Solveeit Logo
Login

Question

Question: A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is \(2...

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 2.0×10−5 km m−1 s−12.0 \times 10 ^ { - 5 } \mathrm {~km} \mathrm {~m} ^ { - 1 } \mathrm {~s} ^ { - 1 } the terminal velocity of the drop will be

(The density of water=103 kg m−3= 10 ^ { 3 } \mathrm {~kg} \mathrm {~m} ^ { - 3 }and g=10)

A

1.0×10−4 m s−11.0 \times 10 ^ { - 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

B

2.0×10−4 m s−12.0 \times 10 ^ { - 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

C

2.5×10−4 m s−12.5 \times 10 ^ { - 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

D

5.0×10−4 m s−15.0 \times 10 ^ { - 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Answer

2.5×10−4 m s−12.5 \times 10 ^ { - 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Explanation

Solution

Here, r=0.0015 mm=0.0015×10−3 m\mathrm { r } = 0.0015 \mathrm {~mm} = 0.0015 \times 10 ^ { - 3 } \mathrm {~m}

η=2.0×10−5 kg m−1 s−1\eta = 2.0 \times 10 ^ { - 5 } \mathrm {~kg} \mathrm {~m} ^ { - 1 } \mathrm {~s} ^ { - 1 }

ρ=1.0×103 kg m−3\rho = 1.0 \times 10 ^ { 3 } \mathrm {~kg} \mathrm {~m} ^ { - 3 }

Neglecting the density of air, the terminal velocity of the water drop is

=2×(0.0015×10−3)2×1.0×103×109×2.0×10−5= \frac { 2 \times \left( 0.0015 \times 10 ^ { - 3 } \right) ^ { 2 } \times 1.0 \times 10 ^ { 3 } \times 10 } { 9 \times 2.0 \times 10 ^ { - 5 } }