Question

Q.A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then calculate the rise of water level (in cm).

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Answer 1

hey here is your answer

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$so \: in \: this \: problem \\ we \: need \: to \: find \: height \: h \: of \: cylindrical \: vessel \: ie \: the \: extent \: of \: rise \: in \: volume \: of \: water \: level \: in \: cylindrical \: vessel$

$so \: for \: a \: sphere \\ diameter(d) = 18 \: cm \\ hence \: radius(r) = d \div 2 \\ = 18 \div 2 \\ = 9 \: cm \\ \\ also \: for \: a \: cylindrical \: vessel \: \\ diameter = 36 \: cm \\ thus \: radius(R) = 18 \: cm$

$so \: as \: when \: sphere \: is \: immersed \: in \: cylindrical \: vessel \: ie \: is \: dropped \: \\ subsequently \: the \: volume \: of \: both \: sphere \: and \: vessel \: will \: get \: equal \: as \: here \: it \: is \: said \: that \: the \: sphere \: is \: partly \: filled \: with \: water \\ this \: is \: because \: of \: law \: of \: buoyancy \: and \: density \: of \: two \: objects \: when \: immersed \: in \: water$

$so \: we \: know \: that\: \\ volume \: of \: sphere = 4 \div 3\pi \: r.cube \\ volume \: of \: cylindrical \: vessel = \pi \: R \: square.h$

$hence \: according \: to \: first \: condition \\ 4 \div 3\pi \: r \: cube = \pi \: R \: square.h \\ ie \: 4 \div 3.r \: cube = R \: square.h \\ so \: 4 \div 3 \times (9)cube = (18)square.h \\ ie \: 4 \div 3 \times 729 = 324h \\ 4 \times 243 = 324h \\ \\ 972 = 324h \\ ie \: h = 3 \: cm$

$hence \: the \: water \: level \: would \: increase \: upto \: an \: extent \: ie \: upto \: a \: height \: of \: 3 \: cm$

Answer 2

Please refer to the attachment