2 Marks Questions - MATHS STD 10 Questions - Vidyadip

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Question 12 Marks

66 cubic cm of silver is drawn into a wire 1mm in diameter. Calculate the length of the wire in metres.

Answer

We have,
Radius of wire, $\text{r}=\frac{1}{2}=0.5\text{mm}=0.05\text{cm}$
Let the length of the wire be l.
As,
Volume of the wire $= 66cm^3$
$\Rightarrow\pi\text{r}^2\text{l}=66$
$\Rightarrow\frac{22}{7}\times0.05\times0.05\times\text{l}=66$
$\Rightarrow\text{l}=66\times\frac{7}{22\times0.05\times0.05}$
$\Rightarrow\text{l}=8400\text{cm}=84\text{m}$
So, the length of the wire is 84 m.

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Question 22 Marks

A solid metallic cuboid of dimension 9m × 8m × 2m is melted and recast into solid cubes of edge 2m. Find the number of cubes so formed.

Answer

Volume of cuboid = Number of cubes × Volume of cubes
$9 \times 8 \times 2 = m \times (2)^3$
$144 = m \times 8$
$\frac{144}{8}=\text{m}$
$18 = m$

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Question 32 Marks

The ratio between the radius of the base and the height of a cylinder is $2 : 3$. If the volume of the cylinder is $12936cm^3$, then find the radius of the base of the cylinder.

Answer

Let the radius of the base and the height of the cylinder be r and h, respectively.
We have,
$r : h = 2 : 3$ i.e. $\frac{\text{r}}{\text{h}}=\frac{2}{3}$
or $\text{h}=\frac{3\text{r}}{2}\ ...(\text{i})$
As,
Volume of the cylinder $= 12936cm^3$
$\Rightarrow\pi\text{r}^2\text{h}=12936$
$\Rightarrow\frac{22}{7}\times\text{r}^2\times\frac{3\text{r}}{2}=12936$ $[\text{Using}(\text{i})]$
$\Rightarrow\frac{33}{7}\times\text{r}^3=12936$
$\Rightarrow\text{r}^3=12936\times\frac{7}{33}$
$\Rightarrow\text{r}^3=2744$
$\Rightarrow\text{r}^3\sqrt[3]{2744}$
$\Rightarrow\text{r}=\sqrt[3]{2744}$
$\therefore\text{r}=14\text{cm}$
So, the radius of the base of the cylinder is 14cm.

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Question 42 Marks

A hemispherical bowl of internal radius 9cm is full of water. This water is to be filled in cylindrical bottles of diameter 3cm and height 4cm. Find the number of bottles needed to fill the whole water of the bowl.

Answer

Radius of hemisphere $=9\text{cm}$
Volume of hemisphere $=\frac{2}{3}\pi\text{r}^3$
$=\Big(\frac{2}{3}\pi\times9\times9\times9\Big)\text{cm}^3$
Radius of each bottle $=\frac{3}{2}\text{cm}$
Height of each bottle $=4\text{cm}$
Volume of each bottle $=\pi\text{r}^2\text{h}$
$=\Big(\pi\times\frac{3}{2}\times\frac{3}{2}\times4\Big)\text{cm}^3$
Number of bottles $=\frac{\text{Volume of the hemisphere}}{\text{Volume of each bottle}}$
$=\frac{2\pi\times9\times9\times9\times2\times2}{3\times\pi\times3\times3\times4}$
$=54$

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Question 52 Marks

The radii of two cylinders are in the ratio of $2 : 3$ and their heights are in the ratio of $5 : 3$. Find the ratio of their volumes.

Answer

Let the radii of the cylinders be $r_1$ and $r_2$; and their heights be $h_1$ and $h_2.$
We have,
$\text{r}_1:\text{r}_2=2:3$ or $\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}\ ....(\text{i})$
and $\text{h}_2:\text{h}_2=5:3$ or $\frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}\ ....(\text{ii})$
Now,
The ratio of the volumes of the cylinders $=\frac{\text{Volume of the first cylinder}}{\text{Volume of the second cylinder}}$
$=\frac{\pi\text{r}_1^2\text{h}_1}{\pi\text{r}_2^2\text{h}_2}$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\times\frac{\text{h}_1}{\text{h}_2}$
$=\Big(\frac{2}{3}\Big)^2\times\frac{2}{3}$ [Using (i) and (ii)]
$=\frac{20}{27}$
$=20:27$
So, the ratio of the volumes of the given cylinders is $20 : 27.$

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Question 62 Marks

The surface areas of a sphere and a cube are equal. Find the ratio of their volumes.

Answer

Surface area of the sphere $=4\pi\text{r}^2$
Surface area of the cube $=6\text{a}^2$
Therefore
$4\pi\text{r}^2=6\text{a}^2$
$\Rightarrow2\pi\text{r}^2=3\text{a}^2$
$\Rightarrow\text{r}^2=\frac{3\text{a}^2}{2\pi}$
$\Rightarrow\text{r}=\sqrt{\frac{3}{2\pi}\text{a}}$
Ratio of their volumes = $=\frac{\frac{4}{3}\pi\text{r}^3}{\text{a}^3}=\frac{4\pi\text{r}^3}{3\text{a}^3}$
$=\frac{4\pi}{3\text{a}^3}\times\frac{3\sqrt{3}\text{a}^3}{2\pi\sqrt{2\pi}}\Big[\sqrt{\frac{3}{2\pi}}\text{a}\Big]$
$=\frac{2\sqrt{3}}{\sqrt{2}\pi}$
$=\frac{2\times\sqrt{3}}{\sqrt{2}\times\frac{\sqrt{22}}{\sqrt{7}}}$
$=\frac{2\times\sqrt{3}\times\sqrt{7}}{\sqrt{2}\times\sqrt{2}\times\sqrt{11}}$
$=\frac{\sqrt{21}}{\sqrt{11}}$
Thus, the ratio of their volumes is $\sqrt{21}:\sqrt{11}.$

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Question 72 Marks

A river 1.5 m deep and 36m wide is flowing at the rate of 3.5km/hr. Find the amount of water (in cubi metres) that runs into the sea per minute.

Answer

We have,
Depth of the river, h = 1.5m,
Width of the river, b = 36m,
Speed of the flowing water, $\text{l}=3.5\text{km/hr}=\frac{3.5\times1000\text{m}}{6=\text{min}}=\frac{175}{3}\text{m}/\text{min}$
Now,
The amount of water that runs into the sea per minute = lbh
$=\frac{175}{3}\times36\times1.5$
$=3150\text{m}^3/\text{min}$

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Question 82 Marks

A spherical shell of lead whose external and internal diameters are 24cm and 18cm, is melted and recast into a right circular cylinder 37cm high. Find the diameter of the base of the cylinder.

Answer

External diameter of the shell = 24cm
External radius of the shell = 12cm
Internal diameter of the shell = 18cm
Internal radius of the shell = 9cm
Volume of the shell $=\frac{4}{3}\pi(12^3-9^3)=\frac{4}{3}\pi(1728-729)$
$=\frac{4}{3}\pi\times(999)=4\pi\times(333)\text{cm}^3$
Height of cylinder = 37cm
Let radius of cylinder be rem.
Volume of cylinder $=\pi\text{r}^2\text{h}=37\pi\text{h}^2\text{cm}^3$
Volume of the shell = Volume of cylinder
Or, $4\pi\times(333)=37\pi\text{r}^2$
$\Rightarrow\text{r}^2=\frac{4\times333}{37}=4\times9$
$\Rightarrow\text{r}\sqrt{4\times9}=\sqrt{36}=6\text{cm}$
So, diameter of the base of the cylinder = 2r = 12cm.

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Question 92 Marks

Metallic spheres of radii $6\ cm, 8\ cm$ and $10\ cm$ respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer

We have,
the radii $r_1 = 6cm, r_2 = 8cm$ and $r_3 = 10cm$
Let the radius of the resulting sphere be R.
As,
Volume of resulting sphere = Volume of three metallic spheres
$\Rightarrow\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\pi\text{r}_1^3+\frac{4}{3}\pi\text{r}_2^3+\frac{4}{3}\pi\text{r}_3^3$
$\Rightarrow\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\pi\big(\text{r}_1^3+\text{r}_2^3+\text{r}_3^3\big)$
$\Rightarrow\text{R}^3=\text{r}_1^3+\text{r}_2^3+\text{r}_3^3$
$\Rightarrow\text{R}^3=6^3+8^3+10^3$
$\Rightarrow\text{R}^3=216+512+1000$
$\Rightarrow\text{R}^3=1728$
$\Rightarrow\text{R}=\sqrt[3]{1728}$
$\Rightarrow\text{R}=12\text{cm}$
So, the radius of the resulting sphere is $12cm.$

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Question 102 Marks

A cone of height 20cm and radius of base 5cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.

Answer

We have,
The base radius of the cone, r = 5cm and
The height of the cone, h = 20cm
Let the radius of the sphere be R.
As,
Volume of sphere = Volume of cone
$\Rightarrow\frac{3}{4}\pi\text{R}^3=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow\text{R}^3=\frac{\pi\text{r}^2\times3}{3\times4\pi}$
$\Rightarrow\text{R}^3=\frac{\text{r}^2\text{h}}{4}$
$\Rightarrow\text{R}^3=\frac{5\times5\times20}{4}$
$\Rightarrow\text{R}^3=125$
$\Rightarrow\text{R}=\sqrt[3]{125}$
$\Rightarrow\text{R}=5\text{cm}$
$\Rightarrow$ Diameter of the sphere = 2R 2 × 5 = 10cm
So, the diameter of the sphere is 10cm.

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Question 112 Marks

The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.

Answer

Let the edges of the cubes be a and b.
As,
$\frac{\text{Volume of the first cube}}{\text{Volume of the second cube}}=\frac{8}{27}$
$\Rightarrow\frac{\text{a}^3}{\text{b}^3}=\frac{8}{27}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\sqrt[3]{\frac{8}{27}}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{2}{3}\ ...(1)$
Now,
The ratio of the surface areas of the cubes $=\frac{\text{Surface area of the first cube}}{\text{Surface area of the second cube}}$
$=\frac{6\text{a}^2}{6\text{b}^2}$
$=\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=\Big(\frac{2}{3}\Big)^3$ [Using (i)]
$=\frac{4}{9}$
$=4:9$
So, the ratio of the surface areas of the given cubes is 4 : 9.

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Question 122 Marks

Three cubes of iron whose edges are $6\ cm, 8\ cm$ and 10cm, respectively are melted and formed into a single cube. Find the edge of the new cube formed.

Answer

We have,
Edges of the cubes : $a_1 = 6cm, a_2 = 8cm$ and $a_3 = 10cm$
Let the edge of the new cube so formed be a.
As,
Volume of the new cube so formed $= a_1^3 + a_2^3 + a_3^3$
$\Rightarrow a^3 = 6^3 + 8^3 + 10^3$
$\Rightarrow a^3 = 216 + 512 + 1000$
$\Rightarrow a^3 1728$
$\Rightarrow\text{a}=\sqrt[3]{1728}$
$\therefore$ $a =12cm$
So, the edge of the new cube so formed is $12cm.$

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Question 132 Marks

A copper rod of diameter 2cm and length 10cm is drawn into a wire of uniform thickness and length 10m. Find the thickness of the Wire.

Answer

We have,
The radius of the copper rod, $\text{R}=\frac{2}{2}=1\text{cm},$
The height of the copper rod, H = 10cm and
The height of the wire, h = 10m = 1000cm
Let the radius of the wire be r.
As,
Volume of the wire = Volume of the rod
$\Rightarrow\pi\text{r}^3\text{h}=\pi\text{R}^2\text{H}$
$\Rightarrow\text{r}^2\text{h}=\text{R}^2\text{H}$
$\Rightarrow\text{r} ^2\times1000=1\times10$
$\Rightarrow\text{r}^2=\frac{1}{100}$
$\Rightarrow\text{r}=\sqrt{\frac{1}{100}}$
$\Rightarrow\text{r}=\frac{1}{10}$
$\Rightarrow\text{r}=0.1\text{cm}$
$\Rightarrow$ The diameter of the wire = 2r = 2 × 0.1 = 0.2cm
$\therefore$ The thickness of the wire = 0.2cm
So, the thickness of the wire is 0.2cm or 2mm.

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Question 142 Marks

How many cubes of 10cm edge can be put in a cubical box of m edge?

Answer

We have,
Edge of the cube, a = 10cm and
dge of the cubical box, l = 1m = 100cm
Now,
The number of cubes that can be put in the box $=\frac{\text{Volume of the cubical box}}{\text{volume of the cube}}$
$=\frac{\text{l}^2}{\text{a}^3}$
$=\frac{100^3}{10^3}$
$=10^3$
$=1000$
So, the number of cubes that can be put in the cubical box is 1000.

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Question 152 Marks

The volume of a cube is $729cm^3$. Find its surface area.

Answer

Let the edge of the cube be a.
As,
Volume of the cube $= 729 cm^3$
$\Rightarrow\text{a}^3=729$
$\Rightarrow\text{a}=\sqrt[3]{729}$
$\Rightarrow486\text{cm} ^2$
Now,
Surface area of the cube $= 6a^2$
$= 6 \times 9 \times 9$
$= 486 cm^2$
So, the surface area of the cube is 486 cm2.

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Question 162 Marks

A solid metallic sphere of diameter 21cm is melted and recast into small cones of diameter 3.5cm and height 3cm each. Find the number of cones so formed.

Answer

Radius of sphere $=\frac{21}{2}\text{cm}$
Volume of the metallic sphere $=\frac{4}{3}\pi\text{r}^3$
$=\Big(\frac{4}{3}\pi\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}\Big)\text{cm}^3$
Radius of cone $=\frac{3.5}{2}\text{cm}$
Height of cone = 3cm
Volume of each small cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\pi\times\frac{35}{20}\times\frac{35}{20}\times3\Big)\text{cm}^3$

Number of cones $=\frac{\text{Volume of the metallic sphere}}{\text{Volume of each cone}}$
$=\frac{4\times\pi\times21\times21\times3\times20\times20}{3\times2\times2\times2\times\pi\times35\times3\times3\times3}$
$=504$

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Question 172 Marks

A solid metal cone with base radius 12cm and height 24cm is melted to form solid spherical balls of diameter 6cm each. Find the number of balls formed.

Answer

Radius of cone = 12cm
Height of cone = 24cm
Volume of the metallic cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\pi\times(12)^2\times24$

Radius of spherical ball $=\frac{6}{2}\text{cm}=3\text{cm}$

Volume of each spherical ball $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\pi\times(3)^3$

Number of balls formed $=\frac{\text{Volume of the metallic cone}}{\text{Volume of each spherical ball}}$
$=\frac{\pi\times12\times12\times24\times3}{3\times4\times\pi\times3\times3\times3}$
$=32$

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Question 182 Marks

A solid metal cone with radius of base 12cm and height 24cm is melted to form solid spherical balls of diameter 6cm each. Find the number of balls thus formed.

Answer

Radius of the cone = 12cm
Height of the cone = 24cm
Volume $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\times12\times12\times24$
$=48\times24\times\pi\text{cm}^3$
Radius of each ball = 3cm
Volume of each ball $=-\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\times3\times3\times3=36\pi\text{cm}^3$
Total number of balls formed by melting the cone $=\frac{\text{volume of cone}}{\text{volume of a ball}}=\frac{48\times24\pi}{36\pi}=32$

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Question 192 Marks

A toy is in the form of a cone mounted on a hemisphere of common base radius 7cm. The total height of the toy is 31cm. Find the total surface area of the toy.

Answer

Radius of the hemisphere = Radius of the cone = 7cm
Height of the cone = (31 - 7)cm = 24
Slant height of the cone, $\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(7)^2+(24)^2}$
$=\sqrt{49+576}$
$=\sqrt{625}$
$=25\text{cm}$
Total surface area of the toy = (Curved surface area of the hemisphere) + (Curved surface area of the cone)
$=2\pi\text{r}^2+\pi\text{r}\text{l}$
$=\pi\times\text{r}\times(2\text{rl})$
$=\frac{22}{7}\times7\times(14+25)\text{cm}^2$
$=858\text{cm}^2$

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Question 202 Marks

The radii of internal and external surfaces of a hollow spherical shell are 3cm and 5cm, respectively. It is melted and recast into a solid cylinder of diameter $14\ cm$. Find the height of the cylinder.

Answer

We have,
The internal base radius of spherical shell, $r_1 = 3cm,$
The external base radius of spherical shell, $r_2 = 5cm$ and
The base radius of solid cylinder, $\text{r}=\frac{14}{2}=7\text{cm}$
Let the height of the cylinder be h.
As,
Volume of solid cylinder= Volume of spherical shell
$\Rightarrow\pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{r}_2^3=\frac{4}{3}\pi\text{r}_1^3$
$\Rightarrow\pi\text{r}^2\text{h}=\frac{4}{3}\pi\big(\text{r}_2^3-\text{r}_1^3\big)$
$\Rightarrow\text{r}^2\text{h}=\frac{4}{3}(\text{r}^3_2-\text{r}_1^3)$
$\Rightarrow7\times7\times\text{h}=\frac{4}{3}(5^3-3^3)$
$\Rightarrow49\times\text{h}=\frac{4}{3}(125-27)$
$\Rightarrow\text{h}=\frac{4}{3}\times\frac{98}{49}$
$\therefore\text{h}=\frac{8}{3}\text{cm}$
So, the height of the cylinder is $\frac{8}{3}$ cm

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Question 212 Marks

The internal and external diameters of a hollow hemispherical shell are 6cm and 10cm. respectively. It is melted and recast into a solid cone of base diameter 14cm. Find the height of the cone so formed.

Answer

Internal diameter of the hemispherical shell = 6cm
Therefore. internal radius of the hemispherical shell = 3cm
External diameter of the hemispherical shell = 10cm
External radius of the hemispherical shell = 5cm
Volume of hemispherical shell $=\frac{2}{3}\pi(5^3-3^3)=\frac{196}{3}\times\frac{22}{7}=\frac{616}{3}\text{cm}^3$
Radius of cone = 7cm
Let the height of the cone be h cm.
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\times\frac{22}{7}\times7\times7\text{h}=\frac{154\text{h}}{3}\text{cm}^3$
The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,
$\frac{154\text{h}}{3}=\frac{616}{3}$
$\Rightarrow\text{h}=\frac{616}{154}=4\text{cm}$

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Question 222 Marks

A drinking glass is in the shape of the frustum of a cone of height 21cm with 6cm and 4cm as the diameters of its two circular ends. Find the capacity of the glass.

Answer

Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h.
Then, $\text{R}\Rightarrow\frac{6}{2}\text{cm}=3\text{cm},\text{r}\Rightarrow\frac{4}{2}\text{cm}=2\text{cm},\text{h}=21\text{cm}$
Capacity of the glass = Capacity of the frustum of the cone
$=\frac{\pi\text{h}}{3}\big[\text{R}^2+\text{r}^2+\text{Rr}\big]$
$=\frac{22}{7}\times\frac{1}{3}\times21\times\big[(3)^2+(2)^2+(3\times2)\big]\text{cm}^3$
$=(22\times19)\text{cm}^3$
$=418\text{cm}^3$

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Question 232 Marks

A hemispherical bowl of internal diameter 30cm is full of a liquid. This liquid is poured into cylindrical bottles of diameter 5cm and height 6cm each. How many bottles are required?

Answer

Radius of hemispherical ball $=\frac{30}{2}=15\text{cm}$
Volume of hemispherical bowl $=\frac{2}{3}\pi\text{r}^3$
$=\Big(\frac{2}{3}\pi\times15\times15\times15\Big)\text{cm}^3$
Radius of each bottle $=\frac{5}{2}\text{cm}$
Height of each bottle $=6\text{cm}$
Volume of each bottle $=\pi\text{r}^2\text{h}$
$=\Big(\pi\times\frac{5}{2}\times\frac{5}{2}\times6\Big)\text{cm}^3$
Number of bottles required $=\frac{\text{Volume of the hemispherical bowl}}{\text{Volume of each bottle}}$
$=\frac{2\times\pi\times15\times15\times15\times2\times2}{3\times\pi\times5\times5\times6}$
$=60$

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Question 242 Marks

A hemispherical bowl of internal diameter 30cm contains some liquid. This liquid is to be filled into cylindrical-shaped bottles each of diameter 5cm and height 6cm find the number of bottles necessary to empty the bowl.

Answer

Inner diameter of the bowl = 30cm
Inner radius of the bowl $=\frac{30\text{cm}}{2}=15\text{cm}$
Inner volume of the bowl = Volume of liquid $=\frac{2}{3}\pi\text{r}^3=\frac{2}{3}\times\pi\times15^3\text{cm}^3$
Radius of each bottle = 2.5cm
Height = 6cm
Volume of each bottle $=\pi\text{r}^2\text{h}=\pi\times\frac{5}{2}\times\frac{5}{2}\times6=\frac{75\pi}{2}\text{cm}^3$
Total number of bottles required $=\frac{\big[\frac{2}{3}\pi\times15\times15\times15\big]}{\frac{75\pi}{2}}=\frac{2\pi\times15\times15\times15\times2}{3\times75\pi}=15\times4=60$

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Question 252 Marks

The diameter of a sphere is 42cm. It is melted and drawn into a cylindrical wire of diameter 2.8cm. Find the length of the wire.

Answer

Radius of the sphere $=\frac{42}{2}=21\text{cm}$ Volume of the sphere $=\frac{4}{3}\pi\text{r}^3$ $=\Big(\frac{4}{3}\pi\times21\times21\times21\Big)\text{cm}^3$ Radius of the wire $=\frac{2.8}{2}=1.4\text{cm}$ Let the length of the wire be h cm. Then, Volume of the wire $=\pi\text{r}^2\text{h}$ $\Big(\pi\times\frac{14}{10}\times\frac{14}{10}\times\text{h}\Big)\text{cm}^3$ Therefore,$\frac{4}{3}\pi\times21\times21\times21=\pi\times\frac{14}{10}\times\frac{14}{10}\times\text{h}$
$\Rightarrow12348=\frac{49}{25}\times\text{h}$
$\Rightarrow\text{h}=\Big(\frac{12348\times25}{49}\Big)$
$\Rightarrow\text{h}=6300\text{cm}$
$\Rightarrow\text{h}=63\text{m}$
Hence, the length of the wire is 63m.

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Question 262 Marks

The slant height of the frustum of a cone is $4\ cm$ and the perimeters (i.e. circumferences) of its circular ends are $18\ cm$ and $6\ cm.$ Find the curved surface area of the frustum.

Answer

Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height be l.
Then,
$2\pi\text{R}=18\Rightarrow\pi\text{R}=9\ ...(\text{i})$
$2\pi\text{r}=6\Rightarrow\pi\text{r}=3\ ....(\text{ii})$
Curved surface area of the frustum $=\pi\text{l}(\text{R}+\text{r})$
$=\text{l}\times(\pi\text{R}+\pi\text{r})$
= 4 × (9 + 3) [Since l = 4cm] (From (i) and (ii))
$= 48cm^2$

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Question 272 Marks

A metallic cone of radius 12cm and height 24cm is melted and made into spheres of radius 2cm each. How many spheres are formed?

Answer

We have, Radius of the metallic cone, r = 12cm, Height of the metallic cone, h = 24cm and Radius of the sphere, R = 2cm Now,$=\frac{\Big(\frac{1}{3}\pi\text{r}^3\text{h}\Big)}{\Big(\frac{4}{3}\pi\text{R}^3\Big)}$
$=\frac{\text{r}^2\text{h}}{4\text{R}^3}$
$=\frac{12\times12\times24}{4\times2\times2\times2}$
$=108$ So, the number of spheres so formed is 108.

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Question 282 Marks

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104cm and the radius of each hemispherical end is 7cm, find the surface area of the solid.

Answer

Radius of the hemispherical end = 7cm
Height of the hemispherical end = 7cm
Height of the cylindrical part = (104 - 2 × 7)cm = 90cm
Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder)
$=\Big[2\big(2\pi\text{r}^2\big)+2\pi\text{rh}\Big]$
$=2\pi\text{r}(2\text{r}+\text{h})$
$=2\times\frac{22}{7}\times7\times7[(2\times7)+90]\text{cm}^2$
$=(44\times104)\text{cm}^2$
$=4576\text{cm}^2$

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Question 292 Marks

Two cubes, each of volume $64cm^3,$ are joined end to end. Find the total surface area of the resulting cuboid.

Answer

Volume of the cube $= a^3$
Therefore,
$a^3 = 64$
$\Rightarrow a^3 = (4)^3$
$\Rightarrow a = 4cm$
Each side of the cube = 4cm
Then,
Length of the cuboid ⇒ (2 × 4)cm = 8cm
Breadth of the cuboid = 4cm
Height of the cuboid = 4cm
Total surface area of the cuboid = 2(lb+bh+lh)
$=2[(8 \times 4) + (4 \times 4) + (8 \times 4)]cm^2$
$=(2 \times 80)cm^2$
$= 160cm^2$

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Question 302 Marks

The radius of the base and the height of a solid right circular cylinder are in the ratio $2 : 3$ and its volume is $1617cm^3$. Find the total surface area of the cylinder. $\Big[\text{Take}\ \pi=\frac{22}{7}.\Big]$

Answer

Let the radius of the cylinder be 2x cm and its height be 3x cm.
Then. volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times(2\text{x})^2\times3\text{x}$
Therefore
$\frac{22}{7}\times(2\text{x})^2\times3\text{x}=1617$
$\Rightarrow\frac{22}{7}\times4\text{x}^2\times3\text{x}=1617$
$\Rightarrow\frac{22}{7}\times12\text{x}^3=1617$
$\Rightarrow\text{x}^3=\Big(1617\times\frac{7}{22}\times12\Big)$
$\Rightarrow\text{x}^3\Big(\frac{7}{2}\times\frac{7}{2}\times\frac{7}{2}\Big)$
$\Rightarrow\text{x}^3=\Big(\frac{7}{2}\Big)^3$
$\Rightarrow\text{x}=\frac{7}{2}$
Now, $\text{r}=7\text{cm}$ and $\text{h}=\frac{21}{2}\text{cm}$
Hence, the total surface area of the cylinder.
$(2\pi\text{rh}+2\pi\text{r})^2$
$=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\times\Big(\frac{21}{2}+7\Big)\text{cm}^2$
$=\Big(2\times\frac{22}{7}\times7\times\frac{35}{2}\Big)\text{cm}^2$
$=770\text{cm}^2$

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Question 312 Marks

The volume of a right circular cylinder with its height equal to the radius is$25\frac{1}{7}\text{cm}^3$ Find the height of the cylinder.

Answer

We have,
Height = Base radius i.e. h = r
As,
Volume of the cylinder $25\frac{1}{7}\text{cm}^3$
$\Rightarrow\pi\text{r}^2\text{h}=\frac{176}{7}$
$\Rightarrow\frac{22}{7}\times\text{h}^2\times\text{h}=\frac{176}{7}$
$\Rightarrow\text{h}^3=\frac{176\times7}{7\times22}$
$\Rightarrow\text{h}^3=8$
$\Rightarrow\text{h}=\sqrt[3]{8}$
$\therefore\text{h}=2\text{cm}$
So, the height of the cylinder is 2cm.

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