**1.** 2 cubes each of volume 64 cm 3 are joined end to end. Find the surface area of the
resulting cuboid.

Given that, $\\$ Volume of cubes = 64 cm 3 $\\$ $(Edge)^3$ = 64 $\\$ Edge = 4 cm $\\$ If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.$\\$ $\therefore$ Surface area of the cuboids = 2(lb + bh+ lh) $\\$ $\qquad \qquad \qquad \qquad =2(4 \times 4 + 4 \times 8 + 4 \times 8)$ $\\$ $\qquad \qquad \qquad \qquad = 2(16 +32 +32)$ $\\$ $\qquad \qquad \qquad \qquad = 2(16 +64)$ $\\$ $\qquad \qquad \qquad \qquad = 2 \times 80 = 160cm^2$

**2.** 2 cubes each of volume 64 cm 3 are joined end to end. Find the surface area of the resulting cuboids.

2 None

Solutions**3.** 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.

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).$\\$ Height of hemispherical part = Radius = 7 cm $\\$ Height of cylindrical part(h)=13-7 =6cm $\\$ inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part $\\$ $2 \pi r h + 2 \pi r^2$ $\\$ Inner surface area of vessel $= 2 \times \frac{22}{7} \times 7 \times 6 + 2 \times \frac{22}{7} \times 7 \times 7$ $\\$ $\qquad \qquad \qquad \qquad \qquad =44(6+7)=444 \times13$ $\\$ $\qquad \qquad \qquad \qquad \qquad =572cm^2$

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

It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).$\\$ Height of hemispherical part = Radius (r) $= 3.5 = \frac{7}{2} cm$ $\\$ Height of conical part (h) = 15.5 - 3.5 =12cm$\\$ Slant height of (l)conical part $= \sqrt{r^2 +h^2}$ $\\$ $\qquad \qquad \qquad \qquad \qquad= \sqrt{\big(\frac{7}{2}\big)^2 +(12)^2}$ $=\sqrt{\frac{49}{4} +144} =\sqrt{\frac{49 + 576}{4}}$ $\\$ $\qquad \qquad \qquad \qquad \qquad=\sqrt{\frac{625}{4}}=\frac{25}{2}$ $\\$ Total surface area of toy = CSA of conical part + CSA of hemispherical part $\\$ =$\pi r l +2 \pi r^2$ $\\$ $= \frac{22}{7} \times \frac{7}{2} \times \frac{25}{2} +2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$ $\\$ $=1375.5 +77 =214.5 cm^2$ $\\$

**5.** 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.

From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.$\\$ Radius (r) of hemispherical part $=\frac{7}{2} = 3.5cm$ $\\$ Total surface area of solid = Surface area of cubical part + CSA of hemispherical part -Area of base of hemispherical part $\\$ $=6(Edge)^2 + 2 \pi r^2 - \pi r^2 = 6(Edge)^2 + \pi r^2$ $\\$ $Total \ surface \ area \ of \ the \ solid = 6(7)^2 +\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$ $\\$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad = 294 +38.5=332.5cm^2$

**6.** A hemispherical depression is cut out from one face of a cubical 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.

Diameter of hemisphere = Edge of cube = $l$ $\\$ Radius of hemisphere = $l/2$ $\\$ Total surface area of solid = Surface area of cubical part + CSA of hemispherical part -Area of base of hemispherical part$\\$ $=6(Edge)^2 + 2 \pi r ^2 -\pi r^2 = 6(Edge)^2 + \pi r^2$ $\\$ $Total \ surface \ area \ of \ solid = 6l^2 + \pi \times \big(\frac{l}{2}\big)^2$ $\\$ $ \qquad \qquad \qquad \qquad \qquad \qquad =6l^2 + \frac{\pi l^2}{4}$ $\\$ $\qquad \qquad \qquad \qquad \qquad \qquad = \frac{1}{4}(24 + \pi)l^2 \ unit^2$

**7.** A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each
of its ends (see Fig. 13.10). The length of
the entire capsule is 14 mm and the diameter
of the capsule is 5 mm. Find its surface area.

It can be observed that $\\$ Radius (r) of cylindrical part = Radius (r) of hemispherical part $\\$ $\qquad \qquad \qquad \qquad \qquad = \frac{Diameter of the Capsule}{2} = \frac{5}{2}$ $\\$ Length of cylindrical part (h) = Length of the entire capsule - 2 × r $\\$ $=14 -5 =9cm$ $\\$ Surafce area of Capsule = 2 $\times$ CSA of hemispherical part + CSA of cylindrical part $\\$ $=2 \times 2 \pi r^2 + 2 \pi r h$ $=4 \pi \big(\frac{5}{2}\big)^2 + 2 \pi \big(\frac{5}{2}\big)(9)$ $=25 \pi + 45 \pi$ $\\$ $= 70 \pi mm^2$ $\\$ $=70 \times \frac{22}{7}$ $\\$ $=220 mm^2$

**8.** A tent is in the shape of a cylinder surmounted by a conical top. If the height and
diameter of the cylindrical part 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 Rs 500 per $m^2$ . (Note that the base of the tent will not
be covered with canvas.)

Given that,$\\$ Height (h) of the cylindrical part = 2.1 m $\\$ Diameter of the cylindrical part = 4 m $\\$ Radius of the cylindrical part = 2 m $\\$ Slant height (l) of conical part = 2.8 m $\\$ Area of canvas used = CSA of conical part +CSA of cylindrical part $\\$ $=\pi r l+ 2\pi r h$ $\\$ $=\pi \times 2 \times 2.8 + 2 \pi \times 2 \times 2.1$ $\\$ $=2 \pi [2.8 + 2 \times 2.1] = 2 \pi [2.8 +4.2] = 2 \times \frac{22}{7} \times 7$ $\\$ $=44 \ m^2$ $\\$ Cost of 1 m 2 canvas = Rs 500 $\\$ Cost of 44 m 2 canvas = 44 × 500 = 22000 $\\$ Therefore, it will cost Rs 22000 for making such a tent.$\\$

**9.** 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 $cm^2$ .

Given that,$\\$ Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm $\\$ Diameter of the cylindrical part = 1.4 cm $\\$ Therefore, radius (r) of the cylindrical part = 0.7 cm $\\$ Slant height of (l) of conical part = $\sqrt{ r^2 +h^2}$ $\\$ $\qquad \qquad \qquad = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76}$ $\\$ $\qquad \qquad \qquad = \sqrt{6.25} = 2.5$ $\\$ Total surface area of the remaining solid will be $\\$ = CSA of cylindrical part + CSA of conical part + Area of cylindrical base $\\$ $=2 \pi r h + \pi r l + \pi r^2$ $\\$ $ = 2 \times \frac{22}{7} \times 0.7 \times 2.4 \times + \frac{22}{7} +\times 0.7 \times 2.5 +\frac{22}{7} \times 0.7 \times 0.7$ $\\$ $=4.4 \times 2.4 +2.2 \times 2.5 +2.2 \times 0.7$ $\\$ $=10.56 + 5.50 + 1.54 =17.60 \ mm^2$ $\\$ The total surface area of the remaining solid to the nearest cm 2 is 18 $cm^2$

**10.** A wooden article was made by scooping
out a hemisphere from each end of a solid
cylinder, as shown in Fig. 13.11. 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.

Given that,$\\$ Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm$\\$ Height of cylindrical part (h) = 10 cm $\\$ Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part $\\$ $=2 \pi r h + 2 \times 2 \pi r^2$ $\\$ $=2 \pi \times 3.5 \times 10 + 2 \times 2 \pi \times 3.5 \times 3.5$ $\\$ $=70 \pi + 49 \pi$ $\\$ $=119 \pi$ $\\$ $=17 \times 22 = 374cm^2$

**11.** A solid is in the shape of a cone standing on a hemisphere with both their radii being
equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid
in terms of $\pi$ .

Given that, $\\$ Height (h) of conical part = Radius(r) of conical part = 1 cm $\\$ Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm$\\$ Volume of solid = Volume of conical part + Volume of hemispherical part $\\$ $=\frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3$ $\\$ $\frac{1}{3} \pi (1)^2 (1) + \frac{2}{3} \pi (1)^3$ $\\$ $\frac{\pi}{3} +\frac{2 \pi}{3} = \frac{3 \pi}{3} = \pi \ cm^3$

**12.** Rachel, an engineering student, was asked to make a model shaped like a cylinder with
two cones attached at its two ends by using a thin aluminium sheet. The diameter of the
model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume
of air contained in the model that Rachel made. (Assume the outer and inner dimensions
of the model to be nearly the same.)

From the figure, it can be observed that $\\$ Height ($h_1$) of each conical part = 2 cm $\\$ Height ($h_2$) of cylindrical part =12 -2 \times Height of the conical part $\\$ $=12-2 \times 2 =8 \ cm$ $\\$ Radius (r) of cylindrical part = Radius of conical part = $\frac{3}{2}$ $\\$ Volume of air present in the model = Volume of cylinder + 2 × Volume of cones $\\$ $=\pi r^2 h_2 + 2 \times \frac{1}{3} \pi r^2 h_1$ $\\$ $=\pi \big(\frac{3}{2}\big)^2 (8) + 2 \times \frac{1}{3} \pi (\frac{3}{2})^2 (2) = \pi \times \frac{9}{4} \times 8 + \frac{2}{3} \times \pi \times \frac{9}{4} \times 2$ $\\$ $=18 \pi + 3 \pi = 21 \pi = 66 \ cm^2$

**13.** A gulab jamun, contains sugar syrup up to about
30% of its volume. Find approximately how much
syrup would be found in 45 gulab jamuns, each
shaped like a cylinder with two hemispherical ends
with length 5 cm and diameter 2.8 cm (see Fig. 13.15).

It can be observed that $\\$ Radius (r) of cylindrical part = Radius (r) of hemispherical part =$\frac{2.8}{2}=1.4 \ cm$ $\\$ Length of each hemispherical part = Radius of hemispherical part = 1.4 cm $\\$ Length (h) of cylindrical part = 5 - 2 $\times$ Length of hemispherical part $= 1.4 \ cm = 2.2 \ cm$ $\\$ Volume of one gulab jamun = Vol. of cylindrical part + 2 $\times$ Vol. of hemispherical part $\\$ $= \pi r^2 h + 2 \times \frac{2}{3} \pi r^3 = \pi r^2 h+ \frac{4}{3} \pi r^3$ $\\$ $=\pi \times (1.4)^2 \times 2.2 + \frac{4}{3} \pi (1.4)^3$ $\\$ $=\frac{22}{7} \times 1.4 \times 1.4 \times 2.2 + \frac{4}{3} \times \frac{22}{7} \times 1.4 \times 1..4 \times 1.4$ $\\$ $=13.552 + 11.498 =25.05 \ cm^3$ $\\$ Volume of 45 gulab jamuns $= 45 \times 25.05 = 1,127.25 \ cm^3$ $\\$ Volume of sugar syrup = 30% of volume $\\$ $=\frac{30}{100} \times 1,127.25$ $\\$ $=338.17 \ cm^3$ $\\$ $\approx 338 \ cm^3$

**14.** A pen stand made of wood is in the shape of a
cuboid with four conical depressions to hold pens.
The dimensions of the cuboid are 15 cm by 10 cm by
3.5 cm. The radius of each of the depressions is 0.5
cm and the depth is 1.4 cm. Find the volume of
wood in the entire stand (see Fig. 13.16).

14 None

SolutionsDepth (h) of each conical depression = 1.4 cm $\\$ Radius (r) of each conical depression = 0.5 cm$\\$ Volume of wood =Volume of cuboid -4 $\times$ Volume of cones $\\$ $l b h - 4 \times \frac{1}{3} \pi r^2 h$ $\\$ $15 \times 10 \times 3.5 - 4 \times \frac{1}{3] \times \frac{22}{7} \times \big(\frac{1}{2}\big)^2 \times 1.4$ $\\$ $=525 -1.47$ $\\$ $=523.53 \ cm^3$