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JEE Physics 2023 Question with Solution

The frequency (ν\nu) of an oscillating liquid drop may depend upon radius (rr) of the drop, density (ρ\rho) of liquid and the surface tension (ss) of the liquid as:

ν=raρbsc\nu = r^a \rho^b s^c

The values of aa, bb, and cc respectively are:

  • A

    (32,−12,12)\left( \frac{3}{2}, -\frac{1}{2}, \frac{1}{2} \right)

  • B

    (32,12,−12)\left( \frac{3}{2}, \frac{1}{2}, -\frac{1}{2} \right)

  • C

    (32,12,12)\left( \frac{3}{2}, \frac{1}{2}, \frac{1}{2} \right)

  • D

    (32,−12,−12)\left( \frac{3}{2}, -\frac{1}{2}, -\frac{1}{2} \right)

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The frequency is assumed to vary as

ν=raρbsc\nu = r^a \rho^b s^c

Find: The values of aa, bb and cc using dimensional analysis.

The dimensional formula of frequency is

[ν]=[T−1][\nu] = [T^{-1}]

The dimensional formulas used in the solution are for radius rr, density ρ\rho and surface tension ss.

Substituting dimensions as shown in the solution,

[T−1]=[L]a[M]b[L−3b][MLT−2]c[T^{-1}] = [L]^a [M]^b [L^{-3b}] [M L T^{-2}]^c

which simplifies to

[T−1]=Mb+c⋅La−3b+c⋅T−2c[T^{-1}] = M^{b+c} \cdot L^{a-3b+c} \cdot T^{-2c}

Equating powers of the fundamental dimensions gives

b+c=0b + c = 0 a−3b+c=0a - 3b + c = 0 −2c=−1-2c = -1

From the time dimension,

c=12c = \frac{1}{2}

Then from b+c=0b+c=0,

b=−12b = -\frac{1}{2}

Now substitute into the length equation:

a−3(−12)+12=0a - 3\left(-\frac{1}{2}\right) + \frac{1}{2} = 0 a+2=0a + 2 = 0 a=−2a = -2

The solution is internally inconsistent: it declares option B, computes a=−32a=-\frac{3}{2} in one place, and the available options all have positive 32\frac{3}{2}. Also, the correct dimensional treatment of surface tension would make the sign of aa negative, so none of the listed options matches the worked result. Therefore the answer cannot be resolved consistently from the provided page.

Consistency Check

Using the standard dimension of surface tension,

[s]=[MT−2][s] = [M T^{-2}]

with

[r]=[L],[ρ]=[ML−3][r] = [L], \qquad [\rho] = [M L^{-3}]

we get

[T−1]=[L]a[ML−3]b[MT−2]c[T^{-1}] = [L]^a [M L^{-3}]^b [M T^{-2}]^c

So,

[T−1]=Mb+cLa−3bT−2c[T^{-1}] = M^{b+c} L^{a-3b} T^{-2c}

Equating powers,

b+c=0,a−3b=0,−2c=−1b+c=0, \qquad a-3b=0, \qquad -2c=-1

which gives

c=12,b=−12,a=−32c=\frac{1}{2}, \qquad b=-\frac{1}{2}, \qquad a=-\frac{3}{2}

Thus the consistent result is

(−32,−12,12)\left(-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}\right)

This value is not present among the listed options, so the source data contains a mismatch between question, options and solution.

Common mistakes

  • Using the wrong dimension of surface tension. Surface tension should be taken as force per unit length, so its dimension is [MT−2][M T^{-2}], not a form containing an extra LL factor. Using the wrong dimension changes the power of rr incorrectly.

  • Not combining the dimensions of density correctly. Since [ρ]=[ML−3][\rho] = [M L^{-3}], raising it to bb gives both mass and length contributions. If the L−3bL^{-3b} term is missed, the equation for aa becomes wrong.

  • Equating only one or two fundamental dimensions instead of all of MM, LL and TT. Dimensional analysis requires matching every independent exponent to get a complete system of equations.

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