NEET Unit And Dimensions MCQs & PYQs

Practice NEET Unit And Dimensions 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]

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

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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]

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

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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]

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

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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]

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

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