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Physics NEET Questions — Free MCQ Practice

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If L, C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for C²LR is

A.

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

B.

$[M^{0} L^{0} T^{3} I^{0}]$

C.

$[M^{- 1} L^{- 2} T^{6} I^{2}]$

D.

$[M^{0} L^{0} T^{2} I^{0}]$

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Explanation

$C^{2} L R = [C^{2} L^{2}] \times [\frac{R}{L}] = [T^{4}] \times [\frac{1}{T}] = [T^{3}]$

As $[\frac{L}{R}] = T$ and $\sqrt{L C} = T$

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If the velocity of light (c), gravitational constant (G) and Planck's constant (h) are chosen as fundamental units, then the dimensions of mass in new system is

A.

$c^{1 / 2} G^{1 / 2} h^{1 / 2}$

B.

$c^{1 / 2} G^{1 / 2} h^{- 1 / 2}$

C.

$c^{1 / 2} G^{- 1 / 2} h^{1 / 2}$

D.

$c^{- 1 / 2} G^{1 / 2} h^{1 / 2}$

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Explanation

$[C] = [M^{0} {LT}^{- 1}] [G] = [M^{- 1} L^{3} T^{- 2}] [h] = \frac{[Ed]}{[C]} = [{ML}^{2} T^{- 1}] Let m \propto C^{x} G^{y} h^{z} m = {KC}^{x} G^{y} h^{z} [{ML}^{0} T^{0}] = K {[{LT}^{- 1}]}^{x} \times {[M^{- 1} L^{3} T^{- 2}]}^{y} \times {[{ML}^{2} T^{- 1}]}^{z} [{ML}^{0} T^{0}] = [M^{- y + z} \cdot L^{x + 3 y + 2 z} \cdot T^{- x - 2 y - z}] Comparing the dimensions : - - y + z = 1 . . . . . . (I) x + 3 y + 2 z = 0 . . . . . . . (II) - x - 2 y - z = 0 \Rightarrow x + 2 y + z = 0 . . . . . . . (III) From eq . (II) - (x + 2 y + z) + y + z = 0 0 + y + z = 0 \Rightarrow Y = - z in eq . (I) - z = \frac{1}{2} y = \frac{- 1}{2} x = \frac{1}{2}$

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Dimensions of charge are

A.

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

B.

MLTA^–1

C.

T^–1A

D.

TA

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Explanation

Charge = Current × Time = $[A T]$

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According to Newton, the viscous force acting between liquid layers of area A and velocity gradient $Δ v / Δ z$ is given by $F = - η A \frac{Δ v}{Δ z}$ where $η$ is constant called coefficient of viscosity. The dimension of $η$ are

A.

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

B.

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

C.

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

D.

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

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Explanation

$F = - η A \frac{Δ v}{Δ z} \Rightarrow [η] = [M L^{- 1} T^{- 1}]$

As $F = [M L T^{- 2}], A = [L^{2}], \frac{Δ v}{Δ z} = [T^{- 1}]$

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Identify the pair whose dimensions are equal

A.

Torque and work

B.

Stress and energy

C.

Force and stress

D.

Force and work

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Explanation

$T o r q u e = F o r c e \times P e r p e n d i c u l a r d i s \tan c e = [M L T^{- 2}] [L] = [M L^{2} T^{- 2}] A n d w o r k = F o r c e \times d i s p l a c e m e n t = [M L T^{- 2}] [L] = [M L^{2} T^{- 2}]$

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The dimensions of pressure is equal to

A.

Force per unit volume

B.

Energy per unit volume

C.

Force

D.

Energy

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Explanation

$\frac{Energy}{Volume} = \frac{M L^{2} T^{- 2}}{L^{3}} = [M L^{- 1} T^{- 2}]$ = Pressure

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Which of the two have same dimensions

A.

Force and strain

B.

Force and stress

C.

Angular velocity and frequency

D.

Energy and strain

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Explanation

$ω = \frac{d θ}{d t} = [T^{- 1}]$ and frequency $[n] = [T^{- 1}]$

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The dimensions of emf in MKS is

A.

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

B.

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

C.

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

D.

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

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Explanation

$V = \frac{W o r k}{C h a r g e} [V] = \frac{M L^{2} T^{- 2}}{Q} = M L^{2} Q^{- 1} T^{- 2}$

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Which of the following quantities is dimensionless

A.

Gravitational constant

B.

Planck's constant

C.

Power of a convex lens

D.

None

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Explanation

$[G] = [M^{- 1} L^{3} T^{- 2}]; [h] = [M L^{2} T^{- 1}]$

Power = $\frac{1}{focal length} = [L^{- 1}]$

All quantities have dimensions

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The dimensional formula for Boltzmann's constant is

A.

$[M L^{2} T^{- 2} θ^{- 1}]$

B.

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

C.

$[M L^{0} T^{- 2} θ^{- 1}]$

D.

$[M L^{- 2} T^{- 1} θ^{- 1}]$

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Explanation

$k = [\frac{R}{N}] = [M L^{2} T^{- 2} θ^{- 1}]$

Physics MCQs for NEET — Practice Questions with Answers

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