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AIEEE 2010 Solved Paper

Section: Physics

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Question : 36 of 89

Marks: +1, -0

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of

30^{\circ}

with each other. When suspended in a liquid of density

0.8 \; \mathrm{g} \; \mathrm{cm}^{-3}

, the angle remains the same. If density of the material of the sphere is

1.6 \; \mathrm{gcm}^{-3}

, the dielectric constant of the liquid is

4
3
2
1

Solution:

\;F_e = T \sin \;15^{\circ}

\;m g = T \cos 15^{\circ}

\; \Rightarrow \tan 15^{\circ} = \frac{F_e}{m g}

......(i)

In liquid,

F_e' = T' \sin \;15^{\circ}

.....(ii)

m g = F_B + T' \cos 15^{\circ}

F_B' = V(d-\rho) g = V(1.6-0.8) g = 0.8 V g

= 0.8 \; \frac{m}{d} g = \frac{0.8 m g}{1.6} = \frac{m g}{2}

\therefore m g = \frac{m g}{2} + T' \cos 15^{\circ}

\Rightarrow \; \frac{m g}{2} = T' \cos 15^{\circ}

From

(A)

and

(B), \tan 15^{\circ} = \frac{2 F_e'}{m g}

.......(2)

From

(1)

and

(2)

\; \frac{F_e}{m g} = \; \frac{2 F_e'}{m g} \Rightarrow F_e = 2 F_e' \Rightarrow F_e' = \frac{F_e}{2}

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