Unit And Dimensions

Practice Questions:

Question 1: difficult

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]

1. \(c^2 G (e^2 / (4\pi\epsilon_0))^{1/2}\)

2. \(\frac{1}{c^2} G (e^2 / (4\pi\epsilon_0))^{1/2}\)

3. \(\frac{1}{c^2} G \frac{e^2}{4\pi\epsilon_0}\)

4. \(\frac{1}{c^2} \left( G \frac{e^2}{4\pi\epsilon_0} \right)^{1/2}\)

View Answer

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: \(\frac{1}{c^2} \left( G \frac{e^2}{4\pi\epsilon_0} \right)^{1/2} = [L^{-2}T^2] \cdot ([M^{-1}L^3T^{-2}] \cdot [ML^3T^{-2}])^{1/2} = [L^{-2}T^2] \cdot ([L^6T^{-4}])^{1/2} = [L^{-2}T^2] \cdot [L^3T^{-2}] = [L]\).

💬 Discuss this Question

Question 2: difficult

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]

1. \(\sqrt{\frac{hc}{G}}\)

2. \(\sqrt{\frac{Gc}{h^{3/2}}}\)

3. \(\frac{\sqrt{hG}}{c^{3/2}}\)

4. \(\frac{\sqrt{hG}}{c^{5/2}}\)

View Answer

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, \([M^{1/2}L^1T^{-1/2}] [M^{-1/2}L^{3/2}T^{-1}] [L^{-3/2}T^{3/2}] = [M^0L^{1+3/2-3/2}T^{-1/2-1+3/2}] = [L]\).

💬 Discuss this Question

Question 3: difficult

If energy \((E)\), velocity \((V)\), and time \((T)\), are chosen as the fundamental quantities, the dimensional formula of surface tension will be:

[2015]

1. \(EV^{-1}T^{-1}\)

2. \(EV^{-2}T^{-2}\)

3. \(E^{-2}V^{-1}T^{-3}\)

4. \(EV^{-2}T^{-1}\)

View Answer

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 \(EV^{-2}T^{-2}\).

💬 Discuss this Question

Question 4: difficult

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]

1. 1, 1, 1

2. 1, -1, -1

3. -1, -1, 1

4. -1, -1, -1

View Answer

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, \(x=1, y=-1, z=-1\).

💬 Discuss this Question