Question

Secondary School Mathematics for
I in the blanks
A vessel is in the form of a hemispherical
bowl mounted by a hollow cylinder. If each
of the diameter and height of the vessel is 2R
then the inner surface area of the vessel is​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{\red{\bf \: CSA_{(Cylinder)} = 2 \: \pi \: r \: h}}$

$\boxed{\red{\bf \:CSA_{(hemisphere)} = 2 \: \pi \: {r}^{2} }}$

$\large\underline{\sf{Solution-}}$

Given that

• A vessel is in the form of hemispherical bowl mounted by a hollow cylinder.

• Diameter of cylinder = Diameter of hemisphere = 2R

• Height of the vessel = 2R.

Since,

• Radius of hemispherical bowl = R

We know,

• Height of hemispherical bowl = Radius of hemispherical bowl

So,

• Height of hemispherical bowl = R

Now,

• Height of vessel = 2R

Thus,

• Height of cylindrical part = 2R - R = R.

Now,

We have,

• Height of cylindrical part, h = R

• Radius of cylindrical part, r = R

• Radius of hemispherical bowl, r = R

So,

$\bf :\longmapsto\:SA_{(bowl)} = CSA_{(hemisphere)} + CSA_{(Cylinder)}$

$\rm :\longmapsto\:SA_{(bowl)} =2 \: \pi \: {R}^{2} + 2 \: \pi \: R \times R$

$\rm :\longmapsto\:SA_{(bowl)} =2 \: \pi \: {R}^{2} + 2 \: \pi \: {R}^{2}$

$\rm :\longmapsto\:SA_{(bowl)} =4 \: \pi \: {R}^{2}$

Additional Information :-

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: CSA{(cuboid)} = 2(l + b) \times h}}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: TSA{(cube)} = {6(edge)}^{2} }}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$