Gujarat Board Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes

 Gujarat Board Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes

Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes

Ex 13.1

Question 1.
2 cubes of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let the side of the cube by a cm.
Volume of a cube = 64 cm3
a3 = 64
⇒ a = 4 cm

Both cubes combined together form a cuboid of length
l = 4 + 4 = 8 cm
b = 4 cm
h = 4 cm
Surface area of cuboid
= 2[lb + bh + lh]
= 2[ 8 x 4 + 4 x 4 + 8 x 4 ]
= 2 [32 + 16 + 32] = 2 [80]
= 160 cm2

Question 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. (CBSE 2013)
Solution:
Diameter of hemisphere = 14 cm

r = 7 cm
Height of cylinder = 13 – 7 = 6 cm
Inner surface area of vessel
= 2πrh + 2πr2
= 2 x 22/7 x (6 + 7)
= 2 x 22/7 x 7(6 + 7)
= 44 x 13 = 572 cm2

Question 3.
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:
Here

Question 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:

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter I of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:

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

Question 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 4m 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:

Question 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 cm2.
Solution:
Height = 2.4 m
and diameter = 1.4 cm
r = 0.7 cm
and radius of conical cavity or base
= 0.7 cm

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the 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:
Here:

h = 10cm
r = 3.5 cm
Total surface area of article
= 2πrh + 2πr2 + 2πr2
= 2πrh + 4πr2 = 2πr (h + 2r)
= 2 x 22/7 x 3.5 (10 + 2 x 3.5)
= 44 x 0.5 (10 + 7)
= 22x 17 = 374 cm2

Ex 13.2

Question 1.
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 it.
Solution:
Let radius be r cm.
Height of cone = Radius
h = r
r = 1 cm
h = 1 cm

Question 2.
Rachel, an engineering student, was asked ta make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum 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 dimension of the model to be nearly the same).
Solution:

Question 3.
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.

 

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depression to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
For cone, h = 8 cm
r = 5 cm

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use ir=3.14)
Solution:
Diameter of cylinder A = 24 cm
So radius r1 = 12 cm
And height h1 = 220cm
Radius of cylinder B = 8 cm
i.e., r2 = 8 cm
and height h2 = 60 cm

Volume of pole
= Volume of cylinder A + Volume of cylinder B
= πr12h1 + πr12h2
= π [r12h1 + r12h2]
= 3.14 [122 x 220 + 82 x 60]
= 3.14 [144 x 220 + 64 x 60]
= 3.14 x 35520
= 11532.8 cm3
Mass of pole = Volume x Density
= 11532.8 x 8
= 11532.8×8/1000
= 892.26 kg

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such ‘ that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of cone h = 120 cm
radius r = 60 cm
Radius of hemisphere = 60 cm

Question 8.
Aspherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Cylindrical neck of vessel has diameter = 2 cm
r = 1 cm
h = 8 cm

Ex 13.3

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radii of solid sphere are
r1 = 6 cm
r2 = 8 cm
and r3 = 10 cm
Let radius of resulting sphere be r.
Volume of three spheres = Volume of resulting sphere

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. [NCERT 20121]
Solution:

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
In case of cylindrical well
Diameter = 3 m

Question 5.
A container shaped like a right circular cylinder having a diameter 12 cm and a height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
In case of right circular cylinder, diameter = 12 cm
r1 = 6 cm
h1 = 15 cm
Volume of ice cream in right circular cylinder
= πr12h1
= π x 62 x 15
In case of cone
Diameter = 6 cm
Radius r2 = 3 cm
Height h2 = 12 cm

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical help of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
In case of cylindrical bucket
Radius (r) of bucket = 18 cm
Height h = 32 cm
Volume of sand filled in cylindrical bucket
= πr2h
= π x 182 x 32
= 10368 π cm3
In case of conical heap
Height (H) = 24 cm
Volume of conical heap = Volume of cylindrical bucket

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
In case of canal
Width of canal (b) = 6 m
Depth of canal (h) = 1.5 m
Speed of flow of water = 10 km/h

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:

Ex 13.4

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Here

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Here,
Slant height l = 4 cm
and let r1 r2 be the radii of frustum and r1 > r2

Perimeter of circular ends = 2πr1
2πr1 = 18 cm
⇒  πr1 = 9 cm ……(1)
and 2πr2 = 6 cm
πr2 = 3 cm …….(2)
Curved surface area of frustum
= π(r1 + r2) l = (πr1 + πr2)l ……(3)
Putting value of eqn. (1) and (2) in eqn. (3),
Curved surface area of frustum = (9 + 3) x 4 = 12 x 4 = 48 cm2

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:
Here
r1 = 10 cm (Radius on the open side)
and r1 = 4 cm (Radius of the upper base)
Surface area of cap = CSA of frustum + Area of upper circular end
= π(r1 + r2) l + πr22

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs  ₹8 per 100 cm2. (Take π = 3.14)
Solution:
Here
h = 16 cm
r1 = 20 cm and r2 = 8 cm

 

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Solution:

 

Ex 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
Solution:
In case of cylinder
Diameter = 10 cm
radius r = 5 cm

Question 2.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of n as found appropriate.)
Solution:
Here
ABC is a right triangle in ∠BAC = 90°

BC2 = AB2 + AC2
= 32 + 42 = 9 + 16
BC2 = 25
BC = 5 cm
Here AD and A’D are radii of two cones which are formed by rotating the right triangle about the hypotenuse BC.
ΔADB – ΔCAB (by AA similarity)

Question 3.
A cistern, internally measuring 150 cm x 120 cm x 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?
Solution:
Volume of cistern = l x b x h
= 150 x 120 x 110
= 1980000 cm3
Volume of water in cistern = 129600 cm3
Volume of cistern to be filled
= Volume of cistern – Volume of water
= 1980000 – 129600
= 1850400 cm3
Let n bricks are needed to fill the cistern up to the brim.
Volume of one brick
= l x b x h
= 22.5 x 7.5 x 6.5
= 1096.875 cm3

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers and each 1072 km long, 75 m wide and 3 m deep.
Solution:

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (See figure).

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols are explained.
Solution:
Let r1 and r2 be the radii of circular base, where r1 > r2 and h be the height of the frustum of a cone.

Let h1 be the height of large cone LPQ
Height of small cone = h1 – h
Let l1 be the slant height of cone LPQ and l be the frustum of cone. The slant height of Cone LRS
= l1 – l
In rt. ΔLMQ and rt. ΔLNS
∠1 = ∠1 (common)
∠M = ∠N (each 900)
∴ ΔLMQ – ΔLNS (by AA similarity)

(i) Lateral surface area or curved surface area of a frustum of cone = CSA of large cone – CSA of small cone
= πr1l1 – πr2(l1 – l)

Question 7.
Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Solution:
Let r1 and r2 be radii of two circular bases where r1 > r2, h be the height of frustum of a cone.

Let h1 be the height of large cone LPD and height of small cone LRS = h1 – h
Let l1, be the slant height of cone and t be the slant height of frustum of a cone.
Slant height of cone LRS = l1 – l
In rt. ΔLMQ and rt. ΔLNS (Common)
∠M = ∠N (Each 900)
ΔLMN – ΔLNS (By AA similarity)

Volume of a frustum of a cone = Volume of large cone LPQ – Volume of small cone LRS