Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2010 · Paper 2 · Physics · Electrostatics

A tiny spherical oil drop carrying a net charge q is balanced in...

A tiny spherical oil drop carrying a net charge qq is balanced in still air with a vertical uniform electric field of strength 81π7×105Vm−1\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}. When the field is switched off, the drop is observed to fall with terminal velocity 2×10−3 ms−12 \times 10^{-3} \mathrm{~ms}^{-1}. Given g=9.8 ms−2g=9.8 \mathrm{~ms}^{-2}, viscosity of the air =1.8×10−5Nsm−2=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2} and the density of oil =900 kg m−3=900 \mathrm{~kg} \mathrm{~m}^{-3}, the magnitude of qq is

Options

  1. A1.6×10−19C1.6 \times 10^{-19} \mathrm{C}
  2. B3.2×10−19C3.2 \times 10^{-19} \mathrm{C}
  3. C4.8×10−19C4.8 \times 10^{-19} \mathrm{C}
  4. D8.0×10−19C8.0 \times 10^{-19} \mathrm{C}

Correct answer

D. 8.0×10−19C8.0 \times 10^{-19} \mathrm{C}

Step-by-step solution

qE=mg6π\etarv=mg43\pir3ρg=mg∴r=(3mg4πρg)1/3\begin{aligned}& q E=m g \\ & \quad 6 \pi \etar v=m g \\ & \frac{4}{3} \pir^3 \rho g=m g \\ & \therefore \quad r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3} \end{aligned}SubstitutingthevalueofSubstituting the value ofrinEq.(ii)weget,6in Eq. (ii) we get, 6\pi \etav(3mg4πρg)1/3=v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=m g or (6πηv)3(3mg4πρg)=(mg)3\quad(6 \pi \eta v)^3\left(\frac{3 m g}{4 \pi \rho g}\right)=(m g)^3 Again substituting mg=qEm g=q E we get, (q E)^2=(34πρg)(\left(\frac{3}{4 \pi \rho g}\right)(6 πη\pi \etav)^3 or qE=(34πρg)1/2(6πηv)3/2\quad q E=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2} \therefore \quad q=1E(34πρg)1/2(\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 πη\pi \etav)^{3 / 2} Substituting the values we get , q=781π×10534π×900×9.8×216π3×(1.8×10−5×2×10−3)3=8.0×10−19C\begin{aligned}q= & \frac{7}{81 \pi \times 10^5} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^3}\\ & \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^3}\\ & =8.0 \times10^{-19} \mathrm{C} \end{aligned}\thereforecorrectoptioniscorrect option is(\mathrm{d})$.
Practice Electrostatics on Quantrex Academy →

More from Electrostatics

Two charges Q_1 = q and Q_2 = mq are placed at the points P_1(a, b) and P_2(ma, mb)… JEE Advanced 2026Consider an electric dipole comprising two charges +q and -q each with mass m , separated… JEE Advanced 2026Two point charges q_1=3\,μ C and q_2=-4\,μ C are placed at points (2 i +3 j +3 k ) and (… JEE Main 2026The electric potential as a function of x, y is given by V = 5(x^2 - y^2) V. The electric… JEE Main 2026A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb.… JEE Main 2026A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an… JEE Main 2026A particle of charge q and mass m is projected from origin with an initial velocity v = (… JEE Main 2026A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an… JEE Main 2026 Full Electrostatics list All JEE Advanced PYQs