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 $0.8 g / c m^{3}$ , the angle remains the same. If the density of the material of the sphere is $1.6 g m / c m^{3}$ , the dielectric constant of the liquid is

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Solution

The correct option is C 2
$t a n θ = \frac{F}{W} = \frac{F}{m g} t a n θ = \frac{F^{'}}{m g - U} ∴ \frac{F^{'}}{m g - U} = \frac{F}{m g} \Rightarrow \frac{F}{F^{'}} = \frac{m g}{m g - U} ∴ k = \frac{/ V . ρ . / g}{/ V (ρ - 0.8) / g} K = \frac{1.6}{16 - 0.8} = \frac{1.6}{0.8} = 2$

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

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Two identical charged spheres are suspended by strings of equal length and the strings make certain angle with each other. When suspended in a liquid of density $400 k g / m^{3}$ , the angle between the threads remains the same. If the density of the material of the sphere is $1600 k g / m^{3}$ , the dielectric constant of the liquid:

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The correct option is C 2tan θ=FW=Fmgtan θ=F′mg−U ∴F′mg−U=Fmg⇒FF′=mgmg−U∴k=/V.ρ./g/V(ρ−0.8)/gK=1.616−0.8=1.60.8=2

The correct option is C 2
$t a n θ = \frac{F}{W} = \frac{F}{m g} t a n θ = \frac{F^{'}}{m g - U} ∴ \frac{F^{'}}{m g - U} = \frac{F}{m g} \Rightarrow \frac{F}{F^{'}} = \frac{m g}{m g - U} ∴ k = \frac{/ V . ρ . / g}{/ V (ρ - 0.8) / g} K = \frac{1.6}{16 - 0.8} = \frac{1.6}{0.8} = 2$