The dimension of \(\frac{1}{2}\epsilon_0 E^2\), where \(\epsilon_0\) is permittivity of free space and \(E\) is electric field, is:
[2010 Pre]
The expression \(\frac{1}{2}\epsilon_0 E^2\) represents electric energy density, which is Energy per unit Volume. \([\text{Energy density}] = [\text{Energy}]/[\text{Volume}] = (ML^2T^{-2})/L^3 = ML^{-1}T^{-2}\).
Which two of the following five physical parameters have the same dimensions?
A. Energy density
B. Refractive index
C. Dielectric constant
D. Young’s modulus
E. Magnetic field
[2008]
Energy density (A) is \(ML^{-1}T^{-2}\). Young's modulus (D) is Stress/Strain, so also \(ML^{-1}T^{-2}\). Therefore, (A) and (D) have the same dimensions.
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be:
[2007]
Resistance \(R = V/I = (W/q)/I = W/(I^2T)\). Work \([W] = ML^2T^{-2}\). So, \([R] = (ML^2T^{-2})/(I^2T) = ML^2T^{-3}I^{-2}\).
The dimensions of universal gravitational constant are:
[2004]
From \(F = G \frac{m_1 m_2}{r^2}\), we get \(G = \frac{Fr^2}{m_1 m_2}\). So, \([G] = \frac{(MLT^{-2})(L^2)}{M^2} = M^{-1}L^3T^{-2}\).
Which pair have not equal dimensions?
[2000]
Force has dimensions \(MLT^{-2}\) and impulse has dimensions \(MLT^{-1}\). These are not equal.
If energy \((E)\), velocity \((V)\), and time \((T)\), are chosen as the fundamental quantities, the dimensional formula of surface tension will be:
[2015]
Surface tension \(gamma\) has dimensions \(MT^{-2}\). Given fundamental quantities are energy \(E = [ML^2T^{-2}]\), velocity \(V = [LT^{-1}]\), and time \(T = [T]\). Let \(\gamma = E^x V^y T^z\).
Equating dimensions: \(MT^{-2} = (ML^2T^{-2})^x (LT^{-1})^y (T)^z = M^x L^{2x+y} T^{-2x-y+z}\). Comparing powers: \(x=1\), \(2x+y=0 ⇒ y=-2\), \(-2x-y+z=-2 ⇒ -2(1)-(-2)+z=-2 ⇒s z=-2\). Thus, surface tension dimensions are \(EV^{-2}T^{-2}\).
The dimensions of impulse are equal to that of:
[1996]
Impulse is defined as change in momentum. Therefore, their dimensions are equal, \(MLT^{-1}\).
If dimension of critical velocity of liquid flowing through a tube are expressed as \(v_c \propto \eta^x \rho^y r^z\) where \(\eta\) and \(\rho\) are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of \(x, y\) and \(z\) are given by:
[2015-Re]
Critical velocity \(v_c = [LT^{-1}]\). Coefficient of viscosity \(\eta = [ML^{-1}T^{-1}]\). Density \(\rho = [ML^{-3}]\). Radius \(r = [L]\). Assume \(v_c \propto \eta^x \rho^y r^z\). Equating dimensions: \([LT^{-1}] = ([ML^{-1}T^{-1}])^x ([ML^{-3}])^y ([L])^z = [M^{x+y} L^{-x-3y+z} T^{-x}]\). Comparing powers: \(x+y = 0\), \(-x-3y+z = 1\), \(-x = -1\). From \(-x = -1\), we get \(x=1\). From \(x+y = 0\), we get \(1+y = 0 ⇒ y=-1\). From \(-x-3y+z = 1\), we get \(-1-3(-1)+z = 1⇒ -1+3+z=1 ⇒ 2+z=1 ⇒ z=-1\). Therefore, \(x=1, y=-1, z=-1\).
Which of the following dimensions will be the same as that of time?
[1996]
The ratio \(L/R\) has the dimensions of time, \(T\).
If Force \((F)\), Velocity \((V)\), and Time \((T)\), are taken as fundamental units, then the dimensions of mass are:
[2014]
Given fundamental units: Force \(F = [MLT^{-2}]\), Velocity \(V = [LT^{-1}]\), Time \(T = [T]\). We want to find dimensions of Mass \(M = F^x V^y T^z\). Equating dimensions: \([M] = [MLT^{-2}]^x [LT^{-1}]^y [T]^z = [M^x L^{x+y} T^{-2x-y+z}]\). Comparing powers: \(x=1\), \(x+y=0 ⇒ 1+y=0 ⇒ y=-1\), \(-2x-y+z=0 ⇒ -2(1)-(-1)+z=0⇒ -2+1+z=0 ⇒ z=1\). Thus, mass dimensions are \(FV^{-1}T\).