A solid sphere with the total surface area of 315 cm2 is dropped into a cubical tank with an edge length of 10 cm that is already filled to m% of its capacity with a liquid. After dropping the sphere into the tank, the height of the liquid in the tank increases by 175% then what is the value of m? (take π = 3.15)

Let's solve the problem step by step to find the value of \( m \).

### Step 1: Find the Radius of the Sphere

The total surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]

Given:
\[ A = 315 \text{ cm}^2 \]
\[ \pi = 3.15 \]

We set up the equation:
\[ 4 \pi r^2 = 315 \]
\[ 4 \times 3.15 \times r^2 = 315 \]
\[ 12.6 r^2 = 315 \]

Solving for \( r^2 \):
\[ r^2 = \frac{315}{12.6} = 25 \]
\[ r = \sqrt{25} = 5 \text{ cm} \]

### Step 2: Calculate the Volume of the Sphere

The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]

Using \( r = 5 \) cm and \( \pi = 3.15 \):
\[ V = \frac{4}{3} \times 3.15 \times 5^3 \]
\[ V = \frac{4}{3} \times 3.15 \times 125 \]
\[ V = \frac{4}{3} \times 393.75 \]
\[ V = 525 \text{ cm}^3 \]

### Step 3: Determine the Initial Volume of Liquid in the Tank

The edge length of the cubical tank is 10 cm, so its total volume \( V_{\text{tank}} \) is:
\[ V_{\text{tank}} = 10^3 = 1000 \text{ cm}^3 \]

Let the initial volume of the liquid be \( V_{\text{liquid}} \), which is \( m\% \) of the tank's capacity:
\[ V_{\text{liquid}} = \frac{m}{100} \times 1000 = 10m \text{ cm}^3 \]

### Step 4: Calculate the Final Volume of the Liquid in the Tank

After dropping the sphere, the volume of the liquid increases by the volume of the sphere:
\[ V_{\text{final}} = 10m + 525 \text{ cm}^3 \]

### Step 5: Calculate the Initial Height of the Liquid in the Tank

The initial height \( h \) of the liquid is given by the volume of the liquid divided by the base area of the tank. Since the tank is a cube with edge length 10 cm, the base area \( A_{\text{base}} \) is:
\[ A_{\text{base}} = 10 \times 10 = 100 \text{ cm}^2 \]

So the initial height \( h \) is:
\[ h = \frac{V_{\text{liquid}}}{A_{\text{base}}} = \frac{10m}{100} = 0.1m \text{ cm} \]

### Step 6: Determine the Final Height Increase

The height increase is 175% of the initial height:
\[ \Delta h = 1.75 \times h = 1.75 \times 0.1m = 0.175m \text{ cm} \]

The final height \( h_{\text{final}} \) is:
\[ h_{\text{final}} = h + \Delta h = 0.1m + 0.175m = 0.275m \text{ cm} \]

### Step 7: Calculate the Final Volume Based on the Height Increase

The final volume of the liquid based on the new height is:
\[ V_{\text{final}} = h_{\text{final}} \times A_{\text{base}} = 0.275m \times 100 \text{ cm}^2 = 27.5m \text{ cm}^3 \]

### Step 8: Set Up the Equation and Solve for \( m \)

We know that the final volume is also given by:
\[ V_{\text{final}} = 10m + 525 \]

Thus, we set up the equation:
\[ 27.5m = 10m + 525 \]

Solving for \( m \):
\[ 27.5m - 10m = 525 \]
\[ 17.5m = 525 \]
\[ m = \frac{525}{17.5} = 30 \]

Thus, the value of \( m \) is \( 30 \).

Would you like further details or have any questions?

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**Relative Questions:**

1. How is the radius of a sphere derived from its surface area?
2. What is the volume formula for a sphere, and how is it applied?
3. How do you determine the volume of a cube?
4. What does the term "175% increase in height" imply in this context?
5. How does the volume of a liquid in a cubical tank relate to the height of the liquid?
6. What are the steps to solve the problem if the tank had a different edge length?
7. How would you calculate the volume change if the sphere were a different shape, such as a cylinder?
8. Why is it important to ensure all units are consistent when performing these calculations?

**Tip:** When solving geometry problems involving volume, always check to ensure the units are consistent throughout the calculation to avoid errors.

Explore how to calculate the increase in liquid volume in a cubical tank after dropping a sphere. Step-by-step solution includes finding sphere radius, calculating liquid volume, and determining final tank liquid height. Suitable for Grades 9-12.

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