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The dimensions of permittivity $ε_{0}$ are

$A^{2} T^{2} M^{- 1} L^{- 3}$

$A^{2} T^{4} M^{- 1} L^{- 3}$

$A^{- 2} T^{- 4} M L^{3}$

$A^{2} T^{- 4} M^{- 1} L^{- 3}$

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Explanation

$F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$\Rightarrow ε_{0} = \frac{| q_{1} | | q_{2} |}{[F] [r^{2}]} = \frac{[A^{2} T^{2}]}{[M L T^{- 2}] [L^{2}]} = [A^{2} T^{4} M^{- 1} L^{- 3}]$

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Dimensions of the following three quantities are the same

Work, energy, force

Velocity, momentum, impulse

Potential energy, kinetic energy, momentum

Pressure, stress, coefficient of elasticity

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Explanation

[Pressure] = [Stress] = [coefficient of elasticity] = $[M L^{- 1} T^{- 2}]$

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The dimensions of Planck's constant and angular momentum are respectively

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

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

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

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

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Explanation

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

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Let $[ε_{0}]$ denotes the dimensional formula of the permittivity of the vacuum and $[μ_{0}]$ that of the permeability of the vacuum. If M = mass, L = length, T = Time and I = electric current, then

$[ε_{0}] = M^{- 1} L^{- 3} T^{2} I$

$[ε_{0}] = M^{- 1} L^{- 3} T^{4} I^{2}$

$[μ_{0}] = M L T^{- 2} I^{- 3}$

$[μ_{0}] = M L^{2} T^{- 1} I$

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Explanation

$F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{R^{2}} \Rightarrow ε_{0} = \frac{q_{1} q_{2}}{4 π F R^{2}} \Rightarrow ε_{0} = \frac{[I T^{2}]}{[M L T^{- 2}] [L^{2}]} \Rightarrow ε_{0} = [M^{- 1} L^{- 3} T^{4} I^{2}]$

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Dimensions of CR are those of

Frequency

Energy

Time period

Current

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Explanation

Capacitance × Resistance = $\frac{Charge}{Potential} \times \frac{Volt}{amp}$

= $\frac{amp \times second \times Volt}{Volt \times amp} = Second$

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The physical quantity that has no dimensions

Angular Velocity

Linear momentum

Angular momentum

Strain

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Explanation

Strain has no dimensions.

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

Stress

Young's Modulus

Pressure

All the above three quantities

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Explanation

All the above three quantities

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Dimensions of magnetic field intensity is

$[M^{0} L^{- 1} T^{0} A^{1}]$

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

$[M L^{0} T^{- 2} A^{- 1}]$

$[M L T^{- 2} A]$

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Explanation

$B = \frac{F}{I L} = \frac{[M L T^{- 2}]}{[A] [L]} = [M T^{- 2} A^{- 1}]$

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The force F on a sphere of radius ‘a’ moving in a medium with velocity ‘v’ is given by $F = 6 π η a v$ . The dimensions of $η$ are

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

$M T^{- 1}$

$M L T^{- 2}$

$M L^{- 3}$

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Explanation

$η = \frac{F}{a v} = \frac{[M L T^{- 2}]}{[L] [L T^{- 1}]} = [M L^{- 1} T^{- 1}]$

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Which physical quantities have the same dimension

Couple of force and work

Force and power

Latent heat and specific heat

Work and power

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Explanation

Couple of force = $| \vec{r} \times \vec{F} | = [M L^{2} T^{- 2}]$

Work = $[\vec{F} . \vec{d}] = [M L^{2} T^{- 2}]$

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