Question

$➝\blue{\underline{\boxed{\sf Question :}}}$
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⟹ Check by the method of dimensions whether the following equation are correct or wrong r = 1/2l √T/M
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Answer 1

Question :

Check by the method of dimensions whether the following equation are correct or wrong $\sf V \ = \ \dfrac {l}{2 l} \times \dfrac {\sqrt {T}}{m}$

$\ \large\underbrace{\underline{\sf{Understanding\;the\;Question}}}$

• The question says that find whether the given formula is correct or not using the method of dimensions. We have to find the dimensions first and then we can easily conclude that the formula is correct or not.

Solution :

: $\implies \sf \dfrac {l}{2 l} \times \dfrac {\sqrt {T}}{m}$

Firstly, write the dimensions.

• LHS = $\sf V \ = \ [T^{-1}] \ = \ [M^0L^0T^{-1}]$
• RHS = $\sf V \ = \ \dfrac {l}{2 l} \times \dfrac {\sqrt {T}}{m}$

${\red{\sf Note : \ \dfrac {1}{2} \ has \ no \ dimensions.}}$

Here,

• Tension is force.
• m is the mass per unit length.

RHS = $\sf \dfrac {l}{L} \sqrt{\dfrac{MLT^{-2}}{ML^{-1}}}$

: $\implies \sf \dfrac {l}{L} \sqrt{L^2T^{-2}} \ = \ \dfrac {l}{L} [LT^{-1}]$

: $\implies \sf [M^0L^0T^{-1}]$

$\therefore$ By dimensionally, LHS = RHS.

$\therefore$ The formula is correct.

Answer 2

Answer:

v = 1/2l √T/m

Write the dimensions on either sides, we have

L.H.S. = v =[T-1] = [M0L0T-1]

R.H.S. = 1/2l = √T/m (As 1/2 has no dimensions)

Tension is force and m is mass per unit length,so

As, L.H.S. = R.H.S., dimensionally.

therefore, formulae is correct.