Exams Question Papers General Knowledge Election Directory
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 π as found appropriate.)
Answer:
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Surface area of double cone = Surface area of cone 1 + Surface area of cone 2
= πrl1 + πrl2
= πr [4 + 3] = 3.14 × 2.4 × 7 = 52.75 cm2
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.
Answer:
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Number of rounds = Height of cylinder/Diameter of wire
= 12/0.3 = 40 rounds
Length of wire required in 1 round = Circumference of base of cylinder
= 2πr = 2π × 5 = 10π
Length of wire in 40 rounds = 40 × 10π
= (400 ×22)/7 = 8800/7
= 1257.14 cm = 12.57 m
Radius of wire = 0.3/2 = 0.15 cm
Volume of wire = Area of cross-section of wire × Length of wire
= π(0.15)2 × 1257.14
= 88.898 cm3
Mass = Volume × Density
= 88.898 × 8.88
= 789.41 gm
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 is drawn into a wire of diameter 1/16 cm, find the length of the wire. [Take π = 22/7]
Answer:
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 Rs.20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm2. [Take π = 3.14]
Answer:
Radius (r1) of upper end of container = 20 cm
Radius (r2) of lower end of container = 8 cm
Height (h) of container = 16 cm
Slant height (l) of frustum
Cost of 1 litre milk = Rs 20
Cost of 10.45 litre milk = 10.45 × 20 = Rs 209
Area of metal sheet used to make the container
Therefore, the cost of the milk which can completely fill the container is Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.
Question 3:
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). 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 use for making it. [Use π = 22/7]
Answer:
Radius (r2) at upper circular end = 4 cm
Radius (r1) at lower circular end = 10 cm
Slant height (l) of frustum = 15 cm
Area of material used for making the fez = CSA of frustum + Area of upper circular end
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.
Answer:
Perimeter of upper circular end of frustum = 18
2πr1 =18
r1 = 9/π
Perimeter of lower end of frustum = 6 cm
2πr2 = 6
r2 = 3/π
Slant height (l) of frustum = 4
CSA of frustum = π (r1 + r2) l
Therefore, the curved surface area of the frustum is 48 cm2.
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. [Use π = 22/7]
Answer:
Radius (r1) of upper base of glass = 4/2 = 2 cm
Radius (r2) of lower base of glass = 2/2 = 1 cm
Capacity of glass = Volume of frustum of cone
Question 9:
A farmer connects a pipe of internal diameter 20 cm form 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?
Answer:
Consider an area of cross-section of pipe as shown in the figure.
Radius (r1) of circular end of pipe = 20/200 = 0.1 m
Area of cross-section
Speed of water = 3 km/h = 3000/60 = 50 metre/min
Volume of water that flows in 1 minute from pipe = 50 × 0.01 π = 0.5 π m3
Volume of water that flows in t minutes from pipe = t × 0.5π m3
Radius (r2) of circular end of cylindrical tank = 10/2 = 5 m
Depth (h2) of cylindrical tank = 2 m
Let the tank be filled completely in t minutes.
Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
Volume of water that flows in t minutes from pipe = Volume of water in tank
t × 0.5π = π ×(r2)2 ×h2
t × 0.5 = 52 ×2
t = 100
Therefore, the cylindrical tank will be filled in 100 minutes.
Question 8:
Water in 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?
Answer:
Consider an area of cross-section of canal as ABCD.
Area of cross-section = 6 × 1.5 = 9 m2
Speed of water = 10 km/h = 10000/60 metre/min
Volume of water that flows in 1 minute from canal = 9 × (10000/60) = 1500 m3
Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m3
Let the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.
Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the reqd. area
45000 = (A × 8)/100
A = 562500 m2
Therefore, area irrigated in 30 minutes is 562500 m2.
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.
Answer:
Height (h1) of cylindrical bucket = 32 cm
Radius (r1) of circular end of bucket = 18 cm
Height (h2) of conical heap = 24 cm
Let the radius of the circular end of conical heap be r2.
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Volume of sand in the cylindrical bucket = Volume of sand in conical heap
Therefore, the radius and slant height of the conical heap are 36 cm and 12√13 cm respectively.
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 × 10 cm × 3.5 cm? [Use π = 22/7]
Answer:
Coins are cylindrical in shape.
Height (h1) of cylindrical coins = 2 mm = 0.2 cm
Radius (r) of circular end of coins = 1.75/2 = 0.875 cm
Let n coins be melted to form the required cuboids.
Volume of n coins = Volume of cuboids
Therefore, the number of coins melted to form such a cuboid is 400.
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.
Answer:
Height (h1) of cylindrical container = 15 cm
Radius (r1) of circular end of container = 12/2 = 6 cm
Radius (r2) of circular end of ice-cream cone = 6/2 = 3 cm
Height (h2) of conical part of ice-cream cone = 12 cm
Let n ice-cream cones be filled with ice-cream of the container.
Volume of ice-cream in cylinder = n × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)
Therefore, 10 ice-cream cones can be filled with the ice-cream in the container.
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.
Answer:
The shape of the well will be cylindrical.
Depth (h1) of well = 14 m
Radius (r1) of the circular end of well = 3/2 m
Width of embankment = 4 m
From the figure, it can be observed that our embankment will be in a cylindrical shape having outer radius (r2) as 4+ (3/2) = 11/2 m and inner radius (r1) as 3/2 m.
Let the height of embankment be h2.
Volume of soil dug from well = Volume of earth used to form embankment
Therefore, the height of the embankment will be 1.125 m.
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. [Use π = 22/7]
Answer:
The shape of the well will be cylindrical.
Depth (h) of well = 20 m
Radius (r) of circular end of well = 7/2 m
Area of platform = Length × Breadth = 22 × 14 m2
Let height of the platform = H
Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.
Volume of soil from well = Volume of soil used to make such platform
π × r2 × h = Area of platform × Height of platform
Therefore, the height of such platform will be 2.5 m.
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.
Answer:
Radius (r1) of 1st sphere = 6 cm
Radius (r2) of 2nd sphere = 8 cm
Radius (r3) of 3rd sphere = 10 cm
Let the radius of the resulting sphere be r.
The object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.
Volume of 3 spheres = Volume of resulting sphere
Therefore, the radius of the sphere so formed will be 12 cm.
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.
Answer:
Radius (r1) of hemisphere = 4.2 cm
Radius (r2) of cylinder = 6 cm
Let the height of the cylinder be h.
The object formed by recasting the hemisphere will be the same in volume.
Volume of sphere = Volume of cylinder
Hence, the height of the cylinder so formed will be 2.74 cm.
Question 8:
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter o 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.
Answer:
Height (h) of cylindrical part = 8 cm
Radius (r2) of cylindrical part = 2/2 = 1 cm
Radius (r1) spherical part = 8.5/2 = 4.25 cm
Volume of vessel = Volume of sphere + Volume of cylinder
Hence, she is wrong.
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. [Use π = 3.14]
Answer:
Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm
Height (h2) of conical part of solid = 120 cm
Height (h1) of cylinder = 180 cm
Radius (r) of cylinder = 60 cm
Volume of water left = Volume of cylinder − Volume of solid
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]
Answer:
From the figure, it can be observed that
Height (h1) of larger cylinder = 220 cm
Radius (r1) of larger cylinder = 24/2 = 12 cm
Height (h2) of smaller cylinder = 60 cm
Radius (r2) of smaller cylinder = 8 cm
Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder
Mass of 1 cm3 iron = 8 g
Mass of 111532.8 cm3 iron = 111532.8 × 8 = 892262.4 g = 892.262 kg
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.
Answer:
Height (h) of conical vessel = 8 cm
Radius (r1) of conical vessel = 5 cm
Radius (r2) of lead shots = 0.5 cm
Let n number of lead shots were dropped in the vessel.
Volume of water spilled = Volume of dropped lead shots
Hence, the number of lead shots dropped in the vessel is 100.