The length of the largest possible rod that can be placed in a cubical roam is `35sqrt3`m. The surface area of the largest possible sphere that fit within the cubical room in sq m is—
(assuming `pi=22/7`)

Options

A. 4250
B. 3500
C. 3850
D. 2450

Answer

To find the surface area of the largest possible sphere that can fit within a cubical room, we can follow these steps:

### Step 1: Understand the relationship between the cube and the sphere
The largest sphere that can fit inside a cube will have a diameter equal to the side length of the cube.

### Step 2: Find the side length of the cube
We are given the length of the diagonal of the cube, which is \(35\sqrt{3}\) m. The formula for the diagonal \(d\) of a cube with side length \(a\) is given by:
\[ d = a\sqrt{3} \]
Setting \(d = 35\sqrt{3}\), we can solve for \(a\):
\[ 35\sqrt{3} = a\sqrt{3} \]
Dividing both sides by \(\sqrt{3}\):
\[ a = 35 \text{ m} \]

### Step 3: Determine the radius of the sphere
The diameter of the sphere is equal to the side length of the cube, which is \(35\) m. Therefore, the radius \(r\) of the sphere is:
\[ r = \frac{diameter}{2} = \frac{35}{2} = 17.5 \text{ m} \]

### Step 4: Calculate the surface area of the sphere
The formula for the surface area \(S\) of a sphere is:
\[ S = 4\pi r^2 \]
Substituting \(r = 17.5\) m and \(\pi = \frac{22}{7}\):
\[ S = 4 \times \frac{22}{7} \times (17.5)^2 \]

### Step 5: Calculate \( (17.5)^2 \)
Calculating \( (17.5)^2 \):
\[ (17.5)^2 = 306.25 \]

### Step 6: Substitute and simplify
Now substituting back into the surface area formula:
\[ S = 4 \times \frac{22}{7} \times 306.25 \]
Calculating \(4 \times 306.25\):
\[ 4 \times 306.25 = 1225 \]
Now substituting this value:
\[ S = \frac{22 \times 1225}{7} \]

### Step 7: Perform the multiplication and division
Calculating \(22 \times 1225\):
\[ 22 \times 1225 = 26950 \]
Now dividing by \(7\):
\[ S = \frac{26950}{7} = 3842.857 \approx 3850 \text{ m}^2 \]

### Final Answer
The surface area of the largest possible sphere that fits within the cubical room is approximately \(3850 \text{ m}^2\).

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