5 Marks Questions

Question 15 Marks

Test if the following equations are dimensionally correct:

$\text{h}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}$
$\text{u}=\sqrt{\frac{\text{P}}{\rho}}$
$\text{V}=\frac{\pi\text{Pr}^4\text{t}}{8\eta l}$
$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{mg}l}{\text{I}}}$

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

Answer

$\text{h}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}$

$\text{LHS}=[\text{L}]$

Surface tension $=\text{S}=\frac{\text{F}}{\text{I}}=\frac{\text{MLT}^{-2}}{\text{L}}=[\text{MT}^{-2}]$

Density $=\rho=\frac{\text{M}}{\text{V}}=[\text{ML}^{-3}\text{T}^{0}]$

Radius $=\text{r}=[\text{L],g = [LT}^{-2}]$

$\text{RHS}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}=\frac{[\text{MT}^{-2}]}{[\text{ML}^{-3}\text{T}^0][\text{L}][\text{LT}^{-2}]}=[\text{M}^0\text{L}^1\text{T}^0]=[\text{L}]$

$\text{LHS = RHS}$

So, the relation is correct

$\text{v}=\sqrt{\frac{\text{P}}{\rho}}$ where v = velocity

$\text{LHS = Dimension of v = [LT}^{-1}]$

Dimension of $\text{p}=\frac{\text{F}}{\text{A}}=[\text{ML}^{-1}\text{T}^{-2}]$

Dimension of $\rho=\frac{\text{m}}{\text{v}} = \text{[ML}^{-2}]$

$\text{RHS}=\sqrt{\frac{\text{p}}{\rho}}=\sqrt{\frac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]}}=[\text{L}^2\text{T}^{-2}]^{\frac{1}{2}}=[\text{LT}^{-1}]$

So, the relation is correct.

$\text{V}=\frac{(\pi\text{pr}^4\text{t})}{(8\eta\text{I})}$

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

Dimension of $\text{p}=[\text{ML}^{-1}\text{T}^{-2}],\text{r}^4=[\text{L}^4],\text{t}=[\text{T}]$

Coefficient of viscosity $=[\text{ML}^{-1}\text{T}^{-1}]$

$\text{RHS}=\frac{\pi\text{pr}^4\text{t}}{8\eta\text{I}}=\frac{[\text{ML}^{-1}\text{T}^{-2}][\text{L}^4][\text{T}]}{[\text{ML}^{-1}\text{T}^{-1}][\text{L}]}$

So, the relation is correct.

$\text{v}=\frac{1}{2\pi}\sqrt{(\text{mgI}/\text{I})}$

$\text{LHS = dimension of v = [T}^{-1}]$

$\text{RHS}=\sqrt{(\text{mgI}/\text{I})}=\sqrt{\frac{[\text{M}][\text{LT}^{-2}][\text{L}]}{[\text{ML}^2]}}=[\text{T}^{-1}]$

$\text{LHS = RHS}$

So, the relation is correct.

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Question 25 Marks

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

Answer

Frequency $\text{f = KL}^{\text{a}}\text{F}^{\text{b}}\text{M}^{\text{c}}$ M = Mass/unit length, L = length, F = tension (force) Dimension of $\text{f}=[\text{T}^{-1}]$ Dimension of right side,$\text{L}^{\text{a}}=[\text{L}^{\text{a}}],\text{F}^{\text{b}}=[\text{MLT}^{-2}]^{\text{b}},\text{M}^{\text{c}}=[\text{ML}^{-1}]^{\text{c}}$
$\therefore[\text{T}^{-1}]=\text{K[L}]^{\text{a}}[\text{MLT}^{-2}]^{\text{b}},[\text{ML}^{-1}]^{\text{c}}$
$\text{M}^0\text{L}^0\text{T}^{-1}=\text{KL}^{\text{b + c}}\text{L}^{\text{a + b}-\text{c}}\text{T}^{-2\text{b}}$
Equating the dimensions of both sides,$\therefore\text{b + c}=0 \ ...(1)$
$-\text{c + a + b} =0 \ ...(2)$
$-2\text{b}=-1\ ...(3)$
Solving the equations we get$\text{a}=-1,\text{b}=\frac{1}{2}$ and $\text{c}=\frac{-1}{2}$
$\therefore$ So frequency $\text{f = KL}^{-1}\text{F}^{\frac{1}{2}}\text{M}^{-\frac{1}{2}}=\frac{\text{K}}{\text{L}}\text{F}^{\frac{1}{2}}\text{M}^{-\frac{1}{2}}=\frac{\text{K}}{\text{L}}=\sqrt{\frac{\text{F}}{\text{M}}}$

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