Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Diameter of hemisphere = 14 cm
Radius of hemisphere \((r)\) = 7 cm
Radius of cylinder will bethe same as the hemisphere = 7 cm
Height of Cylinder \((h)\) = 13-7 = 6 cm
Surface Area of the vessel will be the sum of curved surface Area of a hemisphere and a cylinder
Calculating CSA

$$\require{cancel}\begin{aligned}CSA&=2\pi rh+2\pi r^{2}\\ &=2\pi r\left( r+h\right) \\ &=2\times \dfrac{22}{\cancel{7}}\times \cancel{7}\times \left( 6+7\right) \\ &=2\times 22\times 13\\ &=44\times 13\\ &=572\ \mathrm{cm^2}\end{aligned}$$

Inner surface area of the vessel is \(572\ \mathrm{cm^2}\)

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Radius of the cone \((r)\) = 3.5 cm
Height of the toy = 15.5 cm
Height of the cone \((h)\)= 15.5-3.5 = 12 cm
Total surface Area of the toy = CSA of the cone + CSA of the hemisphere

$$\begin{aligned}TCSA&=\pi rl+2\pi r^{2}\\ &=\pi r\left( l+2r\right) \\\\ l&=\sqrt{r^{2}+h^{2}}\\ &=\sqrt{\left( 3.5\right) ^{2}+12^{2}}\\ &=\sqrt{12.25+144}\\ &=12.50\\\\ TCSA&=\pi \times 3.5\left( 12\cdot 5+2\times 3.5\right) \\ &=\dfrac{22}{7}\times 3.5\left( 12.5+7\right) \\ &=22\times 0.5\left( 19.5\right) \\ &=11\times 19.5\\ &=214.5\ \mathrm{cm^{2}}\end{aligned}$$

Total surface area of the toy is \(214.5\ \mathrm{cm^{2}}\)

Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

Side of cubical block \((a)\) = 7 cm
Greatest diameter of the hemisphere will be 7 cm \(\Rightarrow r=3.5\)
Surface Area of the solid = Surface Area of the cuboid + curved surface area of the hemisphere - Base Area of hemisphere

$$\require{cancel}\begin{aligned}SA&=6a^{2}+2\pi r^{2}-\pi r^{2}\\ &=6a^{2}+\pi r^{2}\\ &=6\cdot 7^{2}+\dfrac{22}{\cancel{7}}\times \cancelto{0.5}{3.5} \times 3.5\\ &=6\times 49+\times 22\times 3.5\times 0.5\\ &=294+11\times 3.5\\ &=294+38.5\\ &=332.5\ \mathrm{cm^{2}}\end{aligned}$$

Surface area of the solid is \(332.5\ \mathrm{cm^{2}}\)

Q5. A hemispherical depression is cut out from one face of a cubic wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Diameter of the hemisphere = \(l\)
Radius of the hemisphere \((r)\)= \(l/2\)
Side of cube = \(l\)
Surface Area of the remaining solid = surface Area of cubic block + CSA of hemisphere- Area of base of hemisphere

$$\begin{aligned}SA&=6l^{2}+2\pi \left( \frac{l}{2}\right) ^{2}-\pi \left( \frac{l}{2}\right) ^{2}\\\\ &=6l^{2}+A\left( \frac{l}{2}\right) ^{2}\\\\ &=6l^{2}+\dfrac{\pi l^{2}}{4}\\\\ &=\dfrac{24l^{2}+\pi l^{2}}{4}\\\\ &=\dfrac{l}{4}\left[ \pi +24\right] \end{aligned}$$

Surface area of the remaining solid \(\dfrac{l}{4}\left[ \pi +24\right] \)

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Diameter of the capsule = 5 mm
Radius of Capsule \((r)\)= 2.5 mm
Entire length of capsule = 14 mm
length of Cylinder \((h)\) = 14-(2.5+2.5) = 9 mm
SA of the capsule = CSA of Cylinder + 2 (CSA of hemisphere )

$$\require{cancel}\begin{aligned}TSA&=2\pi rh+4\pi r^{2}\\ &=2\pi rh+4\pi r^{2}\\ &=2\pi r\left( h+2r\right) \\ &=2\times \dfrac{22}{7}\times 2.5\times \left( 9+5\right) \\ &=2\times \dfrac{22}{\cancel{7}}\times 2.5\times \cancelto{2}{14} \\ &=2\times 22\times 2.5\times 2\\ &=220\ \mathrm{mm^{2}}\end{aligned}$$

Surface area of Capsule is \(=220\ \mathrm{mm^{2}}\)

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical parts are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m. Find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Solution:

Height of cylindrical part \((h)\) 2.1 m
Diameter of cylindrical part = 4 m
Radius of cylindrical part \((r)\) = 4/2 =2 m
Slant-height \((l)\) of conical past =2.8 m
canvas required for the tent = CSA of Cylindrical part + CSA of conical part

$$\begin{aligned}SA&=2\pi rh+\pi rl\\ &=\pi r\left( 2h+l\right) \\ &=\dfrac{22}{7}\times 2\left( 2\times 2.1+2.8\right) \\ &=\dfrac{22}{7}\times 2\left( 4.2+2.8\right) \\ &=\dfrac{22}{7}\times 2\times 7\\ &=44\ \mathrm{m^{2}}\end{aligned}$$

Rate of canvas = ₹500/m² therefore cost of canvas

$$\begin{aligned}&=44\times 500\\ &=\mathbb{\text{₹ }}22000\end{aligned}$$

Cost of the canvas of the tent is \(\mathbb{\text{₹ }}22000\)

Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest \(\mathrm{cm^2}\).

Solution:

Height of cylinder \((h)\) = 2.4 cm
Diameter of cylinder = 1.4 cm
Radius of Cylinder \((r)\) = 1.4/2 = 0.7 cm
Total Surface Area = CSA of cylinder + CSA of Cone + Base Area of cylinder

$$\begin{aligned}TSA&=2\pi rh+\pi rl+\pi r^{2}\\\\ &=\pi r\left[ 2h+l+r\right] \\\\ l&=\sqrt{r^{2}+h^{2}}\\\\ &=\sqrt{\left( 0.7\right) ^{2}+\left( 0.7\right) ^{2}}\\\\ &=\sqrt{0.49+\left( 2.4\right) ^{2}}\\\\ &=2.5\\\\ TSA&=\scriptsize\dfrac{22}{7}\times 0.7\left[ \left( 2\times 2.4\right) +2.5+0.7\right] \\\\ &=2.2\left[ 4.8+3.2\right] \\\\ &=2.2\times 8\\\\ &=17.6\\\\ &\approx 18\ \mathrm{cm^{2}}\end{aligned}$$

Total surface area of the remaining solid is \(\approx 18\ \mathrm{cm^{2}}\)

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

Height of cylinder \(h\) = 10 cm
Radius of the base \(r\)= 3.5 cm
Total surface Area = CSA of Cylinder + 2 (CSA of hemisphere)

$$\begin{aligned}TSA&=2\pi rh+2\left( 2\pi r^{2}\right) \\ &=2\pi rh+4\pi r^{2}\\ &=2\pi r\left[ h+2r\right] \\ &=2\times \dfrac{22}{7}\times 3.5\left( 10+2\times 3\cdot 5\right) \\ &=2\times 22\times 0.5\left( 10+7\right) \\ &=22\times 17\\ &=374\ \mathrm{cm^{2}}\end{aligned}$$

Total surface area of the article is \(374\ \mathrm{cm^{2}}\)