Answer:
6.82 ![mt^3/min](https://tex.z-dn.net/?f=mt%5E3%2Fmin)
Step-by-step explanation:
By congruence of triangles, the radius r of the cone base when its height is 1 meter satisfies the relation
r/1 = 2.5/6
(See picture attached)
So, r = 0.4166 meters when the water is 1 meter high.
The volume of a cone with radius of the base = R is given by
![V=\frac{\pi R^2h}{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%5Cpi%20R%5E2h%7D%7B3%7D)
So, the volume of water when it is 1 meter high is
![V=\frac{\pi* 0.(4166)^2}{3}=0.1817\;mt^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%5Cpi%2A%200.%284166%29%5E2%7D%7B3%7D%3D0.1817%5C%3Bmt%5E3)
One minute later, the height of the water is 1 meter + 30 centimeters = 1.30 mt
The radius now satisfies
r/1.3 = 2.5/6
and now the radius of the base is
r = 0.5416 mt
and the new volume of water is ![0.3071\;mt^3](https://tex.z-dn.net/?f=0.3071%5C%3Bmt%5E3)
So, the water is raising at a speed of
0.3071-0.1871 = 0.12 ![mt^3/min](https://tex.z-dn.net/?f=mt%5E3%2Fmin)
This speed equals the difference between the water being pumped and the water leaking,
So,
<em>0.12 = Speed of water being pumped -6.8 </em>
and
Speed of water being pumped = 6.8+0.12 = 6.82 ![mt^3/min](https://tex.z-dn.net/?f=mt%5E3%2Fmin)