Question 51:
moderateThe dimensional formula of permeability of free space \(\mu_0\) is
[1991]
From Ampere's law, the dimensions of permeability of free space \(\mu_0\) are derived as \(MLT^{-2}A^{-2}\).
Practice NEET Dimensions Questions
Question 52:
moderateAccording to Newton, the viscous force acting between liquid layers of area \(A\) and velocity gradient \( \Delta v / \Delta z \) is given by \(F = -\eta A (\Delta v / \Delta z)\), where \(\eta\) is constant called coefficient of viscosity. The dimensional formula of \(\eta\) is:
[1990]
From the formula \(F = -\eta A (\Delta v / \Delta z)\), the dimension of \(\eta\) is \(F / (A \cdot (\Delta v / \Delta z)) = [MLT^{-2}] / ([L^2] \cdot [T^{-1}]) = [ML^{-1}T^{-1}]\).
Question 53:
moderateWhich of the following quantities, which one has dimensions different from the remaining three?
[1989]
Energy per unit volume, force per unit area, and product of voltage and charge per unit volume all have dimensions \(ML^{-1}T^{-2}\). Angular momentum has dimensions \(ML^2T^{-1}\), which is different.
Question 54:
easyDimensional formula of self inductance is:
[1989]
The energy stored in an inductor is \(U = \frac{1}{2}LI^2\). Therefore, the dimensions of self-inductance \(L\) are \(U/I^2 = [ML^2T^{-2}]/[A^2] = [ML^2T^{-2}A^{-2}]\).
Question 55:
easyThe dimensional formula of torque is:
[1989]
Torque is calculated as Force \(\times\) perpendicular distance. So its dimensions are \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\).
Question 56:
moderateDimensions 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, \(R = [ML^2T^{-2}] / ([IT][I]) = [ML^2T^{-3}I^{-2}]\).
Question 57:
moderateThe 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]).
Question 58:
moderateThe 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.
Question 59:
moderateAn 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, \([a] = [ML^{-1}T^{-2}][L^3]^2 = [ML^{-1}T^{-2}L^6] = [ML^5T^{-2}]\).
Question 60:
easyTurpentine oil is flowing through a tube of length (l) and radius (r). The pressure difference between the two ends of the tube is (P). The viscosity of oil is given by \(\eta = \frac{P(r^2 – x^2)}{4vl}\) where (v) is the velocity of oil at a distance (x) from the axis of the tube. The dimensions of (eta) are:
[1993]
\([P] = [ML^{-1}T^{-2}]). ([r^2 - x^2] = [L^2]\). \([v] = [LT^{-1}]\). \([l] = [L]\). \([\eta] = \frac{[ML^{-1}T^{-2}][L^2]}{[LT^{-1}][L]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]\).