Question 1 1 / -0
A sphere of copper, whose diameter is 18 cm, is melted and converted into a wire of 4 mm diameter. Length of the wire (in meters) is
Question 2 1 / -0
A right circular cone is cut into two pieces by a plane parallel to its base. The base radii of the frustum thus formed are 'a' and 'b' (a > b). If 'h' is the height of the frustum, then find the height of the smaller cone.
Solution
Since ΔOPB and ΔOQD are similar [
QOD =
POB (same angle) and
OQD =
OPB = 90°], so the ratio of their corresponding sides is equal.
Thus,
…………………. (i)
Also, h
1 = h + h
2 ……………….(ii)
By putting (ii) in (i), we get
h
2 a = (h + h
2 )b
h
2 (a – b) = hb
Or h
2 =
is the height of the smaller cone.
Question 3 1 / -0
If both the radii of a frustum are doubled, keeping the height unchanged, then how many times will the new volume of the frustum be as compared to the original volume?
Solution
Let h be the height and r
1 and r
2 be the radii of the ends (r
1 > r
2 ) of the frustum of the cone.
Then, volume of the original frustum =
h (
+
+ r
1 r
2 )
Now, volume of the new frustum =
h [(2r
1 )
2 + (2r
2 )
2 + (2r
1 )(2r
2 )]
= 4 × [
h (
+
+ r
1 r
2 )]
= 4 × (original volume of the frustum)
So, after doubling the radii and keeping the height unchanged, the volume of the frustum becomes 4 times the original volume.
Question 4 1 / -0
A cone with radius 7 m has volume 1232 m3 . What is the cost of painting the curved surface of the cone at the rate of Rs. 20 per m2 ?
Solution
Volume = 1232 m
3 h =
= 24 m
=
= 25 m
CSA of the cone =
=
= 550 m
2 Cost of painting = 550
20 = Rs. 11,000
Question 5 1 / -0
A hollow metallic cylindrical tank has an internal radius of 80 cm and height 200 cm. The thickness of the metal is 1 cm. If the tank is melted and cast into a right circular cone of height 3 m, find the radius of the cone, correct to two decimal places.
Solution
r = Internal radius of the cylindrical tank = 80 cm
R = External radius of the cylindrical tank = (80 + 1) cm = 81 cm
Volume of the hollow cylindrical tank =
=
=
=
=
= 4400
23
= 1,01,200 cm
3 Let the height of the cone be h
1 = 3 m = 300 cm.
Now, volume of the cylindrical tank = volume of the cone
101200 =
h
1 101200 =
r
2 =
r
2 = 322
r =
r = 17.94 cm
Question 6 1 / -0
A hollow cylinder of internal and external radii 8 cm and 10 cm, respectively and height 5 cm is melted is recasted into small cones of base radius 3 cm and height 5 cm. Find the number of cones.
Solution
Volume of hollow cylinder =
= π(10
2 - 8
2 )5
=
(36)
5
= 180
cm
2 Volume of each cone formed =
r
2 h
=
= 15
cm
2 Number of cones =
= 12
Question 7 1 / -0
A solid cylinder is mounted with a hemispherical bowl of radius 5 cm (as shown in the figure). If the height of the whole solid portion is 28 cm. Find the volume of the whole solid. (Answer up to two decimal places.)
Solution
Height of the whole solid = 28 cm
Height of the cylinder = 28 - 5 = 23 cm (Given radius of hemisphere = 5 cm)
Required volume = Volume of the hemisphere + volume of the cylinder
=
=
=
=
=
=
=
= 2069.05 cm
3
Question 8 1 / -0
A hospital room is supposed to accommodate 56 patients. It should be done in such a way that every patient gets 2.2 m2 of floor and 8.8 m3 of space. If the length of the room is 14 m, the breadth and the height of the room respectively are
Solution
Total patients = 56
Let breadth = b m and height = h m
Then, area = (14 m
b m)
Also, area per person = 2.2 m
2 Hence, (14)(b) m
2 = (56)(2.2)
b = 8.8 m
Volume = 8.8 m
3 56
Volume = (14)(8.8)(h) = 8.8
56
h = 4 m
Question 9 1 / -0
The total surface area of a cylinder of height 8 cm is 528 cm2 . It is melted and recast into spheres of radius 2 cm. How many spheres can be obtained?
Solution
Height of the cylinder = 8 cm
Surface area of the cylinder = 528 cm
2 cm
2 r(r + 8) =
r
2 + 8r = 84
r
2 + 8r - 84 = 0
r =
=
=
= 6, - 14
Thus, the radius of the cylinder is 6 cm.
Volume of cylinder = n
volume of spheres
n =
=
Question 10 1 / -0
The volume of one sphere is 8 times that of another sphere. What will be the ratios of their radii and surface areas, respectively?
Solution
Volume of one sphere =
; where r = radius of the sphere
Volume of another sphere =
; where R = radius of the sphere
Given:
= 8
r
3 = 8R
3 Thus, r : R = 2 : 1
Ratio of surface areas of the two spheres =
=
=
Question 11 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 12 1 / -0
From a circle of radius 15 cm, a sector of angle 288° is cut out and its bounding radii are bent so as to form a cone. What is its approximate volume? (Take
= 22/7)
Solution
Given above is the figure based on the data given in the question.
Area of the circle =
r
2 =
(15)
2 cm
2 Since 360°
(15)
2 cm
2 So, 288°
cm
2 The radius of circle becomes the slant height of cone,
.
Let R be the radius of the cone.
R
=
(15)
2 x R x 15 =
x 15 x 15 x
Radius of the cone, R = 12 cm
Slant height of the cone,
= 15 cm
= R
2 + h
2 (15)
2 = (12)
2 + h
2 225 – 144 = h
2 h
2 = 81
h = 9 cm
Volume of the cone =
r
2 h =
x 12 x 12 x 9
= 432
= 432 × (22/7)
= 1357.71 cm
3 = 1358 cm
3 (approximately)
Hence, 1358 cm
3 is the correct answer.
Question 13 1 / -0
A sphere completely fits into a cylinder as shown in the figure. What is the ratio of the curved surface area of cylinder to the surface area of sphere?
Solution
Radius of cylinder = Radius of sphere
Height of cylinder = Diameter of sphere
Ratio of surface areas =
Question 14 1 / -0
A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.
Question 15 1 / -0
A solid right circular cone of base diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, what is the internal diameter of the sphere?
Solution
The radius of the base of cone is 7 cm and the height is 8 cm.
Volume of the cone = (1/3)
r
2 h = (1/3)
(7)
2 x 8 cm
3 External radius of the hollow sphere = 5 cm
Let the internal radius be r cm.
Volume of the hollow sphere = (4/3)
(R
3 - r
3 ) cm
3 = (4/3)
(5
3 - r
3 ) cm
3 Since volume of material used for the cone will be same as the volume of material used to make the hollow sphere,
(1/3)
(7)
2 8 = (4/3)
(125 - r
3 )
98 = 125 - r
3 r
3 = 125 - 98 = 27
r = 3 cm
So, internal diameter of the sphere = 2r = 2 x 3 = 6 cm
Question 16 1 / -0
An ice cream cone is packed full of ice cream and has a shape as depicted in the diagram. If the volume of ice cream inside the cone is the same as the volume of ice cream outside the cone, the height of the cone is
Solution
Radius of cone = Radius of hemisphere = 4 cm
Let height of cone = h cm
Volume of hemisphere = Volume of cone
=
=
= h = 8 cm
Question 17 1 / -0
Water is pumped out through a cylindrical pipe of internal diameter 28 cm. How much water is pumped out in one hour if the flow velocity of water is 4 cm/s?
Solution
Volume of water pumped out in a second =
(r = internal radius of the pipe, h = distance covered by water in one second)
=
cm
3 = 2464 cm
3 Volume of water pumped out in an hour = 2464 × 3600 cm
3 = 88,70,400 cm
3
Question 18 1 / -0
A cylindrical vessel of diameter 9 cm has some water in it. A cylindrical iron piece of diameter 7 cm and height 5 cm is dropped into it. What is the rise in water level after it is completely immersed?
Solution
Volume of cylindrical iron =
r
2 h
Diameter = 7 cm
Radius = 7/2 cm
Height = 5 cm
Volume of cylindrical iron =
(7/2)
2 x h
Let the rise in the water level be H.
Rise in volume of cylindrical vessel =
R
2 H
Diameter = 9 cm
Radius = (9/2) cm
Rise in volume = Volume of cylindrical iron (because the volume would be the same when it is immersed; the only difference would be the rise in water level)
R
2 H =
r
2 h
H = (49 x 5)/81 = 3.02 cm
Question 19 1 / -0
A solid vessel of cylindrical shape, having volume of 33.264 litres, is cast into a conical shape of height 72 cm. What is the cost of painting its CSA at Rs. 12 per sq. m?
Solution
Volume of the cone = 33.264 litres = 33264 cm
3 r
2 =
r
2 = 441
r = 21 cm
=
=
=
= 75 cm
CSA =
r
=
x 21 x 75
= 4950 cm
2 = 0.4950 m
2 Cost of painting per m
2 = Rs. 12
Cost of painting per 0.4950 m
2 = 12 x 0.4950 = Rs. 5.94
Question 20 1 / -0
In the following figure, two cones of the same radius are joined together as shown. Which of the following statements is definitely true regarding the figure?
Solution
Total surface area of the combination is equal to the sum of the curved surface areas of the two cones. So, the statement given in option 2 is definitely true.
Question 21 1 / -0
The ratio of the radius of the base to the height of a cone is 5 : 12. If the cost of smoothening the curved surface area (CSA) is Rs. 115.50 at a rate of 5 paise per sq. cm, then what is the total surface area of the cone?
Question 22 1 / -0
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel, partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution
Radius of the sphere =
= 3 cm
Radius of the cylinder =
= 6 cm
Volume of the sphere =
πr
3 =
× π × 3 × 3 × 3 = 36π cm
3 New volume of the cylindrical vessel - Initial volume of the cylindrical vessel = Volume of the sphere
Initial height = h cm
Final height = H cm
πr
2 H - πr
2 h = 36π
πr
2 (H - h) = 36π
6
2 (H - h) = 36
H - h =
= 1 cm
Question 23 1 / -0
A cylinder, a hemisphere and a cone stand on the same base and have equal heights. The ratio of the areas of their curved surfaces is
Solution
Let r and h be the radius and height of the cylinder, hemisphere and cone.
The height of the hemisphere is the same as its radius.
So, r = h
The cylinder, hemisphere and cone are made on same base, so the radius is same in all. So, r = h for the cylinder as well as the cone.
Ratio of the curved surface areas of the cylinder, hemisphere and the cone standing on the same base and having equal heights =
Slant height of the cone,
Solving, we get the required ratio =
Question 24 1 / -0
A sphere with radius r is cut into two halves by a plane passing through the centre of the sphere. Two hemispheres of same radii are formed. Which of the following relations is true in this case?
Solution
Surface area of sphere = 4
r
2 And curved surface area of hemisphere = 2
r
2 A sphere is cut into two hemispheres.
So, surface area of sphere = C.S.A of the I
st hemisphere + C.S.A of the II
nd hemisphere
4
r
2 = 2
r
2 + 2
r
2 4
r
2 = 4
r
2 L.H.S = R.H.S
Question 25 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 cylinder = 15 m
Radius of cylinder = 3 m
Curved surface area = 2
rh = 90
Radius of cone = 3 m
Height of cone = 4 m (19 – 15 = 4 m)
Slant height,
=
= 5
Curved surface area of cone =
r
= 15
Curved surface area of cone + cylinder
90
+ 15
= 105
= 105 x
= 330 m
2
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