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A capillary tube PQ of length 0.6 m and radius 4 xx 10^(-3)m is connected in series with another capillary tube QS of length 0.45 m and radius 10^(-3)m. The tubes are arranged horizontally. End P is connected to a vessel of water having constant pressure head of 0.8 m. The end S is open to the atmosphere. Find the pressure at the junction Q ?

Answer»


Solution :` V = pi P_1 r_1^(4) //8 ETA l_1 = pi P_2 r_2^(4)//8eta l_2, P_1//P_2= ( r_2//r_1) ^(4)(l_1ll_2) . ` Let P be the pressure at the junctions Q , `P_1 = 0.8- P, P_2 = P= 0 `
` 0.8 - P//P- 0 = ( 10 ^(-3) // 4 xx 10 ^(-3))^(4)xx (0.6 // 0.45 )` Solving P = 0.794 m of WATER


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