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A cubical block of wood, of length 10 cm, floats at the interface between oil of density 800 kg//m^3and water. The lower surface of the block is 1.5 cm below the interface. If the depth of water is 10 cm below the interface and oil is upto 10 cm above the interface then the difference in pressure at the lower and the upper face of the wooden block is (Assume density of waterrho = 1000 kg //m^3) and acceleration of gravity , g = 10 m//s^2) |
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Answer» 850 Pa  density of oil `rho_0 = 800 kg//m^3 ` and density of WATER, `rho_w = 10^3 kg//m^3` If ` V_1` and `V_2`be the volume of wooden block in water and oil respectively, then `V_1 : V_2 = 1.5 : 8.5` ` V_1/V_2 = (1.5)/(8.5) = 3.17` ` therefore V_1 = 3/17 V_2 "" ...(i)` ` therefore V_1 + V_2= V`where is the volume of the block From EQ. (1), we get ` 3/17 V_2 + V_2 = V rArr V_2 = 17/20 V` ` therefore V_1 = 3/20 V` If ` rho ` be the density of the block weight of block ` therefore ` weight of oil displaced+weight of water displaced ` V rho G = V_1 rho_(omega) g + V_2 rho_(oil) g` ` V rho g = 3/20 V rho_w g + 17/20 V rho_(oil) g` ` rho = 3/20 rho_w + 17/20 rho_(oil) = 3/20 xx 10000 + 17/20 xx 800 ` ` = 830 kg//m^3` ` therefore ` PRESSURE difference ` p = rho g h = 830 xx 10 xx 0.1 = 830 Pa` |
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