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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

Dimensional analysis of the equation (velocity) ^ x =...

Dimensional analysis of the equation
(velocity)x=(pressuredifference)3/2×(density)−3/2{\left(velocity\right)}^{x} = {\left(pressuredifference\right)}^{3 / 2} \times {\left(density\right)}^{- 3 / 2},
gives the value of xx as _______.

Correct answer

3

Step-by-step solution

(velocity)x=P3/2×ρ−3/2=(Pρ)2{( velocity )}^{\text{x}} = {\text{P}}^{3 / 2} \times {ρ}^{- 3 / 2} = {\left(\dfrac{\text{P}}{ρ}\right)}^{2}
=[M L−1T−2][M L−3]32= \dfrac{\left[{\text{M L}}^{- 1}{\text{T}}^{- 2}\right]}{\left[{\text{M L}}^{- 3}\right]}\dfrac{3}{2}
=[L2T−2]3/2= {\left[{L}^{2}{T}^{- 2}\right]}^{3 / 2}
=[L3T−3]= \left[{L}^{3}{T}^{- 3}\right]
=[L1T−1]3= {[ {L}^{1}{T}^{- 1} ]}^{3}
But \left[{L}^{1}{T}^{- 1}\right]isthedimensionofvelocity.<br/>is the dimension of velocity.<br />⇒ x = 3
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