On heating water,bubbles formed at the bottom | vedclass

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(B) When the bubble is about to detach,the upward buoyant force is balanced by the downward force due to surface tension.The buoyant force $F_{B}$ act

Vedclass · Accepted Answer

$R^{2} \sqrt{\frac{2 \rho_{w} g}{3 T}}$

Vedclass · Answer

(B) When the bubble is about to detach,the upward buoyant force is balanced by the downward force due to surface tension.The buoyant force $F_{B}$ acting on the bubble is given by $F_{B} = V ho_{w} g = \frac{4}{3} \pi R^{3} ho_{w} g$.The surface tension force $F_{S}$ acts along the circumference of the contact circle of radius $r$. The force is $F_{S} = T 	imes (2 \pi r) 	imes \sin 	heta$,where $	heta$ is the angle the radius makes with the vertical. Since $r Thus,$F_{S} = T 	imes 2 \pi r 	imes \frac{r}{R} = \frac{2 \pi T r^{2}}{R}$.Equating the forces: $\frac{2 \pi T r^{2}}{R} = \frac{4}{3} \pi R^{3} ho_{w} g$.Solving for $r^{2}$: $r^{2} = \frac{4}{3} \pi R^{3} ho_{w} g 	imes \frac{R}{2 \pi T} = \frac{2 R^{4} ho_{w} g}{3 T}$.Therefore,$r = R^{2} \sqrt{\frac{2 ho_{w} g}{3 T}}$.

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