5 Marks Questions

Question 15 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%.

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Question 25 Marks

A cloth having an area of $165\ m^2$ is shaped into the form of a conical tent of radius $5\ m$

How many students can sit in the tent if a student, on an average, occupies $\frac{5}{7}\text{ m}^2$ on the ground?
Find the volume of the cone.

Answer

Given, radius of the base of a conical tent $= 5\ m$

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 $241.7\ m^3$ 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, $\text{number of student}=\frac{\text{Area of the base of a conical tent}}{\text{Area needs to sit a student on ground}}$
$=\frac{\frac{22\times5\times5\text{m}^2}{7}}{\frac{5}{7}}$
$=\frac{22}{7}\times5\times5\times\frac{7}{5}$
$=110$
Hence$, 110$ students can sit in the conical tent. Given, area of the cloth to from a conical tent $= 165\ m^2$

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Question 35 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 $= 64 : 27$
$\frac{\text{V}_1}{\text{V}_2}=\frac{64}{27}$
$\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2{}}=\frac{64}{27}$
$\Rightarrow\ \ \ \Big(\frac{\text{r}_1}{\text{r}_2}\Big)^3$
$=\Big(\frac{4}{3}\Big)^3\ \Big[\because$ 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 $16 : 9.$

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