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The dimensions of K in the equation $W = \frac{1}{2} K x^{2}$ is

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

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

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

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

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Explanation

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

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The physical quantities not having same dimensions are

Speed and ${(μ_{0} ε_{0})}^{- 1 / 2}$

Torque and work

Momentum and Planck's constant

Stress and Young's modules

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Explanation

Momentum $[M L T^{- 1}]$ , Planck's constant $[M L^{2} T^{- 1}]$

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Dimension of R is

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

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

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

None of these

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Explanation

$R = \frac{V}{I}$ $= [\frac{M L^{2} T^{- 3} A^{- 1}}{A}]$ = $[M L^{2} T^{- 3} A^{- 2}]$

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The dimensional formula of relative density is

ML^–3

LT^–1

MLT^–2

Dimensionless

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Explanation

Relative density = $\frac{Density of substance}{density of water}$ $= [M^{0} L^{0} T^{0}]$

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The dimensional formula for young's modulus is

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

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

MLT^–2

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

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Explanation

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

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Frequency is the function of density $(ρ)$ , length (a) and surface tension (T). Then its value is

$\frac{\sqrt{T}}{{kρ}^{1 / 2} a^{3 / 2}}$

$k ρ^{3 / 2} a^{3 / 2} / \sqrt{T}$

$\frac{T^{3 / 2}}{{kρ}^{1 / 2} a^{3 / 2}}$

$\frac{T^{3 / 2}}{{kρ}^{1 / 2} a^{1 / 2}}$

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Explanation

$n = {kp}^{a} a^{b} T^{c} [M^{0} L^{0} T^{- 1}] = {[{ML}^{- 3}]}^{a} {[L]}^{b} {[{MT}^{- 2}]}^{c} [M^{0} L^{0} T^{- 1}] = [M^{a + c} . L^{- 3 a + b} . T^{- 2 c}] \Rightarrow a + c = 0 . . . . . (i) - 3 a + b = 0 . . . . . (ii) - 2 c = - 1 \Rightarrow c = \frac{1}{2} a = - \frac{1}{2} b = - \frac{3}{2} n = k . \frac{\sqrt{T}}{\sqrt{p} . a^{\frac{3}{2}}}$

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The dimensions of electric potential are

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

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

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

$[M L^{2} T^{- 2} Q]$

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Explanation

$V = \frac{W}{Q} = [M L^{2} T^{- 2} Q^{- 1}]$

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Dimensions of potential energy are

MLT^–1

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

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

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

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Explanation

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

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The dimensions of shear modulus are

MLT^–1

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

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

$M L T^{- 2}$

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Explanation

Shear modulus = $\frac{Shearing stress}{Shearing strain}$ $= \frac{F}{A θ} = [M L^{- 1} T^{- 2}]$

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Pressure gradient has the same dimension as that of

Velocity gradient

Potential gradient

Energy gradient

None of these

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Explanation

Velocity gradient $= \frac{v}{x} = \frac{[L T^{- 1}]}{[L]} = [T^{- 1}]$

Potential gradient $= \frac{V}{x} = \frac{[M L^{2} T^{- 3} A^{- 1}]}{[L]}$ $= [M L T^{- 3} A^{- 1}]$

Energy gradient $= \frac{E}{x} = \frac{[M L^{2} T^{2}]}{[L]} = [M L T^{- 2}]$

and pressure gradient $= \frac{P}{x} = \frac{[M L^{- 1} T^{- 2}]}{[L]} = [M L^{- 2} T^{- 2}]$

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