A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by?

Let the height of cylinder = h
and height cone = H
Then, ?r² h = 1/3 ?r² H
? h/H = 1/3 = 1 : 3

Let N cones be needed
Then, 1/3 ?r² h x N = ?r²h
? N = 3

? 4/3 x 22/7 x r³ = 38808
? r³ = (38808 x 7/22 x 3/4)
= (21)³
? r = 21 cm
So, curved surface area of sphere
= 4?r²
= (4 x 22/7 x 21x21) cm²
= 5544 cm²

Let breadth = L meters. Then, height = 5L meters and length = 40L meters
? L x 5L x 40L = 12.8
? L³ = 12.8/200 = 128/2000 = 64/1000
? L = 4/10
Thus, breadth = (4/10) m = (4 x 100)/10 cm = 40 cm

Let original radius = r and original length = h
New radius = r/3
Let new length = H
Then, ? r²h = ? (r/3)² x H = ? r²H/9
? H = 9h
Thus, the length becomes 9 times .

? 4/3 ?r³ = ?r² h
? h = 4/3 r
? Height = 4/3 times its radius.

? Circumference of the girth = 440 cm
? 2?r = 440
? r = 440 / (2 x 7/22) = 70 cm
Thus, Outer radius = 70 cm
Inner radius = (70 - 4) cm = 66 cm
Volume of iron = ?[(70)² - (66)²] x 63 cm³
= (22/7) x 136 x 4 x 63 cm³
= 107712 cm ³

Let the number of spheres be N
Then, N x 4/3 ? x (3)³ = ? x (2)² x 45
? 36N = 180
? N = 180/36 = 5

Radius of sphere = 9 cm
Volume of sphere = [ 4/3 x ? x (9)³] cm³ = 972? cm³
Radius of wire = 0.2 mm = 2/(10 x 10) cm = 1/50 cm
Let the length of wire be = L cm
Then, 972? = ? x (1/50)² x L
? L = (972 x 50 x 50 ) cm
?972? = ? x (1/50)² x L
? L = (972 x 50 x 50) cm
? Length of wire = (972 x 50 x 50)/100 m = 24, 300 m

? 2?rh = 264
? 2 x (22/7) x r x 14 = 264
? r = 3
So, Volume = ?r²h
= (22/7) x 3 x 3 x 14 cm³
= 396 cm³