Physical quantities & their dimensions - Fluid Mechanics

Q.  Match the following physical quantities in Group 1 with their dimensions in Group 2.

1. Work done (Energy) (W) ----------------------------- A. [M L 2 T – 3 ]
2. Power (P) ------------------------------------------------ B. [M L – 1 T   – 1 ]
3. Momentum (M) ---------------------------------------- C. [M L 2 T – 2 ]
4. Modulus of elasticity (E) ----------------------------- D. [M L T – 1 ]
5. Dynamic viscosity (μ) --------------------------------- E. [M L – 1 T – 2 ]

- Published on 05 Aug 15

a. 1-(C), 2-(A), 3-(D), 4-(E), 5-(B)
b. 1-(A), 2-(C), 3-(D), 4-(E), 5-(B)
c. 1-(C), 2-(A), 3-(E), 4-(B), 5-(D)
d. 1-(D), 2-(E), 3-(B), 4-(A), 5-(C)

ANSWER: 1-(C), 2-(A), 3-(D), 4-(E), 5-(B)
 

    Discussion

  • Pankaj   -Posted on 05 Oct 15
    1. Workdone = Force x Displacement

    W = Mass x Acceleration x Displacement

    W = M x (L / T2) x L

    W = [M L 2 T – 2 ]


    2. Power = Workdone / Time

    P = [M x (L / T2) x L] / [T]

    P = [M L 2 T – 3 ]


    3. Momentum = Mass x Velocity

    Momentum = M x [L / T]

    Momentum = [M L T – 1 ]


    4. Modulus of elasticity = Stress / Strain

    E = (Force / Area) / Strain

    Strain is dimensionless term

    E = (Mass x Acceleration) / Area

    E = [M] [L T -2] [L 2]

    E = [M L – 1 T – 2 ]


    5. Dynamic Viscosity = Shear stress / Shear Strain

    μ = (Force / Area) / (du / dy)

    μ = (Mass x Acceleration) / (Area x du/dy)

    μ = [M L T -2] / [L 2] [L T -1] [L -1]

    μ = [M L – 1 T – 1 ]

Post your comment / Share knowledge


Enter the code shown above:
 
(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.)