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Measurement MCQs for NEET — Physics Questions with Answers

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The period of a body under SHM i.e. presented by $T = P^{a} D^{b} S^{c}$ ; where P is pressure, D is density and S is surface tension. The value of a, b and c are

$- \frac{3}{2}, \frac{1}{2}, 1$

$- 1, - 2, 3$

$\frac{1}{2}, - \frac{3}{2}, - \frac{1}{2}$

$1, 2, \frac{1}{3}$

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Explanation

By substituting the dimension of each quantity we get $T = {[M L^{- 1} T^{- 2}]}^{a} {[L^{- 3} M]}^{b} {[M T^{- 2}]}^{c}$

By solving we get a = – 3/2, b = 1/2 and c = 1

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Which of the following pairs of physical quantities has the same dimensions

Work and power

Momentum and energy

Force and power

Work and energy

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Explanation

$D i m e n s i o n o f w o r k = [M^{1} L^{2} T^{- 2}] D i m e n s i o n o f e n e r g y = [M^{1} L^{2} T^{- 2}]$

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The velocity of a freely falling body changes as $g^{p} h^{q}$ where g is acceleration due to gravity and h is the height. The values of p and q are

$1, \frac{1}{2}$

$\frac{1}{2}, \frac{1}{2}$

$\frac{1}{2}, 1$

1, 1

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Explanation

$v \propto g^{p} h^{q}$ (given)

By substituting the dimension of each quantity and comparing the powers in both sides we get $[L T^{- 1}] = {[L T^{- 2}]}^{p} {[L]}^{q}$

$\Rightarrow p + q = 1, - 2 p = - 1, ∴ p = \frac{1}{2}, q = \frac{1}{2}$

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Which one of the following does not have the same dimensions

Work and energy

Angle and strain

Relative density and refractive index

Planck constant and energy

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Explanation

[Planck constant] = $[M L^{2} T^{- 1}]$ and

[Energy] = $[M L^{2} T^{- 2}]$

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Dimensions of frequency are

$M^{0} L^{- 1} T^{0}$

$M^{0} L^{0} T^{- 1}$

$M^{0} L^{0} T$

$M T^{- 2}$

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Explanation

Frequency $= \frac{1}{T} = [M^{0} L^{0} T^{- 1}]$

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Which one has the dimensions different from the remaining three

Power

Work

Torque

Energy

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Explanation

Power = $\frac{Energy}{Time}$

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A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $η$ . After some time the velocity of the ball attains a constant value known as terminal velocity $v_{T}$ . The terminal velocity depends on (i) the mass of the ball m, (ii) $η$ , (iii) r and (iv) acceleration due to gravity g. Which of the following relations is dimensionally correct

$v_{T} \propto \frac{m g}{η r}$

$v_{T} \propto \frac{η r}{m g}$

$v_{T} \propto η r m g$

$v_{T} \propto \frac{m g r}{η}$

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Explanation

By substituting dimension of each quantity in R.H.S. of option we get $[\frac{m g}{η r}] = [\frac{M \times L T^{- 2}}{M L^{- 1} T^{- 1} \times L}]$ = $[L T^{- 1}]$ .

This option gives the dimension of velocity.

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The quantity $X = \frac{ε_{0} L V}{t} : ε_{0}$ is the permittivity of free space, Lis length, V is potential difference and t is time. The dimensions of X are same as that of

Resistance

Charge

Voltage

Current

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Explanation

[ $ε_{0} L$ ] = [C]

$∴ X = \frac{ε_{0} L V}{t} = \frac{C \times V}{t} = \frac{Q}{t}$ = current

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$μ_{0}$ and $ε_{0}$ denote the permeability and permittivity of free space, the dimensions of $μ_{0} ε_{0}$ are

$L T^{- 1}$

$L^{- 2} T^{2}$

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

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

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Explanation

(b) $C = \frac{1}{\sqrt{μ_{0} ε_{0}}}$ $\Rightarrow μ_{0} ε_{0} = (\frac{1}{C^{2}})$ (where C = velocity of light)

$∴ [μ_{0} ε_{0}] = L^{- 2} T^{2}$

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The expression $[M L^{2} T^{- 2}]$ represents

Pressure

Kinetic energy

Momentum

Power

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Explanation

Kinetic energy= $\frac{1}{2} m v^{2} = [M L^{2} T^{- 2}]$

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