LT−2]
Now,
$\frac{2\left[ S \right]\cos \theta}{\left[ \rho \right]\left[ r \right]\left[ g \right]} = \frac{\left[ {MT}^{- 2} \right]}{\left[ {ML}^{- 3} T^0 \right] \left[ L \right] \left[ {LT}^{- 2} \right]} = \left[ M^0 L^1 T^0 \right] = \left[ L \right]$
Since the dimensions of both sides are the same, the equation is dimensionally correct.

Concept: What is Physics?
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 1 Introduction to Physics
Exercise | Q 18.1 | Page 10
Share
Notifications

View all notifications

Forgot password?