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Physics NEET Questions — Free MCQ Practice

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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

A.

$V = \frac{π p r^{4}}{8 η l}$

B.

$V = \frac{π η l}{8 p r^{4}}$

C.

$V = \frac{8 p η l}{π r^{4}}$

D.

$V = \frac{π p η}{8 l r^{4}}$

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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.

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

B.

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

C.

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

D.

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

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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

A.

$M^{0} L^{1} T^{- 1}$ and T^–1

B.

$M^{0} L^{1} T^{0}$ and T^–1

C.

$M^{0} L^{1} T^{- 1}$ and LT^–2

D.

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

A.

[LT^–1]

B.

[L^–1T]

C.

[L^–2T²]

D.

[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

A.

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

B.

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

C.

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

D.

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

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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

A.

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

B.

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

C.

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

D.

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

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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

A.

1%

B.

2%

C.

6%

D.

8%

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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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While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation $g = 4 π^{2} (l / T^{2})$ will be

A.

2%

B.

4%

C.

7%

D.

10%

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Explanation

Percentage error in g = (%error in l) + 2(% error in T) = 1% + 2(3%) = 7%

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The length, breadth, and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm The volume of the block according to the idea of significant figures should be

A.

1 × 10² cm³

B.

2 × 10² cm³

C.

1.763 × 10² cm³

D.

None of these

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Explanation

Volume V = l × b × t

$= 12 \times 6 \times 2 .45$ $= 176 .4 c m^{3}$

V $= 1 .764 \times 10^{2} c m^{3}$

since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, $V = 2 \times 10^{2} c m^{3}$ .

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A physical quantity of the dimensions of length that can be formed out of c,G and $\frac{e^{2}}{4 π ε_{ο}}$ is [c is velocity of light, G is universal constant of gravitation and e is charge]

A.

$\frac{1}{c^{2}} {[G \frac{e^{2}}{4 π ε_{o}}]}^{\frac{1}{2}}$

B.

$c^{2} {[G \frac{e^{2}}{4 π ε_{o}}]}^{\frac{1}{2}}$

C.

$\frac{1}{c^{2}} {[\frac{e^{2}}{G 4 π ε_{o}}]}^{\frac{1}{2}}$

D.

$\frac{1}{c} G \frac{e^{2}}{4 π ε_{0}}$

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Explanation

(a) As force F= $\frac{e^{2}}{4 π ε_{o} r^{2}}$

$\Rightarrow \frac{e^{2}}{4 π ε_{o}} = r^{2} . F$

putting dimensions of r and F, we get

$\Rightarrow [\frac{e^{2}}{4 π ε_{o}}] = [M L^{3} T^{- 2}] . . . (i)$

Also force. F= $\frac{G m^{2}}{r^{2}}$

$\Rightarrow |G| = \frac{[M L T^{- 2}] [L^{2}]}{[M^{2}]}$ ...(ii)

$\Rightarrow |G| = [M^{- 1} L^{3} T^{- 2}]$ ...(iii)

and $[\frac{1}{c^{2}}] = \frac{1}{[L^{2} T^{- 2}]} = [L^{- 2} T^{2}]$

Now, checking optionwise.

= $\frac{1}{c^{2}} {(\frac{G e^{2}}{4 π ε_{o}})}^{\frac{1}{2}}$

= $[L^{- 2} T^{2}] {[L^{6} T^{- 4}]}^{\frac{1}{2}} = |L|$

Physics MCQs for NEET — Practice Questions with Answers

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