MCQ
Surface area of sphere of diameter $14\ cm$ is:
- ✓$616\ cm^2$
- B$400\ cm^2$
- C$516\ cm^2$
- D$2244\ cm^2$
✓
Answer
Correct option: A.
$616\ cm^2$
Surface area of sphere $=4\pi\text{r}^2$
Given that Diametre $= 14\ cm$
Surface area of sphere $=4\pi\text{r}^2$
$=\frac{4\times22}{7\times7^2}$
$= 616\ cm^2$
Given that Diametre $= 14\ cm$
Surface area of sphere $=4\pi\text{r}^2$
$=\frac{4\times22}{7\times7^2}$
$= 616\ cm^2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$y$ co-ordinate is known as:
→$D, E, F$ are the mid-point of the sides $B C, C A$ and $A B$ respectively of $\triangle A B C$. Then $\triangle D E F$ is congruent to triangle
→The distance of the point $P(4, 3)$ from the origin is:
→The runs scored by $11$ members of a cricket term are: $15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0.$ The median score is
→Axioms are assumed:
$(125)^{-1 / 3}=?$
→The Diagonals $AC$ and $BD$ of a Parallelogram $ABCD$ intersect each other at the point $O$ such that $\angle\text{DAC}=30^\circ$ and $\angle\text{AOB}=30^\circ.$ Then, $\angle\text{DBC}\ ?$
→Write the correct answer in the following: In a frequency distribution, the mid value of a class is $10$ and the width of the class is. The lower limit of the class is:
→If $x + 3$ is a factor of $x^2 - ax - 15, $ then $a =$
→In Fig. the quadrilateral $ABCD$ circumscribes a circle with centre $O$. If $\angle\text{AOB} = 115^\circ,$ then find $\angle\text{COD}.$
→