SATHEE: Electrostatics 7 Question 9

Electrostatics 7 Question 9

9. A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 π}{7} \times 10^{5} {Vm}^{- 1}$ . When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{- 3} {ms}^{- 1}$ . Given $g = 9.8 {ms}^{- 2}$ , viscosity of the air $= 1.8 \times 10^{- 5} Ns m^{- 2}$ and the density of oil $= 900 kg m^{- 3}$ , the magnitude of $q$ is

(a) $1.6 \times 10^{- 19} C$

(b) $3.2 \times 10^{- 19} C$

(d) $8.0 \times 10^{- 19} C$

(2010)

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Answer:

Correct Answer: 9. (d)

Solution:

$\begin{aligned} q E & = m g \dots (i) \\ 6 π η r v & = m g \\ \frac{4}{3} π r^{3} ρ g & = m g \dots (i i) \end{aligned}$

$∴ r = {\frac{3 m g}{4 π ρ g}}^{1 / 3} \dots (i i i)$

Substituting the value of $r$ in Eq. (ii) we get,

$\begin{aligned} 6 π η ν {\frac{3 m g}{4 π ρ g}}^{1 / 3} = m g \\ (6 π η v)^{3} \frac{3 m g}{4 π ρ g} = (m g)^{3} \end{aligned}$

Again substituting $m g = q E$ we get,

$\begin{aligned} (q E)^{2} = \frac{3}{4 π ρ g} (6 π η v)^{3} \\ or q E = {\frac{3}{4 π ρ g}}^{1 / 2} (6 π η v)^{3 / 2} \\ ∴ q = \frac{1}{E} {\frac{3}{4 π ρ g}}^{1 / 2} (6 π η v)^{3 / 2} \end{aligned}$

Substituting the values we get,

$\begin{aligned} q & = \frac{7}{81 π \times 10^{5}} \sqrt{\frac{3}{4 π \times 900 \times 9.8} \times 216 π^{3}} \\ \times \sqrt{{(1.8 \times 10^{- 5} \times 2 \times 10^{- 3})}^{3}} \\ = 8.0 \times 10^{- 19} C \end{aligned}$

Electrostatics 7 Question 9

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Electrostatics 7 Question 9

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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