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(B) Let the contact angle be $	heta$. Consider the center of the sphere $O$ and the point of contact $A$ at the edge of the liquid surface. Let $B$ b

Vedclass · Accepted Answer

$\cos^{-1} \left( \frac{R - h}{R} \right)$

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(B) Let the contact angle be $	heta$. Consider the center of the sphere $O$ and the point of contact $A$ at the edge of the liquid surface. Let $B$ be the point on the vertical axis at the same level as the liquid surface.In the right-angled triangle $\Delta OBA$,the hypotenuse $OA = R$ (radius of the sphere).The vertical distance from the center $O$ to the liquid surface is $R - h$.Thus,$OB = R - h$.The angle between the radius $OA$ and the vertical line $OB$ is $(90^{\circ} - 	heta)$,where $	heta$ is the contact angle.In $\Delta OBA$,we have:$\cos(90^{\circ} - 	heta) = \frac{OB}{OA} = \frac{R - h}{R}$Since $\cos(90^{\circ} - 	heta) = \sin 	heta$,we have $\sin 	heta = \frac{R - h}{R}$.However,looking at the geometry,the angle between the tangent to the surface at $A$ and the radius $OA$ is $90^{\circ}$. The angle between the horizontal liquid surface and the radius $OA$ is $	heta$.Therefore,$\cos 	heta = \frac{OB}{OA} = \frac{R - h}{R}$.Thus,$	heta = \cos^{-1} \left( \frac{R - h}{R} ight)$.

$A$ liquid is filled in a spherical container of radius $R$ up to a height $h$. At this position,the liquid surface at the edges is horizontal. The contact angle is:

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