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Properties of Matter question

2010 · Shift 2 · Q41
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Properties of Matter question

2010 · Shift 2 · Q41

JEE AdvancedPhysicsProperties of MatterMCQ+3 / −0.75
When liquid medicine of density ρ\rhoρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When the force becomes smaller than the weight of the drop, the drop gets detached from the dropper. If r = 5 ×\times× 10−4 m, ρ\rhoρ = 103 kg m−3 , g = 10 m/s2 , T = 0.11 Nm−1 , the radius of the drop when it detaches from the dropper is approximately
  1. A
    1.4 ×\times× 10−3 m
  2. B
    3.3 ×\times× 10−3 m
  3. C
    2.0 ×\times× 10−3 m
  4. D
    4.1 ×\times× 10−3 m
View written solutionFree

Correct answer: C

  1. Forces on the hanging drop

A drop hangs from the circular opening of the dropper of radius rrr.

  • Its weight acts downward: W=mg=ρ(43πR3)gW = mg = \rho \left(\frac{4}{3}\pi R^3\right) gW=mg=ρ(34​πR3)g

  • The surface tension acts upward along the rim of the dropper.

The force due to surface tension is FT=2πrTF_T = 2\pi r TFT​=2πrT

This is because surface tension TTT acts along the circumference 2πr2\pi r2πr.

At the point of detachment, the drop is just about to fall, so W=FTW = F_TW=FT​

Hence, ρ(43πR3)g=2πrT\rho \left(\frac{4}{3}\pi R^3\right) g = 2\pi r Tρ(34​πR3)g=2πrT

  1. Simplify the equation

Cancel π\piπ from both sides: 43ρgR3=2rT\frac{4}{3}\rho g R^3 = 2rT34​ρgR3=2rT

So, R3=2rT⋅34ρg=3rT2ρgR^3 = \frac{2rT\cdot 3}{4\rho g} = \frac{3rT}{2\rho g}R3=4ρg2rT⋅3​=2ρg3rT​

  1. Substitute the given values

Given: r=5×10−4 m,ρ=103 kg m−3,g=10 m s−2,T=0.11 N m−1r = 5\times 10^{-4}\,\text{m},\quad \rho = 10^3\,\text{kg m}^{-3},\quad g=10\,\text{m s}^{-2},\quad T=0.11\,\text{N m}^{-1}r=5×10−4m,ρ=103kg m−3,g=10m s−2,T=0.11N m−1

Then R3=3(5×10−4)(0.11)2(103)(10)R^3 = \frac{3(5\times 10^{-4})(0.11)}{2(10^3)(10)}R3=2(103)(10)3(5×10−4)(0.11)​

First calculate numerator: 3×5×0.11×10−4=1.65×10−43\times 5\times 0.11 \times 10^{-4} = 1.65\times 10^{-4}3×5×0.11×10−4=1.65×10−4

Denominator: 2×103×10=2×1042\times 10^3\times 10 = 2\times 10^42×103×10=2×104

Thus, R3=1.65×10−42×104=0.825×10−8=8.25×10−9R^3 = \frac{1.65\times 10^{-4}}{2\times 10^4} = 0.825\times 10^{-8} = 8.25\times 10^{-9}R3=2×1041.65×10−4​=0.825×10−8=8.25×10−9

  1. Find RRR

R=(8.25×10−9)1/3R = (8.25\times 10^{-9})^{1/3}R=(8.25×10−9)1/3

Now,

\qquad (10^{-9})^{1/3} = 10^{-3}$$ So, $$R \approx 2.0\times 10^{-3}\,\text{m}$$ 5. **Match with options** This corresponds to: **Option C: $2.0\times 10^{-3}\,\text{m}$** 6. **Comparison with stored answer** Stored correct answer is **A**, but the calculation clearly gives **C**. So I **disagree** with the stored answer.
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