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Properties of Matter question

2025 · 3 Apr · Shift 2 · Q66
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Properties of Matter question

2025 · 3 Apr · Shift 2 · Q66

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45×10−2 m/s2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}2.45×10−2 m/s while falling under gravity through an oil of density 925 kg m−3925 \mathrm{~kg} \mathrm{~m}^{-3}925 kg m−3. Take density of steel as 7825 kg m−37825 \mathrm{~kg} \mathrm{~m}^{-3}7825 kg m−3 and g as 9.8 m/s29.8 \mathrm{~m} / \mathrm{s}^29.8 m/s2. The viscosity of the oil in SI unit is
  1. A
    2.18
  2. B
    1.68
  3. C
    2.38
  4. D
    1.99
View written solutionFree

Correct answer: D

  1. Use Stokes' law at terminal velocity

For a small sphere falling through a հեղuid, at terminal velocity:

6πηrvt=effective weight=43πr3(ρs−ρo)g6\pi \eta r v_t = \text{effective weight} = \frac{4}{3}\pi r^3 (\rho_s-\rho_o)g6πηrvt​=effective weight=34​πr3(ρs​−ρo​)g

So,

η=43πr3(ρs−ρo)g6πrvt=29 r2(ρs−ρo)gvt\eta = \frac{\frac{4}{3}\pi r^3 (\rho_s-\rho_o)g}{6\pi r v_t} = \frac{2}{9}\,\frac{r^2(\rho_s-\rho_o)g}{v_t}η=6πrvt​34​πr3(ρs​−ρo​)g​=92​vt​r2(ρs​−ρo​)g​
  1. Write the given data
  • Diameter of steel ball: d=3.6 mm=3.6×10−3 md=3.6\text{ mm}=3.6\times 10^{-3}\text{ m}d=3.6 mm=3.6×10−3 m
  • Radius:
r=d2=1.8×10−3 mr=\frac{d}{2}=1.8\times 10^{-3}\text{ m}r=2d​=1.8×10−3 m
  • Terminal velocity:
vt=2.45×10−2 m/sv_t=2.45\times 10^{-2}\text{ m/s}vt​=2.45×10−2 m/s
  • Density of steel:
ρs=7825 kg/m3\rho_s=7825\text{ kg/m}^3ρs​=7825 kg/m3
  • Density of oil:
ρo=925 kg/m3\rho_o=925\text{ kg/m}^3ρo​=925 kg/m3
  • Difference in density:
ρs−ρo=7825−925=6900 kg/m3\rho_s-\rho_o=7825-925=6900\text{ kg/m}^3ρs​−ρo​=7825−925=6900 kg/m3
  • Acceleration due to gravity:
g=9.8 m/s2g=9.8\text{ m/s}^2g=9.8 m/s2
  1. Substitute into the formula
η=29⋅(1.8×10−3)2×6900×9.82.45×10−2\eta = \frac{2}{9}\cdot \frac{(1.8\times 10^{-3})^2\times 6900\times 9.8}{2.45\times 10^{-2}}η=92​⋅2.45×10−2(1.8×10−3)2×6900×9.8​

First compute:

(1.8×10−3)2=3.24×10−6(1.8\times 10^{-3})^2 = 3.24\times 10^{-6}(1.8×10−3)2=3.24×10−6

Then,

3.24×10−6×6900=2.2356×10−23.24\times 10^{-6}\times 6900 = 2.2356\times 10^{-2}3.24×10−6×6900=2.2356×10−2

Next,

2.2356×10−2×9.8=0.21908882.2356\times 10^{-2}\times 9.8 = 0.21908882.2356×10−2×9.8=0.2190888

Now divide by 2.45×10−22.45\times 10^{-2}2.45×10−2:

0.21908880.0245=8.9424\frac{0.2190888}{0.0245} = 8.94240.02450.2190888​=8.9424

Finally multiply by 29\frac{2}{9}92​:

η=29×8.9424=1.9872\eta = \frac{2}{9}\times 8.9424 = 1.9872η=92​×8.9424=1.9872

Thus,

η≈1.99 SI unit=1.99 Pa⋅s\eta \approx 1.99\ \text{SI unit} = 1.99\ \text{Pa·s}η≈1.99 SI unit=1.99 Pa⋅s
  1. Match with options

The correct option is:

D: 1.99\boxed{\text{D: }1.99}D: 1.99​
  1. Comparison with stored answer

Stored correct answer is D, which matches our result.

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