Volume of 1 bullet = 4/3?(2)3 = (32/3 x 22/7) cm3
Volume of the cube = (44)3
? Number of bullets = Volume of solid/Volume of 1 bullet
= (44)3 x 3 x 7 / 32 x 22 = 2541
Let required number of coins be n, Then,
n x 1/3 x ? (1/10)2 x 1 = ? x (3)2 x 5
? n = 9 x 5 x 3 x 100 = 13500
Ratio of curved surface area of cylinder and cone = [2?rh] / [?r?h2 + r2]
= [2 x 6 x 8] / [6 x ?62 + 82]
= 96/(6 x 10) = 96/60 = 8/5 = 8 : 5
Volume of the prism = (Area of the base ) x (Height)
Volume of the pyramid = 1/3 (Area of the base ) x (Height)
Required ratio = [A x H] / [(1/3) x A x H]
Therefore, Ratio of the volumes of the prism and the pyramid = 3 : 1.
Let the sides of the base are 5k, 12k and 13k, respectively.
Given, perimeter of base = 60 cm
? 5k + 12k + 13k = 60
? k = 60/30 = 2
The sides of base are 10 cm, 24 cm 26 cm.
? Volume of prism = (1/2) x 10 x 24 x 50 = 6000 cm3
Volume of prism = Area of base x Height
= (1/2) x 10 x 12 x 20
= 1200 cm2
? Weight of prism = 1200 x 6
= 7200 g
= 7.2 kg
Let r = Radius of cylinder = Radius of sphere, h = Height of the cylinder.
According to the question.
4/3?r3 = ?r2h
? h = 4r/3
? 4r = 3h
? 2r = 3/2h
? 2r/h = 3/2
? Required ration = 3 : 2
Given, 4?r2 = 2?rh
? h = 2r
Now, required ratio
= 4/3?r3 : ?r2h
= 4r : 3h
= 4r : 6r [ ? h = 2r]
= 2 : 3
Volume of the cuboid = 9 x 8 x 6 = 432 cm3
Volume of the cube = 1/2 x 432 = 216 cm3
? Each side of cube = ?216 = 6 cm
Total surface area of the cube = 6 x (Side)2 = 6 x 62
= 6 x 36 = 216 sq cm
Given, l = Length of the cuboid = 5 x 7 = 35 cm
h = Breadth of the cuboid = 5 cm
h = Height of the cuboid = 5 cm
? Surface area = 2(lb + bh + lh) = 2[35 x 5 + 5 x 5 + 35 x 5]
= 2[175 + 25 + 175]
= 2 x 375 = 750 sq cm
1 hec = 10000 m3
Volume of water = Base area x Height
= (10000 x 10)/100 = 1000 m3
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