A cubical block of wood, of length 10 ,cm, fl | vedclass

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(D) Let the side length of the cubical block be $L = 10 \,cm = 0.1 \,m$.The block floats at the interface. Let $h_w = 1.5 \,cm = 0.015 \,m$ be the dep

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$830$

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(D) Let the side length of the cubical block be $L = 10 \,cm = 0.1 \,m$.The block floats at the interface. Let $h_w = 1.5 \,cm = 0.015 \,m$ be the depth of the block in water and $h_o = 10 \,cm - 1.5 \,cm = 8.5 \,cm = 0.085 \,m$ be the height of the block in oil.The pressure at the lower face (in water) is $P_{lower} = P_{interface} + ho_w g h_w$.The pressure at the upper face (in oil) is $P_{upper} = P_{interface} - ho_o g h_o$.The difference in pressure between the lower and upper face is $\Delta P = P_{lower} - P_{upper} = (P_{interface} + ho_w g h_w) - (P_{interface} - ho_o g h_o) = ho_w g h_w + ho_o g h_o$.Substituting the given values:$\Delta P = (1000 \,kg/m^3 	imes 10 \,m/s^2 	imes 0.015 \,m) + (800 \,kg/m^3 	imes 10 \,m/s^2 	imes 0.085 \,m)$$\Delta P = 150 \,Pa + 680 \,Pa = 830 \,Pa$.

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