Measurement NEET Questions — Free MCQ Practice

Measurement MCQs for NEET — Physics Questions with Answers

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In the following list, the only pair which have different dimensions, is

Linear momentum and moment of a force

Planck's constant and angular momentum

Pressure and modulus of elasticity

Torque and potential energy

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Explanation

Linear momentum = Mass × Velocity = $[M L T^{- 1}]$

Moment of a force = Force × Distance = $[M L^{2} T^{- 2}]$

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If R and L represent respectively resistance and self inductance, which of the following combinations has the dimensions of frequency [ Only for droppers]

$\frac{R}{L}$

$\frac{L}{R}$

$\sqrt{\frac{R}{L}}$

$\sqrt{\frac{L}{R}}$

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Explanation

$\frac{R}{L} = \frac{V / I}{V \times T / I} = \frac{1}{T} =$ Frequency

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If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be

$F A^{- 1} v$

$F v^{3} A^{- 2}$

$F v^{2} A^{- 1}$

$F^{2} v^{2} A^{- 1}$

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Explanation

$L \propto v^{x} A^{y} F^{z} \Rightarrow L = k v^{x} A^{y} F^{z}$

Putting the dimensions in the above relation

$[M L^{2} T^{- 1}] = k {[L T^{- 1}]}^{x} {[L T^{- 2}]}^{y} {[M L T^{- 2}]}^{z}$

$\Rightarrow [M L^{2} T^{- 1}] = k [M^{z} L^{x + y + z} T^{- x - 2 y - 2 z}]$

Comparing the powers of M, L and T

$z = 1$ …(i)

$x + y + z = 2$ …(ii)

$- x - 2 y - 2 z = - 1$ …(iii)

On solving (i), (ii) and (iii) $x = 3, y = - 2, z = 1$

So dimension of L in terms of v, A and f

$[L] = [F v^{3} A^{- 2}]$

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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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