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Physics MCQs for NEET — Practice Questions with Answers

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Dimensional formula of velocity of sound is

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

$L T^{0}$

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

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

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Explanation

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

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Dimensional formula of capacitance is

$M^{- 1} L^{- 2} T^{4} A^{2}$

$M L^{2} T^{4} A^{- 2}$

$M L T^{- 4} A^{2}$

$M^{- 1} L^{- 2} T^{- 4} A^{- 2}$

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Explanation

[C] = $(\frac{Q}{V}) = (\frac{Q^{2}}{W}) = [\frac{A^{2} T^{2}}{M L^{2} T^{- 2}}] = [M^{- 1} L^{- 2} T^{4} A^{2}]$

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$M L T^{- 1}$ represents the dimensional formula of

Power

Momentum

Force

Couple

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Explanation

Momentum = mv = $[M L T^{- 1}]$

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Dimensional formula of heat energy is

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

$M L T^{- 1}$

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

None of these

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Explanation

$Q = [M L^{2} T^{- 2}]$ (All energies have same dimension)

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If C and L denote capacitance and inductance respectively, then the dimensions of LC are

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

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

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

$M L T^{2}$

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Explanation

$f = \frac{1}{2 π \sqrt{L C}} \Rightarrow L C = \frac{1}{f^{2}} = [M^{0} L^{0} T^{2}]$

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Which of the following quantities has the same dimensions as that of energy

Power

Force

Momentum

Work

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Explanation

Energy = Work done [Dimensionally]

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The dimensions of "time constant" $\frac{L}{R}$ during growth and decay of current in all inductive circuit is same as that of

Constant

Resistance

Current

Time

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Explanation

$\frac{L}{R} =$ Time constant.

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The period of a body under SHM i.e. presented by $T = P^{a} D^{b} S^{c}$ ; where P is pressure, D is density and S is surface tension. The value of a, b and c are

$- \frac{3}{2}, \frac{1}{2}, 1$

$- 1, - 2, 3$

$\frac{1}{2}, - \frac{3}{2}, - \frac{1}{2}$

$1, 2, \frac{1}{3}$

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Explanation

By substituting the dimension of each quantity we get $T = {[M L^{- 1} T^{- 2}]}^{a} {[L^{- 3} M]}^{b} {[M T^{- 2}]}^{c}$

By solving we get a = – 3/2, b = 1/2 and c = 1

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Which of the following pairs of physical quantities has the same dimensions

Work and power

Momentum and energy

Force and power

Work and energy

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Explanation

$D i m e n s i o n o f w o r k = [M^{1} L^{2} T^{- 2}] D i m e n s i o n o f e n e r g y = [M^{1} L^{2} T^{- 2}]$

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The velocity of a freely falling body changes as $g^{p} h^{q}$ where g is acceleration due to gravity and h is the height. The values of p and q are

$1, \frac{1}{2}$

$\frac{1}{2}, \frac{1}{2}$

$\frac{1}{2}, 1$

1, 1

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Explanation

$v \propto g^{p} h^{q}$ (given)

By substituting the dimension of each quantity and comparing the powers in both sides we get $[L T^{- 1}] = {[L T^{- 2}]}^{p} {[L]}^{q}$

$\Rightarrow p + q = 1, - 2 p = - 1, ∴ p = \frac{1}{2}, q = \frac{1}{2}$

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