A quantity X is given by (IFv2MLL4) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:


Option 1 -

Coefficient of viscosity

Option 2 -

Planck's constant

Option 3 -

Energy density

Option 4 -

Force constant

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 3


    Detailed Solution:

    x = (IFV²)/ (WL? )
    I = [ML²]
    F = [MLT? ²]
    V² = [L² T? ²]
    W = [ML² T? ²]
    Q? = [L? ]
    X = [ML? ¹ T? ²]

Similar Questions for you

R
Raj Pandey

P = α β l o g e ( k t β x )

k t β x = Dimensionless

β = k t x = [ M L 2 T 2 k 1 ] [ k ] [ L ]

α β = dimensionless

a = dimensionless of b

a = MLT-2

V
Vishal Baghel

Using F = MA = m V T

m = FTV1

V
Vishal Baghel

F = ηAdv / dx
η = F (dx/dv) (1/A) = m × a × time × (1/area) = MV/A

A
alok kumar singh

Since, x 2 α k T should be dimensionless.

So, dimension of  α , [ α ] = L 2 M L 2 T 2 = M 1 T 2

Dimension of  α β 2 should be that of W.

So, [ α β 2 ] = M L 2 T 2

[ β 2 ] = M L 2 T 2 M 1 T 2 = M 2 L 2 T 4 [ β ] = M L T 2

A
alok kumar singh

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

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