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Standard XII

Physics

Physical Quantities

What is the d...

Question

What is the dimensional formula of dynamic viscosity?

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

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[m^{1} L^{1} T^{- 1}]

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[m^{1} L^{- 1} T^{- 1}]

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[m^{1} L^{1} T^{1}]

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Solution

The correct option is B $[m^{1} L^{- 1} T^{- 2}]$
Dynamic viscosity is given by:
$μ = τ . \frac{d y}{d c}$
here, $τ =$ shearing stress $\frac{d y}{d c} =$ velocity gradient
$m = [m^{1} L^{- 1} T^{- 2}] [T^{1}] = [m^{1} L^{- 1} T^{- 1}]$ same coefficient of viscosity.

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Similar questions

Q. The dimensional formula for viscosity is

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

. One poise, i.e, C.G.S. unit of viscosity is:

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

is the dimensional formula of

Þ a = 1/2

Now putting a = 1/2 in equation (3)

Þ b = 3/2

On substituting value a, b and c in equation (1) T = k(d)^1/2 (r)^3/2 (s)^_1/2

Hence the correct answer is (c).

Level-3

Illustration 1. A spherical body of mass m and radius r is allowed to fall in a medium of viscosity h. The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity (v) is called time constant (t). Dimensionally t can be represented by (5 min)

(a) (b)

Solution.

The dimensional formulae of mass = [M]

The dimensional formulae of radius = [L]

The dimensional formulae of viscosity = M¹L^_1T¹

So lets check the dimensions of each option

The dimension of is therefore option (a) is not correct.

The dimensions of is therefore option (b) is also not correct.

The dimension of is therefore option (c) is also not correct.

Hence the correct answer is (d).

Illustration 2. The damping force for a body moving through a fluid is proportional to velocity. What will be the dimensional formula for the constant of proportionality ? (3 min)

(a) MLT^_1 (b) ML⁰T^_1

Q. Dimensional formula

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

doesnot represent the physical quantity

Q. Dimensional formula

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

does not represent the physical quantity

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What is the dimensional formula of dynamic viscosity?

The correct option is B [m1L−1T−2]Dynamic viscosity is given by: μ=τ.dydchere, τ= shearing stress dydc= velocity gradientm=[m1L−1T−2][T1]=[m1L−1T−1] same coefficient of viscosity.

The correct option is B $[m^{1} L^{- 1} T^{- 2}]$
Dynamic viscosity is given by:
$μ = τ . \frac{d y}{d c}$
here, $τ =$ shearing stress $\frac{d y}{d c} =$ velocity gradient
$m = [m^{1} L^{- 1} T^{- 2}] [T^{1}] = [m^{1} L^{- 1} T^{- 1}]$ same coefficient of viscosity.