water flows out through a circular pipe whose internal diameter is 2cm at the rate of 0 8m/s into a - Maths - Surface Areas and Volumes

Question

water flows out through a circular pipe whose internal diameter
is 2cm at the rate of 0 8m/s into a cylindrical tank, the radius of
whose base is 40cm by how much will the level rise in one and half
hour?

Priyanka Kedia · Accepted Answer

Given, internal diametre of circular pipe = 2 cm

So, radius of circular pipe = 2/2 cm = 1cm

Since water flows at the rate of 0.8 m/s.

i.e., in 1 second it covers distance = 0.8 m = 80 cm

$S o, v o l u m e o f w a t e r f l o w s i n 1 \sec o n d = π r^{2} d = 3.14 \times 1^{2} \times 80 = 251.2 m^{3} T h u s, v o l u m e o f w a t e r f l o w s i n 1 \frac{1}{2} h r o r (90 \min = 90 \times 60 = 5400 \sec) 5400 \sec o n d = 5400 \times 251.2 m^{3} = 1356480 m^{3}$

Given, radius of cylindrical tank (r) = 40 cm

Let the increase level of water = h cm

$S o, V o l u m e o f c y l i n d r i c a l \tan k = π {(40)}^{2} h$

Since, water flows out through a circular pipe at the rate of 0.8 m/s into a cylindrical tank. So,

$V o l u m e o f w a t e r f l o w s i n 1 \frac{1}{2} h r = V o l u m e o f c y l i n d r i c a l \tan k 1356480 = π {(40)}^{2} h h = \frac{1356480}{{(40)}^{2} \times 3.14} = 270 m$

Hence, water rise by 270 m.

water flows out through a circular pipe whose internal diameter is 2cm at the rate of 0.8m/s into a cylindrical tank, the radius of whose base is 40cm. by how much will the level rise in one and half hour?