Question

Modulus of elasticity is dimensionally equivalent to:
A. Stress
B. Surface tension
C. Strain
D. Coefficient of viscosity.

Answer 1

Hint: The dimension of a physical quantity may be defined as the number of times the fundamental units of mass, length, time, temperature, electric current, luminous intensity and moles appear in the physical quantity.

Complete step by step solution:
$Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{F/A}}{{\Delta l/l}} = \dfrac{{F \times l}}{{A \times \Delta l}}$
Putting the dimensions of the physical quantities
$\therefore $Dimension of$Y = \dfrac{{ML{T^{ - 2}} \times L}}{{{L^2} \times L}}$
$ = M{L^{ - 1}}{T^{ - 2}}$

Now from option –
(a) Stress$ = \dfrac{{Force}}{{area}} = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}$

(b) Surface tension$ = \dfrac{{Force}}{{length}}$
$ = \dfrac{{ML{T^{ - 2}}}}{L} = M{T^{ - 2}}$

(c) Strain$ = \dfrac{{Change{\text{ in length}}}}{{Original{\text{ length}}}}$
$ = \dfrac{L}{L} = 1 = M^\circ L^\circ T^\circ $

(d) $F = 6\pi \eta ru$
$6\pi \to $ is dimensionless
So, dimension of $\eta $
$ = \dfrac{F}{{ru}} = \dfrac{{ML{T^{ - 2}}}}{{L \times L{T^{ - 1}}}}$
$ = M{L^{ - 1}}{T^{ - 1}}$

So option (a) is correct.

Additional information:
1. Dimension of force is$\left[ {ML{T^{ - 2}}} \right]$
2. Dimension of length type physical quantities is$\left[ L \right]$
3. Dimension area is$\left[ {{L^2}} \right]$
4. Dimension of velocity is$\left[ {L{T^{ - 1}}} \right]$

Note: The numerical values such as $2,3,5\pi $, etc. used in equations has no dimensions.