Question

2.1

Answer 1

Answer:

Explanation:

$v = T^a\times \rho^b$

Using dimensional formula on both sides ,

LT⁻¹ = $(ML^{-1}T^{-2})^a(ML^{-3})^b$

= $M^{a+b}L^{-a-3b}T^{-2a}$

equating the power of equal terms

- 2 a = -1

a = 0.5

-0.5 -3b = 1

3b = -1.5

b = -0.5

a + b = 0

Hence

a = 0.5

b = -0.5

$v=\sqrt{\frac{T}{\rho} }$