$(a)\;1\;and\;2\qquad(b)\;1 , 2 \;and\;3\qquad(c)\;2\;and\;3 \qquad(d)\;2 , 3\;and\;4$

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Answer : (b) 1 , 2 and 3

Explanation :

Let the charge on the sphere be q and radius R . Then

$\large\frac{k\;q}{R+5\times10^{-2}}=100----(1)$

$\large\frac{k\;q}{R+10\times10^{-2}}=75----(2)$

$\large\frac{(1)}{(2)}\;we\;get\;\large\frac{R+10^{-1}}{R+5\times10^{-2}}=\large\frac{100}{75}=\large\frac{4}{3}$

$3R+3\times10^{-1}=4R+2\times10^{-1}$

$R=10^{-1}$

$R=10 \;cm$

Putting in equation (1) we get

$\large\frac{9\times10^{9}\times}{15\times10^{-2}}=100$

$q=\large\frac{15\times100\times10^{-11}}{9}$

$q=\large\frac{5}{3}\times10^{-9}\;C$

Potential on it's surface is

$V_{S}=\large\frac{k\;q}{R}=\large\frac{9\times10^{9}\times\large\frac{5}{3}\times10^{-9}}{10^{-1}}$

$V_{S}=150 V$

Electric field on it's surface

$E_{S}=\large\frac{k\;q}{R^2}=\large\frac{9\times10^{9}\times \large\frac{5}{3}\times10^{-9}}{10^{-2}}$

$E_{S}=1500 V/m$

Potential at centre of sphere

$V_{0}=\large\frac{k\;q}{2R^3}\;(3R^2-r^2)|_{r=0}$

$V_{0}=\large\frac{3\;k\;q}{2R}$

$V_{0}=\large\frac{3}{2} \times 150$

$V_{0}=225 V $

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