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Physics NEET Questions — Free MCQ Practice

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If L denotes the inductance of an inductor through which a current i is flowing, the dimensions of Li² are

A.

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

B.

Not expressible in MLT

C.

$M L T^{- 2}$

D.

LT

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Explanation

$\frac{1}{2} L i^{2}$ = Stored energy in an inductor = $[M L^{2} T^{- 2}]$

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Of the following quantities, which one has dimensions different from the remaining three

A.

Energy per unit volume

B.

Force per unit area

C.

Product of voltage and charge per unit volume

D.

Angular momentum per unit mass

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Explanation

Energy per unit volume = $\frac{[M L^{2} T^{- 2}]}{[L^{3}]} = [M L^{- 1} T^{- 2}]$

Force per unit area = $\frac{[M L T^{- 2}]}{[L^{2}]} = [M L^{- 1} T^{- 2}]$

Product of voltage and charge per unit volume

$= \frac{V \times Q}{Volume} = \frac{V I t}{Volume} = \frac{Power \times Time}{Volume}$

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

Angular momentum per unit mass = $\frac{[M L^{2} T^{- 1}]}{[M]} = [L^{2} T^{- 1}]$

So angular momentum per unit mass has different dimension.

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A spherical body of mass m and radius r is allowed to fall in a medium of viscosity $η$ . The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity $(v)$ is called time constant $(τ)$ . Dimensionally $τ$ can be represented by

A.

$\frac{m r^{2}}{6 π η}$

B.

$\sqrt{(\frac{6 π m r η}{g^{2}})}$

C.

$\frac{m}{6 π η r v}$

D.

None of the above

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Explanation

Time constant $τ = [T]$ and Viscosity $η = [M L^{- 1} T^{- 1}]$

For options (1), (2) and (3) dimensions are not matching with time constant.

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The frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a relation of this type $f = C m^{x} K^{y}$ ; where C is a dimensionless quantity. The value of x and y are

A.

$x = \frac{1}{2}, y = \frac{1}{2}$

B.

$x = - \frac{1}{2}, y = - \frac{1}{2}$

C.

$x = \frac{1}{2}, y = - \frac{1}{2}$

D.

$x = - \frac{1}{2}, y = \frac{1}{2}$

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Explanation

By putting the dimensions of each quantity both the sides we get $[T^{- 1}] = {[M]}^{x} {[M T^{- 2}]}^{y}$

Now comparing the dimensions of quantities in both sides we get $x + y = 0 and 2 y = 1$

$∴ x = - \frac{1}{2}, y = \frac{1}{2}$

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The quantities A and B are related by the relation, m = A/B, where m is the linear density and A is the force. The dimensions of B are of

A.

Pressure

B.

Work

C.

Latent heat

D.

None of the above

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Explanation

m = linear density = mass per unit length = $[\frac{M}{L}]$

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

$∴$ [B] = $\frac{[A]}{[m]} = \frac{[M L T^{- 2}]}{[M L^{- 1}]}$ $= [L^{2} T^{- 2}]$

This is same dimension as that of latent heat.

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The velocity of water waves $v$ may depend upon their wavelength $λ$ , the density of water $ρ$ and the acceleration due to gravity g. The method of dimensions gives the relation between these quantities as

A.

$v^{2} \propto g λ^{- 1} ρ^{- 1}$

B.

$v^{2} \propto g λ ρ$

C.

$v^{2} \propto g λ$

D.

$v^{2} \propto g^{- 1} λ^{- 3}$

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Explanation

$V = k . g^{x} . λ^{y} . ρ^{z} [M^{0} {LT}^{- 1}] = {[{LT}^{- 2}]}^{x} . {[L]}^{y} {[{ML}^{- 8}]}^{z} [M^{0} {LT}^{- 1}] = [M^{z} L^{x + y - 3 z} T^{- 2 y}] On comparing - z = 0 . . . . . . (i) x + y - 3 z = 1 \Rightarrow x + y = 1 . . . . . . . . (ii) - 2 y = - 1 y = \frac{1}{2} \Rightarrow x = \frac{1}{2} V = k . \sqrt{gλ} \Rightarrow V^{2} = k . gλ$

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The dimensions of resistivity in terms of M, L, T and Q where Q stands for the dimensions of charge, is

A.

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

B.

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

C.

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

D.

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

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Explanation

$ρ = \frac{R A}{l}$ i.e. dimension of resistivity is $[M L^{3} T^{- 1} Q^{- 2}]$ ........(i)

Using Ohm's law,

V=IR

$R = \frac{V}{I} = \frac{\frac{W o r k}{C h a r g e}}{C u r r e n t} = \frac{\frac{M L^{2} T^{- 2}}{Q}}{\frac{Q}{T}} = \frac{M L^{2} T^{- 1}}{Q^{2}} P u t t i n g t h i s v a l u e o f R i n (i) ρ = \frac{M L^{2} T^{- 1}}{Q^{2}} \times \frac{L^{2}}{L} = \frac{M L^{3} T^{- 1}}{Q^{2}}$

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The dimensions of Farad are , where Q represents electric charge [Only for droppers]

A.

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

B.

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

C.

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

D.

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

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Explanation

Farad is the unit of capacitance and $C = \frac{Q}{V}$ = $\frac{[Q]}{[M L^{2} T^{- 2} Q^{- 1}]}$ = $M^{- 1} L^{- 2} T^{2} Q^{2}$

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The equation of a wave is given by $Y = A \sin ω (\frac{x}{v} - k)$ where $ω$ is the angular velocity, x is length and $v$ is the linear velocity. The dimension of k is

A.

LT

B.

T

C.

$T^{- 1}$

D.

T²

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Explanation

From the principle of homogenity $(\frac{x}{v})$ has dimensions of T.

$[\frac{x}{v}] = [k] [\frac{L}{L T^{- 1}}] = [k] ∴ [k] = T$

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Dimensional formula of stress is

A.

$M^{0} L T^{- 2}$

B.

$M^{0} L^{- 1} T^{- 2}$

C.

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

D.

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

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Explanation

Stress = $\frac{Force}{Area} = \frac{[M L T^{- 2}]}{[L^{2}]} = [M L^{- 1} T^{- 2}]$

Physics MCQs for NEET — Practice Questions with Answers

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