JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Units & Measurement

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If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

(A) $S^{1/2} I^{1/2} h^{0}$ ✓ Correct
(B) $S^{1/2} I^{3/2} h^{- 1}$
(C) $S^{3/2} I^{1/2} h^{0}$
(D) $S^{1/2} I^{1/2} h^{- 1}$

Answer: (A)

Solution

Write momentum as

p = k S^{a} I^{b} h^{c}

and match dimensions. The dimensional formulas are

[p] = M L T^{- 1}

[S] = M T^{- 2}

for surface tension (force per length),

[I] = M L^{2}

for moment of inertia, and

[h] = M L^{2} T^{- 1}

for Planck's constant. Substituting,

M L T^{- 1} = M^{a + b + c} L^{2 b + 2 c} T^{- 2 a - c}

. Matching powers gives

a + b + c = 1

for mass,

2 b + 2 c = 1

for length, and

- 2 a - c = - 1

for time. The length equation says

b + c = \frac{1}{2}

, so the mass equation forces

a = \frac{1}{2}

; then

2 a + c = 1

gives

c = 0

, and finally

b = \frac{1}{2}

. So linear momentum is

S^{1/2} I^{1/2} h^{0}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Units & Measurement

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