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SURFACE AREA AND VOLUME | SHAPE DIMENTION | 3D OBJECTS
Typology: Exercises
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(Class โ X)
Question 1: 2 cubes each of volume 64 cm^3 are joined end to end. Find the surface area of the resulting cuboids.
Answer 1: Given that, Volume of cubes = 64 cm^3 (Edge) 3 = 64 Edge = 4 cm
If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.
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. [Use ๐ = 227 ]
Answer 2:
It can be observed that radius ( r ) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm). Height of hemispherical part = Radius = 7 cm Height of cylindrical part ( h ) = 13 โ7 = 6 cm Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part
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. [Use ๐ = 227 ] Answer 4:
From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cubeโs edge, i.e., 7cm. Radius ( r ) of hemispherical part = 7/2 = 3.5cm Total surface area of solid = Surface area of cubical part + CSA of hemispherical part โ Area of base of hemispherical part
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.
= 6 (Edge ) 2 โ = 6 (Edge) 2 +
Answer 5:
Diameter of hemisphere = Edge of cube = l Radius of hemisphere = l / Total surface area of solid = Surface area of cubical part + CSA of hemispherical part โ Area of base of hemispherical part
Question 6: A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [Use
๐ = 227 ]
= 6 (Edge ) 2 โ = 6 (Edge) 2 +
Answer 7:
Given that, Height ( h ) of the cylindrical part = 2.1 m Diameter of the cylindrical part = 4 m Radius of the cylindrical part = 2 m Slant height ( l ) of conical part = 2.8 m Area of canvas used = CSA of conical part + CSA of cylindrical part
Cost of 1 m^2 canvas = Rs 500 Cost of 44 m^2 canvas = 44 ร 500 = 22000 Therefore, it will cost Rs 22000 for making such a tent.
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 cm^2. [Use ๐ = 227 ]
Answer 8:
Given that, Height ( h ) of the conical part = Height ( h ) of the cylindrical part = 2.4 cm Diameter of the cylindrical part = 1.4 cm Therefore, radius ( r ) of the cylindrical part = 0.7 cm
Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
The total surface area of the remaining solid to the nearest cm^2 is 18 cm^2
(Class โ X)
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 ฯ. Answer 1:
Given that, Height ( h ) of conical part = Radius( r ) of conical part = 1 cm Radius( r ) of hemispherical part = Radius of conical part ( r ) = 1 cm Volume of solid = Volume of conical part + Volume of hemispherical part
=^13 ๐๐^2 โ +^23 ๐๐^3 =^13 ๐. 1^2. 1 +^23 ๐. 1^3 = ๐ ๐๐^3
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.) [Use ๐ = 227 ]
Answer 2:
From the figure, it can be observed that Height ( h 1 ) of each conical part = 2 cm Height ( h 2 ) of cylindrical part = 12 โ 2 ร Height of conical part = 12 โ 2 ร2 = 8 cm Radius ( r ) of cylindrical part = Radius of conical part = 3/2 cm Volume of air present in the model = Volume of cylinder + 2 ร Volume of cones
= ๐๐^2 โ 2 + 2 ร^13 ๐๐^2 โ 1
= ๐ (^32 )
2
. 8 + 2 ร^13 ๐ (^32 )
2
. 2 = 18๐ + 3๐ = 21๐ = 21 ร^227 = 66 ๐๐^3
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 the given figure). [Use ๐ = 227 ]
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 cuboids 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 the following
figure). [Use ๐ = 227 ]
Answer 4:
Depth ( h ) of each conical depression = 1.4 cm Radius ( r ) of each conical depression = 0.5 cm Volume of wood = Volume of cuboid โ 4 ร Volume of cones
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 5:
Height ( h ) of conical vessel = 8 cm Radius ( r 1 ) of conical vessel = 5 cm Radius ( r 2 ) 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.
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 ๐ = 227 ]
Answer 7:
Radius ( r ) of hemispherical part = Radius ( r ) of conical part = 60 cm Height ( h 2 ) of conical part of solid = 120 cm Height ( h 1 ) of cylinder = 180 cm Radius ( r ) of cylinder = 60 cm Volume of water left = Volume of cylinder โ Volume of solid
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 cm^3. Check whether she is correct, taking the above as the inside measurements, and ฯ = 3.14. Answer 8:
Height ( h ) of cylindrical part = 8 cm Radius ( r 2 ) of cylindrical part = 1 cm Radius ( r 1 ) spherical part = 4.25 cm Volume of vessel = Volume of sphere + Volume of cylinder
Hence, she is wrong.
Therefore, the radius of the sphere so formed will be 12 cm.
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 ๐ = 227 ]
Answer 3:
The shape of the well will be cylindrical. Depth ( h ) of well = 20 m Radius ( r ) of circular end of well = 7/2 cm Area of platform = Length ร Breadth = 22 ร 14 m^2 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
Therefore, the height of such platform will be 2.5 m.
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 4:
The shape of the well will be cylindrical. Depth ( h 1 ) of well = 14 m Radius ( r 1 ) 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 ( r 2 ) as 4+ (3/2) = 11/2 m and inner radius ( r 1 ) as 3/2 m. Let the height of embankment be h 2. Volume of soil dug from well = Volume of earth used to form embankment
