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(D) Let the pressure at the interface of mercury and glycerin be $P_1$ and at the interface of mercury and oil be $P_2$. Since the upper surfaces of t

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$9.6$

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(D) Let the pressure at the interface of mercury and glycerin be $P_1$ and at the interface of mercury and oil be $P_2$. Since the upper surfaces of the oil and glycerin are at the same level,and the mercury levels in the two arms are at different heights,we equate the pressures at the level of the lower mercury interface.Let $h$ be the length of the oil column.The pressure at the level of the mercury-glycerin interface in the left arm is $P_0 + ho_{	ext{glycerin}} g (10 \ cm)$.The pressure at the same horizontal level in the right arm is $P_0 + ho_{	ext{oil}} g h + ho_{	ext{mercury}} g (10 - h)$.Equating the two pressures:$P_0 + 1.3 	imes g 	imes 10 = P_0 + 0.8 	imes g 	imes h + 13.6 	imes g 	imes (10 - h)$$13 = 0.8h + 136 - 13.6h$$12.8h = 136 - 13$$12.8h = 123$$h = \frac{123}{12.8} \approx 9.6 \ cm$.

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