True False[1 Marks ]

Question 11 Mark

The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r.

Answer

True.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.

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Question 21 Mark

If the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged.

Answer

False.Solution:
Let the original radius of the cone be r and height be h.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
$\text{V}=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\times2\text{h}=\frac{1}{3}\pi\times\frac{\text{r}^2}{4}\times2\text{h}=\frac{1}{2}\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)$
We see that the volume becomes half of the original volume. Hence, the given statement is false.

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Question 31 Mark

In a right circular cone, height, radius and slant height do not always be sides of a right triangle.

Answer

True.Solution:
On rotating a right-angled triangular lamina AOB about OA, it generates a cone. The point A is the vertex of a cone. Its base is a circle with centre O and radius OB. The length OA is the height of the cone and the length AB is called its slant height.

Clearly, $\angle\text{AOB}=90^\circ$ Let the radius of the base = r unit, height = h units and slant height = l unit, then $\text{l}^2=\text{h}^2+\text{r}^2\Rightarrow\text{l}=\sqrt{\text{h}^2+\text{r}^2}$

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Question 41 Mark

The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

Answer

True.Solution:
Let the radius of the sphere and cylinder be r. Given, height of the cylinder = diameter of the base ⇒ h = 2r According to the given condition, Volume of sphere $=\Big(\frac{2}{3}\Big)\times$ Volume of cylinder $\frac{4}{3}\pi\text{r}^3=\Big(\frac{4}{3}\Big)\times\pi\text{r}^2\times2\text{r}$ $\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\text{r}^3$ Hence, the volume of a sphere is equal to two-third of the volume of a cylinder.

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Question 51 Mark

If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

Answer

True.Solution:
Let radius of hemisphere is r.
Volume of a cone, $\text{V}_1=\frac{1}{3}\pi\text{r}^2\text{h}$
$\text{V}_1=\frac{1}{3}\pi\text{r}^2(\text{r})$
$\Rightarrow =\frac{1}{3}\pi\text{r}^2$
Now, Volume of a hemisphere, $\text{V}_2=\frac{2}{3}\pi\text{r}^3$and Volume of cylinder, $\text{V}_3=\pi\text{r}^2\text{h}=\pi\text{r}^2\times\text{r}=\pi\text{r}^3$
Therefore $\text{V}_1:\text{V}_2:\text{V}_3=\frac{1}{3}\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3:\pi\text{r}^3=1:2:3$Hence, the ratio of their volumes is 1 : 2 : 3.

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Question 61 Mark

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.

Answer

True.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.

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Question 71 Mark

If the length of the diagonal of a cube is $6\sqrt{3}\text{cm},$ then the length of the edge of the cube is 3cm.

Answer

False.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.

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Question 81 Mark

If the radius of a cylinder is doubled and height is halved, the volume will be doubled.

Answer

True.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.

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Question 91 Mark

A cylinder and a right circular cone are having the same base and same height. The volume of the cylinder is three times the volume of the cone.

Answer

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.

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Question 101 Mark

If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be $6:\pi.$

Answer

True.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 $6:\pi.$

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MATHS True False[1 Marks ]

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