5 Marks Questions

Question 15 Marks

The volumes of the two spheres are in the ratio $64 : 27$. Find the ratio of their surface areas.

Answer

Let the radius of two spheres be $r_1$ and $r_2$. Given, the ratio of the volume of two spheres = $64 : 27$
$\frac{\text{V}_1}{\text{V}_2}=\frac{64}{27}\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2{}}=\frac{64}{27}$
$\Rightarrow\ \ \ \Big(\frac{\text{r}_1}{\text{r}_2}\Big)^3=\Big(\frac{4}{3}\Big)^3\ \Big[\because\text{volume of sphere}=\frac{4}{3}\pi\text{r}^3\Big]$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{3}$
Let the surface areas of the two spheres be $S_1$ and $S_2$.
$\therefore\ \ \ \frac{\text{S}_1}{\text{S}_2}=\frac{4\pi\text{r}^2_1}{4\pi\text{r}^2_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\ \ \ \ \Rightarrow\text{S}_1:\text{S}_2=\Big(\frac{4}{3}\Big)^2=\frac{16}{9}$
$\Rightarrow\ \ \ \text{S}_1:\text{S}_2=16:9$
Hence, the ratio of the their surface areas is $16 : 9.$

View full question & answer→

Question 25 Marks

A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceed its height$?$

Answer

Let the radius of sphere $= r =$ Radius of a right circular cylinderAccording to the question,
Volume of cylider $=$ Volume of a sphere
$\Rightarrow\ \ \pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{r}^3 \ \ \ \Rightarrow\ \ \ \text{h}=\frac{4}{3}\text{r}$
$\because$ Diameter of the cylinder $= 2r$
$\therefore$ Increased diameter from hight of the cylinder $=2\text{r}-\frac{4\text{r}}{3}=\frac{2\text{r}}{3}$
Now, percentage increase in diameter of the cylinder $=\frac{\frac{2\text{r}}{3}\times100}{\frac{4}{3}\text{r}}50\%$
Hence, the diameter of the cylinder exceeds its height by $50\%.$

View full question & answer→

Question 35 Marks

A cloth having an area of $165\ m^2$ is shaped into the form of a conical tent of radius $5\ m$
$i.$ How many students can sit in the tent if a student, on an average, occupies $\frac{5}{7}\text{m}^2$ on the ground?
$ii.$ Find the volume of the cone.

Answer

$i.$ Given, radius of the base of a conical tent $= 5\ m$
And area needs to sit a student on the ground $=\frac{5}{7}\text{m}^2$
$\therefore$ Area of the base of a conical tent $=\pi\text{r}^2$
$=\frac{22}{7}\times5\times5\text{ m}^2$
Now, number of student$=\frac{\text{Area of the base of a conical tent}}{\text{Area needs to sit a student on ground}}$
$=\frac{\frac{22\times5\times5\text{m}^2}{7}}{\frac{5}{7}}=\frac{22}{7}\times5\times5\times\frac{7}{5}=110$
Hence, $110$ students can sit in the conical tent.
$ii.$ Given, area of the cloth to from a conical tent $= 165\ m^2$
Radius of the base of a conical tent, $r = 5\ m$
Curved surface area of a conical tent $=$ Area of cloth to from a conical tent
$\Rightarrow\pi\text{rl}=165$
$\Rightarrow\frac{22}{7}\times5\times\text{l}=165$
$\therefore \text{l}=\frac{165\times7}{22\times5}=\frac{33\times7}{22}=10.5\text{ m}$
Now, height of a conical tent $=\sqrt{\text{l}^2-\text{r}^2}=\sqrt{(10.5)^2-(5)^2}$
$=\sqrt{110.25-25}=\sqrt{8525}=9.23\text{ m}$
Volume of a cone (conical tent) $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\times\frac{22}{7}\times5\times5\times923$
$=\frac{1}{3}\times\frac{1550\times923}{7}=\frac{50765}{7\times3}=241.7\text{ m}^3$
Hence, the volume of the cone $($conical tent$)$ is $241.7\ m^3$

View full question & answer→

Maths 5 Marks Questions

Test yourself on this topic