Dimensions - NEET Physics Questions
Question 51: moderate

The dimensional formula of permeability of free space \(\mu_0\) is

[1991]

1. \(MLT^{-2}A^{-2}\)
2. \(M^0L^{-1}T\)
3. \(M^0LT^{-1}A^2\)
4. None of these
View Answer

From Ampere's law, the dimensions of permeability of free space \(\mu_0\) are derived as \(MLT^{-2}A^{-2}\).

Question 52: moderate

According 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]

1. \(ML^{-2}T^{-2}\)
2. \(ML^0T^0\)
3. \(ML^2T^{-1}\)
4. \(ML^{-1}T^{-1}\)
View Answer

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: moderate

Which of the following quantities, which one has dimensions different from the remaining three?

[1989]

1. Energy per unit volume
2. Force per unit area
3. Product of voltage and charge per unit volume
4. Angular momentum
View Answer

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: easy

Dimensional formula of self inductance is:

[1989]

1. \(MLT^{-2}A^{-2}\)
2. \(ML^2T^{-1}A^{-2}\)
3. \(ML^2T^{-2}A^{-2}\)
4. \(ML^2T^2A^{-1}\)
View Answer

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: easy

The dimensional formula of torque is:

[1989]

1. \(ML^2T^{-2}\)
2. \(MLT^{-2}\)
3. \(ML^{-1}T^{-2}\)
4. \(ML^{-2}T^{-2}\)
View Answer

Torque is calculated as Force \(\times\) perpendicular distance. So its dimensions are \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\).

Question 56: moderate

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]

1. \(ML^2T^{-2}I^{-2}\)
2. \(ML^2T^{-1}I^{-1}\)
3. \(ML^2T^{-3}I^{-2}\)
4. \(ML^2T^{-3}I^{-1}\)
View Answer

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: moderate

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]

1. \((LT^{-2}), (L) and (T)\)
2. \((L), (T) and (LT^2)\)
3. \((L^2T^{-2}), (LT) and (L)\)
4. \((L), (LT) and (T^2)\)
View Answer

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: moderate

The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of:

[2005]

1. Frequency
2. Velocity
3. Angular momentum
4. Time
View Answer

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: moderate

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:

1. \(ML^{-5}T^{-1}\)
2. \(ML^5T^{-1}\)
3. \(ML^5T^{-2}\)
4. \(M^{-1}L^5T^2\)
View Answer

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: easy

Turpentine 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]

1. \(M^1L^0T^0\)
2. \(MLT^{-1}\)
3. \(ML^{-2}T^{-1}\)
4. \(ML^{-1}T^{-1}\)
View Answer

\([P] = [ML^{-1}T^{-2}]). ([r^2 - x^2] = [L^2]\). \([v] = [LT^{-1}]\). \([l] = [L]\). \([\eta] = \frac{[ML^{-1}T^{-2}][L^2]}{[LT^{-1}][L]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]\).