Taking the earth to be a spherical conductor of diameter 12.8 × 10^3 ~km . Its capacity will be

Correct answer: A. 711 μ F Here : Diameter of earth aligned & =12.8 × 10^3 km So, radius R & =(12.8)/(2) × 10^3 aligned aligned & =6.4 × 10^3 km=6.4 × 10^6 m Capacity C & =4 π _0 R =(1)/(9 × 10^9) × 6.4 × 10^6 & =711 × 10^ -6 F=711 μ F aligned

JIPMER 2004

Exam: JIPMER
Year: 2004
Subject: Physics
Chapter: Capacitance

JIPMER 2004 · Physics · Capacitance

Taking the earth to be a spherical conductor of diameter 12.8 ×...

Taking the earth to be a spherical conductor of diameter

12.8 \times 10^3 \mathrm{~km}

. Its capacity will be

Options

A $711 \mu \mathrm{F}$
B $611 \mu \mathrm{F}$
C $811 \mu \mathrm{F}$
D $511 \mu \mathrm{F}$

Correct answer

A. $711 \mu \mathrm{F}$

Step-by-step solution

Here : Diameter of earth

\begin{aligned} & =12.8 \times 10^3 \mathrm{ km} \\ \text{So, radius } R & =\frac{12.8}{2} \times 10^3 \end{aligned}

\begin{aligned} & =6.4 \times 10^3 \mathrm{ km}=6.4 \times 10^6 \mathrm{ m} \\ \text{ Capacity } C & =4 \pi \varepsilon_0 R =\frac{1}{9 \times 10^9} \times 6.4 \times 10^6 \\ & =711 \times 10^{-6} \mathrm{ F}=711 \mu \mathrm{F}\end{aligned}

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