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Dimensional Analysis

Question

The dimensions of a modulus of elasticity are:
[ $M L T^{- 2}$ ]
[ $M L^{- 1} T^{- 2}$ ]
[ $M L^{- 2} T^{- 2}$ ]
[ $M^{2} L^{- 1} T^{- 2}$ ]

A

[

M^{2} L^{- 1} T^{- 2}

]

B

[

M L^{- 1} T^{- 2}

]

C

[

M L^{- 2} T^{- 2}

]

D

[

M L T^{- 2}

]

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Solution

Verified by Toppr

Young modulus, $γ = \frac{s t r e s s}{s t r a i n} = \frac{F o r c e / A r e a}{Δ L / L} = \frac{[M L T^{- 2}] / [M^{o} L^{2} T^{o}]}{[M^{o} L T^{o}] / [M^{o} L T^{o}]} = [M L^{- 1} T^{- 2}]$

Hence, Dimension is $[M L^{- 1} T^{- 2}]$ .

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Similar Questions

Q1

The dimensions

[M L^{- 1} T^{- 2}]

may be of

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Q2

Match List I with List II and select the correct answer.

\begin{matrix} List I(Physical quantity) & List II (Dimensions) A. Force & {1. T}^{- 1} B. Angular velocity & {2. MLT}^{- 2} C. Torque & {3. ML}^{- 1} T^{- 2} D. Stress & {4. ML}^{- 1} T^{- 1} {5. ML}^{2} T^{- 2} \end{matrix}

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Q3

A physical quantity

X

is expressed as velocity

= \sqrt{(\frac{X}{density})}

. The dimensions of

X

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Q4

The dimension ML⁻¹T⁻² can correspond to
(a) moment of a force
(b) surface tension
(c) modulus of elasticity
(d) coefficient of viscosity

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Q5

Dimensional formula

M L^{- 1} T^{- 2}

represents the physical quantity of strain.

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