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प्रश्न
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?
बेरीज
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उत्तर
Radius of the circular pipe = 0 . 01 m
Length of the water column in 1 sec = 6 m
\[\text{ Volume of the water flowing in 1 s }= \pi r^2 h = \pi(0 . 01 )^2 (6) m^3 \]
\[\text{ Volume of the water flowing in 30 mins }= \pi(0 . 01 )^2 (6) \times 30 \times 60 m^3 \]
Let h m be the rise in the level of water in the cylindrical tank .
\[\text{ Volume of the cylindrical tank in which water is being flown }= \pi(0 . 6 )^2 \times h\]
Volume of water flowing in 30 mins = Volume of the cylindrical tank in which water is being flown
\[\pi(0 . 01 )^2 (6) \times 30 \times 60 = \pi(0 . 6 )^2 \times h\]
\[h = \frac{6(0 . 01 )^2 \times 30 \times 60}{0 . 6 \times 0 . 6}\]
h = 3 m
Length of the water column in 1 sec = 6 m
\[\text{ Volume of the water flowing in 1 s }= \pi r^2 h = \pi(0 . 01 )^2 (6) m^3 \]
\[\text{ Volume of the water flowing in 30 mins }= \pi(0 . 01 )^2 (6) \times 30 \times 60 m^3 \]
Let h m be the rise in the level of water in the cylindrical tank .
\[\text{ Volume of the cylindrical tank in which water is being flown }= \pi(0 . 6 )^2 \times h\]
Volume of water flowing in 30 mins = Volume of the cylindrical tank in which water is being flown
\[\pi(0 . 01 )^2 (6) \times 30 \times 60 = \pi(0 . 6 )^2 \times h\]
\[h = \frac{6(0 . 01 )^2 \times 30 \times 60}{0 . 6 \times 0 . 6}\]
h = 3 m
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