Answer :
Certainly! Let's break down each part of the problem step by step.
### Given:
- Cylindrical Tank Dimensions:
- Radius, [tex]\( r_{\text{tank}} = 14 \)[/tex] cm
- Height, [tex]\( h_{\text{tank}} = 40 \)[/tex] cm
- Conical Pile Dimensions:
- Height, [tex]\( h_{\text{cone}} = 30 \)[/tex] cm
- Weight of cement per cubic centimeter:
- Weight per [tex]\( \text{cm}^3 = 832.5 \)[/tex] grams
## a) Finding the radius of the conical shaped pile of cement
First, we need to find the volume of the cylindrical tank, which equals the volume of the conical pile.
### Volume of the Cylindrical Tank:
The formula for the volume of a cylinder is:
[tex]\[ V_{\text{cylinder}} = \pi r_{\text{tank}}^2 h_{\text{tank}} \][/tex]
### Volume of the Cone:
The formula for the volume of a cone is:
[tex]\[ V_{\text{cone}} = \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}} \][/tex]
Since the volume of the conical pile equals the volume of the cylindrical tank:
[tex]\[ \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}} = \pi r_{\text{tank}}^2 h_{\text{tank}} \][/tex]
We can solve for [tex]\( r_{\text{cone}} \)[/tex]:
[tex]\[ r_{\text{cone}}^2 = \frac{3 r_{\text{tank}}^2 h_{\text{tank}}}{h_{\text{cone}}} \][/tex]
By calculating these values:
[tex]\[ r_{\text{cone}} = \sqrt{\frac{3 \times 14^2 \times 40}{30}} \][/tex]
[tex]\[ r_{\text{cone}} \approx 28 \text{ cm} \][/tex]
So, the radius of the conical pile of cement is approximately [tex]\( 28 \)[/tex] cm.
## b) Finding the surface area of the pile of cement excluding the base
The surface area of a cone (excluding the base) is given by:
[tex]\[ A_{\text{cone}} = \pi r_{\text{cone}} l \][/tex]
where [tex]\( l \)[/tex] is the slant height of the cone.
### Slant Height [tex]\( l \)[/tex]:
The slant height can be found using the Pythagorean theorem:
[tex]\[ l = \sqrt{r_{\text{cone}}^2 + h_{\text{cone}}^2} \][/tex]
Substitute the known values:
[tex]\[ l = \sqrt{28^2 + 30^2} \][/tex]
[tex]\[ l \approx 41.23 \text{ cm} \][/tex]
Now, calculate the surface area:
[tex]\[ A_{\text{cone}} = \pi \times 28 \times 41.23 \][/tex]
[tex]\[ A_{\text{cone}} \approx 3610 \text{ cm}^2 \][/tex]
So, the surface area of the conical pile of cement (excluding the base) is approximately [tex]\( 3610 \text{ cm}^2 \)[/tex].
## c) Finding the total weight of the cement
To find the total weight of the cement, we first calculate the volume of the cylindrical tank and then use the given weight per cubic centimeter.
### Volume of the Cylindrical Tank:
[tex]\[ V_{\text{cylinder}} = \pi \times 14^2 \times 40 \][/tex]
The total weight is then:
[tex]\[ \text{Total Weight} = V_{\text{cylinder}} \times 832.5 \][/tex]
Using the values calculated:
[tex]\[ \text{Total Weight} \approx 20504547 \text{ grams} \][/tex]
So, the total weight of the cement is approximately [tex]\( 20504547 \)[/tex] grams or [tex]\( 20504.547 \)[/tex] kilograms.
### Summary of Answers:
a) The radius of the conical shaped pile of cement is approximately [tex]\( 28 \)[/tex] cm.
b) The surface area of the pile of cement, excluding the base, is approximately [tex]\( 3610 \text{ cm}^2 \)[/tex].
c) The total weight of the cement is approximately [tex]\( 20504547 \text{ grams} \)[/tex] or [tex]\( 20504.547 \)[/tex] kilograms.
### Given:
- Cylindrical Tank Dimensions:
- Radius, [tex]\( r_{\text{tank}} = 14 \)[/tex] cm
- Height, [tex]\( h_{\text{tank}} = 40 \)[/tex] cm
- Conical Pile Dimensions:
- Height, [tex]\( h_{\text{cone}} = 30 \)[/tex] cm
- Weight of cement per cubic centimeter:
- Weight per [tex]\( \text{cm}^3 = 832.5 \)[/tex] grams
## a) Finding the radius of the conical shaped pile of cement
First, we need to find the volume of the cylindrical tank, which equals the volume of the conical pile.
### Volume of the Cylindrical Tank:
The formula for the volume of a cylinder is:
[tex]\[ V_{\text{cylinder}} = \pi r_{\text{tank}}^2 h_{\text{tank}} \][/tex]
### Volume of the Cone:
The formula for the volume of a cone is:
[tex]\[ V_{\text{cone}} = \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}} \][/tex]
Since the volume of the conical pile equals the volume of the cylindrical tank:
[tex]\[ \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}} = \pi r_{\text{tank}}^2 h_{\text{tank}} \][/tex]
We can solve for [tex]\( r_{\text{cone}} \)[/tex]:
[tex]\[ r_{\text{cone}}^2 = \frac{3 r_{\text{tank}}^2 h_{\text{tank}}}{h_{\text{cone}}} \][/tex]
By calculating these values:
[tex]\[ r_{\text{cone}} = \sqrt{\frac{3 \times 14^2 \times 40}{30}} \][/tex]
[tex]\[ r_{\text{cone}} \approx 28 \text{ cm} \][/tex]
So, the radius of the conical pile of cement is approximately [tex]\( 28 \)[/tex] cm.
## b) Finding the surface area of the pile of cement excluding the base
The surface area of a cone (excluding the base) is given by:
[tex]\[ A_{\text{cone}} = \pi r_{\text{cone}} l \][/tex]
where [tex]\( l \)[/tex] is the slant height of the cone.
### Slant Height [tex]\( l \)[/tex]:
The slant height can be found using the Pythagorean theorem:
[tex]\[ l = \sqrt{r_{\text{cone}}^2 + h_{\text{cone}}^2} \][/tex]
Substitute the known values:
[tex]\[ l = \sqrt{28^2 + 30^2} \][/tex]
[tex]\[ l \approx 41.23 \text{ cm} \][/tex]
Now, calculate the surface area:
[tex]\[ A_{\text{cone}} = \pi \times 28 \times 41.23 \][/tex]
[tex]\[ A_{\text{cone}} \approx 3610 \text{ cm}^2 \][/tex]
So, the surface area of the conical pile of cement (excluding the base) is approximately [tex]\( 3610 \text{ cm}^2 \)[/tex].
## c) Finding the total weight of the cement
To find the total weight of the cement, we first calculate the volume of the cylindrical tank and then use the given weight per cubic centimeter.
### Volume of the Cylindrical Tank:
[tex]\[ V_{\text{cylinder}} = \pi \times 14^2 \times 40 \][/tex]
The total weight is then:
[tex]\[ \text{Total Weight} = V_{\text{cylinder}} \times 832.5 \][/tex]
Using the values calculated:
[tex]\[ \text{Total Weight} \approx 20504547 \text{ grams} \][/tex]
So, the total weight of the cement is approximately [tex]\( 20504547 \)[/tex] grams or [tex]\( 20504.547 \)[/tex] kilograms.
### Summary of Answers:
a) The radius of the conical shaped pile of cement is approximately [tex]\( 28 \)[/tex] cm.
b) The surface area of the pile of cement, excluding the base, is approximately [tex]\( 3610 \text{ cm}^2 \)[/tex].
c) The total weight of the cement is approximately [tex]\( 20504547 \text{ grams} \)[/tex] or [tex]\( 20504.547 \)[/tex] kilograms.
