Maths M.C.Q

4. 1 : 3

Solution:

Let the height of a circular cylinder and a right circular cone be h cm and H cm respectively.

Since a right circular and a right circular cone have the same radius and the same volume,

$\Rightarrow\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^2\text{H}$

$\Rightarrow\text{h}=\frac{1}{3}\text{H}$

$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{1}{3}$

⇒ Ratio of the height is 1 : 3.

4. $ 10\sqrt{3}\text{cm}$

Solution:

The length of the longest rod = length of the diagonal

$=\sqrt{3}\text{a}$

$=\sqrt{3}\times10$

$=10\sqrt{3}\text{cm}$

3. 220m

Solution:

The amount of cloth required to make a tent is equal to the curved surface area of a cone.

$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$

$\text{l}=\sqrt{7^2+24^2}$

$\text{l}=\sqrt{49+576}$

$\text{l}=25\text{m}$

Curved surface area of the cone $=\pi\text{rl}$

$=\frac{22}{7}\times7\times25$

$=550\text{m}^2$

Length of the cloth $=\frac{\text{area}}{\text{width}}$

$=\frac{550}{2.5}=220\text{m}$

2. 2.1cm

Solution:

Let the radius of the sphere be r cm.

Since the cone is melted and recast into a sphere,

Volume of the sphere = volume of the cone

$\Rightarrow\frac{4}{3}\pi\times\text{r}^3=\frac{1}{3}\pi\times(2.1)^2\times8.4$

$\Rightarrow\text{r}^3=2.1\times2.1\times2.1$

$\Rightarrow\text{r}=2.1\text{cm}$

1. 6.3cm

Solution:

Let the radius of the base be r cm.

Since the sphere is melted and cast into a cone,

Volume of the sphere = volume of the cone

$\Rightarrow\frac{4}{3}\pi(6.3)^3=\frac{1}{3}\pi\text{r}^2(25.2)$

$\Rightarrow4(6.3)^3=\text{r}^2(25.2)$

$\Rightarrow\frac{4\times(6.3)^3}{(25.2)}=\text{r}^2$

$\Rightarrow\text{r}=6.3\text{cm}$

2. $\sqrt2:1$

Solution:

Let their heights be x cm and 2x cm respectively and let their radii be R₁ and R₂ respectively.

$\Rightarrow\pi\text{R}_1^2\text{h}=\pi\text{R}_2^2\text{h}$

$\Rightarrow\pi\times\text{R}_1^2\times\text{x}=\pi\times\text{R}_2^2\times2\text{x}$

$\Rightarrow\text{R}_1^2=\text{R}_2^2\times2$

$\Rightarrow\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2=2$

$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac{\sqrt2}{1}$

⇒ Ratio of their radii is $\sqrt2:1$

4. $(144\pi)\text{cm}^3$

Solution:

The height of the cone is 12cm and the radius of its base is 6cm.

Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$

$=\frac{1}{3}\times\pi\times6\times6\times12$

$=144\pi\text{cm}^3$

4. 8 times.

Solution:

The volume of a cone of height h and radius r $=\frac{1}{3}\pi\text{r}^2\text{h}=\text{v}$

Since the height and the radius of cone are doubled,

New height = 2h and new radius = 2r

⇒ New volume $=\frac{1}{3}\pi(2\text{r})^2\times2\text{h}$

$=\frac{1}{3}\pi\times4\text{r}^2\times2\text{h}$

$=8\times\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)$

$=8\text{v}$

3. 1 : 9

Solution:

$\frac{\text{Volume}\ \text{of}\ \text{cube}\ 1}{\text{Volume}\ \text{of}\ \text{cube}\ 2}=\frac{\text{a}^3}{\text{b}^3}$

$=\frac{1}{27}$

$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^3=\Big(\frac{1}{3}\Big)^3$

$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{1}{3}$

$\frac{\text{a}^2}{\text{b}^2}=\Big(\frac{\text{a}}{\text{b}}\Big)^2$

$=\Big(\frac{1}{3}\Big)^2$

$=\frac{1}{9}$

Surface area $=\frac{6\text{a}^2}{6\text{b}^2}$

$=\frac{6\times1}{6\times9}$

$=\frac{1}{9}$

$\therefore$ Ratio of their surface areas = 1 : 9.

3. 10 : 9

Solution:

Let r₁ and r₂ be the radii and h₁ and h₂ be the height of two cylinders.

$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}\ \text{and}\ \frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$

Ratio of the surface area $=\frac{2\pi\text{r}_1\text{h}_1}{2\pi\text{r}_2\text{h}_2}$

$=\frac{\text{r}_1}{\text{r}_2}\times\frac{\text{h}_1}{\text{h}_2} $

$=\frac{2}{3}\times\frac{5}{3}$

$=\frac{10}{9}$

$=10:9$

3. 3 unit

Solution:

Volume of the sphere and the surface area of the sphere are numerically the same,

$\Rightarrow\frac{4}{3}\pi\text{r}^3=4\pi\text{r}^2$

$\Rightarrow\text{r}=3\ \text{units}$

1. 15m

​​​​​​​Solution:

Length of the longest rod = length of the diagonal

$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$

$=\sqrt{10^2+10^2+5^2}$

$=\sqrt{100+100+25}$

$=\sqrt{225}$

$=15\text{m}$

1. 5544cm²

Solution:

Volume of sphere = 38808cm³

$\Rightarrow\frac{4}{3}\pi\text{r}^3=38808$

$\Rightarrow\frac{4}{3}\times\frac{22}{7}\times\text{r}^3=38808$

$\Rightarrow\text{r}^3=\frac{38808\times7\times3}{88}$

$\Rightarrow\text{r}^3=441\times21$

$\Rightarrow\text{r}^3=(21)^3$

$\Rightarrow\text{r}=21\text{cm}$

Curved surface area of a sphere $=4\pi\text{r}^2$

$=4\times\frac{22}{7}\times21\times21$

$=4\times22\times3\times21$

$=5544\text{cm}^2$

1. 100%

Solution:

Let the original height of the cone be h and the radius be r.

Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$

New height is 2h and the radius is the same

So, the new volume of the cone $=\frac{1}{3}\pi\text{r}^2(2\text{h})=\frac{2}{3}\pi\text{r}^2\text{h}$

Increase in the volume $=\frac{2}{3}\pi\text{r}^2\text{h}-\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^2\text{h}$

Increase $\%=\frac{\text{Increase}}{\text{Original}\ \text{volume}}\times100$

⇒ Increase $=\frac{\frac{1}{3}\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}\times100$

⇒ Increase $\%=100$

1. $\frac{32\pi\text{r}^3}{3}$

Solution:

Volume of sphere $=\frac{4}{3}\pi\text{r}^3$

$=\frac{4}{3}\pi\times(2\text{r})^3$

$=\frac{4}{3}\pi\times8\text{r}^3$

$=\frac{32\pi\text{r}^3}{3}$

3. $2\pi\text{rh}$

Solution:

The lateral surface area of a cylinder is given to be $2\pi\text{rh}.$

3. 2.5cm

Solution:

Let the radius of the third ball be r cm.

Volume of the spherical ball = sum of the volume of the three balls

$\Rightarrow\frac{4}{3}\pi(3)^3=\frac{4}{3}\pi(1.5)^3+\frac{4}{3}\pi(2)^3+\frac{4}{3}\pi\text{r}^3$

$\Rightarrow(3)^3=(1.5)^3+(2)^3+\text{r}^3$

$\Rightarrow\text{r}^3=15.625$

$\Rightarrow\text{r}=2.5\text{cm}$

4. 25 : 64

Solution:

Let the radii of the cones be 4x cm and 5x cm respectively.

Let their be h cm and H cm respectively.

⇒ Ratio of the volume of the cones $=\frac{\frac{1}{3}\pi\times(4\text{x}^2\times\text{h})}{\frac{1}{3}\pi\times(5\text{x})^2\times\text{H}}$

$\Rightarrow\frac{\text{V}}{4\text{v}}=\frac{\frac{1}{3}\pi\times(4\text{x})^2\times\text{h}}{\frac{1}{3}\pi\times(5\text{x})^2\times\text{H}}$

$\Rightarrow\frac{1}{4}=\frac{16\text{h}}{25\text{H}}$

$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{1}{4}\times\frac{25}{16}$

$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{25}{64}$

$\Rightarrow\text{h}:\text{H}=25:64$

3. 36m

Solution:

Given that the diameter of the sphere is 6cm.

So, the radius of the sphere is 3cm.

Diameter of the wire = 2mm = 1mm = 0.1cm

Since the sphere is drawn into a wire,

Volume of the sphere = volume of the wire

$\frac{4}{3}\pi(3)^3=\pi(0.1)^2\text{h}$

$\Rightarrow36\pi=\pi(0.1)^2\text{h}$

$\Rightarrow\text{h}=3600\text{cm}=36\text{m}$

4. 9 times.

Solution:

Let the radius of the wire be r and the height be h.

So, the new radius $=\frac{1}{3}\text{r}$

Let the new height be H.

Note that the height of the wire is the same as the length of the wire.

Given that the volume remains the same.

So, $\pi\text{r}^2\text{h}=\pi\Big(\frac{1}{3}\text{r}\Big)^2\text{H}$

$\Rightarrow\text{r}^2\text{h}=\frac{1}{9}\text{r}^2\text{H}$

$\Rightarrow\text{h}=\frac{1}{9}\text{H}$

$\Rightarrow\text{H}=9\text{h}$

Thus, the length will become 9 times the original length.

4. 11.2cm

​​​​​​​Solution:

Given: $\sqrt{5}=2.24$

Required length $=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$

$=\sqrt{8^2+6^2+5^2}$

$=\sqrt{64+36+25}$

$=\sqrt{125}$

$=5\sqrt{5}$

$=2\times2.24$

$=11.2\text{cm}$