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(B) The total volume of $100$ drops is $V_{total} = 100 	imes \frac{4}{3} \pi r^3$.Given $r = \left(\frac{3}{40 \pi}ight)^{1/3} 	imes 10^{-3} \, c

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) The total volume of $100$ drops is $V_{total} = 100 	imes \frac{4}{3} \pi r^3$.Given $r = \left(\frac{3}{40 \pi}ight)^{1/3} 	imes 10^{-3} \, cm$,so $r^3 = \frac{3}{40 \pi} 	imes 10^{-9} \, cm^3$.$V_{total} = 100 	imes \frac{4}{3} \pi 	imes \frac{3}{40 \pi} 	imes 10^{-9} = 10^{-8} \, cm^3$.The thickness of the film $t_T$ is given by $Area 	imes t_T = V_{total}$.$4 \, cm^2 	imes t_T = 10^{-8} \, cm^3 \implies t_T = 0.25 	imes 10^{-8} = 25 	imes 10^{-10} \, cm$.Converting to meters: $t_T = 25 	imes 10^{-12} \, m$.The concentration of oleic acid is $0.01$,so the thickness of the oleic acid layer $t_0 = 0.01 	imes t_T$.$t_0 = 0.01 	imes 25 	imes 10^{-12} \, m = 25 	imes 10^{-14} \, m$.Comparing with $x 	imes 10^{-14} \, m$,we get $x = 25$.

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