Test if the following equations are dimensionally correct-a h -2 S cos- prgb u - P - rh 0-c V - Pr 4 t -8 n ld v -1-2 - mg - l -where h - height- S - surface tension- - density- P - pressure- V - volume- - coefficient of viscosity- v - frequency and I - moment of inertia-

(a) h = 2S cos θ/ρ rg LHS = h = [L] Surface Tension, S = F/L = MT^ 2/L = [MT^ 2] Density, ρ = M/V = [ML^ 3 T^0] Radius, r = [L], g = [LT^ 2] Now, RHS = 2S cos ...

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Standard XII

Physics

Rotational Power

Test if the f...

Question

Test if the following equations are dimensionally correct :

(a) $h = \frac{2 S c o s θ}{ρ r g}$ ,

(b) $u = \sqrt{\frac{P}{r h o}}$ ,

(c) $V = \frac{π P r^{4} t}{8 η l}$ ,

(d) $v = \frac{1}{2 π \sqrt{\frac{m g l}{I}}}$ ;

where h = height, S = surface tension, $ρ$ = density, P = pressure, V = volume, $η$ = coefficient of viscosity, v = frequency and I = moment of inertia.

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Solution

(a) $h = \frac{2 S c o s θ}{ρ r g}$

LHS = h = [L]

Surface Tension, $S = \frac{F}{L} = \frac{M T^{- 2}}{L} = [M T^{- 2}]$

Density, $ρ = \frac{M}{V} = [M L^{- 3} T^{0}]$

Radius, $r = [L], g = [L T^{- 2}]$

Now, RHS = $\frac{2 S c o s θ}{ρ r g}$

$= \frac{[M L^{- 2}]}{[M L^{- 3} T^{0}] [L] [L T^{- 2}]}$

$= [M^{0} L^{1} T^{0}] = [L]$

Hence the relation is correct

(b) Velocity, $v = \sqrt{(\frac{P}{ρ})}$

LHS = Demension of v

$= [L T^{- 1}]$

Dimension of $P = \frac{F}{A} = [M L^{- 1} T^{- 2}]$

Dimension of $ρ = \frac{m}{V} = [M L^{- 3}]$

RHS = $\sqrt{\frac{P}{ρ}}$ .

$= {[\frac{[M L^{- 1} T^{- 2}]}{[M L^{- 3}]}]}^{\frac{1}{2}}$

$= [L^{2} T^{- 2}]^{\frac{1}{2}}$

$= [L T^{- 1}]$ = LHS

Hence the relation is correct.

(c) $V = \frac{π P r^{4} t}{(8 η l)}$

LHS = Dimension of V = $[L^{3}]$

Dimension of P = $[M L^{- 1} T^{- 2}]$ ,

$r^{4} = [L^{4}], t = [T]$

Coefficient of viscosity

$= [M L^{- 1} T^{- 1}]$

Dimensions of RHS $= \frac{(π P r^{4} t)}{(8 η l)}$

$= \frac{[M L^{- 1} T^{- 2}] [L^{4}] [T]}{[M L^{- 1} T^{- 1}] [L]}$

$= L^{3} =$ LHS

Hence the relation is correct.

(d) $v = \frac{1}{2 π} \sqrt{(\frac{m g l}{I})}$

RHS = dimension of $v = [T^{- 1}]$

RHS = $\sqrt{(\frac{m g l}{I})}$

$= \sqrt{\frac{[M] [L T^{- 2}] [L]}{M L^{2}}}$

$= {[\frac{[M L^{2} T^{- 2}]}{M L^{2}}]}^{\frac{1}{2}}$

= $[T^{- 1}]$ = LHS

Hence the relation is correct.

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Similar questions

Q. Test if the following equations are dimensionally correct:
(a)

h = \frac{2 S cosθ}{ρ r g},

(b)

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Test if the following equations are dimensionally correct : (a) h=2Scosθρrg, (b) u=√Prho, (c) V=πPr4t8ηl, (d) v=12π√mglI; where h = height, S = surface tension, ρ = density, P = pressure, V = volume, η = coefficient of viscosity, v = frequency and I = moment of inertia.