Question 177: Properties of Matter

Question

A thin uniform tube is bent into a circle of radius $$r$$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $${ho _1}$$ and $${ho _2}$$ $$\left( {{ho _1} > {ho _2}} ight),$$ fill half the circle. The angle $$	heta $$ between the radius vector passing through the common interface and the vertical is :

Accepted Answer

As system is in equilibrium so the pressuse at A from both side  of the liquid must be equal .
(r cos $$	heta $$ + r sin $$	heta $$) $$ho $$
2
g =  (r cos $$	heta $$ $$-$$ r sin $$	heta $$) $$ho $$
1
g
$$ \Rightarrow $$$$\,\,\,$$ $${{{ho _1}} \over {{ho _2}}} = {{\sin 	heta  + \cos 	heta } \over {\cos 	heta  - \sin 	heta }} = {{1 + 	an 	heta } \over {1 - 	an 	heta }}$$
$$ \Rightarrow $$$$\,\,\,$$ $$ho $$
1
$$-$$ $$ho $$
1
tan$$	heta $$ = $$ho $$
2
+ $$ho $$
2
tan$$	heta $$
$$ \Rightarrow $$$$\,\,\,$$ ($$ho $$
1
+ $$ho $$
2
) tan$$	heta $$ = $$ho $$
1
$$-$$ $$ho $$
2
$$ \Rightarrow $$$$\,\,\,$$ tan$$	heta $$ =  $${{{ho _1} - {ho _2}} \over {{ho _1} + {ho _2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $$	heta $$ = tan
$$-$$1
$$\left( {{{{ho _1} - {ho _2}} \over {{ho _1} + {ho _2}}}} ight)$$

Answer

$$	heta  = {	an ^{ - 1}}\pi \left( {{{{ho _1}} \over {{ho _2}}}} ight)$$

Answer

$$	heta  = {	an ^{ - 1}}{\pi  \over 2}\left( {{{{ho _1}} \over {{ho _2}}}} ight)$$

Answer

$$	heta  = {	an ^{ - 1}}\left( {{{{ho _1} - {ho _2}} \over {{ho _1} + {ho _2}}}} ight)$$

Answer

$$	heta  = {	an ^{ - 1}}{\pi  \over 2}\left( {{{{ho _1} + {ho _2}} \over {{ho _1} - {ho _2}}}} ight)$$

Properties of Matter

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