Physics NEET Questions — Free MCQ Practice

Physics MCQs for NEET — Practice Questions with Answers

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Dimensions of time in power are

$T^{- 1}$

$T^{- 2}$

$T^{- 3}$

T⁰

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Explanation

Dimensions of power is $[M L^{2} T^{- 3}]$

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Dimensional formula for torque is

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

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

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

$L M T^{- 2}$

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Explanation

Torque = force × distance = $[M L^{2} T^{- 2}]$

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Dimensions of coefficient of viscosity are

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

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

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

MLT

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Explanation

$F = - η . A \frac{d v}{d x} \Rightarrow [η] = [M L^{- 1} T^{- 1}]$

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The dimension of quantity (L/RCV) is

[A]

[A²]

$[A^{- 1}]$

None of these

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Explanation

$\frac{L}{R C V} = [\frac{L}{R}] \frac{1}{C V} = \frac{T}{Q} = [A^{- 1}]$

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The dimension of the ratio of angular to linear momentum is

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

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

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

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

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Explanation

$\frac{Angular momentum}{Linear momentum} = \frac{m v r}{m v} = r = [M^{0} L^{1} T^{0}]$

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The pair having the same dimensions is

Angular momentum, work

Work, torque

Potential energy, linear momentum

Kinetic energy, velocity

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Explanation

Dimension of work and torque = $[M L^{2} T^{- 2}]$

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The dimensions of surface tension are

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

$M L T^{- 2}$

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

$M T^{- 2}$

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Explanation

Surface tension = $\frac{Force}{Length} = \frac{[M L T^{- 2}]}{L} = [M T^{- 2}]$

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In the following list, the only pair which have different dimensions, is

Linear momentum and moment of a force

Planck's constant and angular momentum

Pressure and modulus of elasticity

Torque and potential energy

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Explanation

Linear momentum = Mass × Velocity = $[M L T^{- 1}]$

Moment of a force = Force × Distance = $[M L^{2} T^{- 2}]$

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If R and L represent respectively resistance and self inductance, which of the following combinations has the dimensions of frequency [ Only for droppers]

$\frac{R}{L}$

$\frac{L}{R}$

$\sqrt{\frac{R}{L}}$

$\sqrt{\frac{L}{R}}$

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Explanation

$\frac{R}{L} = \frac{V / I}{V \times T / I} = \frac{1}{T} =$ Frequency

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If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be

$F A^{- 1} v$

$F v^{3} A^{- 2}$

$F v^{2} A^{- 1}$

$F^{2} v^{2} A^{- 1}$

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Explanation

$L \propto v^{x} A^{y} F^{z} \Rightarrow L = k v^{x} A^{y} F^{z}$

Putting the dimensions in the above relation

$[M L^{2} T^{- 1}] = k {[L T^{- 1}]}^{x} {[L T^{- 2}]}^{y} {[M L T^{- 2}]}^{z}$

$\Rightarrow [M L^{2} T^{- 1}] = k [M^{z} L^{x + y + z} T^{- x - 2 y - 2 z}]$

Comparing the powers of M, L and T

$z = 1$ …(i)

$x + y + z = 2$ …(ii)

$- x - 2 y - 2 z = - 1$ …(iii)

On solving (i), (ii) and (iii) $x = 3, y = - 2, z = 1$

So dimension of L in terms of v, A and f

$[L] = [F v^{3} A^{- 2}]$

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