Chapter 13 Surface Areas And Volumes

Let the initial radius of the cylinder be $r_1$ and the initial height be $h_1$.

The formula for the volume of a cylinder with radius $r$ and height $h$ is given by:

$V = \pi r^2 h$

The initial volume of the cylinder is therefore:

$V_1 = \pi r_1^2 h_1$

According to the problem statement, the radius is halved and the height is doubled.

So, the new radius, $r_2$, is half of the initial radius:

$r_2 = \frac{r_1}{2}$

And the new height, $h_2$, is double the initial height:

$h_2 = 2h_1$

The new volume of the cylinder, $V_2$, is calculated using the new radius and height:

$V_2 = \pi r_2^2 h_2$

Substitute the expressions for $r_2$ and $h_2$ in terms of $r_1$ and $h_1$ into this equation:

$V_2 = \pi \left(\frac{r_1}{2}\right)^2 (2h_1)$

$V_2 = \pi \left(\frac{r_1^2}{4}\right) (2h_1)$

Rearrange the terms:

$V_2 = \pi r_1^2 h_1 \left(\frac{2}{4}\right)$

$V_2 = \pi r_1^2 h_1 \left(\frac{1}{2}\right)$

Now, observe that the term $\pi r_1^2 h_1$ is the initial volume, $V_1$. So, we can write:

Equation (i) shows that the new volume ($V_2$) is half of the initial volume ($V_1$).

Thus, if the radius is halved and the height is doubled, the volume of the cylinder will be halved.

The correct option is (C) halved.

Given:

Radius of the sphere = $2r$

To Find:

The volume of the sphere.

Solution:

The formula for the volume of a sphere with radius $R$ is given by:

$V = \frac{4}{3}\pi R^3$

In this problem, the radius of the sphere is given as $2r$. So, we substitute $R = 2r$ into the volume formula.

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

Now, we simplify the term $(2r)^3$:

$(2r)^3 = 2^3 \times r^3 = 8r^3$

Substitute this back into the volume formula:

$V = \frac{4}{3}\pi (8r^3)$

Multiply the terms:

$V = \frac{4 \times 8}{3}\pi r^3$

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

Comparing this result with the given options, we find that the volume of the sphere with radius $2r$ is $\frac{32}{3}\pi r^3$.

The correct option is (D) $\frac{32}{3}\pi r^3$.

Given:

Total surface area (TSA) of the cube = $96$ cm$^2$.

To Find:

The volume of the cube.

Solution:

Let the side length of the cube be $a$ cm.

The formula for the total surface area (TSA) of a cube is $6a^2$.

We are given that the TSA is $96$ cm$^2$. So, we can write the equation:

To find the side length $a$, we first isolate $a^2$:

$a^2 = \frac{96}{6}$

Divide $96$ by $6$:

$a^2 = 16$

Now, take the square root of both sides to find $a$:

$a = \sqrt{16}$

Since the side length must be a positive value:

$a = 4$ cm

So, the side length of the cube is $4$ cm.

The formula for the volume of a cube with side length $a$ is $V = a^3$.

Substitute the value of $a = 4$ cm into the volume formula:

$V = (4 \text{ cm})^3$

$V = 4 \times 4 \times 4 \text{ cm}^3$

$V = 16 \times 4 \text{ cm}^3$

$V = 64 \text{ cm}^3$

Thus, the volume of the cube is $64$ cm$^3$.

Comparing this result with the given options, we find that the correct volume is $64$ cm$^3$.

The correct option is (C) 64 cm³.

Given:

Height of the cone, $h = 8.4$ cm

Radius of the base of the cone, $r = 2.1$ cm

To Find:

The radius of the sphere formed by melting the cone.

Solution:

The volume of a cone with radius $r$ and height $h$ is given by the formula:

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

Substitute the given values of the radius and height of the cone:

$V_{\text{cone}} = \frac{1}{3}\pi (2.1 \text{ cm})^2 (8.4 \text{ cm})$

$V_{\text{cone}} = \frac{1}{3}\pi (4.41 \text{ cm}^2) (8.4 \text{ cm})$

Now, we calculate the numerical value:

$V_{\text{cone}} = \frac{1}{3} \pi (4.41 \times 8.4) \text{ cm}^3$

$V_{\text{cone}} = \frac{1}{3} \pi (37.044) \text{ cm}^3$

$V_{\text{cone}} = 12.348\pi \text{ cm}^3$

When the cone is melted and recast into a sphere, the volume of the material remains the same.

Let the radius of the sphere be $R$ cm.

The volume of a sphere with radius $R$ is given by the formula:

$V_{\text{sphere}} = \frac{4}{3}\pi R^3$

Since the volume is conserved during the transformation:

Equate the expressions for the volumes:

$\frac{4}{3}\pi R^3 = 12.348\pi$

Divide both sides of the equation by $\pi$ (since $\pi \neq 0$):

$\frac{4}{3} R^3 = 12.348$

Now, solve for $R^3$ by multiplying both sides by $\frac{3}{4}$:

$R^3 = 12.348 \times \frac{3}{4}$

$R^3 = \frac{37.044}{4}$

$R^3 = 9.261$

To find the radius $R$, we take the cube root of $9.261$:

$R = \sqrt[3]{9.261}$

By calculating or recognizing the cube root:

$2.1 \times 2.1 = 4.41$

$4.41 \times 2.1 = 9.261$

So,

The radius of the sphere is $2.1$ cm.

Comparing our result with the given options, the correct radius is $2.1$ cm.

The correct option is (B) 2.1 cm.

Given:

A cylinder with an initial radius and height.

To Find:

How the curved surface area (CSA) changes when the radius is doubled and the height is halved.

Solution:

Let the initial radius of the cylinder be $r_1$ and the initial height be $h_1$.

The formula for the curved surface area (CSA) of a cylinder with radius $r$ and height $h$ is given by:

CSA $= 2\pi r h$

The initial curved surface area of the cylinder is therefore:

CSA$_1 = 2\pi r_1 h_1$

According to the problem statement, the radius is doubled and the height is halved.

So, the new radius, $r_2$, is double the initial radius:

$r_2 = 2r_1$

And the new height, $h_2$, is half of the initial height:

$h_2 = \frac{h_1}{2}$

The new curved surface area of the cylinder, CSA$_2$, is calculated using the new radius and height:

CSA$_2 = 2\pi r_2 h_2$

Substitute the expressions for $r_2$ and $h_2$ in terms of $r_1$ and $h_1$ into this equation:

CSA$_2 = 2\pi (2r_1) \left(\frac{h_1}{2}\right)$

Rearrange the terms and perform the multiplication:

CSA$_2 = 2\pi (2 \times \frac{1}{2}) r_1 h_1$

CSA$_2 = 2\pi (1) r_1 h_1$

CSA$_2 = 2\pi r_1 h_1$

Now, observe that the expression $2\pi r_1 h_1$ is the initial curved surface area, CSA$_1$. So, we can write:

This shows that the new curved surface area (CSA$_2$) is equal to the initial curved surface area (CSA$_1$).

Thus, if the radius is doubled and the height is halved, the curved surface area of the cylinder will be same.

The correct option is (C) same.

Given:

Radius of the cone, $R = \frac{r}{2}$

Slant height of the cone, $L = 2l$

To Find:

The total surface area of the cone.

Solution:

The formula for the total surface area (TSA) of a cone with radius $R$ and slant height $L$ is given by:

TSA $= \pi R L + \pi R^2$

Substitute the given values of the radius and slant height into the formula:

TSA $= \pi \left(\frac{r}{2}\right) (2l) + \pi \left(\frac{r}{2}\right)^2$

Now, simplify the terms:

First term: $\pi \left(\frac{r}{2}\right) (2l) = \pi \left(\frac{r \times 2l}{2}\right) = \pi (rl)$

Second term: $\pi \left(\frac{r}{2}\right)^2 = \pi \left(\frac{r^2}{2^2}\right) = \pi \left(\frac{r^2}{4}\right)$

So, the total surface area is:

TSA $= \pi rl + \pi \frac{r^2}{4}$

To match the options, we can factor out $\pi r$ from the expression:

TSA $= \pi r \left(l + \frac{r}{4}\right)$

Comparing this result with the given options, we find that the total surface area of the cone is $\pi r \left(l + \frac{r}{4}\right)$.

The correct option is (B) $\pi r\left( l + \frac{r}{4} \right)$.

Given:

Ratio of radii of two cylinders = $2:3$

Ratio of heights of two cylinders = $5:3$

To Find:

The ratio of their volumes.

Solution:

Let the radii of the two cylinders be $r_1$ and $r_2$, and their heights be $h_1$ and $h_2$.

According to the given information:

$\frac{r_1}{r_2} = \frac{2}{3}$

$\frac{h_1}{h_2} = \frac{5}{3}$

The formula for the volume of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$.

The volume of the first cylinder is $V_1 = \pi r_1^2 h_1$.

The volume of the second cylinder is $V_2 = \pi r_2^2 h_2$.

We need to find the ratio of their volumes, which is $\frac{V_1}{V_2}$.

$\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}$

Cancel out $\pi$ from the numerator and denominator:

$\frac{V_1}{V_2} = \frac{r_1^2 h_1}{r_2^2 h_2}$

We can rewrite this as:

$\frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$

Substitute the given ratios into this equation:

$\frac{V_1}{V_2} = \left(\frac{2}{3}\right)^2 \times \left(\frac{5}{3}\right)$

Calculate the square of the radius ratio:

$\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$

Now substitute this back into the volume ratio equation:

$\frac{V_1}{V_2} = \frac{4}{9} \times \frac{5}{3}$

Multiply the fractions:

$\frac{V_1}{V_2} = \frac{4 \times 5}{9 \times 3}$

$\frac{V_1}{V_2} = \frac{20}{27}$

The ratio of their volumes is $20:27$.

Comparing our result with the given options, the correct ratio is $20:27$.

The correct option is (B) 20 : 27.

Given:

Lateral surface area (LSA) of the cube = $256$ m$^2$.

To Find:

The volume of the cube.

Solution:

Let the side length of the cube be $a$ meters.

The lateral surface area (LSA) of a cube is the sum of the areas of its four side faces. The area of each face is $a^2$.

The formula for the lateral surface area of a cube is $4a^2$.

We are given that the LSA is $256$ m$^2$. So, we can set up the equation:

To find the side length $a$, we first isolate $a^2$ by dividing both sides by 4:

$a^2 = \frac{256}{4}$

Performing the division:

$a^2 = 64$

Now, take the square root of both sides to find the value of $a$. Since $a$ represents a length, it must be positive.

$a = \sqrt{64}$

$a = 8$ meters

So, the side length of the cube is $8$ meters.

The formula for the volume of a cube with side length $a$ is $V = a^3$.

Substitute the value of $a = 8$ meters into the volume formula:

$V = (8 \text{ m})^3$

$V = 8 \times 8 \times 8 \text{ m}^3$

$V = 64 \times 8 \text{ m}^3$

$V = 512 \text{ m}^3$

Thus, the volume of the cube is $512$ m$^3$.

Comparing this result with the given options, we find that the correct volume is $512$ m$^3$.

The correct option is (A) 512 m³.

Given:

Dimensions of the pit: Length = $16$ m, Width = $12$ m, Depth = $4$ m.

Dimensions of each plank: Length = $4$ m, Width = $50$ cm, Thickness = $20$ cm.

To Find:

The number of planks that can be stored in the pit.

Solution:

First, we need to ensure all dimensions are in the same unit. Let's convert the dimensions of the plank from centimeters to meters. ($1$ m = $100$ cm)

Width of plank = $50$ cm $= \frac{50}{100}$ m $= 0.5$ m

Thickness of plank = $20$ cm $= \frac{20}{100}$ m $= 0.2$ m

Dimensions of the pit are: Length = $16$ m, Width = $12$ m, Depth = $4$ m.

Dimensions of each plank are: Length = $4$ m, Width = $0.5$ m, Thickness = $0.2$ m.

The pit and the planks are in the shape of cuboids. The volume of a cuboid is given by the formula:

$V = \text{Length} \times \text{Width} \times \text{Height}$

Calculate the volume of the pit ($V_{\text{pit}}$):

$V_{\text{pit}} = 16 \text{ m} \times 12 \text{ m} \times 4 \text{ m}$

$V_{\text{pit}} = (16 \times 12 \times 4) \text{ m}^3$

$16 \times 12 = 192$

$192 \times 4 = 768$

Calculate the volume of one plank ($V_{\text{plank}}$):

$V_{\text{plank}} = 4 \text{ m} \times 0.5 \text{ m} \times 0.2 \text{ m}$

$V_{\text{plank}} = (4 \times 0.5 \times 0.2) \text{ m}^3$

$4 \times 0.5 = 2.0$

$2.0 \times 0.2 = 0.4$

To find the number of planks that can be stored in the pit, divide the volume of the pit by the volume of one plank:

Number of planks $= \frac{V_{\text{pit}}}{V_{\text{plank}}}

Number of planks $= \frac{768 \text{ m}^3}{0.4 \text{ m}^3}$

To perform the division, we can remove the decimal by multiplying the numerator and denominator by 10:

Number of planks $= \frac{768 \times 10}{0.4 \times 10} = \frac{7680}{4}$

Now, perform the division:

$\frac{7680}{4} = 1920$

Number of planks $= 1920$

Thus, 1920 planks of the given dimensions can be stored in the pit.

Comparing our result with the given options, the correct number of planks is 1920.

The correct option is (B) 1920.

Given:

Dimensions of the room:

Length ($l$) = $10$ m

Width ($w$) = $10$ m

Height ($h$) = $5$ m

To Find:

The length of the longest pole that can be put in the room.

Solution:

The room is in the shape of a cuboid with length $l$, width $w$, and height $h$. The longest pole that can be placed in a cuboid room is along its space diagonal.

The formula for the length of the space diagonal ($d$) of a cuboid is given by:

$d = \sqrt{l^2 + w^2 + h^2}$

Substitute the given dimensions ($l=10$, $w=10$, $h=5$) into the formula:

$d = \sqrt{(10 \text{ m})^2 + (10 \text{ m})^2 + (5 \text{ m})^2}$

$d = \sqrt{100 \text{ m}^2 + 100 \text{ m}^2 + 25 \text{ m}^2}$

$d = \sqrt{(100 + 100 + 25) \text{ m}^2}$

$d = \sqrt{225 \text{ m}^2}$

Calculate the square root of 225:

$d = 15$ m

Thus, the length of the longest pole that can be put in the room is $15$ meters.

Comparing our result with the given options, the correct length is $15$ m.

The correct option is (A) 15 m.

Given:

Initial radius of the hemispherical balloon, $r_1 = 6$ cm

Final radius of the hemispherical balloon, $r_2 = 12$ cm

To Find:

The ratio of the surface areas of the balloon in the two cases (initial and final).

Solution:

The surface area of a hemispherical balloon, when inflated, refers to its outer surface. This outer surface is the curved surface area of a hemisphere.

The formula for the curved surface area (CSA) of a hemisphere with radius $r$ is given by:

CSA $= 2\pi r^2$

Calculate the initial curved surface area (CSA$_1$) when the radius is $r_1 = 6$ cm:

CSA$_1 = 2\pi (6 \text{ cm})^2$

CSA$_1 = 2\pi (36 \text{ cm}^2)$

CSA$_1 = 72\pi \text{ cm}^2$

Calculate the final curved surface area (CSA$_2$) when the radius is $r_2 = 12$ cm:

CSA$_2 = 2\pi (12 \text{ cm})^2$

CSA$_2 = 2\pi (144 \text{ cm}^2)$

CSA$_2 = 288\pi \text{ cm}^2$

Now, find the ratio of the initial surface area to the final surface area:

Ratio $= \frac{\text{CSA}_1}{\text{CSA}_2}$

Ratio $= \frac{72\pi \text{ cm}^2}{288\pi \text{ cm}^2}$

Cancel out $\pi$ and the unit cm$^2$ from the numerator and denominator:

Ratio $= \frac{72}{288}$

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 72:

$\frac{\cancel{72}^{1}}{\cancel{288}_{4}}$

Ratio $= \frac{1}{4}$

The ratio of the surface areas of the balloon in the two cases is $1:4$.

Comparing our result with the given options, the correct ratio is $1:4$.

The correct option is (A) 1 : 4.

The statement is True.

Justification:

Let the radius of the sphere be $r$.

The formula for the surface area of a sphere with radius $r$ is given by:

$A_{\text{sphere}} = 4\pi r^2$

Since the right circular cylinder just encloses the sphere, the dimensions of the cylinder are determined by the sphere's dimensions.

The height of the cylinder ($h$) must be equal to the diameter of the sphere, which is $2r$.

$h = 2r$

The radius of the base of the cylinder ($R$) must be equal to the radius of the sphere.

$R = r$

The formula for the curved surface area (CSA) of a cylinder with radius $R$ and height $h$ is given by:

CSA$_{\text{cylinder}} = 2\pi R h$

Substitute the dimensions of the cylinder in terms of $r$:

CSA$_{\text{cylinder}} = 2\pi (r) (2r)$

CSA$_{\text{cylinder}} = 2\pi (2r^2)$

CSA$_{\text{cylinder}} = 4\pi r^2$

Comparing the surface area of the sphere and the curved surface area of the cylinder:

$A_{\text{sphere}} = 4\pi r^2$

CSA$_{\text{cylinder}} = 4\pi r^2$

We see that $A_{\text{sphere}} = \text{CSA}_{\text{cylinder}}$.

Therefore, the surface area of the sphere is indeed equal to the curved surface area of the cylinder when the cylinder just encloses the sphere.

The statement is False.

Justification:

Let the edge of the cube measure $r$ cm.

For the largest possible right circular cone to be cut out of this cube, the cone's base must be inscribed in one face of the cube, and its vertex must be at the center of the opposite face.

This means:

• The diameter of the base of the cone is equal to the edge length of the cube, which is $r$.
• The radius of the base of the cone, $R_{\text{cone}}$, is half of the diameter: $R_{\text{cone}} = \frac{r}{2}$.
• The height of the cone, $H_{\text{cone}}$, is equal to the edge length of the cube, which is $r$.

The formula for the volume of a right circular cone with radius $R$ and height $H$ is given by:

$V = \frac{1}{3}\pi R^2 H$

Substitute the dimensions of the cone ($R_{\text{cone}} = \frac{r}{2}$ and $H_{\text{cone}} = r$) into the volume formula:

$V_{\text{cone}} = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 (r)$

$V_{\text{cone}} = \frac{1}{3}\pi \left(\frac{r^2}{4}\right) (r)$

$V_{\text{cone}} = \frac{1}{3} \pi \frac{r^2 \times r}{4}$

$V_{\text{cone}} = \frac{1}{3} \pi \frac{r^3}{4}$

$V_{\text{cone}} = \frac{1 \times \pi \times r^3}{3 \times 4}$

$V_{\text{cone}} = \frac{\pi r^3}{12}$

So, the volume of the largest possible cone cut out of the cube is $\frac{1}{12}\pi r^3$ cm$^3$.

The statement claims the volume is $\frac{1}{6}\pi r^3$. Since $\frac{1}{12}\pi r^3 \neq \frac{1}{6}\pi r^3$, the statement is false.

The statement is True.

Justification:

Let the radius of the sphere be $r$.

The diameter of the sphere is $2r$.

The volume of the sphere is given by the formula:

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

Consider a cylinder whose height and diameter are equal to the diameter of the sphere.

Height of the cylinder, $H = \text{Diameter of sphere} = 2r$.

Diameter of the base of the cylinder, $D = \text{Diameter of sphere} = 2r$.

Radius of the base of the cylinder, $R = \frac{D}{2} = \frac{2r}{2} = r$.

The volume of this cylinder is given by the formula:

$V_{\text{cylinder}} = \pi R^2 H$

Substitute the dimensions of the cylinder in terms of $r$:

$V_{\text{cylinder}} = \pi (r)^2 (2r)$

$V_{\text{cylinder}} = \pi r^2 (2r)$

$V_{\text{cylinder}} = 2\pi r^3$

Now, let's check if the volume of the sphere is equal to two-third of the volume of this cylinder.

We need to verify if $V_{\text{sphere}} = \frac{2}{3} V_{\text{cylinder}}$.

Substitute the expressions for the volumes:

Simplify the right side of the equation:

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

So, the equation becomes:

Since both sides are equal, the statement is true.

This relationship was first discovered by Archimedes.

The statement is False.

Justification:

Let the initial radius of the right circular cone be $r_1$ and the initial height be $h_1$.

The formula for the volume of a cone with radius $r$ and height $h$ is given by:

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

The initial volume of the cone is therefore:

$V_1 = \frac{1}{3}\pi r_1^2 h_1$

According to the statement, the radius is halved and the height is doubled.

So, the new radius, $r_2$, is half of the initial radius:

$r_2 = \frac{r_1}{2}$

And the new height, $h_2$, is double the initial height:

$h_2 = 2h_1$

The new volume of the cone, $V_2$, is calculated using the new radius and height:

$V_2 = \frac{1}{3}\pi r_2^2 h_2$

Substitute the expressions for $r_2$ and $h_2$ in terms of $r_1$ and $h_1$ into this equation:

$V_2 = \frac{1}{3}\pi \left(\frac{r_1}{2}\right)^2 (2h_1)$

Simplify the term $\left(\frac{r_1}{2}\right)^2$:

$\left(\frac{r_1}{2}\right)^2 = \frac{r_1^2}{2^2} = \frac{r_1^2}{4}$

Substitute this back into the $V_2$ equation:

$V_2 = \frac{1}{3}\pi \left(\frac{r_1^2}{4}\right) (2h_1)$

Combine the terms:

$V_2 = \frac{1}{3} \pi \frac{r_1^2 \times 2h_1}{4}$

$V_2 = \frac{2}{12} \pi r_1^2 h_1$

$V_2 = \frac{1}{6} \pi r_1^2 h_1$

Now, compare $V_2$ with the initial volume $V_1 = \frac{1}{3}\pi r_1^2 h_1$.

We can see that:

$V_2 = \frac{1}{6} \pi r_1^2 h_1 = \frac{1}{2} \times \left(\frac{1}{3} \pi r_1^2 h_1\right) = \frac{1}{2} V_1$

The new volume $V_2$ is half of the initial volume $V_1$.

Since the volume changes (it is halved), the statement that the volume will remain unchanged is false.

The statement is False.

Justification:

Consider a right circular cone.

• The height ($h$) is the perpendicular distance from the vertex to the center of the base.
• The radius ($r$) is the radius of the circular base.
• The slant height ($l$) is the distance from the vertex to any point on the circumference of the base.

In a right circular cone, the height, the radius, and the slant height form a right-angled triangle. This right triangle is formed by:

• One leg being the height ($h$).
• The other leg being the radius ($r$).
• The hypotenuse being the slant height ($l$).

This geometrical relationship is inherent to the definition of a right circular cone (where the axis is perpendicular to the base).

According to the Pythagorean theorem, for this right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

Equation (i) shows the relationship between the radius, height, and slant height in any right circular cone. This equation is only possible because these three lengths form the sides of a right triangle.

Therefore, in a right circular cone, the height, radius, and slant height always form the sides of a right triangle.

The statement claims they "do not always be sides of a right triangle", which contradicts this fundamental property of a right circular cone.

The statement is True.

Justification:

Let the initial radius of the cylinder be $r_1$ and the initial height be $h_1$.

The formula for the curved surface area (CSA) of a cylinder with radius $r$ and height $h$ is given by:

CSA $= 2\pi r h$

The initial curved surface area of the cylinder is:

According to the statement, the radius is doubled, and the curved surface area is not changed.

So, the new radius, $r_2$, is double the initial radius:

$r_2 = 2r_1$

Let the new height be $h_2$. The new curved surface area (CSA$_2$) is:

CSA$_2 = 2\pi r_2 h_2$

We are given that the curved surface area is not changed, which means:

Substitute the expressions for CSA$_1$ and CSA$_2$:

$2\pi r_2 h_2 = 2\pi r_1 h_1$

Cancel out the common term $2\pi$ from both sides (since $2\pi \neq 0$):

$r_2 h_2 = r_1 h_1$

Now substitute $r_2 = 2r_1$ into this equation:

$(2r_1) h_2 = r_1 h_1$

$2r_1 h_2 = r_1 h_1$

Since $r_1$ is the initial radius, it is a positive value, so we can divide both sides by $r_1$:

$2h_2 = h_1$

Solve for $h_2$:

This result shows that the new height ($h_2$) must be half of the initial height ($h_1$) if the radius is doubled and the curved surface area remains unchanged.

Thus, the statement is true.

The statement is True.

Justification:

Let the edge of the cube be $a$. According to the problem, the edge of the cube is $2r$.

So, $a = 2r$.

For the largest right circular cone that can be fitted in this cube:

• The diameter of the base of the cone will be equal to the edge length of the cube, $a$.
• The radius of the base of the cone, $R_{\text{cone}}$, will be half of the diameter: $R_{\text{cone}} = \frac{a}{2} = \frac{2r}{2} = r$.
• The height of the cone, $H_{\text{cone}}$, will be equal to the edge length of the cube: $H_{\text{cone}} = a = 2r$.

The formula for the volume of a right circular cone is $V = \frac{1}{3}\pi R^2 H$.

Substitute the dimensions of the largest cone ($R_{\text{cone}} = r$ and $H_{\text{cone}} = 2r$):

$V_{\text{cone}} = \frac{1}{3}\pi (r)^2 (2r)$

$V_{\text{cone}} = \frac{1}{3}\pi (r^2)(2r)$

$V_{\text{cone}} = \frac{1}{3}\pi (2r^3)$

Now, consider a hemisphere of radius $r$.

The formula for the volume of a sphere with radius $r$ is $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.

The volume of a hemisphere is half the volume of a sphere with the same radius.

Volume of hemisphere, $V_{\text{hemisphere}} = \frac{1}{2} \times V_{\text{sphere}}$

$V_{\text{hemisphere}} = \frac{1}{2} \times \left(\frac{4}{3}\pi r^3\right)$

$V_{\text{hemisphere}} = \frac{4}{6}\pi r^3$

Comparing the volume of the largest cone fitted in the cube and the volume of the hemisphere of radius $r$:

$V_{\text{cone}} = \frac{2}{3}\pi r^3$

$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$

Since $V_{\text{cone}} = V_{\text{hemisphere}}$, the volume of the largest right circular cone that can be fitted in a cube whose edge is $2r$ is equal to the volume of a hemisphere of radius $r$.

The statement is True.

Justification:

Let the radius of the common base of the cylinder and the right circular cone be $r$.

Let their common height be $h$.

The formula for the volume of a cylinder with radius $r$ and height $h$ is given by:

$V_{\text{cylinder}} = \pi r^2 h$

The formula for the volume of a right circular cone with radius $r$ and height $h$ is given by:

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

Now, let's compare the two volumes. We can express the volume of the cylinder in terms of the volume of the cone:

$V_{\text{cylinder}} = \pi r^2 h$

We can see that $\pi r^2 h$ is 3 times $\frac{1}{3}\pi r^2 h$.

$V_{\text{cylinder}} = 3 \times \left(\frac{1}{3}\pi r^2 h\right)$

Since $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$, we can substitute this into the equation:

This equation shows that the volume of the cylinder is three times the volume of the cone when they have the same base radius and the same height.

Therefore, the statement is true.

The statement is True.

Justification:

Let the radius of the equal bases of the cone and cylinder be $r$. The hemisphere also stands on a base with radius $r$.

Let the common height of the cone and cylinder be $h$.

For a hemisphere to have the same height as its base radius, its height must be equal to its radius. Since the hemisphere has the same height ($h$) as the cone and cylinder, its radius must also be $h$. As the base radius of the hemisphere is $r$, this implies that the common height $h$ must be equal to the base radius $r$.

So, for the cone, cylinder, and hemisphere:

• Radius = $r$
• Height = $h$
• Condition from the problem: $h = r$

The formulas for the volumes of these shapes are:

Volume of Cone ($V_{\text{cone}}$) with radius $r$ and height $h$: $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$

Volume of Hemisphere ($V_{\text{hemisphere}}$) with radius $r$: $V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$

Volume of Cylinder ($V_{\text{cylinder}}$) with radius $r$ and height $h$: $V_{\text{cylinder}} = \pi r^2 h$

Now, we apply the condition $h = r$ to the volumes of the cone and cylinder:

$V_{\text{cone}} = \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3$

$V_{\text{cylinder}} = \pi r^2 (r) = \pi r^3$

The volume of the hemisphere remains $V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$.

Now, we find the ratio of their volumes: $V_{\text{cone}} : V_{\text{hemisphere}} : V_{\text{cylinder}}$

Ratio $= \frac{1}{3}\pi r^3 : \frac{2}{3}\pi r^3 : \pi r^3$

Divide all parts of the ratio by the common factor $\pi r^3$ (assuming $r \neq 0$):

Ratio $= \frac{1}{3} : \frac{2}{3} : 1$

To express the ratio in simple integers, multiply all parts by 3:

Ratio $= \left(\frac{1}{3} \times 3\right) : \left(\frac{2}{3} \times 3\right) : (1 \times 3)$

Ratio $= 1 : 2 : 3$

The calculated ratio of the volumes of the cone, hemisphere, and cylinder is $1:2:3$, which matches the ratio given in the statement.

Therefore, the statement is true.

The statement is False.

Justification:

Let the length of the edge of the cube be $a$ cm.

The formula for the length of the space diagonal ($d$) of a cube with edge length $a$ is given by:

$d = a\sqrt{3}$

We are given that the length of the diagonal is $6\sqrt{3}$ cm.

So, we can set up the equation:

To find the length of the edge $a$, divide both sides of the equation by $\sqrt{3}$:

$a = \frac{6\sqrt{3}}{\sqrt{3}}$

Cancel out the common term $\sqrt{3}$:

$a = 6$

The length of the edge of the cube is $6$ cm.

The statement claims that the length of the edge of the cube is 3 cm.

Since $6 \text{ cm} \neq 3 \text{ cm}$, the statement is false.

The statement is True.

Justification:

Let the edge length of the cube be $a$.

Let the radius of the sphere inscribed in the cube be $r$.

When a sphere is inscribed in a cube, the diameter of the sphere is equal to the edge length of the cube.

So, the diameter of the sphere is $2r$, which is equal to the edge length $a$.

The volume of the cube with edge length $a$ is given by:

$V_{\text{cube}} = a^3$

Substitute $a = 2r$ into the formula for the volume of the cube:

$V_{\text{cube}} = (2r)^3$

$V_{\text{cube}} = 2^3 r^3$

The volume of the sphere with radius $r$ is given by the formula:

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

Now, we need to find the ratio of the volume of the cube to the volume of the sphere:

Ratio $= \frac{V_{\text{cube}}}{V_{\text{sphere}}}$

Substitute the expressions for the volumes:

Ratio $= \frac{8r^3}{\frac{4}{3}\pi r^3}$

Simplify the expression:

Ratio $= 8r^3 \times \frac{3}{4\pi r^3}$

Cancel out the common term $r^3$ (assuming $r \neq 0$):

Ratio $= 8 \times \frac{3}{4\pi}$

Cancel out the common numerical factor 4:

Ratio $= \cancel{8}^2 \times \frac{3}{\cancel{4}^1\pi}$

Ratio $= \frac{2 \times 3}{1 \times \pi}$

Ratio $= \frac{6}{\pi}$

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

This matches the ratio given in the statement. Therefore, the statement is true.

The statement is True.

Justification:

Let the initial radius of the cylinder be $r_1$ and the initial height be $h_1$.

The formula for the volume of a cylinder with radius $r$ and height $h$ is given by:

$V = \pi r^2 h$

The initial volume of the cylinder is therefore:

According to the statement, the radius is doubled and the height is halved.

So, the new radius, $r_2$, is double the initial radius:

$r_2 = 2r_1$

And the new height, $h_2$, is half of the initial height:

$h_2 = \frac{h_1}{2}$

The new volume of the cylinder, $V_2$, is calculated using the new radius and height:

$V_2 = \pi r_2^2 h_2$

Substitute the expressions for $r_2$ and $h_2$ in terms of $r_1$ and $h_1$ into this equation:

$V_2 = \pi (2r_1)^2 \left(\frac{h_1}{2}\right)$

Simplify the terms:

$V_2 = \pi (4r_1^2) \left(\frac{h_1}{2}\right)$

$V_2 = \pi \frac{4r_1^2 h_1}{2}$

$V_2 = 2\pi r_1^2 h_1$

Now, compare the new volume $V_2$ with the initial volume $V_1 = \pi r_1^2 h_1$.

We can see that:

The new volume $V_2$ is double the initial volume $V_1$.

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

The statement is true.

Given:

Radius of the sphere, $r_s = 5$ cm

Radius of the cone, $r_c = 4$ cm

Surface area of sphere = $5 \times$ Curved surface area of cone

Value of $\pi = \frac{22}{7}$

To Find:

The height ($h_c$) and volume ($V_c$) of the cone.

Solution:

First, calculate the surface area of the sphere. The formula for the surface area of a sphere with radius $r$ is $4\pi r^2$.

Surface area of sphere = $4\pi r_s^2$

Substitute $r_s = 5$ cm:

Surface area of sphere = $4 \times \pi \times (5 \text{ cm})^2$

Surface area of sphere = $4 \times \pi \times 25 \text{ cm}^2$

Surface area of sphere = $100\pi \text{ cm}^2$

Next, consider the curved surface area of the cone. The formula for the curved surface area of a cone with radius $r_c$ and slant height $l_c$ is $\pi r_c l_c$.

Curved surface area of cone (CSA$_c$) = $\pi r_c l_c$

Substitute $r_c = 4$ cm:

CSA$_c = \pi (4) l_c = 4\pi l_c \text{ cm}^2$

According to the given condition, the surface area of the sphere is five times the curved surface area of the cone:

Surface area of sphere $= 5 \times$ CSA$_c$

$100\pi = 5 \times (4\pi l_c)$

$100\pi = 20\pi l_c$

To find the slant height $l_c$, divide both sides by $20\pi$:

$l_c = \frac{100\pi}{20\pi}$

$l_c = 5$ cm

The slant height of the cone is $5$ cm.

Now, we can find the height of the cone using the relationship between the radius, height, and slant height in a right circular cone, which is given by the Pythagorean theorem: $l_c^2 = r_c^2 + h_c^2$.

Substitute $l_c = 5$ cm and $r_c = 4$ cm:

$(5)^2 = (4)^2 + h_c^2$

$25 = 16 + h_c^2$

$h_c^2 = 25 - 16$

$h_c^2 = 9$

Take the square root of both sides to find $h_c$ (height must be positive):

$h_c = \sqrt{9}$

The height of the cone is $3$ cm.

Finally, calculate the volume of the cone. The formula for the volume of a cone with radius $r_c$ and height $h_c$ is $V_c = \frac{1}{3}\pi r_c^2 h_c$.

Substitute $r_c = 4$ cm, $h_c = 3$ cm, and $\pi = \frac{22}{7}$:

$V_c = \frac{1}{3} \times \frac{22}{7} \times (4 \text{ cm})^2 \times (3 \text{ cm})$

$V_c = \frac{1}{3} \times \frac{22}{7} \times (16 \text{ cm}^2) \times (3 \text{ cm})$

Cancel the common factor of 3 in the numerator and denominator:

$V_c = \frac{1}{\cancel{3}^1} \times \frac{22}{7} \times 16 \times \cancel{3}^1 \text{ cm}^3$

$V_c = \frac{22}{7} \times 16 \text{ cm}^3$

$V_c = \frac{22 \times 16}{7} \text{ cm}^3$

$22 \times 16 = 352$

The volume of the cone is $\frac{352}{7}$ cm$^3$. This can also be expressed as a mixed number or a decimal: $\frac{352}{7} = 50 \frac{2}{7}$ cm$^3$ or approximately $50.29$ cm$^3$.

The height of the cone is 3 cm and the volume of the cone is $\frac{352}{7}$ cm³.

Given:

Initial radius of a sphere is $r_1$.

The radius is increased by $10\%$.

To Prove:

The volume of the sphere will be increased by $33.1\%$ approximately.

Solution:

Let the initial radius of the sphere be $r_1$.

The initial volume of the sphere is given by the formula:

$V_1 = \frac{4}{3}\pi r_1^3$

The radius is increased by $10\%$. The increase in radius is $10\%$ of $r_1$, which is $0.10 r_1$.

The new radius, $r_2$, is the initial radius plus the increase:

$r_2 = r_1 + 10\% \text{ of } r_1$

$r_2 = r_1 + \frac{10}{100} r_1$

$r_2 = r_1 + 0.1 r_1$

The new volume of the sphere, $V_2$, with radius $r_2$ is:

$V_2 = \frac{4}{3}\pi r_2^3$

Substitute the value of $r_2$ from equation (i) into this formula:

$V_2 = \frac{4}{3}\pi (1.1 r_1)^3$

Calculate $(1.1)^3$:

$(1.1)^3 = 1.1 \times 1.1 \times 1.1 = 1.21 \times 1.1$

Using multiplication:

$\begin{array}{cc}& & 1 & . & 2 & 1 \\ \times & & 1 & . & 1 & \\ \hline && 1 & 2 & 1 \\ & 1 & 2 & 1 & \times \\ \hline & 1 & . & 3 & 3 & 1 \\ \hline \end{array}$

So, $(1.1)^3 = 1.331$

Substitute this back into the $V_2$ equation:

$V_2 = \frac{4}{3}\pi (1.331 r_1^3)$

Rearrange the terms:

$V_2 = 1.331 \times \left(\frac{4}{3}\pi r_1^3\right)$

We know that the initial volume $V_1 = \frac{4}{3}\pi r_1^3$. Substitute $V_1$ into the equation for $V_2$:

The increase in volume is the difference between the new volume and the initial volume:

Increase in volume $= V_2 - V_1$

Substitute the value of $V_2$ from equation (ii):

Increase in volume $= 1.331 V_1 - V_1$

Increase in volume $= (1.331 - 1) V_1$

Increase in volume $= 0.331 V_1$

To express this increase as a percentage of the initial volume, $V_1$:

Percentage increase $= \left(\frac{\text{Increase in volume}}{V_1}\right) \times 100\%$

Percentage increase $= \left(\frac{0.331 V_1}{V_1}\right) \times 100\%$

Percentage increase $= 0.331 \times 100\%$

Percentage increase $= 33.1\%$

Thus, if the radius of a sphere is increased by $10\%$, its volume will be increased by $33.1\%$.

Given:

Radius of each metal sphere, $r = 2$ cm.

Internal dimensions of the rectangular box: Length ($L$) = $16$ cm, Width ($W$) = $8$ cm, Height ($H$) = $8$ cm.

Number of spheres packed = $16$.

Value of $\pi = 3.14$.

To Find:

The volume of the preservative liquid in the box, to the nearest integer.

Solution:

The volume of the preservative liquid is the space inside the box that is not occupied by the spheres.

Volume of liquid = Volume of box - Volume of 16 spheres

Step 1: Calculate the volume of the rectangular box.

The formula for the volume of a rectangular box (cuboid) is $V = L \times W \times H$.

$V_{\text{box}} = 16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$

$V_{\text{box}} = (16 \times 8 \times 8) \text{ cm}^3$

$V_{\text{box}} = 16 \times 64 \text{ cm}^3$

Let's perform the multiplication:

$\begin{array}{cc}& & 1 & 6 \\ \times & & 6 & 4 \\ \hline && 6 & 4 \\ & 9 & 6 & \times \\ \hline 1 & 0 & 2 & 4 \\ \hline \end{array}$

Step 2: Calculate the volume of one metal sphere.

The formula for the volume of a sphere with radius $r$ is $V = \frac{4}{3}\pi r^3$.

$V_{\text{sphere}} = \frac{4}{3} \times 3.14 \times (2 \text{ cm})^3$

$V_{\text{sphere}} = \frac{4}{3} \times 3.14 \times 8 \text{ cm}^3$

$V_{\text{sphere}} = \frac{32}{3} \times 3.14 \text{ cm}^3$

Approximate value of $\frac{32}{3} \approx 10.6667$.

$V_{\text{sphere}} \approx 10.6667 \times 3.14 \text{ cm}^3$

Let's perform $32 \times 3.14$ and then divide by 3:

$32 \times 3.14 = 100.48$

$V_{\text{sphere}} = \frac{100.48}{3} \text{ cm}^3$

Performing the division:

$\begin{array}{r} 33.493... \\ 3{\overline{\smash{\big)}\,100.480\dots}} \\ \underline{-~\phantom{(}9}\phantom{00.480} \\ 10\phantom{0.480} \\ \underline{-~\phantom{()}9}\phantom{.480} \\ 14\phantom{80} \\ \underline{-~12}\phantom{80} \\ 28\phantom{0} \\ \underline{-~27}\phantom{0} \\ 10\dots \end{array}$

$V_{\text{sphere}} \approx 33.493$ cm$^3$

Step 3: Calculate the total volume of 16 spheres.

Total volume of spheres $= 16 \times V_{\text{sphere}}$

Total volume of spheres $= 16 \times \frac{100.48}{3}$

Total volume of spheres $= \frac{16 \times 100.48}{3}$

$16 \times 100.48 = 1607.68$

Total volume of spheres $= \frac{1607.68}{3}$

Performing the division:

$\begin{array}{r} 535.893... \\ 3{\overline{\smash{\big)}\,1607.680\dots}} \\ \underline{-~15}\phantom{07.680} \\ 10\phantom{7.680} \\ \underline{-~\phantom{1}9}\phantom{.680} \\ 17\phantom{.680} \\ \underline{-~15}\phantom{.680} \\ 26\phantom{80} \\ \underline{-~24}\phantom{80} \\ 28\phantom{0} \\ \underline{-~27}\phantom{0} \\ 10\dots \end{array}$

Total volume of spheres $\approx 535.893$ cm$^3$

Step 4: Calculate the volume of the preservative liquid.

$V_{\text{liquid}} = V_{\text{box}} - \text{Total volume of spheres}$

$V_{\text{liquid}} = 1024 \text{ cm}^3 - 535.893 \text{ cm}^3$

Let's perform the subtraction:

$\begin{array}{cccccc}& 10 & 2 & 4 & . & 000 \\ - & \phantom{0}5 & 3 & 5 & . & 893 \\ \hline & \phantom{0}4 & 8 & 8 & . & 107 \\ \hline \end{array}$

$V_{\text{liquid}} \approx 488.107$ cm$^3$

Step 5: Round the volume of the liquid to the nearest integer.

The volume is approximately $488.107$ cm$^3$. The decimal part is $0.107$, which is less than $0.5$.

Rounding to the nearest integer, we get $488$.

The volume of the preservative liquid is approximately 488 cm$^3$.

Given:

The storage tank is in the form of a cube.

Volume of water when the tank is full, $V_{\text{full}} = 15.625$ m$^3$.

Present depth of water, $h = 1.3$ m.

To Find:

The volume of water already used from the tank.

Solution:

Step 1: Find the side length of the cubic tank.

Since the tank is a cube and is full of water, the volume of the tank is equal to the volume of water when full.

Let the side length of the cube be $a$ meters.

The volume of a cube is given by the formula $V = a^3$.

So, $V_{\text{full}} = a^3$

To find the side length $a$, we take the cube root of $15.625$:

$a = \sqrt[3]{15.625}$

We know that $2.5 \times 2.5 = 6.25$ and $6.25 \times 2.5 = 15.625$.

So,

The side length (edge) of the cubic tank is $2.5$ meters. This means the length, width, and full height of the tank are all $2.5$ m.

Step 2: Find the volume of water currently in the tank.

The water currently in the tank forms a cuboid with dimensions equal to the tank's length, width, and the present depth of the water.

Length of water = side of cube = $2.5$ m

Width of water = side of cube = $2.5$ m

Height (depth) of water = $1.3$ m (Given)

The volume of a cuboid is given by the formula $V = \text{length} \times \text{width} \times \text{height}$.

Volume of water present, $V_{\text{present}} = 2.5 \text{ m} \times 2.5 \text{ m} \times 1.3 \text{ m}$

$V_{\text{present}} = (2.5 \times 2.5) \times 1.3 \text{ m}^3$

$2.5 \times 2.5 = 6.25$

$V_{\text{present}} = 6.25 \times 1.3 \text{ m}^3$

Perform the multiplication:

$\begin{array}{cc}& & 6 & . & 2 & 5 \\ \times & & 1 & . & 3 & \\ \hline && 1 & 8 & 7 & 5 \\ & 6 & 2 & 5 & \times \\ \hline & 8 & . & 1 & 2 & 5 \\ \hline \end{array}$

Step 3: Find the volume of water used.

The volume of water used is the difference between the initial full volume and the current volume of water.

Volume of water used $= V_{\text{full}} - V_{\text{present}}$

Volume of water used $= 15.625 \text{ m}^3 - 8.125 \text{ m}^3$

Perform the subtraction:

$\begin{array}{cc}& 1 & 5 & . & 6 & 2 & 5 \\ - & \phantom{1}8 & . & 1 & 2 & 5 \\ \hline & \phantom{1}7 & . & 5 & 0 & 0 \\ \hline \end{array}$

The volume of water already used from the tank is 7.5 m³.

Given:

Diameter of the solid spherical ball = $4.2$ cm.

To Find:

The amount of water displaced by the spherical ball when completely immersed.

Solution:

According to Archimedes' principle, when a solid object is completely immersed in a liquid, the volume of the liquid displaced is equal to the volume of the object.

Therefore, the amount of water displaced by the spherical ball is equal to the volume of the spherical ball.

First, find the radius of the spherical ball. The radius ($r$) is half of the diameter.

$r = \frac{\text{Diameter}}{2}$

$r = \frac{4.2 \text{ cm}}{2}$

$r = 2.1$ cm

The formula for the volume of a sphere with radius $r$ is given by:

$V = \frac{4}{3}\pi r^3$

Substitute the radius $r = 2.1$ cm into the formula. We can use $\pi = \frac{22}{7}$ for calculation.

$V = \frac{4}{3} \times \frac{22}{7} \times (2.1 \text{ cm})^3$

$V = \frac{4}{3} \times \frac{22}{7} \times (2.1 \times 2.1 \times 2.1) \text{ cm}^3$

It can be helpful to write $2.1$ as a fraction: $2.1 = \frac{21}{10}$.

$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{10}\right)^3 \text{ cm}^3$

$V = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} \text{ cm}^3$

Now, perform the cancellation:

$V = \frac{4}{\cancel{3}^1} \times \frac{22}{\cancel{7}^1} \times \frac{\cancel{21}^7}{10} \times \frac{\cancel{21}^3}{10} \times \frac{21}{10} \text{ cm}^3$

Wait, cancellation should be: $\frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10}$

Cancel 3 with one of the 21s, and 7 with another 21:

$V = \frac{4}{\cancel{3}^1} \times \frac{22}{\cancel{7}^1} \times \frac{\cancel{21}^3}{10} \times \frac{\cancel{21}^{2.1 \rightarrow 3/7}}{10} \times \frac{21}{10} \text{ cm}^3$ - Incorrect Cancellation

Let's do it step-by-step:

$V = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10}$

$V = \frac{4}{3} \times \frac{22}{7} \times \frac{\cancel{21}^7}{10} \times \frac{21}{10} \times \frac{21}{10}$ (Cancel 3 with 21, leaving 7)

$V = \frac{4}{1} \times \frac{22}{7} \times \frac{7}{10} \times \frac{21}{10} \times \frac{21}{10}$

$V = 4 \times \frac{22}{\cancel{7}^1} \times \frac{\cancel{7}^1}{10} \times \frac{21}{10} \times \frac{21}{10}$ (Cancel 7 with 7)

$V = 4 \times 22 \times \frac{1}{10} \times \frac{21}{10} \times \frac{21}{10}$

$V = \frac{4 \times 22 \times 21 \times 21}{10 \times 10 \times 10}$

$V = \frac{88 \times 441}{1000}$

Calculate the numerator $88 \times 441$:

$\begin{array}{cc}& & 4 & 4 & 1 \\ \times & & & 8 & 8 \\ \hline && 3 & 5 & 2 & 8 \\ & 3 & 5 & 2 & 8 & \times \\ \hline 3 & 8 & 8 & 0 & 8 \\ \hline \end{array}$

$V = \frac{38808}{1000}$

$V = 38.808$

The volume of the spherical ball is $38.808$ cm$^3$.

Since the volume of water displaced is equal to the volume of the spherical ball when completely immersed:

Volume of water displaced = $38.808$ cm$^3$.

The amount of water displaced by the solid spherical ball is 38.808 cm³.

Given:

Height of the conical tent, $h = 3.5$ m.

Radius of the base of the cone, $r = 12$ m.

To Find:

The amount of canvas required for the conical tent (which is its curved surface area).

Solution:

The amount of canvas required for a conical tent is equal to its curved surface area (CSA).

The formula for the curved surface area of a cone is CSA $= \pi r l$, where $r$ is the radius of the base and $l$ is the slant height.

We are given the radius ($r$) and the height ($h$), but not the slant height ($l$). We can find the slant height using the Pythagorean theorem, as the radius, height, and slant height form a right-angled triangle:

Substitute the given values of $r = 12$ m and $h = 3.5$ m into equation (i):

$l^2 = (12 \text{ m})^2 + (3.5 \text{ m})^2$

$l^2 = 144 \text{ m}^2 + 12.25 \text{ m}^2$

$l^2 = 156.25 \text{ m}^2$

Take the square root of both sides to find $l$. Since $l$ represents a length, it must be positive.

$l = \sqrt{156.25 \text{ m}^2}$

$l = 12.5$ m

The slant height of the cone is $12.5$ m.

Now, we can calculate the curved surface area using the formula CSA $= \pi r l$.

Substitute the values of $r = 12$ m and $l = 12.5$ m:

CSA $= \pi \times 12 \text{ m} \times 12.5 \text{ m}$

CSA $= \pi \times (12 \times 12.5) \text{ m}^2$

Calculate the product $12 \times 12.5$:

$12 \times 12.5 = 12 \times \frac{25}{2} = \cancel{12}^6 \times 25 = 6 \times 25 = 150$

So,

CSA $= 150\pi \text{ m}^2$

The problem does not specify the value of $\pi$. However, if we assume $\pi \approx 3.14$ or $\pi = \frac{22}{7}$, we can get a numerical value. Using $\pi \approx 3.14$ often gives simpler decimal results in such cases.

Using $\pi \approx 3.14$:

CSA $\approx 150 \times 3.14 \text{ m}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 3 & . & 1 & 4 \\ \times & & 1 & 5 & 0 \\ \hline &&& 0 & 0 & 0 \\ & 1 & 5 & 7 & 0 & \times \\ 3 & 1 & 4 & \times & \times \\ \hline 4 & 7 & 1 & . & 0 & 0 \\ \hline \end{array}$

CSA $\approx 471 \text{ m}^2$

The amount of canvas required for the conical tent is approximately 471 square metres (using $\pi \approx 3.14$). If the question intends the answer in terms of $\pi$, it would be $150\pi$ m$^2$. Based on common practice in such problems without specified $\pi$ and the ease of calculation with $3.14$, $471$ m$^2$ is a likely intended answer.

Given:

Weight of the larger sphere, $W_1 = 5920$ g.

Weight of the smaller sphere, $W_2 = 740$ g.

Diameter of the smaller sphere = $5$ cm.

The spheres are made of the same metal, implying they have the same density ($\rho$).

To Find:

The radius of the larger sphere.

Solution:

Since the spheres are made of the same metal, their density is constant. The weight of a sphere is related to its mass and volume by the formula: Weight $\propto$ Mass = Density $\times$ Volume.

Therefore, the ratio of the weights of the two spheres is equal to the ratio of their volumes:

Let the radius of the larger sphere be $r_1$ and the radius of the smaller sphere be $r_2$.

The volume of a sphere with radius $r$ is given by the formula $V = \frac{4}{3}\pi r^3$.

Volume of larger sphere, $V_1 = \frac{4}{3}\pi r_1^3$

Volume of smaller sphere, $V_2 = \frac{4}{3}\pi r_2^3$

The ratio of their volumes is:

$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$

Now, equate the ratio of weights to the ratio of volumes:

$\frac{W_1}{W_2} = \left(\frac{r_1}{r_2}\right)^3$

We are given the diameter of the smaller sphere is 5 cm, so its radius $r_2$ is:

$r_2 = \frac{\text{Diameter}}{2} = \frac{5 \text{ cm}}{2} = 2.5$ cm

Substitute the given weights and the radius of the smaller sphere into the equation:

$\frac{5920 \text{ g}}{740 \text{ g}} = \left(\frac{r_1}{2.5 \text{ cm}}\right)^3$

Simplify the left side of the equation:

$\frac{5920}{740} = \frac{592}{74}$

Performing the division $592 \div 74$:

We can test multiples of 74. $74 \times 8 = (70 + 4) \times 8 = 560 + 32 = 592$.

So, $\frac{592}{74} = 8$.

The equation becomes:

To find the ratio $\frac{r_1}{2.5}$, take the cube root of both sides:

$\sqrt[3]{8} = \sqrt[3]{\left(\frac{r_1}{2.5}\right)^3}$

$2 = \frac{r_1}{2.5}$

Now, solve for $r_1$ by multiplying both sides by $2.5$:

$r_1 = 2 \times 2.5$ cm

The radius of the larger sphere is $5$ cm.

The radius of the larger sphere is 5 cm.

Given:

Diameter of the cylindrical glass = $7$ cm.

Height of milk in the glass, $h = 12$ cm.

Number of students = $1600$.

To Find:

Total volume of milk needed to serve 1600 students, in liters.

Solution:

Step 1: Find the radius of the cylindrical glass.

The radius ($r$) is half of the diameter.

$r = \frac{\text{Diameter}}{2} = \frac{7 \text{ cm}}{2} = 3.5 \text{ cm}$

We can also write this as a fraction: $r = \frac{7}{2}$ cm.

Step 2: Calculate the volume of milk in one glass.

The volume of a cylinder with radius $r$ and height $h$ is given by the formula:

$V = \pi r^2 h$

Substitute the values of $r = \frac{7}{2}$ cm and $h = 12$ cm. We will use $\pi = \frac{22}{7}$.

$V_{\text{glass}} = \frac{22}{7} \times \left(\frac{7}{2} \text{ cm}\right)^2 \times 12 \text{ cm}$

$V_{\text{glass}} = \frac{22}{7} \times \frac{7^2}{2^2} \times 12 \text{ cm}^3$

$V_{\text{glass}} = \frac{22}{7} \times \frac{49}{4} \times 12 \text{ cm}^3$

Now, perform cancellation and multiplication:

$V_{\text{glass}} = \frac{22}{\cancel{7}^1} \times \frac{\cancel{49}^7}{\cancel{4}^1} \times \cancel{12}^3 \text{ cm}^3$ (Cancel 7 with 49, leaving 7; Cancel 4 with 12, leaving 3; Cancel 22/2 with 11. Wait, cancellation order changed. Let's restart cancellation with the correct fractions.)

$V_{\text{glass}} = \frac{22}{7} \times \frac{49}{4} \times 12$

$V_{\text{glass}} = \frac{\cancel{22}^{11}}{\cancel{7}^1} \times \frac{\cancel{49}^7}{\cancel{4}^1} \times \cancel{12}^3$

$V_{\text{glass}} = 11 \times 7 \times 3$

$V_{\text{glass}} = 77 \times 3$

$V_{\text{glass}} = 231 \times 2 = 462$

$V_{\text{glass}} = \frac{22}{7} \times \frac{49}{4} \times 12$

$V_{\text{glass}} = \frac{22}{7} \times 49 \times \frac{12}{4}$

$V_{\text{glass}} = \frac{22}{\cancel{7}^1} \times \cancel{49}^7 \times \cancel{3}^3$

$V_{\text{glass}} = 22 \times 7 \times 3 \text{ cm}^3$

$V_{\text{glass}} = 154 \times 3 \text{ cm}^3$

$V_{\text{glass}} = 462 \text{ cm}^3$

The volume of milk in one glass is $462$ cm$^3$.

Step 3: Calculate the total volume of milk needed for 1600 students.

Total volume = Volume per glass $\times$ Number of students

$V_{\text{total}} = 462 \text{ cm}^3 \times 1600$

$V_{\text{total}} = (462 \times 1600) \text{ cm}^3$

Calculate $462 \times 16$:

$\begin{array}{cc}& & 4 & 6 & 2 \\ \times & & & 1 & 6 \\ \hline && 2 & 7 & 7 & 2 \\ & 4 & 6 & 2 & \times \\ \hline & 7 & 3 & 9 & 2 \\ \hline \end{array}$

$462 \times 16 = 7392$. Adding the two zeros from 1600:

$V_{\text{total}} = 739200 \text{ cm}^3$

Step 4: Convert the total volume from cubic centimeters to liters.

We know that $1$ liter = $1000$ cubic centimeters.

To convert cm$^3$ to liters, divide by 1000.

$V_{\text{total}} (\text{liters}) = \frac{739200 \text{ cm}^3}{1000 \text{ cm}^3/\text{liter}}$

$V_{\text{total}} (\text{liters}) = \frac{739200}{1000}$ liters

$V_{\text{total}} (\text{liters}) = 739.2$ liters

The amount of milk needed to serve 1600 students is 739.2 liters.

Given:

Length of the cylindrical roller, $h = 2.5$ m

Radius of the base of the cylindrical roller, $r = 1.75$ m

Total area covered by the roller = $5500$ m$^2$

To Find:

The number of revolutions the roller made.

Solution:

When a cylindrical roller is rolled on a road, the area covered in one revolution is equal to its curved surface area.

The formula for the curved surface area (CSA) of a cylinder with radius $r$ and height $h$ is given by:

CSA $= 2\pi r h$

Substitute the given values of $r = 1.75$ m and $h = 2.5$ m into the formula. It's convenient to work with fractions: $1.75 = 1\frac{3}{4} = \frac{7}{4}$ and $2.5 = 2\frac{1}{2} = \frac{5}{2}$. We will use $\pi = \frac{22}{7}$.

CSA $= 2 \times \frac{22}{7} \times \frac{7}{4} \text{ m} \times \frac{5}{2} \text{ m}$

CSA $= 2 \times \frac{22}{7} \times \frac{7}{4} \times \frac{5}{2} \text{ m}^2$

Perform the multiplication and cancellation:

CSA $= \cancel{2}^1 \times \frac{\cancel{22}^{11}}{\cancel{7}^1} \times \frac{\cancel{7}^1}{\cancel{4}^2} \times \frac{5}{\cancel{2}^1}$ (Cancelling 2 and 4, and 7 and 7)

CSA $= \frac{1 \times 11 \times 1 \times 5}{1 \times 2 \times 1} \text{ m}^2$

CSA $= \frac{55}{2} \text{ m}^2$

CSA $= 27.5 \text{ m}^2$

The area covered by the roller in one revolution is $27.5$ m$^2$.

The total area covered by the roller is the product of the area covered in one revolution and the number of revolutions made.

Total area covered = Number of revolutions $\times$ CSA

Number of revolutions $= \frac{\text{Total area covered}}{\text{CSA}}$

Number of revolutions $= \frac{5500 \text{ m}^2}{27.5 \text{ m}^2}$

To simplify the division, multiply the numerator and the denominator by 10 to remove the decimal:

Number of revolutions $= \frac{5500 \times 10}{27.5 \times 10} = \frac{55000}{275}$

Now, perform the division:

Number of revolutions $= \frac{55000}{275}$

We can recognize that $550 = 2 \times 275$.

So, $\frac{55000}{275} = \frac{550 \times 100}{275} = \frac{(275 \times 2) \times 100}{275} = 2 \times 100 = 200$.

Number of revolutions $= 200$

The cylindrical roller made 200 revolutions to cover an area of 5500 m$^2$.

Given:

Population of the village = $5000$ people.

Water requirement per head per day = $75$ litres.

Dimensions of the overhead tank: Length ($L$) = $40$ m, Width ($W$) = $25$ m, Height ($H$) = $15$ m.

To Find:

The number of days the water in the tank will last.

Solution:

Step 1: Calculate the total daily water requirement for the village.

Daily requirement = Water per head per day $\times$ Population

Daily requirement = $75 \text{ litres/person} \times 5000 \text{ people}$

Daily requirement = $(75 \times 5000)$ litres

$75 \times 5000 = 375000$

Step 2: Calculate the volume of the overhead tank.

The tank is a rectangular shape (cuboid). The formula for the volume of a cuboid is $V = L \times W \times H$.

$V_{\text{tank}} = 40 \text{ m} \times 25 \text{ m} \times 15 \text{ m}$

$V_{\text{tank}} = (40 \times 25 \times 15) \text{ m}^3$

$40 \times 25 = 1000$

$V_{\text{tank}} = 1000 \times 15 \text{ m}^3$

Step 3: Convert the tank volume to liters.

We know that $1$ cubic meter ($1 \text{ m}^3$) is equal to $1000$ litres.

$V_{\text{tank}} (\text{liters}) = V_{\text{tank}} (\text{m}^3) \times 1000 \text{ litres/m}^3$

$V_{\text{tank}} (\text{liters}) = 15000 \text{ m}^3 \times 1000 \text{ litres/m}^3$

$V_{\text{tank}} (\text{liters}) = 15000000$ litres

Step 4: Calculate the number of days the water will last.

Number of days = $\frac{\text{Total volume of water in tank}}{\text{Daily water requirement}}$

Number of days $= \frac{15000000 \text{ litres}}{375000 \text{ litres/day}}$

Cancel out the zeros:

Number of days $= \frac{15000}{375}$

We can simplify this fraction. Divide both numerator and denominator by 25:

$15000 \div 25 = 600$

$375 \div 25 = 15$

Number of days $= \frac{600}{15}$

Now, divide 600 by 15:

$600 \div 15 = 40$

The water in the overhead tank will last for 40 days.

Given:

Radius of the large spherical laddoo, $R = 5$ cm.

Radius of each small spherical laddoo, $r = 2.5$ cm.

The same amount of material is used.

To Find:

The number of small laddoos that can be made.

Solution:

The volume of a sphere with radius $x$ is given by the formula $V = \frac{4}{3}\pi x^3$.

Step 1: Calculate the volume of the large laddoo.

Volume of the large laddoo, $V_L = \frac{4}{3}\pi R^3$

Substitute $R = 5$ cm:

$V_L = \frac{4}{3}\pi (5 \text{ cm})^3$

$V_L = \frac{4}{3}\pi (125) \text{ cm}^3$

Step 2: Calculate the volume of one small laddoo.

Volume of one small laddoo, $V_S = \frac{4}{3}\pi r^3$

Substitute $r = 2.5$ cm. It's helpful to note that $2.5 = \frac{5}{2}$.

$V_S = \frac{4}{3}\pi (2.5 \text{ cm})^3$

$V_S = \frac{4}{3}\pi \left(\frac{5}{2} \text{ cm}\right)^3$

$V_S = \frac{4}{3}\pi \left(\frac{5^3}{2^3}\right) \text{ cm}^3$

$V_S = \frac{4}{3}\pi \left(\frac{125}{8}\right) \text{ cm}^3$

$V_S = \frac{4 \times 125}{3 \times 8}\pi \text{ cm}^3$

Step 3: Determine the number of small laddoos.

Let $n$ be the number of small laddoos that can be made. Since the same amount of material is used, the total volume of the small laddoos is equal to the volume of the large laddoo.

Total volume of small laddoos $= n \times V_S$

We have the equation:

Substitute the volumes we calculated:

$n \times \frac{125}{6}\pi = \frac{500}{3}\pi$

Divide both sides by $\pi$ (assuming $\pi \neq 0$):

$n \times \frac{125}{6} = \frac{500}{3}$

Solve for $n$ by multiplying both sides by the reciprocal of $\frac{125}{6}$, which is $\frac{6}{125}$:

$n = \frac{500}{3} \times \frac{6}{125}$

$n = \frac{\cancel{500}^{4}}{\cancel{3}^{1}} \times \frac{\cancel{6}^{2}}{\cancel{125}^{1}}$

$n = 4 \times 2$

Thus, 8 small laddoos of radius 2.5 cm can be made from the same amount of material as the large laddoo of radius 5 cm.

The number of small laddoos that can be made is 8.

Given:

A right triangle with sides 6 cm, 8 cm, and 10 cm.

The triangle is revolved about the side measuring 8 cm.

To Find:

The volume and the curved surface area of the solid formed.

Solution:

When a right triangle is revolved about one of its legs, the solid formed is a right circular cone.

In the given right triangle, the sides are 6 cm, 8 cm, and 10 cm. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$, the sides 6 cm and 8 cm are the legs, and 10 cm is the hypotenuse.

The triangle is revolved about the side measuring 8 cm. This side becomes the height of the cone.

• Height of the cone ($h$) = Side of revolution = $8$ cm.
• Radius of the base of the cone ($r$) = The other leg of the right triangle = $6$ cm.
• Slant height of the cone ($l$) = The hypotenuse of the right triangle = $10$ cm.

Step 1: Calculate the volume of the cone.

The formula for the volume of a cone with radius $r$ and height $h$ is $V = \frac{1}{3}\pi r^2 h$.

Substitute the values $r = 6$ cm and $h = 8$ cm:

$V = \frac{1}{3}\pi (6 \text{ cm})^2 (8 \text{ cm})$

$V = \frac{1}{3}\pi (36 \text{ cm}^2) (8 \text{ cm})$

$V = \frac{1}{3} \times 36 \times 8 \times \pi \text{ cm}^3$

Cancel 3 from the denominator with 36 in the numerator:

$V = \cancel{36}^{12} \times \frac{1}{\cancel{3}^1} \times 8 \times \pi \text{ cm}^3$

$V = 12 \times 8 \times \pi \text{ cm}^3$

Step 2: Calculate the curved surface area of the cone.

The formula for the curved surface area (CSA) of a cone with radius $r$ and slant height $l$ is CSA $= \pi r l$.

Substitute the values $r = 6$ cm and $l = 10$ cm:

CSA $= \pi (6 \text{ cm}) (10 \text{ cm})$

The volume of the solid formed is $96\pi$ cm³.

The curved surface area of the solid formed is $60\pi$ cm².

Given:

Dimensions of the rectangular surface: Length ($L$) = $6$ m, Breadth ($B$) = $4$ m.

Rainfall height, $h_{\text{rain}} = 1$ cm.

Internal radius of the cylindrical vessel, $r_c = 20$ cm.

Value of $\pi = 3.14$.

To Find:

The height of water in the cylindrical vessel, rounded to the nearest integer.

Solution:

The volume of rainwater that falls on the rectangular surface is equal to the volume of water collected in the cylindrical vessel.

First, let's ensure all dimensions are in the same unit. We will convert all measurements to centimeters.

Length of rectangular surface, $L = 6$ m $= 6 \times 100$ cm $= 600$ cm.

Breadth of rectangular surface, $B = 4$ m $= 4 \times 100$ cm $= 400$ cm.

Rainfall height, $h_{\text{rain}} = 1$ cm (already in cm).

Radius of cylindrical vessel, $r_c = 20$ cm (already in cm).

Step 1: Calculate the volume of rainwater collected.

The volume of rainwater forms a cuboid with dimensions $L \times B \times h_{\text{rain}}$.

$V_{\text{rain}} = L \times B \times h_{\text{rain}}$

$V_{\text{rain}} = 600 \text{ cm} \times 400 \text{ cm} \times 1 \text{ cm}$

$V_{\text{rain}} = (600 \times 400 \times 1) \text{ cm}^3$

Step 2: Set up the volume of water in the cylindrical vessel.

Let the height of water in the cylindrical vessel be $h_c$. The volume of water in the cylinder is given by the formula $V = \pi r_c^2 h_c$.

$V_{\text{cylinder}} = \pi r_c^2 h_c$

Substitute the values $r_c = 20$ cm and $\pi = 3.14$:

$V_{\text{cylinder}} = 3.14 \times (20 \text{ cm})^2 \times h_c$

$V_{\text{cylinder}} = 3.14 \times 400 \text{ cm}^2 \times h_c$

Calculate $3.14 \times 400$:

$3.14 \times 400 = 314 \times 4 = 1256$

Step 3: Equate the volumes and solve for $h_c$.

Since the volume of rainwater is transferred to the cylindrical vessel:

$240000 \text{ cm}^3 = 1256 h_c \text{ cm}^3$

Solve for $h_c$:

$h_c = \frac{240000}{1256}$ cm

Let's perform the division. We can simplify the fraction first by dividing both numerator and denominator by 8:

$240000 \div 8 = 30000$

$1256 \div 8 = 157$

$h_c = \frac{30000}{157}$ cm

Now, perform the long division:

$\begin{array}{r} 191.08... \\ 157{\overline{\smash{\big)}\,30000.00\dots}} \\ \underline{-~157}\phantom{00.00} \\ 1430\phantom{0.00} \\ \underline{-~1413}\phantom{0.00} \\ 170\phantom{.00} \\ \underline{-~\phantom{1}0}\phantom{.00} \\ 1700\phantom{0} \\ \underline{-~1570}\phantom{0} \\ 1300\phantom{} \\ \underline{-~1256}\phantom{} \\ 44 \dots \end{array}$

$h_c \approx 191.08$ cm

Step 4: Round the height to the nearest integer.

The calculated height is approximately $191.08$ cm. To round to the nearest integer, look at the first decimal digit. It is 0, which is less than 5. Therefore, we round down.

$h_c \approx 191$ cm

The height of water in the cylindrical vessel will be approximately 191 cm.

Given:

The cylindrical tube is open at both ends.

Thickness of the iron sheet = $2$ cm.

Outer diameter of the tube = $16$ cm.

Length of the tube (height), $h = 100$ cm.

To Find:

The volume of iron used in making the tube, in cubic centimeters.

Solution:

The tube is a hollow cylinder. The volume of material used in a hollow cylinder is the difference between the volume of the outer cylinder and the volume of the inner cylinder.

Volume of iron = Volume of outer cylinder - Volume of inner cylinder.

Let the outer radius be $R$ and the inner radius be $r$. The height of both the outer and inner cylinders is the length of the tube, $h$.

Step 1: Determine the outer radius $R$ and inner radius $r$.

The outer diameter is given as $16$ cm.

Outer radius, $R = \frac{\text{Outer Diameter}}{2} = \frac{16 \text{ cm}}{2} = 8$ cm.

The thickness of the iron sheet is $2$ cm.

Inner radius, $r = \text{Outer radius} - \text{Thickness}$

$r = 8 \text{ cm} - 2 \text{ cm} = 6$ cm.

The height of the tube is $h = 100$ cm.

Step 2: Calculate the volume of the outer cylinder.

The formula for the volume of a cylinder is $V = \pi \times (\text{radius})^2 \times \text{height}$.

Volume of outer cylinder, $V_{\text{outer}} = \pi R^2 h$

$V_{\text{outer}} = \pi \times (8 \text{ cm})^2 \times (100 \text{ cm})$

$V_{\text{outer}} = \pi \times 64 \text{ cm}^2 \times 100 \text{ cm}$

$V_{\text{outer}} = 6400\pi \text{ cm}^3$

Step 3: Calculate the volume of the inner cylinder.

Volume of inner cylinder, $V_{\text{inner}} = \pi r^2 h$

$V_{\text{inner}} = \pi \times (6 \text{ cm})^2 \times (100 \text{ cm})$

$V_{\text{inner}} = \pi \times 36 \text{ cm}^2 \times 100 \text{ cm}$

$V_{\text{inner}} = 3600\pi \text{ cm}^3$

Step 4: Calculate the volume of iron used.

$V_{\text{iron}} = V_{\text{outer}} - V_{\text{inner}}$

$V_{\text{iron}} = 6400\pi \text{ cm}^3 - 3600\pi \text{ cm}^3$

$V_{\text{iron}} = (6400 - 3600)\pi \text{ cm}^3$

$V_{\text{iron}} = 2800\pi \text{ cm}^3$

The volume of iron used in making the tube is $2800\pi$ cubic centimeters. (Note: If a numerical value for $\pi$ was provided, we would substitute it here. Since none was given, the answer is left in terms of $\pi$).

Given:

A semi-circular sheet of metal with diameter 28 cm.

The sheet is bent to form an open conical cup.

To Find:

The capacity (volume) of the conical cup.

Solution:

When a semi-circular sheet is bent into an open conical cup, the following transformations occur:

1. The radius of the semi-circular sheet becomes the slant height of the cone.
2. The circumference of the semi-circular sheet becomes the circumference of the base of the cone.

Step 1: Find the radius of the semi-circular sheet.

Diameter of the semi-circular sheet = 28 cm.

Radius of the semi-circular sheet, $R_{\text{sheet}} = \frac{\text{Diameter}}{2} = \frac{28 \text{ cm}}{2} = 14$ cm.

This radius is the slant height ($l$) of the formed cone.

Step 2: Find the radius of the base of the cone.

The circumference of the semi-circular sheet is half the circumference of a full circle with radius $R_{\text{sheet}}$.

Circumference of semi-circle = $\frac{1}{2} \times (2\pi R_{\text{sheet}}) = \pi R_{\text{sheet}}$

Substitute $R_{\text{sheet}} = 14$ cm:

Circumference of semi-circle = $\pi \times 14 = 14\pi$ cm.

This circumference is equal to the circumference of the base of the cone. Let the radius of the cone's base be $r$.

Circumference of cone base = $2\pi r$

Divide both sides by $2\pi$:

Step 3: Find the height of the cone.

In a right circular cone, the radius ($r$), height ($h$), and slant height ($l$) are related by the Pythagorean theorem:

Substitute the values $l = 14$ cm and $r = 7$ cm into equation (i):

$(14)^2 = (7)^2 + h^2$

$196 = 49 + h^2$

Solve for $h^2$:

$h^2 = 196 - 49$

$h^2 = 147$

Take the square root of both sides to find $h$. Since $h$ is a height, it must be positive.

$h = \sqrt{147}$

Factorize 147: $147 = 3 \times 49 = 3 \times 7^2$.

$h = \sqrt{7^2 \times 3} = 7\sqrt{3}$ cm.

Step 4: Calculate the capacity (volume) of the cone.

The formula for the volume of a cone with radius $r$ and height $h$ is given by:

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

Substitute the values $r = 7$ cm and $h = 7\sqrt{3}$ cm:

$V = \frac{1}{3}\pi (7 \text{ cm})^2 (7\sqrt{3} \text{ cm})$

$V = \frac{1}{3}\pi (49 \text{ cm}^2) (7\sqrt{3} \text{ cm})$

$V = \frac{1}{3} \pi \times 49 \times 7\sqrt{3} \text{ cm}^3$

$V = \frac{343\sqrt{3}}{3}\pi \text{ cm}^3$

The capacity of the conical cup is $\frac{343\sqrt{3}}{3}\pi$ cm³.

Given:

Area of cloth used for the tent = $165$ m$^2$. This area forms the curved surface area (CSA) of the conical tent.

Radius of the base of the conical tent, $r = 5$ m.

Area occupied by one student on the ground = $\frac{5}{7}$ m$^2$.

To Find:

(i) The number of students that can sit in the tent.

(ii) The volume of the cone.

Solution:

(i) Number of students:

The students sit on the ground inside the tent, which is the base of the cone. The base of the cone is a circle with radius $r = 5$ m.

The area of the base of the cone is given by the formula $A_{\text{base}} = \pi r^2$.

Since the area per student is given in terms of sevenths, it is appropriate to use $\pi = \frac{22}{7}$.

$A_{\text{base}} = \frac{22}{7} \times (5 \text{ m})^2$

$A_{\text{base}} = \frac{22}{7} \times 25 \text{ m}^2$

$A_{\text{base}} = \frac{22 \times 25}{7} \text{ m}^2 = \frac{550}{7} \text{ m}^2$

The area occupied by one student on the ground is $\frac{5}{7}$ m$^2$.

The number of students that can sit in the tent is given by:

Number of students $= \frac{\text{Area of the base of the cone}}{\text{Area occupied by one student}}$

Number of students $= \frac{\frac{550}{7} \text{ m}^2}{\frac{5}{7} \text{ m}^2}$

Number of students $= \frac{550/7}{5/7} = \frac{550}{7} \times \frac{7}{5}$

Number of students $= \frac{550 \times \cancel{7}^1}{\cancel{7}^1 \times 5} = \frac{550}{5}$

Number of students $= 110$

(ii) Volume of the cone:

The formula for the volume of a cone is $V = \frac{1}{3}\pi r^2 h$, where $r$ is the radius and $h$ is the height.

We know $r = 5$ m. We need to find the height $h$.

The curved surface area (CSA) of the cone is given as $165$ m$^2$. The formula for CSA is $\text{CSA} = \pi r l$, where $l$ is the slant height.

So, $\pi r l = 165$

Substitute $r = 5$ m and use $\pi = \frac{22}{7}$:

$\frac{22}{7} \times 5 \times l = 165$

$\frac{110}{7} l = 165$

Solve for $l$:

$l = 165 \times \frac{7}{110}$

$l = \frac{\cancel{165}^{15} \times 7}{\cancel{110}^{10}}$ (Dividing by 11)

$l = \frac{\cancel{15}^{3} \times 7}{\cancel{10}^{2}}$ (Dividing by 5)

$l = \frac{3 \times 7}{2} = \frac{21}{2} = 10.5$ m

The slant height of the cone is $l = 10.5$ m.

Now, we find the height $h$ using the relationship $l^2 = r^2 + h^2$ (Pythagorean theorem):

$h^2 = l^2 - r^2$

$h^2 = (10.5 \text{ m})^2 - (5 \text{ m})^2$

$h^2 = 110.25 \text{ m}^2 - 25 \text{ m}^2$

$h^2 = 85.25 \text{ m}^2$

$h = \sqrt{85.25} \text{ m}$

We can write $85.25$ as a fraction: $85.25 = \frac{8525}{100}$. Simplify the fraction:

$\frac{8525}{100} = \frac{1705}{20} = \frac{341}{4}$

So, $h = \sqrt{\frac{341}{4}} \text{ m} = \frac{\sqrt{341}}{\sqrt{4}} \text{ m} = \frac{\sqrt{341}}{2} \text{ m}$.

Now, calculate the volume of the cone using $V = \frac{1}{3}\pi r^2 h$ with $r=5$ m, $h=\frac{\sqrt{341}}{2}$ m, and $\pi = \frac{22}{7}$.

$V = \frac{1}{3} \times \frac{22}{7} \times (5 \text{ m})^2 \times \left(\frac{\sqrt{341}}{2} \text{ m}\right)$

$V = \frac{1}{3} \times \frac{22}{7} \times 25 \times \frac{\sqrt{341}}{2} \text{ m}^3$

$V = \frac{22 \times 25 \times \sqrt{341}}{3 \times 7 \times 2} \text{ m}^3$

Cancel common factors:

$V = \frac{\cancel{22}^{11} \times 25 \times \sqrt{341}}{3 \times 7 \times \cancel{2}^1} \text{ m}^3$

$V = \frac{11 \times 25 \times \sqrt{341}}{21} \text{ m}^3$

(i) The number of students that can sit in the tent is 110.

(ii) The volume of the cone is $\frac{275\sqrt{341}}{21}$ m³.

Given:

The tank is hemispherical.

Internal diameter of the tank = $14$ m.

Volume of water initially in the tank = $50$ kilolitres (kL).

To Find:

The volume of water pumped into the tank to fill it to its capacity.

Solution:

Step 1: Calculate the radius of the hemispherical tank.

The radius ($r$) is half of the internal diameter.

$r = \frac{\text{Diameter}}{2} = \frac{14 \text{ m}}{2} = 7$ m

Step 2: Calculate the total capacity (volume) of the hemispherical tank.

The formula for the volume of a hemisphere with radius $r$ is $V = \frac{2}{3}\pi r^3$.

Substitute the radius $r = 7$ m. We can use $\pi = \frac{22}{7}$ for easier calculation.

$V_{\text{tank}} = \frac{2}{3} \times \frac{22}{7} \times (7 \text{ m})^3$

$V_{\text{tank}} = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \text{ m}^3$

Cancel out one factor of 7:

$V_{\text{tank}} = \frac{2}{3} \times 22 \times \cancel{7}^1 \times 7 \times 7 \text{ m}^3$

$V_{\text{tank}} = \frac{2}{3} \times 22 \times 49 \text{ m}^3$

$V_{\text{tank}} = \frac{44 \times 49}{3} \text{ m}^3$

Calculate $44 \times 49$:

$\begin{array}{cc}& & 4 & 4 \\ \times & & 4 & 9 \\ \hline & 3 & 9 & 6 \\ 1 & 7 & 6 & \times \\ \hline 2 & 1 & 5 & 6 \\ \hline \end{array}$

$V_{\text{tank}} = \frac{2156}{3}$ m$^3$

Step 3: Convert the initial volume of water to cubic meters.

We are given that the tank contains 50 kilolitres of water. We need to convert this to cubic meters to match the unit of the tank's volume.

We know that $1$ litre (L) = $1000$ cubic centimeters (cm$^3$).

We also know that $1$ cubic meter (m$^3$) = $100 \text{ cm} \times 100 \text{ cm} \times 100 \text{ cm} = 1000000$ cm$^3$.

So, $1 \text{ m}^3 = 1000000 \text{ cm}^3$.

Also, $1000 \text{ L} = 1 \text{ m}^3$. This means $1$ kilolitre ($1000$ L) is equal to $1$ cubic meter.

$1$ kL = $1$ m$^3$

Initial volume of water = $50$ kL $= 50$ m$^3$.

Step 4: Calculate the volume of water to be pumped.

The volume of water to be pumped is the difference between the total capacity of the tank and the volume of water already in the tank.

Volume to be pumped = Total capacity of tank - Initial volume of water

Volume to be pumped $= V_{\text{tank}} - V_{\text{initial}}$

Volume to be pumped $= \frac{2156}{3} \text{ m}^3 - 50 \text{ m}^3$

To subtract, find a common denominator:

Volume to be pumped $= \frac{2156}{3} - \frac{50 \times 3}{3}$

Volume to be pumped $= \frac{2156 - 150}{3}$

Perform the subtraction in the numerator:

$\begin{array}{cccc}& 2 & 1 & 5 & 6 \\ - & \phantom{2} & 1 & 5 & 0 \\ \hline & 2 & 0 & 0 & 6 \\ \hline \end{array}$

Volume to be pumped $= \frac{2006}{3}$ m$^3$

This value can also be expressed as a mixed number or a decimal. $\frac{2006}{3} = 668$ with a remainder of 2, so it is $668 \frac{2}{3}$ m$^3$, or approximately $668.67$ m$^3$. Since the question asks to calculate the volume and doesn't specify the format or rounding, the exact fraction is a valid answer.

The volume of water pumped into the tank is $\frac{2006}{3}$ m³ (or $668\frac{2}{3}$ m³ or approximately 668.67 m³).

Given:

The ratio of the volumes of two spheres is $64 : 27$.

To Find:

The ratio of their surface areas.

Solution:

Let the radii of the two spheres be $r_1$ and $r_2$.

The formula for the volume of a sphere with radius $r$ is $V = \frac{4}{3}\pi r^3$.

The volumes of the two spheres are $V_1 = \frac{4}{3}\pi r_1^3$ and $V_2 = \frac{4}{3}\pi r_2^3$.

The given ratio of their volumes is:

$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$

We are given that $\frac{V_1}{V_2} = \frac{64}{27}$.

So, we have:

To find the ratio of the radii, we take the cube root of both sides of equation (i):

$\frac{r_1}{r_2} = \sqrt[3]{\frac{64}{27}}$

$\frac{r_1}{r_2} = \frac{\sqrt[3]{64}}{\sqrt[3]{27}}$

Now, we consider the surface area of the spheres. The formula for the surface area of a sphere with radius $r$ is $A = 4\pi r^2$.

The surface areas of the two spheres are $A_1 = 4\pi r_1^2$ and $A_2 = 4\pi r_2^2$.

The ratio of their surface areas is:

$\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2$

Substitute the ratio of radii from equation (ii) into this expression:

$\frac{A_1}{A_2} = \left(\frac{4}{3}\right)^2$

$\frac{A_1}{A_2} = \frac{4^2}{3^2}$

The ratio of their surface areas is $16 : 9$.

The ratio of the surface areas of the two spheres is 16 : 9.

Given:

Side length of the cube, $a = 4$ cm.

A sphere is inscribed in the cube, touching its sides.

To Find:

The volume of the gap between the cube and the sphere.

Solution:

The volume of the gap is the difference between the volume of the cube and the volume of the sphere.

Volume of gap = Volume of cube - Volume of sphere

Step 1: Calculate the volume of the cube.

The formula for the volume of a cube with side length $a$ is $V_{\text{cube}} = a^3$.

Substitute the given side length $a = 4$ cm:

$V_{\text{cube}} = (4 \text{ cm})^3$

$V_{\text{cube}} = 4 \times 4 \times 4 \text{ cm}^3$

Step 2: Determine the radius of the inscribed sphere and calculate its volume.

When a sphere is inscribed in a cube and touches all its sides, the diameter of the sphere is equal to the edge length of the cube.

Diameter of sphere = Side of cube = $a = 4$ cm.

The radius of the sphere, $r$, is half of its diameter.

$r = \frac{\text{Diameter}}{2} = \frac{4 \text{ cm}}{2} = 2$ cm.

The formula for the volume of a sphere with radius $r$ is $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.

Substitute the radius $r = 2$ cm:

$V_{\text{sphere}} = \frac{4}{3}\pi (2 \text{ cm})^3$

$V_{\text{sphere}} = \frac{4}{3}\pi (8) \text{ cm}^3$

Step 3: Calculate the volume of the gap.

$V_{\text{gap}} = V_{\text{cube}} - V_{\text{sphere}}$

Substitute the calculated volumes:

$V_{\text{gap}} = 64 \text{ cm}^3 - \frac{32}{3}\pi \text{ cm}^3$

We can factor out the common term 32:

The volume of the gap in between the cube and the sphere is $\frac{32(6 - \pi)}{3}$ cm³.

Given:

A sphere and a right circular cylinder have the same radius.

The volume of the sphere is equal to the volume of the cylinder.

To Find:

The percentage by which the diameter of the cylinder exceeds its height.

Solution:

Let the common radius of the sphere and the cylinder be $r$.

Let the height of the cylinder be $h$.

The formula for the volume of a sphere with radius $r$ is given by:

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

The formula for the volume of a right circular cylinder with radius $r$ and height $h$ is given by:

$V_{\text{cylinder}} = \pi r^2 h$

According to the problem statement, the volumes of the sphere and the cylinder are equal:

Substitute the respective volume formulas into the equation:

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

Assuming the radius $r$ is not zero (as it's a geometrical object), we can divide both sides of the equation by $\pi r^2$:

$\frac{4}{3} r = h$

This equation gives us the relationship between the height ($h$) and the radius ($r$) of the cylinder.

The diameter of the cylinder is $D = 2r$.

We need to find the percentage by which the diameter exceeds the height. The excess amount is the difference between the diameter and the height:

Excess $= \text{Diameter} - \text{Height}$

Excess $= 2r - h$

Substitute the expression for $h$ in terms of $r$ ($h = \frac{4}{3}r$) into the excess calculation:

Excess $= 2r - \frac{4}{3}r$

Combine the terms by finding a common denominator:

Excess $= \left(\frac{6}{3} - \frac{4}{3}\right)r$

Excess $= \frac{2}{3}r$

To find the percentage by which the diameter exceeds the height, we calculate the ratio of the excess to the height and multiply by 100%:

Percentage excess $= \left(\frac{\text{Excess}}{\text{Height}}\right) \times 100\%$

Substitute the expressions for Excess and Height:

Percentage excess $= \left(\frac{\frac{2}{3}r}{\frac{4}{3}r}\right) \times 100\%$

Cancel the common term $r$ (assuming $r \neq 0$):

Percentage excess $= \left(\frac{2/3}{4/3}\right) \times 100\%$

Percentage excess $= \left(\frac{2}{3} \times \frac{3}{4}\right) \times 100\%$

Percentage excess $= \frac{2}{4} \times 100\%$

Percentage excess $= \frac{1}{2} \times 100\%$

Percentage excess $= 50\%$

The diameter of the cylinder exceeds its height by 50%.

Given:

Number of circular plates = $30$.

Radius of each plate, $r_{\text{plate}} = 14$ cm.

Thickness of each plate, $t_{\text{plate}} = 3$ cm.

The plates are placed one above another to form a cylindrical solid.

To Find:

(i) The total surface area of the cylinder so formed.

(ii) The volume of the cylinder so formed.

Solution:

When 30 circular plates of radius 14 cm and thickness 3 cm are stacked one above another, they form a cylinder.

The radius of the base of the cylinder formed is the radius of a single plate.

Radius of cylinder, $r = r_{\text{plate}} = 14$ cm.

The height of the cylinder formed is the total thickness of all 30 plates.

Height of cylinder, $h = \text{Number of plates} \times \text{Thickness of each plate}

$h = 30 \times 3 \text{ cm} = 90$ cm.

(i) Total Surface Area of the cylinder:

The formula for the total surface area (TSA) of a closed cylinder with radius $r$ and height $h$ is:

TSA $= 2\pi r (r + h)$

Substitute the values $r = 14$ cm and $h = 90$ cm. We will use $\pi = \frac{22}{7}$.

TSA $= 2 \times \frac{22}{7} \times 14 \text{ cm} \times (14 \text{ cm} + 90 \text{ cm})$

TSA $= 2 \times \frac{22}{7} \times 14 \text{ cm} \times 104 \text{ cm}$

Cancel 7 from the denominator with 14 in the numerator:

TSA $= 2 \times 22 \times \cancel{14}^2 \times 104 \text{ cm}^2$

TSA $= 2 \times 22 \times 2 \times 104 \text{ cm}^2$

TSA $= 44 \times 2 \times 104 \text{ cm}^2$

TSA $= 88 \times 104 \text{ cm}^2$

Calculate $88 \times 104$:

$\begin{array}{cc}& & 1 & 0 & 4 \\ \times & & & 8 & 8 \\ \hline &&& 8 & 3 & 2 \\ & 8 & 3 & 2 & \times \\ \hline & 9 & 1 & 5 & 2 \\ \hline \end{array}$

(ii) Volume of the cylinder:

The formula for the volume of a cylinder with radius $r$ and height $h$ is:

$V = \pi r^2 h$

Substitute the values $r = 14$ cm and $h = 90$ cm. Use $\pi = \frac{22}{7}$.

$V = \frac{22}{7} \times (14 \text{ cm})^2 \times 90 \text{ cm}$

$V = \frac{22}{7} \times (14 \times 14) \text{ cm}^2 \times 90 \text{ cm}$

$V = \frac{22}{7} \times 196 \times 90 \text{ cm}^3$

Cancel 7 from the denominator with 196 in the numerator ($196 \div 7 = 28$):

$V = 22 \times \cancel{196}^{28} \times 90 \text{ cm}^3$

$V = 22 \times 28 \times 90 \text{ cm}^3$

Calculate $22 \times 28$:

$\begin{array}{cc}& & 2 & 2 \\ \times & & 2 & 8 \\ \hline & 1 & 7 & 6 \\ & 4 & 4 & \times \\ \hline & 6 & 1 & 6 \\ \hline \end{array}$

$V = 616 \times 90 \text{ cm}^3$

$V = 616 \times 9 \times 10 \text{ cm}^3$

Calculate $616 \times 9$:

$\begin{array}{cc}& & 6 & 1 & 6 \\ \times & & & & 9 \\ \hline & 5 & 5 & 4 & 4 \\ \hline \end{array}$

$V = 5544 \times 10 \text{ cm}^3$

(i) The total surface area of the cylinder so formed is 9152 cm².

(ii) The volume of the cylinder so formed is 55440 cm³.