Chapter 13 – Surface Areas and Volumes
Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.
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Nov
Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed. 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed. Three metallic solid cubes [...]
An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.
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Nov
An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of [...]
The curved surface area of frustum of a cone is π/(r1+r2), where l = h^2 + (r1+r2)^2−−√,r1 and r2 are the radii of two ends of the frustum and h s the vertical height.
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Nov
The curved surface area of frustum of a cone is π/(r1+r2), where l = h^2 + (r1+r2)^2−−√,r1 and r2 are the radii of two ends of the frustum and h s the vertical height. r1 and r2 are the radii of two ends of the frustum and h s the vertical height. The curved surface [...]
The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is 3/r^2 (3h−2r)
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Nov
The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is 3/r^2 (3h−2r) The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is 3/r^2 (3h−2r) November 25, 2020 Category: Chapter 13 - [...]
The volume of the frustum of a cone is 1/3πh[r^21+r^22−r1r2], where h is vertical height of the frustum and r1,r2 are the radii of the ends.
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Nov
The volume of the frustum of a cone is 1/3πh[r^21+r^22−r1r2], where h is vertical height of the frustum and r1,r2 are the radii of the ends. r2 are the radii of the ends. The volume of the frustum of a cone is 1/3πh[r^21+r^22−r1r2] where h is vertical height of the frustum and r1 November 25, [...]
A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4/3 πa^3.
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Nov
A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4/3 πa^3. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4/3 πa^3. November 25, 2020 Category: Chapter 13 - Surface Areas and Volumes , Maths [...]
A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone The total surface area of the combined solid is pie r [root r^2 + h^2 +3r + 2h].
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Nov
A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone The total surface area of the combined solid is pie r [root r^2 + h^2 +3r + 2h]. A cylindrical pencil sharpened at one edge is the combination of [...]
A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr^2.
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Nov
A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr^2. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area [...]
Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr^2.
25
Nov
Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr^2. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr^2. November 25, 2020 Category: Chapter [...]
Volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is
25
Nov
Volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is then the ratio of their surface areas is Volumes of two spheres are in the ratio 64 : 27 November 25, 2020 Category: Chapter 13 - Surface Areas and Volumes , Maths , NCERT Class 10 [...]