Question 12: Properties of Matter

Question

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\_\_\_\_$ $\%$.

Accepted Answer

The height (h) to which a liquid rises in a capillary tube is given by:
$$ \mathrm{h}=\frac{2 \mathrm{~S} \cos 	heta}{\mathrm{r} ho \mathrm{~g}} $$
Where :
S is the surface tension of the liquid.
$	heta$ is the angle of contact.
$\mathrm{r}$ is the inner radius of the capillary tube.
$ho$ is the density of the liquid.
$g$ is the acceleration due to gravity.
Assuming the angle of contact ( ÃÂ¸) and gravity ( $g$ ) remain constant, the height $h$ depends on variables $S, r$, and $ho$ as,
$$ h \propto \frac{S}{r ho} $$
Taking the natural logarithm of both sides :
$$ \ln (h)=\ln (S)-\ln (r)-\ln (ho)+	ext { constant } $$
Differentiating this equation,
$$ \frac{\Delta \mathrm{h}}{\mathrm{~h}}=\frac{\Delta \mathrm{S}}{\mathrm{~S}}-\frac{\Delta \mathrm{r}}{\mathrm{r}}-\frac{\Delta ho}{ho} $$
So, the percentage change in height is,
$$ \% \Delta \mathrm{~h}=\% \Delta \mathrm{~S}-\% \Delta \mathrm{r}-\% \Delta ho $$
It is given that all three quantities (radius, density, and surface tension) decrease by $1 \%$. Therefore, their percentage changes are negative :
Change in surface tension, $\% \Delta \mathrm{~S}=-1 \%$
Change in inner radius, $\% \Delta \mathrm{r}=-1 \%$
Change in density, $\% \Delta ho=-1 \%$
Substituting these values into our percentage change equation :
$$ \% \Delta \mathrm{~h}=(-1 \%)-(-1 \%)-(-1 \%) $$
$$\Rightarrow $$ $$ \% \Delta h=-1 \%+1 \%+1 \%=1 \% $$
The positive sign indicates that the height of the liquid in the tube will increase by $1 \%$. Hence, the correct option is (B).

Answer

+3

Answer

+1

Answer

-1

Answer

-3

Properties of Matter

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