The bob of a pendulum of mass 8 μ g carries an electric charge of 39.2 × 10-10 coulomb in an electric field of 20 × 103 volt/metre and it is at rest. The angle made by the pendulum with the vertical will be

Question

The bob of a pendulum of mass 8 μ g carries an electric charge of 39.2 × 10-10 coulomb in an electric field of 20 × 103 volt/metre and it is at rest. The angle made by the pendulum with the vertical will be (A) 27° (B) 45° (C) 87° (D) 127°

Tardigrade Admin · Accepted Answer

T sin θ=q E, T cos θ=m g <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a423f6dce8a2a3d6f2408cfa6c719b5f-.png" /> ∴ tan θ=(q E/m g)=(39.2 × 10-10 × 20 × 103/8 × 10-6 × 9.8)=1 ⇒ θ=45°

Q. The bob of a pendulum of mass $8 μg$ carries an electric charge of $39.2 \times 1 0^{- 10}$ coulomb in an electric field of $20 \times 1 0^{3}$ volt/metre and it is at rest. The angle made by the pendulum with the vertical will be

$2 7^{\circ}$

$4 5^{\circ}$

$8 7^{\circ}$

$12 7^{\circ}$

$T sin θ = qE, T cos θ = m g$

$∴ tan θ = \frac{qE}{m g} = \frac{39.2 \times 1 0 ^{- 10} \times 20 \times 1 0 ^{3}}{8 \times 1 0 ^{- 6} \times 9.8} = 1$
$\Rightarrow θ = 4 5^{\circ}$

Q. The bob of a pendulum of mass 8μg carries an electric charge of 39.2×10−10 coulomb in an electric field of 20×103 volt/metre and it is at rest. The angle made by the pendulum with the vertical will be

27∘

45∘

87∘

127∘