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Dimensional formula for angular momentum is

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

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

$M L T^{- 1}$

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

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Explanation

Angular momentum = mvr

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

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The dimensional formula for impulse is

$M L T^{- 2}$

$M L T^{- 1}$

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

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

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Explanation

Impulse = Force × Time = $[M L T^{- 2}] [T] = [M L T^{- 1}]$

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The dimensional formula for the modulus of rigidity is

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

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

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

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

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Explanation

Modulus of rigidity = $\frac{Shear stress}{Shear strain} =$ $[M L^{- 1} T^{- 2}]$

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The dimensional formula for r.m.s. (root mean square) velocity is

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

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

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

$M L T^{- 3}$

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Explanation

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

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The dimensional formula for Planck's constant (h) is

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

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

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

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

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Explanation

$E = h v \Rightarrow [M L^{2} T^{- 2}] = [h] [T^{- 1}] \Rightarrow [h] = [M L^{2} T^{- 1}]$

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The dimensional formula for impulse is same as the dimensional formula for

Momentum

Force

Rate of change of momentum

Torque

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Explanation

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

Impulse = Force × Time = $[M L T^{- 2}] \times [T] = [M L T^{- 1}]$

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Which of the following is dimensionally correct

Pressure = Energy per unit area

Pressure = Energy per unit volume

Pressure = Force per unit volume

Pressure = Momentum per unit volume per unit time

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Explanation

Pressure = $\frac{Force}{Area} = \frac{Energy}{Volume}$ = $M L^{- 1} T^{- 2}$

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Planck's constant has the dimensions (unit) of

Energy

Linear momentum

Work

Angular momentum

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Explanation

$[h] = [Angular momentum] = [M L^{2} T^{- 1}]$

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The equation of state of some gases can be expressed as $(P + \frac{a}{V^{2}}) (V - b) = R T$ . Here P is the pressure, V is the volume, T is the absolute temperature and a, b, R are constants. The dimensions of ‘a’ are

$M L^{5} T^{- 2}$

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

$M^{0} L^{3} T^{0}$

$M^{0} L^{6} T^{0}$

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Explanation

By principle of dimensional homogenity $[\frac{a}{V^{2}}] = [P]$

$∴$ $[a] = [P] [V^{2}] = [M L^{- 1} T^{- 2}] \times [L^{6}]$ = $[M L^{5} T^{- 2}]$

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If V denotes the potential difference across the plates of a capacitor of capacitance C, the dimensions of CV² are [Only for Dropper and XII batch]

Not expressible in MLT

$M L T^{- 2}$

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

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

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Explanation

$\frac{1}{2} C V^{2} =$ Stored energy in a capacitor = $[M L^{2} T^{- 2}]$

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