**Maths Class 10
Exercise 13.1**

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

**Solution:** According to question, volume of the
cube = 64 cm^{3}

Volume of a cube = *a*^{3}

⇒ 64 = *a*^{3}

⇒ (*a*)^{3}
= 4^{3}

⇒ Side = 4 cm

For the resulting
cuboid, length (*l*) = 4 + 4 = 8 cm, breadth (*b*) = 4 cm and
height (*h*) =
4 cm.

Surface
area of the resulting cuboid = 2(*lb*
+ *bh* + *hl*)

= 2(8 × 4 + 4 × 4 + 4 × 8)

= 2(32 + 16 + 32)

= 2 × 80

= 160 cm^{2}

^{ }

**2. 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:** Given, the diameter of the hollow hemisphere = 14
cm

Therefore, the radius of the hollow hemisphere = 14/2 = 7 cm

Total
height of the vessel = 13 cm

Therefore,
the height of
the hollow cylinder (*h*) = 13 – 7 = 6
cm

So, the inner surface area of the
vessel = Inner surface area of the hollow hemisphere + Inner curved surface
area of the hollow cylinder

=
2Ï€(7)^{2} + 2Ï€(7)(6)

= 98Ï€ + 84Ï€

= 182Ï€

=
182 × 22/7

=
26 × 22

=
572 cm^{2}

**3. 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.**

**Solution: **Here, the radius of the cone = 3.5 cm

Therefore, the
radius of the hemisphere = 3.5 cm

Total
height of the toy = 15.5 cm

So, the height
of the cone = 15.5 – 3.5 = 12 cm

Slant
height of the cone (*l*) =

=
12.5 cm

Now,
the total
surface area of the toy = Curved surface area of the hemisphere + Curved
surface area of the cone

=
2Ï€*r*^{2}
+Ï€*rl*

= 2Ï€(3.5)^{2} + Ï€(3.5)(12.5)

=
24.5Ï€ + 43.75Ï€

= 68.25Ï€

= 68.25
× 22/7

= 214.5
cm^{2}

**4. 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**: The greatest diameter of the hemisphere = Edge
of the cubical block = 7 cm

Now,
the total surface
area of the solid = External surface area of the cubical block + Curved surface
area of the hemisphere

=
[6(7)^{2} – Ï€(7/2)^{2}] + 2Ï€(7/2)^{2}

=
294 – 49Ï€/4 + 49Ï€/2

=
294 + 49Ï€/4

=
294 + 49/4 × 22/7

=
294 + 77/2

=
294 + 38.5

=
332.5 cm^{2}

**5. 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.**

**Solution**: Given, the diameter of the hemisphere = *l*

Therefore, the
radius of the hemisphere = *l*/2

Also,
the length of the edge of the cube = *l*

Now,
the total surface
area of the remaining solid = External surface area of the cubical wooden block
+ Inner curved surface area of the hemisphere

=
6*l*^{2} – Ï€(*l*/2)^{2} + 2Ï€(*l*/2)^{2}

=
Ï€(*l*/2)^{2 }+^{ }6*l*^{2 }

=
Ï€*l*^{2}/4 +^{ }6*l*^{2}

=
¼ *l*^{2}(Ï€ + 24)

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

**Solution:** Given, the diameter of the capsule = 5 mm

Then,
radius of the hemisphere and cylinder (r) =
2.5 mm

Cylindrical
height (h) = Total height – 2 × Radius of hemisphere = 14 – 2 × 2.5 = 9 mm

Surface area of
the capsule = Curved surface area of the cylinder + 2 × Surface area of the
hemisphere

=
2Ï€rh + 2(2Ï€r^{2})

=
2Ï€(2.5)(9) + 2[2Ï€(2.5)^{2}]

=
45Ï€ + 25Ï€

=
70Ï€

=
70 × 22/7

= 220 mm^{2}

**7. 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.)**

**Solution**: Given, the diameter of the cylindrical part = 4 m

Therefore, the radius of the cylindrical part (*r*) = 2 m

Height
of the cylindrical part (*h*) = 2.1 m

Slant
height of the conical part (*l*) = 2.8
m

Total surface
area of the tent = Curved surface area of the cylindrical part + Curved surface
area of the conical part

=
2Ï€*rh* + Ï€*rl*

=
2Ï€(2)(2.1) + Ï€(2)(2.8)

= 8.4Ï€ + 5.6 Ï€

= 14Ï€

=
14 × 22/7

=
44 m^{2}

Therefore,
the cost of the canvas
of the tent at the rate of Rs. 500 per m^{2}

= 44 × 500

= Rs. 22000

**8. 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}.**

**Solution**: Given, the diameter of the solid cylinder = 1.4 cm

Therefore, the radius of the solid cylinder (r) = 0.7
cm

Now, the radius
of the base of the conical cavity = 0.7 cm

Height
of the solid cylinder (h) = 2.4 cm

Therefore, the height of the conical cavity = 2.4 cm

Therefore, the slant height of the conical cavity =

Then, the
total surface area of the remaining solid = 2Ï€(0.7)(2.4) + Ï€(0.7)^{2} + Ï€(0.7)(2.5)

= 3.36Ï€ + 0.49Ï€ + 1.75Ï€

= 5.6Ï€

=
5.6 × 22/7

=
17.6 cm^{2}

= 18 cm^{2 }(to
the nearest cm^{2})

**9. A wooden article was made by scooping out a
hemisphere from each end of a solid cylinder as shown in figure. 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**: Given, the height of the
solid cylinder (h) = 10 cm and the base radius of the cylinder (r) = 3.5 cm.

Therefore,
the radius of the hemisphere = 3.5 cm

Now, the total surface area of the article
= 2Ï€rh + 2(2Ï€r^{2})

=
2Ï€(3.5)(10) + 2[2Ï€(3.5)^{2}]

=
70Ï€ + 49Ï€

= 119Ï€

=
119 × 22/7

= 374
cm^{2}