The question asks us to find the surface area of a ball (which is a sphere) given its radius. We are also provided with a specific value to use for pi ($\pi$).
To calculate the surface area of a sphere, we use the standard mathematical formula:
$$A = 4 \pi r^2$$
Where:
Now, let's substitute the given values into the formula:
$$A = 4 \times \frac{22}{7} \times (21 \text{ mm})^2$$
$$(21 \text{ mm})^2 = 21 \times 21 \text{ mm}^2 = 441 \text{ mm}^2$$
$$A = 4 \times \frac{22}{7} \times 441 \text{ mm}^2$$
$$A = 4 \times \frac{22}{7} \times 21 \times 21 \text{ mm}^2$$
$$A = 4 \times 22 \times \frac{21}{7} \times 21 \text{ mm}^2$$
$$A = 4 \times 22 \times 3 \times 21 \text{ mm}^2$$
$$A = 88 \times 63 \text{ mm}^2$$
Let's calculate $88 \times 63$:
$88 \times 60 = 5280$
$88 \times 3 = 264$
$5280 + 264 = 5544$
Therefore,
$$A = 5544 \text{ mm}^2$$
The calculated surface area is $5544 \text{ mm}^2$. Comparing this result with the given options, we find that it matches one of the choices.
Result: The surface area of the ball is $5544 \text{ mm}^2$.
A hemispherical bowl has a 21 cm radius. It is to be painted inside as well as outside. The cost of painting it at the rate of ₹0.05 per $cm^2$ and assuming that the thickness of the bowl is negligible, is: (Use $\pi = \frac{22}{7}$)
A solid cube is painted yellow, blue and black such that opposite faces are of same colour. The cube is then cut into 36 cubes of two different sizes such that 32 cubes are small and the other four cubes are Big. None of the faces of the bigger cubes is painted blue. How many cubes have only one face painted?
A and B are two heavy steel blocks. If B is placed on the top of A, the weight increases by 60%. How much weight will reduce with respect to the total weight of A and B, if B is removed from the top of A?
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The centroid of an equilateral triangle ABC is G. If AB is 6 cms, the length of AG is