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...

Chapter 14 Statistics NCERT Class 10th solution of Exercise 14.1 NCERT Class 10th solution of Exercise 14.2 NCERT Class 10th Maths Projects Exercise 14.3 Q1. The following frequency distribution gives the monthly consumption of electricity of `68` consumers of a locality. Find the median, mean, and mode of the data and compare them. Sol. : `text{For Median}` `text{Median } = n/2 = 68/2 = 34^text(th) text{ term}` `text{Median class is (125—145)}` `text{lower limit } l = 125` `text{class-size } h = 145—125 = 20` `cf = 22` `text{Median}` `= l + ((n/2 - cf)/f) times h` `text{Median}` `= 125 + ((68/2 - 22)/ 20) times 20` `text{Median}` `= 125 + (34 - 22)` `text{Median}` `= 125 + 12` `text{Median}` `=137 text{ units}` `text{For Mean}` `text{Assume mean }` `a = 135` `text{class-size }` `h = 20` `text{Mean }overline x = a +((∑f_iu_i)/(∑f_i)) times h` `overline x = 135 +((7 times 20)/68)` `overline x = 135 + 140/68` `overline x = 135 + 2.06` `overline x = 137.06 text{ units (approx.)}` `text{Fo...

Chapter 14 Statistics NCERT Class 10th solution of Exercise 14.1 NCERT Class 10th solution of Exercise 14.3 NCERT Class 10th Maths Projects Exercise 14.2 Q1. The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Sol. : `text{For Mode}` `text{maximum class frequency is 23}` `text{modal class = 35—45}` `text{lower limit l = 35}` `text{class size h = 10}` `text{frequency of modal class}` `f_1 = 23` `text{frequency just before modal class}` `f_0 = 21` `text{frequency just after modal class}` `f_2 = 14` `text{Mode Z}` `= l + ((f_1 - f_0)/(2f_1-f_0-f_2))``times text{ h}` `text{Mode Z}` `= 35 + ((23 - 21)/(2 times 23 - 21 - 14)) times 10` `text{Mode Z}` `= 35 + ((2)/(46-35)) times 10` `text{Mode Z}` `= 35 + (20)/(11)` `text{Mode Z}` `= 35 + 1.82` `text{Mode Z}` `= 36.82` `text{For Mean}` `text{Mean } overline x = (sum f_i x_i)/(sum f_i)` `overline ...