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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 4 April · Morning Shift · Physics · Electrostatics

Two small spherical balls of mass 10 g each with charges -2 μ C and...

Two small spherical balls of mass 10 g each with charges −2μC-2 \mu \mathrm{C} and 2μC2 \mu \mathrm{C}, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of 100μC/m2100 \mu \mathrm{C} / \mathrm{m}^2 such that length of rod makes an angle of 30∘30^{\circ} with electric field generated by charge sheet. Net torque acting on the rod is:
(Take ε0:8.85×10−12C2/Nm2\varepsilon_0: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2)Two small spherical balls of mass 10 g each with charges -2 μ C and 2 μ C , are attached to two ends of very light

Options

  1. A112 Nm
  2. B1.12 Nm
  3. C2.24 Nm
  4. D11.2 Nm

Correct answer

B. 1.12 Nm

Step-by-step solution

Two small spherical balls of mass 10 g each with charges -2 μ C and 2 μ C , are attached to two ends of very light
E=σ2ε0τ=PEsin⁡θ\begin{aligned} \mathrm{E} & =\frac{\sigma}{2 \varepsilon_0} \\ \tau & =\mathrm{PE} \sin \theta\end{aligned}
=[(2×10−6)(210)][100×10−62×8.85×10−12](12)=108.85=1.12Nm\begin{aligned} & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm}\end{aligned}
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