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Properties of Matter question

2022 · 28 Jun · Shift 2 · Q56
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  5. /2022 · 28 Jun · Shift 2 · Q56

Properties of Matter question

2022 · 28 Jun · Shift 2 · Q56

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A water drop of radius 1 μ\muμ m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 ×\times× 10 −-− 5 Nsm −-− 2 and its density is negligible as compared to that of water 106 gm −-− 3. Terminal velocity of the water drop is : (Take acceleration due to gravity = 10 ms −-− 2)
  1. A
    145.4 ×\times× 10 −-− 6 ms −-− 1
  2. B
    118.0 ×\times× 10 −-− 6 ms −-− 1
  3. C
    132.6 ×\times× 10 −-− 6 ms −-− 1
  4. D
    123.4 ×\times× 10 −-− 6 ms −-− 1
View written solutionFree

Correct answer: D

  1. Use Stokes' law for terminal velocity

For a small spherical drop falling through air, neglecting buoyancy, terminal velocity is

vt=2r2ρg9ηv_t = \frac{2 r^2 \rho g}{9\eta}vt​=9η2r2ρg​

where:

  • r=1 μm=1×10−6 mr = 1\,\mu m = 1 \times 10^{-6}\,mr=1μm=1×10−6m
  • ρ=106 g m−3=103 kg m−3\rho = 10^6\,g\,m^{-3} = 10^3\,kg\,m^{-3}ρ=106gm−3=103kgm−3 (density of water)
  • g=10 m s−2g = 10\,m\,s^{-2}g=10ms−2
  • η=1.8×10−5 N s m−2\eta = 1.8 \times 10^{-5}\,N\,s\,m^{-2}η=1.8×10−5Nsm−2
  1. Substitute the values

vt=2(1×10−6)2(103)(10)9(1.8×10−5)v_t = \frac{2(1\times10^{-6})^2(10^3)(10)}{9(1.8\times10^{-5})}vt​=9(1.8×10−5)2(1×10−6)2(103)(10)​

  1. Simplify numerator

(1×10−6)2=10−12(1\times10^{-6})^2 = 10^{-12}(1×10−6)2=10−12

So numerator becomes

2×10−12×103×10=2×10−82 \times 10^{-12} \times 10^3 \times 10 = 2 \times 10^{-8}2×10−12×103×10=2×10−8

  1. Simplify denominator

9×1.8×10−5=16.2×10−5=1.62×10−49 \times 1.8 \times 10^{-5} = 16.2 \times 10^{-5} = 1.62 \times 10^{-4}9×1.8×10−5=16.2×10−5=1.62×10−4

  1. Compute terminal velocity

vt=2×10−81.62×10−4v_t = \frac{2\times10^{-8}}{1.62\times10^{-4}}vt​=1.62×10−42×10−8​

vt=21.62×10−4v_t = \frac{2}{1.62} \times 10^{-4}vt​=1.622​×10−4

vt≈1.234×10−4 m s−1v_t \approx 1.234 \times 10^{-4}\,m\,s^{-1}vt​≈1.234×10−4ms−1

vt=123.4×10−6 m s−1v_t = 123.4 \times 10^{-6}\,m\,s^{-1}vt​=123.4×10−6ms−1

  1. Match with options

This matches:

Option D: 123.4×10−6 m s−1123.4 \times 10^{-6}\,m\,s^{-1}123.4×10−6ms−1

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