Unless stated otherwise, take π = 22/7
13.1 Class 10 Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of one cube = 64 cm3
Let edge of one cube = a
Volume of the cube = (edge)3
a3= 64 ⇒ a = 4 cm
Similarly, edge of the another cube = 4 cm.
Now, both cubes are joined together and a cuboid is formed as shown in the figure.
Now, length of the cuboid (l) = 8 cm
breadth of the cuboid (b) = 4 cm
height of the cuboid (h) = 4 cm
Surface area of the cuboid so formed = 2 (lb + bh + hl)
= 2(8 x 4 + 4 x 4 + 4 x 8)
= 2(32 + 16 + 32) = 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.
Solution:
Given: diameter of the hemisphere = 14 cm
Radius =14/2= 7 cm
Curved surface area of the hemisphere = 2πr² = 2 x 22/7 x 7 x 7 cm²
= 14 x 22 cm² = 308 cm²
Here, total height of the vessel = 13 cm
Height of the cylinder = Total height – Height of the hemisphere = 13 cm – 7 cm = 6 cm
and radius of the cylinder = radius of the hemisphere = 7 cm
Inner surface area of the cylinder = 2πrh = 2 x 22/7 x 7 x 6
= 2 x 22 x 6 = 264 cm²
Inner surface area of the vessel = Inner surface area of the cylinder + curved surface area of the hemisphere
= 264 cm² + 308 cm² = 572 cm²
Question 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.
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 l 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 (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:
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 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 ? 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:
Question 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:
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 n.
Solution:
Radius of the hemisphere = 1 cm
Volume of the hemisphere = 2/3 πr³ = 2/3 π(1)³ = 23πcm³
Radius of base of the cone = 1 cm
Height of the cone = 1 cm
Volume of the cone = 1/3πr²h = 1/3πr² x 1 = 1/3π cm³
Total volume of the solid = Volume of the hemisphere + Volume of the cone
= 2/3π cm³ + 1/3π cm³ = π cm³
Question 2.
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.)
Solution:
Volume of air contained in the model = Total volume of the solid
Diameter of base of each cone = 3 cm
∴ Radius of base of each cone = 3/2
Height of each cone = 2 cm
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 (see figure).
Solutions:
Volume of one piece of gulab jamun
= Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8
Radius of each hemispherical portion = 2.8/2 = 1.4 cm
Volume of both hemispherical ends = 2 x 5.74 cm³ = 11.48 cm³
Height of the cylindrical portion = (total height) - (radius of both hemispherical ends)
= 5 cm -2(1.4) cm = 5 cm - 2.8 cm = 2.2 cm
Radius of the cylindrical portion = 1.4 cm
Volume of the cylindrical portion of gulab jamun = πr²h
= 22/7 x (1.4)² x 2.2cm³
=22x1.4x1.4x2.2/7 cm³ = 13.55 cm³
Total volume of one gulab jamun = Volume of the two hemispherical ends + Volume of the cylindrical portion
= 11.48 cm³ + 13.55 cm³ = 25.03 cm³
Volume of sugar syrup = 30% of volume of gulab jamun
=30/100 x 25.03 cm³ = 7.50 cm³
∴ Volume of sugar syrup in 45 gulab jamuns
= 45 (volume of sugar syrup in one gulab jamun)
= 45 x 7.50 cm³ = 337.5 cm³ = 338 cm³ approx.
Question 4.
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 figure).
Solution:
Radius of one conical depression = 0.5 cm
Depth of one conical depression = 1.4 cm
Volume of cuboidal box = l x b x h
= 15 x 10 x 3.5 cm³
= 525 cm³
Remaining volume of box = Volume of cubical box - Volume of four conical depressions
= 525 cm³ - 1.464 cm³ = 523.5 cm³
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:
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 of 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 π = 3.14)
Solution:
Given: radius of 1st cylinder = 12 cm
and height of 1st cylinder = 220 cm
∴ Volume of 1st cylinder = πr²h
= π(12)² (220) cm³
= 144 x 220π cm³
= 144 x 220 x 3.14 cm³
= 99475.2 cm³ … (i)
Given: radius of 2nd cylinder = 8 cm
and height of 2nd cylinder = 60 cm
∴ Volume of 2nd cylinder = πr²h
= π(8)² (60) cm³ = 64 x 60π cm³
= 64 x 60 x 3.14 cm³
= 12057.6 cm³ … (ii)
Total volume of solid = Volume of 1st cylinder + Volume of 2nd cylinder
= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³
Given: mass of 1 cm³ of iron = 8 g
∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g
= 892262.4 g = 892.262 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:
Radius of hemisphere = 60 cm
When solid (hemisphere + conical) is kept in cylindrical solid, then volume of water left in cylinder
= Volume of cylinder - (Volume of hemisphere + Volume of cone)
= [π(60)² x 180 – π(60)² x 80] cm³
= π(60)² [180 – 80]cm³ =π x 3600 x 100 cm³ = 1130400 cm³ = 1.130 m³
Question 8.
A spherical 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:
Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)
Radius of the cylindrical part = 2/2 = 1 cm
Height of the cylindrical part = 8 cm
∴ Volume of the cylindrical part = πr²h
= 3.14 x (1)² x 8 cm³
Diameter of the spherical part Radius = 8.5 cm
∴ Radius = 8.5/2 cm =85/20
Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
= 25.12 cm³ + 321.39 cm³ = 346.51 cm³
Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.
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:
Given: radius of metallic sphere = 4.2 cm
∴ Volume =4/3 π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere
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:
Radius of 1st metallic sphere = 6 cm
∴ Volume of 1st metallic sphere =4/3π(6)³ cm³
Radius of 2nd metallic sphere = 8 cm
∴ Volume of 2nd metallic sphere = 4/3π(8)³ cm³
Radius of 3rd metallic sphere = 10 cm
∴ Volume of 3rd metallic sphere =4/3π(10)³ cm³
Volume of all three metallic spheres =4/3π(6³+8³+10³) cm³
∵ 3 spheres are melted and recast into a new metallic sphere of radius r.
∴ Volume of new metallic sphere = 43πr³
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.
Solution:
Given: diameter of the well = 7 m Radius =7/2 m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²
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:
Given: diameter of the well = 3 m
⇒ Radius =3/2 m
Depth of the well = 14 m
Volume of the earth taken out from the well = πr²h
= π(3/2)² x 14 =πx9x14/4 =63/2πm³
∵ Earth taken out from the well evenly spread to form an embankment having height h and width of embankment around the well is 4 m.
∴ External radius (R) = radius of well + width of the embankment
=32m + 4m =11/2 m
Internal radius =3/2 m = radius of well
Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and 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:
Let total number of ice cream cones are n.
∴ All ice cream cones are filled from ice cream in the cylinder.
Total volume of n number of ice cream cones = Volume of ice cream in the cylinder
n x π x 54 = π(36)15
⇒ n x 54 = 36 x 15
⇒ n = 36 x 15/54 = 10
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:
Given: diameter of each coin = 1.75 cm ⇒ radius =1.75/2 cm
and thickness of each coin = 2 mm
Let n number of coins are melted to form a cuboid.
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 heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm
∴ Volume of the cylindrical bucket = π²h = π(18)² x 32 cm³
Let radius of the conical heap = r cm
Given: height of the conical heap = 24 cm
∴ Volume of the conical heap =1/3 π(r²) 24cm³
According to question,
Volume of the cylinderical bucket = Volume of the conical heap
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:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water 10 km/h
Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, 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:
Given: diameter of the pipe = 20 cm ⇒ radius of the pipe = 10 cm
Water flowing from the pipe at rate = 3 km
Let it filled the tank in ‘t’ hours.
Volume of the water flowing in ‘t’ hours.
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:
Given: upper diameter = 4 cm ⇒ upper radius =1/2= 2 cm = R
lower diameter = 2 cm ⇒ lower radius =2/2= 1 cm = r
height of glass = 14 cm
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:
Given: upper circumference of the frustum = 18 cm
Slant height (l) = 4 cm
We have C.S.A of the frustum = π (r1 + r2)l
Putting values from equation (i) and (ii), we get
Curved surface area = (πr1+πr2)l = (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 (see figure). 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:
Radius of open side (r1) = 10 cm
Radius of upper base (r2) = 4 cm
Slant height (l) = 15 cm
Total surface area of the cap = C.S.A. of the frustum + Area of upper base
= 660 cm² + 50.28 cm² = 710.28 cm²
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:
Radius of the lower end (r1) = 8 cm
Radius of the upper end (r2) = 20 cm
Height of the frustum (h) = 16 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:
Let ADC is a cone with vertical angle 600.
Now, cone is cut into two parts, parallel to its base at height 10 cm.
Radius of larger end of the frustum = R1
A wire be formed having diameter 1/16 cm and length be H cm
Volume of wire so obtained = πr²H
