When the size of a spherical nanoparticle decreases from $30 nm$ to $10 nm$ . The ratio surface area/volume becomes:

1 / 3

of the original

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3

times the original

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1 / 9

of the original

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9

times the original

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Solution

The correct option is B $3$ times the original
From the given information.
$\frac{(s u r f a c e a r e a)}{v o l u m e} = \frac{(π d^{2})}{(\frac{π d^{2}}{6})} = \frac{6}{d}$
$d_{1} = 30 n m, d_{2} = 10 n m$
$∴$ $\frac{{(\frac{s u r f a c e a r e a}{v o l u m e})}_{2}}{{(\frac{s u r f a c e a r e a}{v o l u m e})}_{1}} = \frac{(\frac{6}{d_{2}})}{(\frac{6}{d_{1}})} = \frac{d_{1}}{d_{2}}$ = $\frac{30}{10}$ = $3$

Hence, the correct answer is option $B$ .

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When the size of a spherical nanoparticle decreases from 30 nm to 10 nm. The ratio surface area/volume becomes:

The correct option is B 3 times the originalFrom the given information.(surface area)volume=(πd2)(πd26)=6dd1= 30nm , d2=10nm∴ (surface areavolume)2(surface areavolume)1=(6d2)(6d1) = d1d2 = 3010 = 3Hence, the correct answer is option B.

When the size of a spherical nanoparticle decreases from $30 nm$ to $10 nm$ . The ratio surface area/volume becomes: