A horizontally oriented bent thin tube ABC of length 1.4 m rotates with a constant angular velocity $ω$ about a stationary vertical axis OO’ passing through the end A as shown in figure. The portion BC of tube is parallel to OO’ and its length is equal 0.4 m. At t = 0, AB part of tube is filled with an ideal fluid. The end A of tube is open, the closed end C has a very small orifice. Find the maximum height(in m) reached by fluid relative to end C. (take $g = 10 m / s^{2}$ and $ω = 10 r a d / s$ )

Question

A horizontally oriented bent thin tube ABC of length 1.4 m rotates with a constant angular velocity $ω$ about a stationary vertical axis OO’ passing through the end A as shown in figure. The portion BC of tube is parallel to OO’ and its length is equal 0.4 m. At t = 0, AB part of tube is filled with an ideal fluid. The end A of tube is open, the closed end C has a very small orifice. Find the maximum height(in m) reached by fluid relative to end C. (take $g = 10 m / s^{2}$ and $ω = 10 r a d / s$ )

Accepted Answer

tan30=vxvy=ugt=13⇒t=3ug      .....(1) x=ut=3u2g          ....(2) s=uyt+12t2=0+12gt2 s=12g3u2g2=3u22g          ....(3)Now  y=h−s=h−3u22g               ......(4)Velocity of  =u=2gh1=2g(10−h)                 ....(5)From the triangle  tan300=yx ∴y=x tan300=3u2g×13=u2g          ....(6)Combing (4), (5) and (6) u2g=h-32g[2g(10-h)] ⇒2g(10−h)g=h−3(10−h) ⇒20−2h=h−30+3h ⇒20+30=6h ⇒h=506=106×x ⇒x=5

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