Water is flowing at the rate of 2.52km/h through a cylindrical pipe into a cylindrical tank the radius of whose base is 40cm. If the increase in the level of water in the tank in half an hour is 3.15m. Find the internal diameter of the pipe.
Answers
Answered by
23
ANSWER:
METHOD 1
GIVEN:
After solving, we'll get that the internal diameter of the pipe is four (4)
Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = πr2 h
= π x 402 x 315
= 5,04,000 π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 *30 / 60 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = π x (d/2)2 x 126000 cm3
We know
Volume of the water that flows through the pipe in half an hour = Volume of water
that falls in the tank in half an hour
⇒ π x (d/2)2 x 126000 =504000π
⇒(d/2)2 =4
⇒d2=16
⇒d=4 cm
Thus, the internal diameter of the pipe is 4 cm.
METHOD 2
Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = π r2 h
= π x 402 x 315
= 5,04,000 π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 *30 / 60 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = π x (d/2)2 x 126000 cm3
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour
⇒ π x (d/2)2 x 126000 =504000π
⇒(d/2)2 =4
⇒d2=16
⇒d=4 cm
Thus, the internal diameter of the pipe is 4 cm.
METHOD 1
GIVEN:
After solving, we'll get that the internal diameter of the pipe is four (4)
Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = πr2 h
= π x 402 x 315
= 5,04,000 π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 *30 / 60 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = π x (d/2)2 x 126000 cm3
We know
Volume of the water that flows through the pipe in half an hour = Volume of water
that falls in the tank in half an hour
⇒ π x (d/2)2 x 126000 =504000π
⇒(d/2)2 =4
⇒d2=16
⇒d=4 cm
Thus, the internal diameter of the pipe is 4 cm.
METHOD 2
Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = π r2 h
= π x 402 x 315
= 5,04,000 π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 *30 / 60 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = π x (d/2)2 x 126000 cm3
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour
⇒ π x (d/2)2 x 126000 =504000π
⇒(d/2)2 =4
⇒d2=16
⇒d=4 cm
Thus, the internal diameter of the pipe is 4 cm.
wilcypsam:
hi
plz mark me as brainliest if you like my answer
Answered by
15
Answer:
Internal Diameter of Pipe =4cm
Step-by-step explanation:✪ ✫ ✬ ✭ ✮ ✯
Increase in level of water in tank in half hour =3.15m
⇒Increase in level of water in 1hour =6.3m
Or increase in level of water in cylindrical tank =6.3m/h
So Volume of water in tank in 1hour =πr²h
given that radius of cylindrical tank =40cm =0.4m
Volume of water =(22/7)(0.4)²×6.3
=22×(0.16)×(0.9)
=22×1.44
=3.168m³
It means 3.168m³ Volume of water increases per hour.
So Volume of water flowing thorough Pipe =3.168m³/h
Or (2.52km/h)×cross sectional area of pipe =3.168m³/h
Or (2520m/h)×πr² =3.168m³/h
Or 360×22r²=3.168
Or r² =3.168/7920
Or r²=0.0004m²
⇒ r=+0.02m=2cm (radius is always +ve)
D=2r=4cm ✪ ✫ ✬ ✭ ✮ ✯
Similar questions