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Drift Velocity

The dimension...

Question

The dimensional formula of mobility is __________.

A

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

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B

M^{1} L^{0} T^{- 2} A^{- 1}

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C

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

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D

M^{- 1} L^{0} T^{2} A^{1}

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Solution

The correct option is C $M^{- 1} L^{0} T^{2} A^{1}$
Mobility of the charged particle is defined as the drift velocity of the particle per unit external electric field applied.
Mobility $μ = \frac{V_{d}}{E}$
Dimensional formula of mobility $μ = [M^{- 1} L^{0} T^{2} A^{1}]$
Correct answer is option D.

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Similar questions

tan A = \frac{1}{\sqrt{5}}

, If the value of

\frac{{cosec}^{2} A - {sec}^{2} A}{{cosec}^{2} A + {sec}^{2} A} i s \frac{1}{m}

m

Þ a = 1/2

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

Þ

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

Þ

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)

(c) (d) None of these

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. 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

[M L T^{- 1}]

is dimensional formula of

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