NCERT Class 10th solution of Exercise 13.1 NCERT Class 10th solution of Exercise 13.2 NCERT Class 10th Maths Projects Exercise 13.3 Take `pi = 22/7` unless stated otherwise. Q1. A metallic sphere of radius `text{4.2 cm}` is melted and recast into the shape of a cylinder of radius `text{6 cm}`. Find the height of the cylinder. `text{Sol. :}` `text{Given :}` `text{Radius of Sphere }R = 4.2 text{ cm}` `text{Radius of Cylinder }r = 6 text{ cm}` `text{To find :}` `text{Height of the Cylinder}` `text{Solve :}` `text{According to question}` `text{Volume of the Cylinder = Volume of the Sphere}` `pir^2h = 4/3piR^3` `h = (4piR^3)/(3pir^2)` `h = (4R^3)/(3r^2)` `h = (4times(4.2)^3)/(3(6)^2)` `h = (4times4.2times4.2times4.2)/(3times6times6)` `h = (4times4.2times0.7times0.7)/3` `h = (8.232)/3` `h = 2.744text{ cm}` `text{Answer :}` `text{The height of the cylinder 2.744 cm.` ☝  Like,  Share,  and  Subscribe.  Q2. Metallic spheres of radii `text{6 cm, 8 cm}` and `text{10 cm,}` ...

NCERT Class 10th solution of Exercise 13.1 NCERT Class 10th solution of Exercise 13.3 NCERT Class 10th Maths Projects Chapter 13 Surface Areas And Volume Exercise 13.2 Unless stated otherwise, take `pi = 22/7`. Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to `text{1 cm}` and the height of the cone is equal to its radius. Find the volume of the solid in terms of `pi`. `text{Sol. :}` `text{Given :}` `text{Radius of Cone & hemisphere } r = 1text{ cm}` `text{Height of Cone & hemisphere } h = r =1text{ cm}` `text{To find :}` `text{Volume of the solid}` `text{Solve :}` `text{Volume of Solid = Volume of hemisphere}` `text{                           +  Volume of Cone}` `text{V} = 2/3pir^3 + 1/3pir^2h` `text{    }= pir^2(2/3r + 1/3h)` `text{    }= pi(1)^2(2/3times1 + 1/3 times 1)` `text{    }= pi(2/3 + 1/3)` `text{    }= pi...

NCERT Class 10th solution of Exercise 13.2 NCERT Class 10th solution of Exercise 13.3 NCERT Class 10th Maths Projects Exercise 13.1 Q1. `2` cubes each of volume `text{64 cm}^3` are joined end to end. Find the surface area of the resulting cuboid. `text{Sol. :}` `text{Given :}` `text{Volume of cube 64 cm}^3` `text{To find :}` `text{The Surface area of the resulting cuboid.}` `text{Solve :}` `text{Volume of cube = (side)}^3` `64 = text{(side)}^3` `(4)^3 = text{(side)}^3` `text{side = 4 cm}` After join `2` cubes the resulting cuboid `text{Height = 4cm, Length = 8 cm, Breadth = 4 cm}` `text{Surface Area of Cuboid = 2(LB + BH + HL)}` `text{S A of Cuboid = 2}(8 times 4 + 4 times 4 + 4 times 8)` `text{S A of Cuboid = 2}(32 + 16 + 32)` `text{S A of Cuboid = 2}(80)` `text{S A of Cuboid = 160 cm}^2` `text{Answer:}` `text{The Surface Area of resulting Cuboid is 160 cm}^2.`  ☝  Like, Share, and Subscribe.  Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinde...