Question

the potential energy of a particle varies with distance x from origin as U=
$a \sqrt{x } \div x {}^{2} + b$
where a & b are dimensional constants then dimensional formula for a&b​

Answer 1

$\sf \bold{Given\::}$

$\sf U\:=\:\dfrac{a\sqrt{x} }{x^{2} } + b$

$\sf \bold{To\:find\::}$

Dimensions of a & b

$\sf \bold{Method\:used}:}$

$\bullet$Only those quantities can be added whose dimensions are same

$\sf \bold{Dimension\::}$

$\sf \implies \:\: Potential\: energy\:(U)\:=\: [ML^{2} T^{-2} ]\\\\\sf \implies\:\: Distance\:(s)\:=\:[L]$

$\sf \bold{Solution\::}$

✪ $\sf U\:=\:\dfrac{a\sqrt{x} }{x^{2} } + b$

$\sf \implies \: Dimension\:of\:U = \: Dimension\:of\:\:\dfrac{a\sqrt{x} }{x^{2} } + b$

✪ ALSO,as only quantity with same dimensions can be added,

$\sf \implies \: Dimension\:of\: \dfrac{a\sqrt{x} }{x^{2} } = \: Dimension\:of\: b$

✪ Dimension of U = Dimension of b

$\sf \implies Dimension\: of\:\bold{b} = [ML^2T^{-2}]$

✪ Dimension of U = Dimension of $\sf \dfrac{a\sqrt{x} }{x^{2} }$

Let, Dimension of a = $\sf [M^{p}L^qT^r]$

$\sf \implies [ML^2T^{-2}] = \dfrac{[M^pL^qT^r]\times [L]^{1/2}}{[L]^{2}} \\\\\\\sf \implies [ML^2T^{-2}] = [M^{p}\times L^{(q+\frac{1}{2}-2)}\times T^r]$

Comparing LHS & RHS

$\sf \bullet\: M^1 \:=\:M^q\\\sf \implies q = 1\\\\\sf \bullet\: L^2 \:=\:L^{(q-\frac{3}{2})}\\\\\sf \implies 2 \:=\:q-\frac{3}{2}\\\\\sf \implies q\:=2+\frac{3}{2}\\\\\sf \implies q = \frac{7}{2}\\\\\sf \bullet T^{-2} = T^r\\\sf \implies r = -2$

Therefore , Dimension of a = $\sf [ML^{\frac{7}{2}} T^{-2}]$

$\sf \bold{ANSWER\::}$

Dimensions of a = $\sf \bold{ [ML^{\frac{7}{2}} T^{-2}]}$

Dimensions of b = $\sf \bold{[ML^2T^{-2}]}$

Answer 2

Given:

the potential energy of a particle varies with distance x from origin as :

$\rm \: U = \dfrac{a \sqrt{x} }{ {x}^{2} } + b$

To find:

Dimensional formula of "a" and "b" in terms of basic physical quantities ?

Calculation:

We need to use the following rules in dimensional analysis :

• Only quantities with same dimensions can be added.

• Dimensions of quantities on either side of an equation has to be same.

Following these rules , we can say that dimensions of "b" is same as that of "U":

$\therefore \: \bigg \{b \bigg \} = \bigg \{U \bigg \}$

$\boxed{ \implies\: \bigg \{b \bigg \} = \bigg \{M{L}^{2} {T}^{ - 2} \bigg \}}$

Again , dimensions of "a√x/x²" is same as that of "U":

$\therefore \: \bigg \{ \dfrac{a \sqrt{x} }{ {x}^{2} } \bigg \} = \bigg \{U \bigg \}$

$\implies\: \bigg \{ \dfrac{a }{ {x}^{ \frac{3}{2} } } \bigg \} = \bigg \{U \bigg \}$

$\implies\: \bigg \{ \dfrac{a }{ {L}^{ \frac{3}{2} } } \bigg \} = \bigg \{M{L}^{2} {T}^{ - 2} \bigg \}$

$\implies\: \bigg \{ a \bigg \} = \bigg \{M{L}^{(2 + \frac{3}{2} )} {T}^{ - 2} \bigg \}$

$\boxed{ \implies\: \bigg \{ a \bigg \} = \bigg \{M{L}^{ \frac{7}{2}} {T}^{ - 2} \bigg \}}$

Hope It Helps.