Maths True False[1 Marks ]

Solution:

Let radius of hemisphere is r.

Volume of a cone, $\text{V}_1=\frac{1}{3}\pi\text{r}^2\text{h}$

$\text{V}_1=\frac{1}{3}\pi\text{r}^2(\text{r})\ \ [\therefore\text{h=r}]$

$=\frac{1}{3}\pi\text{r}^3$

Volume of a hemisphere, $\text{V}_2=\frac{2}{3}\pi\text{r}^3$

Volume of cylinder, $\text{V}_3=\pi\text{r}^2\text{h}=\pi\text{r}^2\times\text{r}=\pi\text{r}^3\ \ [\therefore\text{h}=\text{r}]$

$\text{V}_1:\text{V}_2:\text{V}_3=\frac{1}{2}\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3=1:2:3$

Hence, the ratio of their volumes is 1 : 2 : 3.

Solution:

The height of the largest cone is 2r that can be fitted in a cube whose edge is 2r.

Its volume $=\frac{1}{3}\pi\text{r}^2(2\text{r})=\frac{2}{3}\pi\text{r}^3$

But $\frac{2}{3}\pi\text{r}^3$ is the volume of a hemisphere of radius r.

Hence, the given statement is true.

Solution:

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

Now, when radius of a height circular cone is halved and height is doubled, than

$\text{V}=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\times2\text{h}=\frac{1}{3}\pi\times\frac{\text{r}^2}{4}\times2\text{h}=\frac{1}{2}\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)$

Solution:

Solution:

Solution:

Let radius of hemisphere is r.

Volume of a cone, $\text{V}_1=\frac{1}{3}\pi\text{r}^2\text{h}$

$\text{V}_1=\frac{1}{3}\pi\text{r}^2(\text{r})$

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

and Volume of cylinder, $\text{V}_3=\pi\text{r}^2\text{h}=\pi\text{r}^2\times\text{r}=\pi\text{r}^3$

Hence, the ratio of their volumes is 1 : 2 : 3.

Solution:

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

$\therefore$ Original volume $(\text{V}_1)=\pi\text{r}^2\text{h}.$

When radius of cylinder is doubled and height is halved

$\text{(V)}_2=\pi(2\text{r})^2\times\frac{\text{h}}{2}=\pi\times4\text{r}^2\times\frac{\text{h}}{2}=2\pi\text{r}^2\text{h}=2\text{V}_1$

Hence, the given statement is true.

Solution:

Let r the radius and h be hight of the cylinder and a right circular cone.

Now, volume of cylinder $=\pi\text{r}^2\text{h}$

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

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

$\Rightarrow3\pi\text{r}^2\text{h}=\pi\text{r}^2\text{h}$

Therefore, volume of cylinder = 3(volume of cone)

Hence, the given statement is true.

Solution:

Let the length of the edge of the cube be a

Then the diagonal of the cube is $\sqrt{3\text{a}}$

Also, the length of the diagonal of a cube is $6\sqrt{3}.$

$\therefore\sqrt{3\text{a}}=6\sqrt{3}\Rightarrow\text{a}=6\text{cm}$

So. the length of the edge of the cube is 6cm.

Hence, the given statement is false.

Solution:

Volume of cube $(\text{V}_1)=(\text{Side})^ 3$

Radius of sphere $(\text{r})=\frac{\text{Side}}{2}$

Volume of sphere $(\text{V}_1)=\frac{4}{3}\pi(\text{r}^3)$

$=\frac{4}{3}\pi\Big(\frac{\text{Side}}{2}\Big)^3=\frac{4}{3}\pi\times\frac{(\text{Side})^3}{8}$

$\frac{\text{V}_1}{\text{V}_2}=\frac{(\text{Side})^3}{\frac{4}{3}\pi\frac{\text{(Side)}^3}{8}}=\frac{6}{\pi}\text{ or }6:\pi$

Hence, the ratio of the volume of the cube to the volume of the sphere is $6:\pi.$