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The dimensions of calorie are

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

$M L T^{- 2}$

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

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

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Explanation

Calorie is the unit of heat i.e., energy.

So dimensions of energy $= M L^{2} T^{- 2}$

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If L and R are respectively the inductance and resistance, then the dimensions of $\frac{L}{R}$ will be

[Only for Dropper and XII batch]

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

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

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

Cannot be represented in terms of M, L and T

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Explanation

$V = \frac{L i}{t} L = \frac{V t}{i} R = \frac{V}{i} T h e n, \frac{L}{R} = \frac{\frac{V t}{i}}{\frac{V}{i}} = t$

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

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Which pair has the same dimensions

Work and power

Density and relative density

Momentum and impulse

Stress and strain

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Explanation

Impulse = change in momentum so dimensions of both quantities will be same and equal to MLT^–1

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If C and R represent capacitance and resistance respectively, then the dimensions of RC are

[Only for droppers]

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

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

$M L^{- 1}$

None of the above

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Explanation

$R C = T$

âˆµ $[R] = [M L^{2} T^{- 3} I^{- 2}]$ and $[C] = [M^{- 1} L^{- 2} T^{4} I^{2}]$

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Dimensions of one or more pairs are same. Identify the pairs

Torque and work

Angular momentum and work

Energy and Young's modulus

Year and wavelength

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Explanation

[Torque] = [work] = [ML²T^–2]

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Dimensional formula for volume elasticity is

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

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

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

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

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Explanation

Volume elasticity = $\frac{Force/Area}{Volume strain}$

Strain is dimensionless, so

[Elasticity] = [Stress]

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The dimensions of universal gravitational constant are

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

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

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

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

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Explanation

$F = \frac{G m_{1} m_{2}}{d^{2}} \Rightarrow G = \frac{F d^{2}}{m_{1} m_{2}}$

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

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The dimensional formula of angular velocity is

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

$M L T^{- 1}$

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

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

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Explanation

Angular velocity = $\frac{θ}{t},$ $[ω] = \frac{[M^{0} L^{0} T^{0}]}{[T]} = [T^{- 1}]$

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The dimensions of power are

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

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

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

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

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Explanation

Power = $\frac{Work done}{Time} = [\frac{M L^{2} T^{- 2}}{T}] = [M L^{2} T^{- 3}]$

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The dimensions of couple are

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

$M L T^{- 2}$

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

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

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Explanation

Couple = Force × Arm length = $[M L T^{- 2}] [L] = [M L^{2} T^{- 2}]$

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