Sphere Volume Numerically Equals Surface Area
This problem asks for the radius and surface area of a sphere where its volume is numerically equal to its surface area, using \(\pi = \frac{22}{7}\) and rounding to 3 decimal places.
Sphere Formulas and Given Values
- Volume (\(V\)) of a sphere: \(\small V = \frac{4}{3}\pi r^3\)
- Surface Area (\(A\)) of a sphere: \(\small A = 4\pi r^2\)
- Given value of pi: \(\pi = \frac{22}{7}\)
- Condition: Volume = Surface Area (numerically)
Equating Sphere Volume and Surface Area
The condition \(V = A\) translates to:
\( \frac{4}{3}\pi r^3 = 4\pi r^2 \)
Sphere Radius Calculation
To solve for the radius (\(r\)), we simplify the equation. Assuming \(r > 0\) (since it's a sphere):
Divide both sides by \(4\pi r^2\):
\( \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{4\pi r^2}{4\pi r^2} \)
This simplifies to:
\( \frac{1}{3}r = 1 \)
Solving for \(r\) yields:
\( r = 3 \text{ units} \)
Sphere Surface Area Calculation
Using the calculated radius (\(r = 3\)) and the given \(\pi = \frac{22}{7}\), calculate the surface area (\(A\)):
\( A = 4\pi r^2 \)
\( A = 4 \times \frac{22}{7} \times (3)^2 \)
\( A = 4 \times \frac{22}{7} \times 9 \)
\( A = \frac{4 \times 22 \times 9}{7} \)
\( A = \frac{792}{7} \)
Converting to a decimal and rounding to three decimal places:
\( A \approx 113.143 \text{ unit}^2 \)
Final Results
The radius of the sphere is 3 units, and its surface area is approximately 113.143 unit\(^2\).
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Important Questions from Mensuration
In a circular garden of radius 15 m, a path of 2 m wide has to be made inside the garden at the rate of ₹ 24 per sq. m. The cost of making the path is: (Take π = \(\frac{22}{7}\) )
A hollow cylinder with outer radius 4 cm and height 2 cm is made up of 1 cm thick metal sheet. What is the volume of metal used? (Take π = \(\frac{22}{7}\))
The sides of a triangular park are 60 m, 297 m and 303 m. Its area is equal to the area of a square-shaped garden. What is the perimeter (in m) of the garden?
The length and breadth of a rectangular field are in the ratio 4 : 3. If the cost of cultivating the field at ₹2 per m 2is ₹600, then the length of the field is:
A cylindrical tank has a capacity of 5632 m3. If the diameter of its base is 8 m, what is the depth of the cylindrical tank? (Use π = \(\frac{22}{7}\))
