The dimensional formula of angular momentum is:
[1988]
Angular momentum is defined as \(L = r \times p\), where \(r\) is distance and \(p\) is linear momentum. Its dimensions are \([L][MLT^{-1}] = [ML^2T^{-1}]\).
If force \([F]\), acceleration \([A]\) and time \([T]\) are chosen as the fundamental physical quantities. Find the dimensions of energy.
[2021]
We know \(E = ML^2T^{-2}\), \(F = MLT^{-2}\), \(A = LT^{-2}\), \(T = T\). From \(F = MA\), \(M = F/A\). Substitute \(M\) into \(E\): \(E = (F/A)L^2T^{-2}\). Also, \(L = AT^2\). So, \(E = (F/A)(AT^2)^2T^{-2} = (F/A) A^2 T^4 T^{-2} = F A T^2\).
A physical quantity of the dimensions of length can be formed out of \(c\), \(G\) and \(e^2 / (4\pi\epsilon_0)\), where \(c\) is velocity of light, \(G\) is universal constant of gravitation and \(e\) is charge.
[2017, Delhi]
Dimensions: \(c = [LT^{-1}]\), \(G = [M^{-1}L^3T^{-2}]\), and \(e^2/(4\pi\epsilon_0) = [ML^3T^{-2}]\). Let's check option d: \(\frac{1}{c^2} \left( G \frac{e^2}{4\pi\epsilon_0} \right)^{1/2} = [L^{-2}T^2] \cdot ([M^{-1}L^3T^{-2}] \cdot [ML^3T^{-2}])^{1/2} = [L^{-2}T^2] \cdot ([L^6T^{-4}])^{1/2} = [L^{-2}T^2] \cdot [L^3T^{-2}] = [L]\).
The dimension of Planck constant equals to that of:
[2001]
Planck's constant \(h\) has dimensions \(ML^2T^{-1}\). Angular momentum also has dimensions \(ML^2T^{-1}\).
Planck’s constant \(h\), speed of light in vacuum \(c\), and Newton’s gravitational constant \(G\), are three fundamental constants. Which of the following combinations of these has the dimension of length?
[2016-II]
The Planck length formula is \(l_P = \sqrt{\frac{hG}{c^3}}\). Checking option c: \(\frac{\sqrt{hG}}{c^{3/2}} = h^{1/2}G^{1/2}c^{-3/2}\). \(h=[ML^2T^{-1}]\), \(G=[M^{-1}L^3T^{-2}]\), \(c=[LT^{-1}]\). Thus, \([M^{1/2}L^1T^{-1/2}] [M^{-1/2}L^{3/2}T^{-1}] [L^{-3/2}T^{3/2}] = [M^0L^{1+3/2-3/2}T^{-1/2-1+3/2}] = [L]\).
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, \(T\).