Question 1 1 / -0
Let 'a' and 'b' respectively be the radius of the base and the height of a right circular cylinder. Which of the following expressions represents the ratio of its curved surface area to its total surface area?
Solution
Radius of the base of the cylinder = a
Height of the cylinder = b
Question 2 1 / -0
A cylinder of height 7 cm and diameter 44 cm is melted to form cubes, each of side 2 cm. Calculate the number of cubes formed.
Solution
Number of cubes formed =
=
= 11
11
11 = 1331
Question 3 1 / -0
If the total surface area of a right circular cone with base radius 6 cm is 96
cm
2 , then its altitude is
Solution
Total surface area of the cone =
r(r +
)
So, 96
=
6(6 +
)
16 = 6 +
= 16 - 6
= 10 cm
Now,
h =
=
=
=
=
= 8 cm
Question 4 1 / -0
A sphere is melted and recast into smaller right cones of height 7 centimeters and radius 7 centimeters. If the radius of the sphere is 7 meters, then how many cones can be made?
Solution
Radius of the sphere, r = 7 meters
Radius of the cone, r
2 = 7 centimeters = 0.07 meters
Height of the cone, h = 7 centimeters = 0.07 meters
Volume of the sphere =
=
...(1)
Volume of the cone =
=
...(2)
Number of cones =
Solving,
Number of cones = 4 × 10
6
Question 5 1 / -0
Two right circular cones have equal slant height. If the ratio of their base radii is 5 : 3, then the ratio of their curved surface areas is
Solution
Given:
(
are the slant heights of the first and second cones, respectively.)
And
(r
1 and r
2 are the base radii of the first and second cones, respectively.)
Also, curved surface area of a cone =
Ratio of curved surface areas of the two cones = Curved surface area of the first cone/Curved surface area of the second cone
=
=
=
[
]
Ratio = 5 : 3
Question 6 1 / -0
A hemisphere, a cylinder and a cone are joined as shown in the figure.
If C
1 , C
2 and C
3 respectively are the curved surface areas of the hemisphere, the cylinder and the cone, then which of the following expressions represents the total surface area of the figure?
Solution
If C
1 , C
2 and C
3 respectively are the curved surface areas of the hemisphere, the cylinder and the cone, then the total surface area of the given solid is C
1 + C
2 + C
3 .
Question 7 1 / -0
A circus tent is in the form of a cone over a cylinder. The diameter of the base is 6 m, the height of the cylindrical part is 15 m and the total height of the tent is 19 m. The canvas required for the tent is
Solution
Height of the cylinder = 15 m
Height of the cone, 'h' = 19 - 15 = 4 m
Radius =
= 3 m
Slant height of the cone =
= 5 m
Canvas required = C.S.A. of (cone + cylinder)
=
r
+ 2
rH
=
r(5 + 2 × 15)
=
= 330 m
2
Question 8 1 / -0
A metallic right circular cone of height 9 cm and base radius 7 cm is melted to form a cuboid with two sides measuring 11 cm and 6 cm. The third side of the cuboid measures
Solution
Let the length of the third side of the cuboid or h
2 be x cm.
h
1 = 9 cm,
Now, according to the question,
Volume of cone = Volume of new cuboid
=
⇒ 66x = 462
⇒ x = 7 cm
Question 9 1 / -0
If a square is rotated along one of its diagonals, then find the curved surface area of the shape formed by it, given that one of its side is 9 metres.
Solution
According to the question, rotating square along a diagonal makes two right-circular cones.
As the diagonal makes a right-angled triangle by the sides of square, so
D =
m (Using Pythagoras theorem)
r =
m
So, curved surface area of the cone =2
r l
= 2
×
× 9 =
m
2
Question 10 1 / -0
The height of a frustum made from a cone is 8 cm. If the base radii of this frustum are 6 cm and 2 cm, then find the height of the original cone.
Solution
Height of the frustum (h) = 8 cm
Radii of the base of the frustum are:
r
1 = 6 cm and r
2 = 2 cm
Height of the original cone = h
1 = 8 + h
2 … (i)
∵
OPB and
OQD are similar
[
QOD =
POB (same angle) and
OQD =
OPB = 90°]
So, the ratio of their corresponding sides is equal.
Thus,
=
=
or h
1 = 3h
2 Putting the value h
1 = 3h
2 into equation (i), we get
3h
2 = 8 + h
2 or 2h
2 = 8
h
2 = 4 cm
Again putting h
2 = 4 into (i), we get
h
1 = 8 + 4 = 12 cm
So, height of the original cone = 12 cm
Question 11 1 / -0
A spherical ball of radius 7 cm is melted and recast into 8 small balls of the same radius. The radius of each smaller ball is
Solution
Radius of the bigger ball r
1 = 7 cm
Let the radius of the smaller ball be r
2 cm.
Now,
Volume of bigger ball =
Volume of smaller ball =
Since the sum of the volumes of the small balls is equal to the volume of the bigger ball.
Therefore,
Volume of bigger ball = 8 × Volume of smaller ball
Hence, the radius of each smaller ball is 3.5 cm.
Question 12 1 / -0
The height of a cone is 40 cm. A small cone is cut off from the top by a plane parallel to its base. If the volume of the frustum left out is
of the volume of the original cone, find the height of the cone cut off from it.
Solution
So,
Or
, where 'r' and 'h' are the radius and the height of the smaller cone, and R and H are the radius and the height of the larger cone, respectively.
As we know,
So,
Or h = 10 cm
Question 13 1 / -0
The sector of a circle with radius 10 cm and angle 120° is rolled up such that the two bounding radii are joined together to form a cone. Find the volume of the cone.
Solution
Length of arc AXB = Circumference of the base of the cone
Circumference =
× 2 × π × R
=
× 2 ×
× 10
= 20.95 cm
Let the radius of the base be r.
So, 2πr = 20.95
2 ×
× r = 20.95
× r = 20.95
44r = 146.65
r =
r = 3.33 cm
Now, to find out height we can use Pythagorean theorem.
Let the height of the cone be h.
So, h
2 = R
2 - r
2 h
2 = 100 - 11.08
h =
h = 9.42 cm
Now, volume of the cone =
×
× 11.08 × 9.42
=
= 109.3 cm
3
Question 14 1 / -0
The length, breadth and height of a cuboid are 10 m, 6 m and 5 m, respectively. It is melted and converted into 10 equal cubes. In this process, 10% of the material is lost. What is the length of the edge of each cube?
Solution
Volume of the original cuboid = 10 × 6 × 5 = 300 m3 3 3 Now, according to the question,3
Question 15 1 / -0
Let
be the slant height of a frustum. If a and b represent the two base radii, which of the following expressions represents the ratio of the total surface area to the curved surface area of the frustum?
Solution
(i) Curved surface area of the frustum of the cone =
(r
1 + r
2 )
, where
=
(ii) Total surface area of the frustum of the cone =
(r
1 + r
2 ) +
+
, where
=
Let h be the height.
Given:
= slant height
'a' and 'b' are the two base radii of the frustum of the cone.
Then,
=
= 1 +
Question 16 1 / -0
A toy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The base radius of the cone is 3.5 m and its volume is (2/3)rd of the volume of the hemisphere. Calculate its height.
Solution
Radius of the hemisphere = Base radius of the cone = 3.5 m
Volume of the hemisphere =
=
m
3 Volume of the cone =
m
3 Let the height of the cone be h metres.
Volume =
R
2 h =
m
3 h = 4.67 m
Question 17 1 / -0
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m, while the vertex of the cone is 16 m above the ground. What is the area of the curved surface for the conical portion?
Solution
Radius of cone = 12 m
Height of cone = 5 m
Hence, slant height =
m
= 13 m
Curved surface area of cone =
=
m
2 =
Question 18 1 / -0
Ram cuts a semicircle of radius 8 cm from a piece of paper and folds it in such a way that it forms a right circular cone. What is the volume of this cone?
Solution
Let 'R' be the radius of the semicircular piece of paper.
Let the radius of the base of the cone be r.
Let the height of the cone be h.
Circumference of base of cone = 2
r =
R
So, r = R/2
Now, the slant height of the cone so formed = R
Applying Pythagoras theorem, we get
h =
Volume of the cone, V =
Putting the values of r and h, we get
Putting the value of R, we get V =
cm
3
Question 19 1 / -0
An oil tanker is in the form of a right circular cylinder of outer radius 10 cm and depth 25 cm. The thickness of the tanker's outer layer is 2 cm. Find the cost of the oil filling in the tanker, if it costs 5 paisa per cubic cm.
Solution
Radius of cylinder = (10 - 2) cm = 8 cm
Volume of cylinder =
r
2 H =
× 64 × 25 cm
3 = 1600
cm
3 Cost = Rs.
× 1600
= Rs. 251.2
Question 20 1 / -0
If one cubic cm of cast iron weighs 21 g, then find the weight of a pipe made of cast iron having length 1 m, internal diameter 3 cm and thickness 1 cm.
Solution
Volume of the metallic pipe =
= 400π cm
3 Weight of the metallic pipe = 21 × 400 × 22/7 = 26.4 kg
Question 21 1 / -0
A solid metal sphere is melted and smaller spheres, all with the same radius, are formed. 20% of the material is lost in this process. The radius of each smaller sphere is
the radius of the original sphere. If 20 litres of paint was needed to paint the original sphere, then how many litres of paint would be required to paint all the smaller spheres?
Solution
Let the radius of the original sphere be R and that of each smaller one be r.
⇒ r =
Then, volumes of the larger and smaller spheres are
πR
3 and
π(
)
3 , respectively.
Volume lost due to melting =
πR
3 ×
=
Volume left =
πR
3 ×
=
Number of smaller spheres formed =
=
= 8
3 × 0.8
Surface area of original sphere = 4πR
2 Surface area of each smaller sphere = 4πr
2 = 4π(
)
2 =
Surface area of all smaller spheres = Number of small spheres × Surface area of each smaller sphere = 8
3 × 0.8 ×
= 6.4 × (4πR
2 )
Therefore, ratio of the total surface area of all smaller spheres to the surface area of original sphere =
=
Volume of paint required = 6.4 × Volume of paint used on original sphere = 6.4 × 20 = 128 litres
Question 22 1 / -0
Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes, if 8 cm of standing water is required?
Solution
Width of canal = 30 dm = 3 m
Depth of canal = 12 dm = 1.2 m
Velocity of water = 100 km/hr = 1,00,000 m/hr
Length of water column in 1 hour = 1,00,000 m
So, length of water column in 30 minutes =
= 50,000 m
Volume of water in 30 minutes = 3 × 1.2 × 50,000 = 1,80,000 m
3 Required height of standing water = 8 cm = 0.08 m
So, area irrigated in 30 minutes =
=
= 22,50,000 m
2
Question 23 1 / -0
A sphere of radius 3 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cross-section of the vessel is 12 cm. After the sphere is completely submerged in water, the rise in the water level in the vessel is
Solution
Let r be the radius of cross-section of the vessel, r
1 be the radius of the sphere; h and H are initial heights and increased height, respectively, as shown in the figure above.
Now, volume of water rise after sphere is submerged in it = volume of sphere
36 (H - h) cm
3 =
cm
3 H - h = 1 cm
So, increase in water level = 1 cm
Question 24 1 / -0
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 5 m and 8 m, respectively, and the slant height of the top is 1.4 m, find the cost of the canvas of the tent at the rate of Rs. 200 per m2 . It is to be noted that the tent has no base.
Solution
Radius of cylinder = 4 m, height = 5 m and slant height of conical top = 1.4 m
Curved Surface Area of cylinder
=
=
=
Curved Surface Area of cone
=
=
=
Total CSA
Cost of canvas = Rate × Surface Area = Rs. 200 × 143.3 = Rs. 28,660
Thus, answer option 3 is correct.
Question 25 1 / -0
The height of a right circular cone and the radius of its circular base are 9 cm and 3 cm, respectively. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e. the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking
=
) is
Solution
Let the radius of the upper circular surface of the frustum be 'r' cm and the height of the smaller cone be 'h' cm.
Now,
h = 3r
Then, by putting the value of h, we get the ratio of volumes as given below:
=
...(i)
Now, Volume of smaller cone = Volume of larger cone - Volume of frustum
= [
(3)
2 × 9 - 44] cm
3 =
cm
3 ...(ii)
Volume of larger cone =
(3)
2 × 9 cm
3 =
cm
3 ...(iii)
Using (i), (ii) and (iii),
=
r
3 = 13
r =
cm
×
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