A charged cork of mass m suspended by a light string is placed in uniform electric filed of strength E=(i^+j^)×105 NC−1 as shown in the figure. If in equilibrium position tension in the string is 2mg/(1+3) then angle α with the vertical can be

Ex=Ey=105 NC−1qEx=Tsin⁡αqEy=mg−Tcos⁡αDividing ExEy=T sin ⁡αmg−T cos ⁡αor mg−T cos ⁡α=T sin⁡ αor mg=2mg1+3(cos⁡ α+sin ⁡α) (given T=2mg1+3) or cos ⁡α+sin⁡ α=3+12,⇒α=30∘ or 60∘

A charged cork of mass m suspended by a light string is placed in uniform electric filed of strength E=(i^+j^)×105 NC−1 as shown in the figure. If in equilibrium position tension in the string is 2mg/(1+3) then angle α with the vertical can be

Ex=Ey=105 NC−1qEx=Tsin⁡αqEy=mg−Tcos⁡αDividing ExEy=T sin ⁡αmg−T cos ⁡αor mg−T cos ⁡α=T sin⁡ αor mg=2mg1+3(cos⁡ α+sin ⁡α) (given T=2mg1+3) or cos ⁡α+sin⁡ α=3+12,⇒α=30∘ or 60∘

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A charged cork of mass m suspended by a light string is placed in uniform electric filed of strength $E = (\hat{i} + \hat{j}) \times 10^{5} {NC}^{- 1}$ as shown in the figure. If in equilibrium position tension in the string is $2 mg / (1 + \sqrt{3})$ then angle $α$ with the vertical can be

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60°

30°

45°

18°

answer is A.

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Detailed Solution

$\begin{array}{l} E_{x} = E_{y} = 10^{5} {NC}^{- 1} \\ {qE}_{x} = Tsin α \\ {qE}_{y} = mg - Tcos α \end{array}$

Dividing $\frac{E_{x}}{E_{y}} = \frac{T \sin α}{mg - T \cos α}$
or    $mg - T \cos α = T \sin α$
or    $mg = \frac{2 mg}{1 + \sqrt{3}} (\cos α + \sin α)$ (given $T = \frac{2 mg}{1 + \sqrt{3}}$ )
or    $\cos α + \sin α = \frac{\sqrt{3} + 1}{2}, \Rightarrow α = 30^{\circ} or 60^{\circ}$