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(D) The pressure inside the soap bubble at $B$ is given by $P_{in} = P_0 + \frac{4T}{r}$,where $P_0$ is the atmospheric pressure,$T$ is the surface te

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$\frac{r \rho gh}{4}$

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(D) The pressure inside the soap bubble at $B$ is given by $P_{in} = P_0 + \frac{4T}{r}$,where $P_0$ is the atmospheric pressure,$T$ is the surface tension,and $r$ is the radius of the bubble.The pressure at the same horizontal level in the manometer limbs is equal. Let the pressure at the liquid surface in the open limb be $P_0$. Then the pressure at the level corresponding to the lower limb $D$ is $P_0 + \rho gh$.Equating the pressures at the level of the bubble $B$ and the manometer limb $D$ (assuming $B$ is at the same level as $D$):$P_0 + \frac{4T}{r} = P_0 + \rho gh$Simplifying the equation:$\frac{4T}{r} = \rho gh$Solving for $T$:$T = \frac{r \rho gh}{4}$

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