Question

A cylindrical glass with diameter 20cm has water to a height of 9cm. A small cylindrical metal of radius 5cm and height 4cm is immersed it completely. calculate the raise of the water in the glass ?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Rise\:in\:height=1\:cm}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline\bold{Given : }} \\ \tt{ : \implies Radius \: of \: cylinder( r_{1} )=10 \: cm} \\ \\ \tt{ : \implies Height \: of \: cylinder( h_{2})= 9 \: cm} \\ \\ \tt{ : \implies Radius \: of \: cylinder( r_{2}) = 5 \: cm} \\ \\ \tt{ : \implies Height \: of \: cylinder ( h_{2}) = 4 \: cm} \\ \\ \red{ \underline\bold{To \: Find : }} \\ \tt{: \implies Raise \: of \: water \: in \: glass = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Volume \: of \: cylinder \: glass= \pi { r_{1} }^{2} h_{1} } \\ \\ \tt{: \implies Volume \: of \: cylinder \: glass =3.14 \times {10}^{2} \times 9} \\ \\ \green{\tt{: \implies Volume \: of \: cylinder \: glass =2826 {cm}^{3} }} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies Volume \: of \: cylindrical \: metal = \pi { r_{2} }^{2} h_{2} } \\ \\ \tt{: \implies Volume \: of \: cylindrical \: metal =3.14 \times {5}^{2} \times 4 } \\ \\ \green{\tt{: \implies Volume \: of \: cylindrical \: metal = 314 {cm}^{3} }} \\ \\ \bold{Total \: Volume \:after \: adding \: cylindrical \: metal} \\ \tt{: \implies New \: volume = \pi { r_{1} }^{2} h_{3}} \\ \\ \tt{: \implies 2826 + 314 = 3.14 \times 10 \times 10 \times h_{3}} \\ \\ \tt{: \implies h_{3} = \frac{3140}{314}} \\ \\ \green{\tt{: \implies h_{3} =10 \: cm }} \\ \\ \bold{Rise \: height : } \\ \tt{: \implies Height \: rise = 10 - 9} \\ \\ \green{\tt{: \implies Height \: rise =1 \: cm }}$

Answer 2

Answer:

$\large{\underline{\boxed{\mathfrak\blue{Height \: of \: water \: rises = 1 \: cm }}}}$

Given:-

• Diameter of cylindrical glass(d1)=20 cm
• Height of cylindrical glass (h1) = 9 cm
• Radius of cylindrical metal (r2) = 5 cm
• Height of cylindrical metal (h2) = 4 cm

To find:-

• Height of water rise in the glass if cylindrical metal is immersed into cylindrical glass = ?

$\rm We \: know, \\\\\star \: {\boxed{\rm{Volume \: of \: cylinder = \pi r^2 h }}}$

$\rm At \: first, \\\\\Rightarrow\rm Volume \: of \: cylindrical \: glass = \pi (\dfrac{20}{2})^2 \times 9 \\\\\Rightarrow\rm Volume \:of \:cylindrical\: glass = 3.14 \times (10)^2 \times 9 \\\\\Rightarrow\rm Volume\: of \:cylindrical\: glass = 3.14 \times 100 \times 9 \\\\\Rightarrow\rm Volume\: of\: cylindrical\: glass ={\boxed{\rm{2826 \: cm^3 }}}$

$\rm Secondly, \\\\\Rightarrow\rm Volume\: of\: cylindrical \:metal = 3.14 \times (5)^2 \times 4 \\\\\Rightarrow\rm Volume \:of\: the\: cylindrical metal = 3.14 \times 25 \times 4 \\\\\Rightarrow\rm Volume \:of\: the \:cylindrical\: metal ={\boxed{\rm{314 \: cm^3}}}$

$\rule{200}{1}$

$\rm New \:volume \: of cylindrical \: glass:- \\\\\Rightarrow\rm New \: volume = 3.14 \times (10)^2 \times h_3 \\\\\Rightarrow\rm \: {Volume}_{cylindrical \: glass} + {Volume}_{cylindrical \: metal} = 314h \\\\\Rightarrow\rm 2826 + 314 = 314h \\\\\Rightarrow\rm 3140 = 314h \\\\\Rightarrow\rm h_3 = 3140/314 \\\\\Rightarrow\rm h_3 ={\boxed{\rm{10\: cm }}}$

$\because\rm Old\:height \: of \: cylindrical \: glass = 9 \: cm$

$\rm And, New \: height \: of \: cylindrical \: glass = 10 \: cm$

$\therefore\rm Height \: rises = (10 -9)\: cm \\\\\therefore\rm Height \: rises = {\boxed{\rm{1\: cm }}}$

${\underline{\underline{\therefore{\text{Height \: rises \: 1 \: cm \: in \: the \: glass }}}}}$