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1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^{- 4} N / C$ $2.55 \times 10^{- 4} N / C$ (Millikan’s oil drop experiment). The density of the oil is $1.26 gm / {cm}^{3}$ $1.26 gm / {cm}^{3}$ . Estimate the radius of the drop.
(g = 9.81 $m / s^{2}$ $m / s^{2}$ ; e = $1.60 \times 10^{- 19} C$ $1.60 \times 10^{- 19} C$ ).

Let the radius of the oil drop =r

Force (F) due to electric field E is equal to the weight of the oil drop (W).

$F_{e} = W$ $F_{e} = W$
$qE = mg$ $qE = mg$
$neE = mg = \frac{4}{3} {πr}^{3} ρg$ $neE = mg = \frac{4}{3} {πr}^{3} ρg$

$\begin{array}{l} r = {[\frac{3 neE}{4 πρg}]}^{\frac{1}{3}} = {[\frac{3 \times 2 .55 \times 10^{4} \times 12 \times 1 .6 \times 10^{- 19}}{4 \times 3 .14 \times 1 .26 \times 10^{3} \times 9 .81}]}^{\frac{1}{3}} \\ = {[946 .09 \times 10^{- 21}]}^{\frac{1}{3}} = 9 .82 \times 10^{- 7} mm \end{array}$ $\begin{array}{l} r = {[\frac{3 neE}{4 πρg}]}^{\frac{1}{3}} = {[\frac{3 \times 2 .55 \times 10^{4} \times 12 \times 1 .6 \times 10^{- 19}}{4 \times 3 .14 \times 1 .26 \times 10^{3} \times 9 .81}]}^{\frac{1}{3}} \\ = {[946 .09 \times 10^{- 21}]}^{\frac{1}{3}} = 9 .82 \times 10^{- 7} mm \end{array}$

The radius of the oil drop is 9.82 x $10^{- 7}$ $10^{- 7}$ mm.

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