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Measurement MCQs for NEET — Physics Questions with Answers

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What are the units of $K = 1 / 4 π ε_{0}$

C²N^–1m^–2

Nm²C^–2

Nm²C²

Unitless

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Explanation

Unit of $ε_{0} = C^{2} / N - m^{2}$

∴ Unit of K = Nm²C^–2

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The SI unit of surface tension is

Dyne/cm

Newton/cm

Newton/metre

Newton-metre

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Explanation

Newton/metre

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E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the dimension of $\frac{E l^{2}}{m^{5} G^{2}}$ are

Angle

Length

Mass

Time

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Explanation

$[E] = [M L^{2} T^{- 2}], [m] = [M], [l] = [M L^{2} T^{- 1}]$ and $[G] = [M^{- 1} L^{3} T^{- 2}]$ Substituting the dimension of above quantities in the given formula :

$\frac{E l^{2}}{m^{5} G^{2}}$ = $\frac{[M L^{2} T^{- 2}] {[M L^{2} T^{- 1}]}^{2}}{[M^{5}] {[M^{- 1} L^{3} T^{- 2}]}^{2}} = \frac{M^{3} L^{6} T^{- 4}}{M^{3} L^{6} T^{- 4}} = [M^{0} L^{0} T^{0}]$

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A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity η flowing per second through a tube of radius r and length l and having a pressure difference p across its end, is

$V = \frac{π p r^{4}}{8 η l}$

$V = \frac{π η l}{8 p r^{4}}$

$V = \frac{8 p η l}{π r^{4}}$

$V = \frac{π p η}{8 l r^{4}}$

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Explanation

Formula for viscosity $η = \frac{π p r^{4}}{8 V l} \Rightarrow V = \frac{π p r^{4}}{8 η l}$

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The velocity v (in cm/sec) of a particle is given in terms of time t (in sec) by the relation $v = a t + \frac{b}{t + c}$ ; the dimensions of a, b and c are

$a = L^{2}, b = T, c = L T^{2}$

$a = L T^{2}, b = L T, c = L$

$a = L T^{- 2}, b = L, c = T$

$a = L, b = L T, c = T^{2}$

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Explanation

From the principle of dimensional homogenity $[v] = [a t] \Rightarrow [a] = [L T^{- 2}]$ . Similarly [b] = [L] and [c] = [T]

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The position of a particle at time t is given by the relation $l = \frac{v_{o}}{α} (1 - e^{α t})$ , where v₀ is a constant and α > 0. The dimensions of v₀ and α are respectively

$M^{0} L^{1} T^{- 1}$ and T^–1

$M^{0} L^{1} T^{0}$ and T^–1

$M^{0} L^{1} T^{- 1}$ and LT^–2

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

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Explanation

Power of exponential quantity needs to be dimensionless. So

Dimension of $α t$ = $[M^{0} L^{0} T^{0}]$

∴ [α] = $[T^{- 1}]$

Also $1 - e^{α t} i s d i m e n s i o n l e s s$

Again $[\frac{v_{0}}{α}] = [L]$ so $[v_{0}] = [L T^{- 1}]$

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Dimensions of $\frac{1}{μ_{0} ε_{0}}$ , where symbols have their usual meaning, are

[LT^–1]

[L^–1T]

[L^–2T²]

[L²T^–2]

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Explanation

$C = \frac{1}{\sqrt{μ_{0} ε_{0}}} \Rightarrow \frac{1}{μ_{0} ε_{0}} = c^{2} = [L^{2} T^{- 2}]$

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The dimensions of $e^{2} / 4 π ε_{0} h c$ , where $e, ε_{0}, h$ and c are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively

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

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

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

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

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Explanation

$[e] = [A T],$ $\in_{0} = [M^{- 1} L^{- 3} T^{4} A^{2}],$ $[h] = [M L^{2} T^{- 1}]$ and $[c] = [L T^{- 1}]$

$∴ [\frac{e^{2}}{4 π \in_{0} h c}] = [\frac{A^{2} T^{2}}{M^{- 1} L^{- 3} T^{4} A^{2} \times M L^{2} T^{- 1} \times L T^{- 1}}]$

$= [M^{0} L^{0} T^{0}]$

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If radius of the sphere is (5.3 ± 0.1)cm. Then percentage error in its volume will be

$3 + 6 .01 \times \frac{100}{5 .3}$

$\frac{1}{3} \times 0 .01 \times \frac{100}{5 .3}$

$(\frac{3 \times 0 .1}{5 .3}) \times 100$

$\frac{0 .1}{5 .3} \times 100$

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Explanation

Volume of sphere $(V) = \frac{4}{3} π r^{3}$

% error in volume $= 3 \times \frac{Δ r}{r} \times 100$ $= (3 \times \frac{0 .1}{5 .3}) \times 100$

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The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is

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Explanation

$P = \frac{F}{A} = \frac{F}{l^{2}},$ so maximum error in pressure (P)

${(\frac{Δ P}{P} \times 100)}_{\max} = \frac{Δ F}{F} \times 100 + 2 \frac{Δ l}{l} \times 100$

= 4% + 2 × 2% = 8%

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