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H Harsh Kankaria
Given, The radius of the wooden spheres =   The surface area of a single sphere =  Again, the Radius of the cylinder support =  Height of the support =   The base area of the cylinder =  Now, Cost of painting silver =   Cost of painting 1 wooden sphere = Cost of painting silver =  Now, Curved surface area of the cylindrical support =  Now, Cost of painting  black =   Cost of painting 1 such...

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H Harsh Kankaria
Let the radius of the sphere be  Diameter of the sphere =  According to question, Diameter is decreased by  So, the new diameter =  So, the new radius =  New surface area =  Decrease in surface area =   Percentage decrease in the surface area = 

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H Harsh Kankaria
External dimension od the book shelf = (Note: There is no front face) The external surface area of the shelf = We know, each stripe on the front surface is also to be polished. which is 5 cm stretch. Area of front face = Area to be polished = Cost of polishing  area = Cost of polishing  area =  Now, Dimension of inner part =  Area to be painted in 3 rows = Cost of painting  area =...

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H Harsh Kankaria
Given, The radius of the spherical capsule = The volume of the capsule =  Therefore,  of medicine is needed to fill the capsule.

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H Harsh Kankaria
Given, The radius of a small sphere =  The surface area of a small sphere =  The radius of the bigger sphere =  The surface area of the bigger sphere =  And,  We know, the surface area of a sphere =  The ratio of their surface areas =  Therefore, the required ratio is     

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H Harsh Kankaria
Given, The radius of a small sphere =  The radius of the bigger sphere =  The volume of each small sphere=  And, Volume of the big sphere of radius =  According to question,

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H Harsh Kankaria
Let the radius of the hemisphere be  Inside the surface area of the dome =  We know, Surface area of a hemisphere =    The volume of the hemisphere = 

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H Harsh Kankaria
Given,  is the cost of white-washing  of the inside area  is the cost of white-washing  of inside area (i) Therefore, the surface area of the inside of the dome is       

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H Harsh Kankaria
Given, The surface area of the sphere =  We know, Surface area of a sphere =    The volume of the sphere = 

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H Harsh Kankaria
Given, Inner radius of the hemispherical tank =  Thickness of the tank =   Outer radius = Internal radius + thickness =  We know, Volume of a hemisphere =   Volume of the iron used = Outer volume - Inner volume 

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H Harsh Kankaria
The radius of the hemispherical bowl =  We know, Volume of a hemisphere =  The volume of the given hemispherical bowl =  The capacity of the hemispherical bowl =     

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H Harsh Kankaria
Given, Let  be the diameters of Earth  The diameter of the Moon =  We know, Volume of a sphere =    The ratio of the volumes =  Therefore, the required ratio of the volume of the moon to the  volume of the earth is 

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H Harsh Kankaria
Given, The radius of the metallic sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Now, the density of the metal is  per ,which means, Mass of  of the metallic sphere =  Mass of  of the metallic sphere = 

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H Harsh Kankaria
The solid spherical ball will displace water equal to its volume.  Given, The radius of the sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Therefore, amount of water displaced will be 

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H Harsh Kankaria
The solid spherical ball will displace water equal to its volume.  Given, The radius of the sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Therefore, the amount of water displaced will be 

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H Harsh Kankaria
Given, The radius of the sphere =  We know, Volume of a sphere =  The required volume of the sphere = 

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H Harsh Kankaria
Given, The radius of the sphere =  We know, Volume of a sphere =  The required volume of the sphere = 

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H Harsh Kankaria
Given, Height of the conical heap =   Base radius of the cone =  We know,  The volume of a cone =  The required volume of the cone formed =  Now, The slant height of the cone =  We know, the curved surface area of a cone =  The required area of the canvas to cover the heap  =   

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H Harsh Kankaria
When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose, Height of the cone = Length of the axis=  Base radius of the cone =  And, Slant height of the cone =  We know,  The volume of a cone =  The required volume of the cone formed =  Now, Ratio of the volumes of the two solids =  Therefore, the required ratio is 

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H Harsh Kankaria
When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose, Height of the cone = Length of the axis=  Base radius of the cone =  And, Slant height of the cone =  We know,  The volume of a cone =  The required volume of the cone formed =  Therefore, the volume of the solid cone obtained is 
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