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, \([M^{1/2}L^1T^{-1/2}] [M^{-1/2}L^{3/2}T^{-1}] [L^{-3/2}T^{3/2}] = [M^0L^{1+3/2-3/2}T^{-1/2-1+3/2}] = [L]\).
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 \(R = V/I = W/(QI)\). (W) is work done, (Q) is charge. \(W = [ML^2T^{-2}]\), \(Q = [IT]\). So, \(R = [ML^2T^{-2}] / ([IT][I]) = [ML^2T^{-3}I^{-2}]\).
The velocity (v) of a particle at time (t) is given by \[v = at + \frac{b}{t+c}\] where (a), (b) and (c) are constants. The dimensions of (a), (b) and (c) are respectively:
[2006]
From dimensional homogeneity: ([c] = [t] = [T]). ([at] = [v]) so ([a] = [v]/[t] = [LT^{-1}]/[T] = [LT^{-2}]). ([b/(t+c)] = [v]) so ([b] = [v][t] = [LT^{-1}][T] = [L]).
The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of:
[2005]
Planck's constant (h) has dimensions of angular momentum, \([ML^2T^{-1}]\). Moment of inertia (I) has dimensions \([ML^2]\). The ratio \(h/I = [ML^2T^{-1}]/[ML^2] = [T^{-1}]\). \([T^{-1}]\) is the dimension of frequency.
An equation is given here \[\left(P + \frac{a}{V^2}\right) = b\frac{\theta}{V}\] where P = Pressure, V = Volume and \(\theta =\) Absolute temperature. If (a) and (b) are constants, the dimensions of (a) will be:
From dimensional homogeneity, \([a/V^2] = [P]\). \([a] = [P][V^2]\). Pressure \(P = [ML^{-1}T^{-2}]\), Volume \(V = [L^3]\). So, \([a] = [ML^{-1}T^{-2}][L^3]^2 = [ML^{-1}T^{-2}L^6] = [ML^5T^{-2}]\).
The error in measurement of radius of a sphere is 0.1% then error in its volume is:
Volume of a sphere \(V = \frac{4}{3}pi r^3\). The percentage error in volume is (3) times the percentage error in radius. Given \(\Delta r/r times 100% = 0.1%\). So, Percentage error in \(V = 3 \times 0.1% = 0.3% \).
The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and lengths are 3% and 2% respectively, the maximum error in the measurement of density would be:
[1996]
Density of a cube \(\rho = M/L^3\). Percentage error in \( \rho = (\Delta M/M) + 3(\Delta L/L)\). Given \(\Delta M/M \times 100% = 3%\) and \(\Delta L/L \times 100% = 2%\). So, Percentage error in \(\rho \) = 3% + 3(2%) = 3% + 6% = 9% .