3 Marks Question

Question 13 Marks

The trunk of a tree is cylindrical and its circumference is $176\ cm$. If the length of the tree is $3\ m$. Find the volume of the timber that can be obtained from the trunk.

Answer

We know that, circumference $=2\pi\text{r}$
$\Rightarrow176=2\pi\text{r}$
$\Rightarrow\text{r}=\frac{176}{\text{2}\pi}$
$\Rightarrow\text{r}=\frac{176\times7}{2\times22}$
$\Rightarrow\text{r}=28\text{cm}$
Here, height $(h) = 3m = 300\ cm$
Volume of timber $= r^2\times h$
$=\frac{22}{7}\times28\times28\times300$
$=44\times8400=739200\text{cm}^3 $
$\text{(or)}\ 0.7392\text{m}^3$

View full question & answer→

Question 23 Marks

The inner diameter of a cylindrical wooden pipe is $24\ cm$ and its outer diameter is $28\ cm$. The length of the pipe is $35\ cm$. Find the mass of the pipe, if $1cm^3$ of wood has a mass of $0.6\ gm$.

Answer

Given, Inner radius ($r_1$) of a cylindrical pipe $=\frac{24}{2}=12\text{cm}$
Outer radius ($r_2$) of a cylindrical pipe $=\frac{24}{2}=14\text{cm}$
Height of pipe $(h)$ = length of pipe $= 35\ cm$ Mass of pipe = Volume $\times $ density
$=\pi\big(\text{r}_2^2-\text{r}_1^2\big)\text{h}$
$=\frac{22}{7}(142 - 122)35 = 5720\text{cm}^3$
Mass of $1cm^3$ wood $= 0.6\ gm$
Therefore, mass of $5720cm^3$
wood $= 5720 \times 0.6 = 3432gm = 3.432kg$

View full question & answer→

Question 33 Marks

The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius $20\ cm$ and height $10\ m$. how much concrete mixture would be required to build $14$ such pillars?

Answer

Given,
The concrete mixture is used to build up the pillars is required for the entire space of the pillar i.e, we need to find the volume of the cylinders.
Radius of the base of a cylinder $= 20\ cm$
Volume of the cylindrical pillar $=\pi\text{R}^2\text{H}$
$=\big(\frac{22}{7}\times20^2\times1000\Big)\text{cm}^3$
$=\frac{8800000}{7}\text{cm}^3$
$=\frac{8.8}{7}\text{m}^3\ [1\text{m}=100\text{cm}]$
Therefore, volume of $14$ pillars $=\frac{8.8}{7}\times14\text{m}^3=17.6\text{m}^3$

View full question & answer→

Question 43 Marks

The cost of painting the total outside surface of a closed cylindrical oil tank at $50$ paise per square decimetre is $₹ 198$. The height of the tank is $6$ times the radius of the base of the tank. Find the volume corrected to $2$ decimal places.

Answer

Let the radius of the tank be rdm
Then, height $=6 \mathrm{rdm}$
Cost of painting for $50$ paisa per $\mathrm{dm}^2=₹ 198$
$\Rightarrow 2 \pi \mathrm{r}(\mathrm{r}+\mathrm{h})=198$
$\Rightarrow 2 \times \frac{22}{7} \times \mathrm{r}(\mathrm{r}+6 \mathrm{r}) \times \frac{1}{2}=198$
$\Rightarrow \mathrm{r}=3 \mathrm{dm}$
Therefore, $\mathrm{h}=(6 \times 3) \mathrm{dm}=18 \mathrm{dm}$
Volume of the tank $=\pi \mathrm{r}^2 \mathrm{~h}=\frac{22}{7} \times 9 \times 18=509.14 \mathrm{dm}^3$

View full question & answer→

Question 53 Marks

The capacity of a closed cylindrical vessel of height $1m$ is $15.4$ liters. How many square meters of the metal sheet would be needed to make it?

Answer

Given, Height of the cylindrical vessel $= 15.4$ litres $= 0.0154m^3 [1m^3 = 1000$ litres$]$
Let the radius of the circular ends of the cylinders be $'r'$ $\pi\text{r}^2\text{h}=0.0154\text{m}^3$
$\text{r}=0.07\text{m}$
$[\pi=31.4, \text{h}=1\text{m}]$
Total surface area of a vessel $=2\pi\text{r(r}+\text{h})$
$= 2(3.14(0.07)(0.07 +1))m^2= 0.4703m^2$

View full question & answer→

Question 63 Marks

A rectangular sheet of paper, $44\ cm \times 20\ cm$, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.

Answer

Given, the dimensions of the sheet are $44\ cm \times 20\ cm$
Here, length $= 44\ cm$ Height $= 20\ cm$ $2\pi\text{r}=44$
$\text{r}=\frac{44}{2}\pi$ $\text{r}=\frac{44}{2}\pi$
$\text{r}=\frac{44\times7}{2\times22}$
$\text{r}=7\text{cm}$ Volume of cylinder $= r^2\times h$
$=\frac{22}{7}\times7\times7\times20$
$=154\times20=3080\text{cm}^3$

View full question & answer→

Question 73 Marks

If the lateral surface of a cylinder is $94.2 \ cm^2$ and its height is $5\ cm$ , find: $[$Use $\pi=3.141]$
$i.$ radius of its base.
$ii.$ volume of the cylinder.

Answer

$i.$ Given,
Height of the cylinder $= 5\ cm$
Let radius of cylinder be $'r\ '$
Curved surface of the cylinder $= 94.2\ cm^2$
$2\pi\text{rh}=94.2\text{ cm}^2$
$\text{r}=3\text{ cm}\ [\pi=31.4,\ \text{h}=5\text{ cm}]$
$ii.$ Volume of the cylinder $=\pi\text{r}^2\text{h} $
$=(3.14\times32\times5)\text{ cm}^3$
$=141.3\text{ cm}^3$

View full question & answer→

Question 83 Marks

Write the ratio of total surface area to the curved surface area of a cylinder of radius $r$ and height $h$.

Answer

Total Surface Area of a cylinder $=2\pi\text{rh}+2\pi\text{r}^2$
Curved Surface Area of a cylinder $=2\pi\text{rh}$
Ratio of Total Surface Area $(TSA)$ to Curved Surface Area $(CSA)$ is given by,
$\frac{\text{TSA}}{\text{CSA}}=\frac{2\pi\text{rh}+2\pi\text{r}^2}{2\pi\text{rh}}$
$=\frac{2\pi\text{r}(\text{h}+\text{r})}{2\pi\text{rh}}$
$=\frac{\text{h}+\text{r}}{\text{h}}$
The ratio of Total Surface Area to Curved Surface Area is $(h + r) : h$

View full question & answer→

Question 93 Marks

If the radii of two cylinder are in the ratio $2 : 3$ and their heights are in the ratio $5 : 3$, then find the ratio of their volumes.

Answer

Let $r_1$ and $r_2$ be the radii of the two cylinders respectively and $h_1$and $h_2$ are the heights of the two cylinders respectively.
It is given that $r_1 : r_2 = 2 : 3$ and $h_1 : h_2 = 5 : 3$
We are asked to find the ratio of the volumes of the two cylinders
Now, $\frac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}}=\frac{\pi\text{r}^2_1\text{h}_1}{\pi\text{r}^2_2\text{h}_2}$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$
$=\Big(\frac{2}{3}\Big)^2\Big(\frac{5}{3}\Big)$
$=\frac{20}{27}$ Therefore the ratio of the volumes of the two cylinders is $20 : 27$.

View full question & answer→

Maths 3 Marks Question

Test yourself on this topic