Question

$\huge{\frak{\underline{Bonjour!♡}}}$
⤷ Two identical charged metallic spheres are hung by strings. The strings make 60° with each other in vacuum when the whole arrangement is kept in a liquid of density
0.5 g/CC , the angle remains same. Calculate dielectric constant of fluid given density of sphere is 1.5 g/CC .
[Ans:-- 1.5]
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# JEE Question ​

Answer 1

$\large{\underline{\frak{Given:-}}}$

• Density of sphere= 1.5 g/CC
• Density of liquid= 0.5 g/CC
• angle between the two strings= 60°

$\large{\underline{\frak{To\:Find-}}}$

• Dielectric constant or relative permittivity

$\large{\underline{\frak{solution:-}}}$

• refer to the diagram 1.

→∅= 30°

→ Tsin ∅= Fo (repulsive force)....(1)

→ T cos∅= mg........(2)

dividing (1) by (2)

→ tan∅= Fo/mg

as m = μ V [ μ= density of sphere]

→ tan∅= Fo/μVg ....(3)

• Now, as per second diagram

→ ∅= 30° ( given)

→ T' sin∅= Fm......(4) [T' = new tension developed due to liquid]

→ T' cos∅ + U = mg[ U= upthrust exerted by liquid]

→ T'cos∅= mg- U ....(5)

dividing (4) by (5)

→ tan∅= Fm/mg-U

as U = Vpg (where p is density of liquid)

→ tan∅= Fm/μVg- Vpg ....(6)

now, comparing eq( 3) and (6)

→ Fo/μVg= Fm/μVg-Vpg

→Fo/μ = Fm/μ-p

putting the given values

→ Fo/1.5= Fm/1.5-0.5

→ Fo/Fm= 1.5

which is dielectric constant or relative permittivity.

⤷ hope it helps you shubhu ❤️

Answer 2

Answer:

Given:−

Density of sphere= 1.5 g/CC

Density of liquid= 0.5 g/CC

angle between the two strings= 60°

\large{\underline{\frak{To\:Find-}}}

ToFind−

Dielectric constant or relative permittivity

\large{\underline{\frak{solution:-}}}

solution:−

refer to the diagram 1.

→∅= 30°

→ Tsin ∅= Fo (repulsive force)....(1)

→ T cos∅= mg........(2)

dividing (1) by (2)

→ tan∅= Fo/mg

as m = μ V [ μ= density of sphere]

→ tan∅= Fo/μVg ....(3)

Now, as per second diagram

→ ∅= 30° ( given)

→ T' sin∅= Fm......(4) [T' = new tension developed due to liquid]

→ T' cos∅ + U = mg[ U= upthrust exerted by liquid]

→ T'cos∅= mg- U ....(5)

dividing (4) by (5)

→ tan∅= Fm/mg-U

as U = Vpg (where p is density of liquid)

→ tan∅= Fm/μVg- Vpg ....(6)

now, comparing eq( 3) and (6)

→ Fo/μVg= Fm/μVg-Vpg

→Fo/μ = Fm/μ-p

putting the given values

→ Fo/1.5= Fm/1.5-0.5

→ Fo/Fm= 1.5

which is dielectric constant or relative permittivity.

⤷ hope it helps you shubhu