If V is the volume of cuboid whose length is a , breadth is b and height is c and S is its total surface area, then prove that 1 V = 2 S 1 a + 1 b + 1 c .

Given: ' V ' is the Volume of the cuboid, ' S ' is the Surface area of the cuboid and a × b × c are the dimensions of the cuboid. To prove: 1 V = 2 S 1 a + 1 b + 1 c . We know that, S = 2 a b + b c + c a and V = a b c 1 V = 1 a b c Multiply the numerator and denominator by S . ⇒ 1 V = S S a b c ⇒ 1 V = 2 a b + b c + c a S a b c {Since, S = 2 a b + b c + c a } ⇒ 1 V = 2 S a b a b c + b c a b c + c a a b c ⇒ 1 V = 2 S 1 c + 1 a + 1 b Re-arranging the terms, ⇒ 1 V = 2 S 1 a + 1 b + 1 c . Hence, proved.

A hemispherical tank full of water is emptied by a pipe at the rate of 3 4 7 litters per second. How much time will it take to empty half the tank, if it is 3 m in diameter.(Take π = 22 7 )

Radius of base of tank = D 2 = 3 2 m Volume of tank = 2 3 πr 3 = 2 3 × 22 7 × 3 2 3 = 99 14 m 3 = 99 14 × 1000 l i t r e s = 99000 14 l i t r e s Volume of water to be emptied = 1 2 × volume of tank = 1 2 × 99000 14 l i t r e s = 99000 28 l i t r e s Time taken to empty 3 4 7 = 25 7 l i t r e s = 1 s e c Time taken to empty 1 l i t r e s = 7 25 s e c Time taken to empty 99000 28 l i t r e s = 99000 28 × 7 25 = 69300 700 = 990 s e c = 990 60 m i n = 16 . 5 m i n Time taken to empty half tank is 16 minute and 5 second.

The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm respectively, find the volume.

The volume of frustum of cone, V = 1 3 π H R 2 + R r + r 2 The height of frustum be H , R as the radius of the larger end of frustum of cone and r is the radius of smaller end of frustum of cone. Given H = 45 cm, R = 28 cm and r = 7 cm respectively. Therefore, Volume of frustum of cone V = 1 3 × 22 7 × 45 × 28 2 + 28 × 7 + 7 2 V = 22 7 × 15 × 1029 V = 48510 cubic centimetre.

Two cubes each of 15 cm sides are joined end to end. Find the total surface area of the resulting cuboid.

When the two cubes each of 15 cm edge are joined end to end, we get a cuboid of dimensions Length l = 15 cm + 15 cm = 30 cm Breadth b = 15 cm Width h = 15 cm Surface area of resulting cuboid is: = 2 ( lb + bh + hl ) = 2 ( 30 × 15 + 15 × 15 + 15 × 30 ) cm 2 = 2 ( 1125 ) cm 2 = 2250 cm 2 Hence the surface area of the resulting cuboid is 2250 cm 2

A iron sphere of radius 8 cm is melted then recasted into small spheres each of radius 1 cm . Find the number of small spheres.

Radius of given iron sphere R = 8 cm Volume of sphere = 4 3 π r 3 , where r is the radius of sphere. Volume of big sphere = 4 3 π R 3 = 4 3 π 8 3 = 4 3 π × 512 cm 3 Radius of recasted small sphere r = 1 cm Volume of small sphere = 4 3 π r 3 = 4 3 π 1 3 = 4 3 π cm 3 Number of small spheres = Volume of big sphere Volume of small sphere Number of small spheres = 4 3 π × 512 4 3 π = 512 Hence, 512 small spheres can be recasted.