Question

If power
$p = a - {x}^{2} \div b$
where x represents displacement find the dimensionof a and b.

Please give correct answer and uncopided from net.. ​

Answer 1

Given :

• P = a - x²/b

To Find :

• Dimensions of a and b

Solution :

We're given the equation as :

$\bigstar \: \boxed{\sf{P \: = \: \dfrac{a \: - \: x^2}{b}}}$

Where,

• P is power
• a and b are some quantities
• x is displacement

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Dimension of Power is $\sf{[ML^2 T^{-3}]}$

And, dimension formula of displacement is $\sf{[M^0 L T^0]}$

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Dimensional formula of a is equal to the dimensional formula of x² . Because only same quantities can be added or subtracted. So, dimensional formula of a is $\sf{[M^0 L^2 T^0]}$

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And, dimensional formula of b can be calculated by :

$\implies \sf{P \: = \: \dfrac{1}{b}} \\ \\ \implies \sf{b \: = \: \dfrac{1}{P}} \\ \\ \implies \sf{b \: = \: \dfrac{1}{[M L^2 T^{-3}]}} \\ \\ \implies \sf{b \: = \: [M^{-1} L^{-2} T^3]}$

Dimensional formula of b is $\sf{[M ^{-1} L^{-2} T^3]}$

Answer 2

Given :

• P = a - x²/b

To Find :

We have to find the dimensions of a and b ?

Solution :

We know the dimension formula of the power.

⇒ Power = M¹L²T ⁻³

→ M¹L²T⁻³ = a/b - x²/b

⊕ According to the principle of homoginity

→ a = x²

→ a = [M⁰L²T⁰]

∴ Dimension formula of a is M⁰L²T⁰.

$\rule{200}{2}$

→ P = 1/b

→  M¹L²T ⁻³ = 1/b

→ b = 1/ M¹L²T ⁻³

→ b = [M⁻¹L⁻²T³]