Advertisements
Advertisements
प्रश्न
Water in a canal, 5·4 m wide and 1·8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?
Advertisements
उत्तर
Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = `25000/60=1250/3` m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal x `1250/3`
`= 5.4 xx 1.8 xx 1250/3`
=4050 m3
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m3 = 162000 m3
Now,
Area of irrigation = `"Volume of water flowing out from canal in 40 min"/"Height of the standing water needed for irrigation"`
`= 162000/0.1`
`= 1620000 m^2`
= 162 hectare (∵ 1 hectare = 10000 m2)
Thus, the area irrigated in 40 minutes is 162 hectare.
APPEARS IN
संबंधित प्रश्न
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area.
[Use π=22/7]
In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use `pi=22/7` and `sqrt5=2.236`)
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. [Use `pi = 22/7`]
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [Use π = `22/7`]
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2.
(Note that the base of the tent will not be covered with canvas.) [Use `pi = 22/7`]
In Figure 4, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region.\[[Use\pi = 3 . 14]\]
A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is
Find the ratio of the volume of a cube to that of a sphere which will fit inside it.
There are two identical solid cubical boxes of side 7 cm. From the top face of the first cube a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. Find
- the ratio of the total surface area of the two new solids formed
- volume of each new solid formed.
