3 Marks Question - MATHS STD 10 Questions - Vidyadip

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Question 13 Marks

If the heights of two right circular cones are in the ratio $1 : 2$ and the perimeters of their bases are in the ratio $3 : 4$, what is the ratio of their volumes?

Answer

Given that
ratio of height of right circular cones
$h_1: h_2 = 1 : 2$
ratio of base of perimeter
$2\pi\text{r}_1:2\pi\text{r}_2=3:4$
$\Rightarrow\text{r}_1:\text{r}_2=3:4$
therefore,
the ratio of volume of their cones
$=\text{v}_1:\text{v}_2=\frac{1}{3}\pi\text{r}^2_1\text{h}_1:\frac{1}{3}\pi\text{r}^2_2\text{h}_2$
$=\frac{\frac{1}{3}\pi\text{r}^2_1\text{h}_1}{\frac{1}{3}\pi\text{r}^2_2\text{h}_2}$
$=\frac{\text{r}^2_1\text{h}_1}{\text{r}^2_2}=\Big(\frac{3}{4}^2\Big)\times\frac{1}{2}$
$=\frac{9}{32}$
$\Rightarrow\text{V}_1:\text{V}_2=9:32$
Hence, the ratio of their volumes are 9:32

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Question 23 Marks

A well of diameter 3m is dug $14\ m$ deep. The earth taken out of it has been spread evenly all around it to a width of $4\ m$ to form an embankment. Find the height of the embankment.

Answer

The inner radius of the well is $\frac{3}{2}\text{m}$ and the height is 14m. Therefore, the volume of the Earth taken out of it is
$\text{V}_1=\pi\times\Big(\frac{3}{2}\Big)^2\times14\text{m}^3$
The inner and outer radii of the embankment are $\frac{3}{2}\text{m}$ and $4+\frac{3}{2}=\frac{11}{2}$ respectively. Let the height of the embankment be h. Therefore, the volume of the embankment is
$\text{V}_2=\pi\times\Big\{\Big(\frac{11}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2\Big\}\times\text{h m}^3$
Since, the volume of the well is same as the volume of the embankment; we have
$V_1 = V_2$
$\Rightarrow\pi\times\Big(\frac{3}{2}\Big)^2\times14=\pi\times\Big\{\Big(\frac{11}{2}\Big)-\Big(\frac{3}{2}\Big)^2\Big\}\times\text{h}$
$\Rightarrow\text{h}=\frac{9\times14}{112}$
$\Rightarrow\text{h}=\frac{9}{8}\text{m} $
Hence, the height of the embankment is $\text{h}=\frac{9}{8}\text{m}$

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Question 33 Marks

A cylindrical tank full of water is emptied by a pipe at the rate of 225 litres per minute. How much time will it take to empty half the tank, if the diameter of its base is 3m and its height is 3.5m?

Answer

Diameter of cylindrical tank = 3m
height (h) = 3.5m = $\frac{7}{2}\text{m}$
$\therefore$ Radius (r) = $\frac{3}{2}\text{m}$
Volume of water filled in it
$=\pi\text{r}^2\text{h}=\frac{22}{7}\times\frac{3}{2}\times\frac{3}{2}\times\frac{7}{2}\text{m}^3$
$=\frac{99}{4}\text{m}^3$
Volume of water in half the tank
$=\frac{99}{4\times2}\times1000=\frac{9900}{8}\text{l}$
water flow at the rate of 225 l per min.
$\therefore$ Total time taken to empting the bank
$=\frac{99000}{8\times225}$
$= 55\ \text{minutes}$

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Question 43 Marks

What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

Answer

Given that:
Eact solid has the same diameter and height
Therefore,
$r_1 = r_2 = r_3= r = h_{1 =}h_2$
The volume of cylinder $\text{V}_1=\pi\text{r}_1^2\text{h}_1$
The volume of cone $\text{V}_2=\pi\text{r}_2^2\text{h}_2$
The volume of sphere $\text{V}_3=\frac{4}{3}\pi\text{r}^3$
The ratio of their volumes
$\text{V}_1:\text{V}_2:\text{V}_3=\pi\text{r}^2\text{h}:\pi\text{r}^2\text{h}:\frac{4}{3}\pi\text{r}^3$
$=1:\frac{1}{3}:\frac{4}{3}$
$\text{V}_1:\text{V}_2:\text{V}_3=3:1:4$
Hence, the required ratio are $3 : 1 : 4$

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Question 53 Marks

The height of a solid cylinder is $15\ cm$ and the diameter of its base is $7\ cm$. Two equal conical holes each of radius $3\ cm$, and height $4\ cm$ are cut off. Find the volume of the remaining solid.

Answer

Diameter of the base of a cylinder $= 7cm$
$\therefore$ Radius $(r_1) =\Big(\frac{7}{2}\Big)\text{cm}$
Height of cylinder $(h_1) = 15cm$
$\therefore$ Volume of cylinder $=\pi\text{r}_1^2\text{h}_1$
$=\frac{22}{7}\Big(\frac{7}{2}\Big)^2\times15\text{cm}^3$
$=\frac{22}{7}\times\frac{49}{4}\times15=\frac{1155}{2}\text{cm}^2$
Radius of each conical hole $(r_2) = 3cm$
anf height $(h_2) = 4cm$
Volume of 2 such conical holes
$=2\times\frac{1}{3}\pi\text{r}_2^2\text{h}_2$
$=\frac{2}{3}\times\frac{22}{7}\times(3)^2\times4=\frac{2}{3}\times\frac{22}{7}\times9\times4\text{cm}^3$
$=\frac{528}{7}\text{cm}^3$
$\therefore$ Volume of remaining solid $=\frac{1155}{2}-\frac{528}{7}$
$=\frac{8085-1056}{14}=\frac{7029}{14}\text{cm}^3$
$=502.07=502.1\text{cm}^3$

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Question 63 Marks

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9cm.

Answer

Edge of cube = 9cm
$\therefore$ Diameter of cone = 9cm

And radius (r) $=\frac{9}{2}\text{cm}$
Hieght(h) = 9cm
$\therefore\text{volume}=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times\frac{9}{2}\times\frac{9}{2}\times9=190.93\text{cm}^3$

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Question 73 Marks

A cone of radius 4cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of two parts.

Answer

Let h be the height of the given cone. One dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following figure:

In two similar triagles OAB and DCB, we have $\frac{\text{OA}}{\text{CD}}=\frac{\text{OB}}{\text{BD}}$ This implies $\frac{4}{\text{r}}=\frac{\text{h}}{\frac{\text{h}}{2}}$
Therefore, r = 2
Therefore,
$\frac{\text{volume of the smaller cone}}{\text{Volume of the frustum of the cone}}$
$=\frac{\frac{1}{3}\pi\times(2)^2\times\Big(\frac{\text{h}}{2}\Big)}{\frac{1}{3}\pi\times\Big(\frac{\text{h}}{2}\Big)[4^2+2^2+4\times2]}=\frac{1}{7}$
Therefore the ratio of volume of the smaller cone to the volume of the frustum of the cone is 1:7

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Question 83 Marks

The surface area of a sphere is $616cm^2$. Find its radius.

Answer

The surface area of sphere $= 616cm^2$
We know that
$4\pi\text{r}^2=616$
$\text{r}^2=\frac{616}{4\pi}$
Taking square root both the side
$\sqrt{\text{r}^2}=\sqrt{\frac{616}{4\pi}}$
$\text{r}=7\text{cm}$

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Question 93 Marks

Three cubes of a metal whose edges are in the ratio $3 : 4 : 5$ are melted and converted into a single cube whose diagonal is $12\sqrt{3}\text{cm}.$ Find the edges of the three cubes.

Answer

Let the edges of three cubes (in cm) be 3x, 4x and 5x, respectively.
Volume of the cubes after melting is $= (3x)^3 + (4x)^3 + (5x)^3 = 216 \times 3cm^3$
Let a be the side of new cube so formed after melting. Therefore, $a^3 = 216x^3$
So, a = 6x, $\text{Diagonal}=\sqrt{\text{a}+\text{a}^2+\text{a}^2}=\text{a}\sqrt{3}$
But it is given that diagonal of that diagonal of the new cube is $12\sqrt{3}\text{cm}.$ Therefore, $\text{a}\sqrt{3}=12\sqrt{3},$
i.e. $a = 12$
This gives $x = 2$
Therefore, edges of the three cubes are 6cm, 8cm and 10cm respectively.

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Question 103 Marks

$500$ persons have to dip in a rectangular tank which is $80\ m$ long and $50\ m$ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $0.04m^3?$

Answer

The average displacement of water by a person is 0.04 cubic m. Hence, the total displacement of water in the rectangular tank by 500 persons is $V = 500 \times 0.04 = 20$ Cubic m.
The length and width of the rectangular tank are 80m and 50m respectively. Upon dipping in the tank, let the height of the raised water is be h m. Therefore, the volume of the raised
water is $V_1 = 80 \times 50 \times h$
= 4000h cubic m
Since, the volume of the raised water is same as the volume of the water displaced by $500$ persons, we have
$V_1 = V$
$\Rightarrow 4000h = 20$
$\Rightarrow\text{h}=\frac{20}{4000}$
$\Rightarrow 0.005$
Therefore, the water will be raised by $0.005m$ or $0.5cm$

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Question 113 Marks

The radii of two cylinders are in the ratio $3 : 5$ and their heights are in the ratio $2 : 3$. What is the ratio of their curved surface areas?

Answer

Given that:
Ratio of radii of two cylinder
$r_1 : r_2 = 3 : 5$
Ratioof height of two cylinder
$h_1 : h_2 = 2 : 3$
Now, the ratio of their curved surface area
$\text{S}_1:\text{S}_2=2\pi\text{r}_1\text{h}_1:2\pi\text{r}_2\text{h}_2$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}_1\text{h}_1}{2\pi\text{r}_2\text{h}_2}=\frac{3}{5}\times\frac{2}{3}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{6}{15}=\frac{2}{5}$
$\Rightarrow S_1 : S_2 = 2 : 5$
Hence, the ratio of their curved surface area are $2 : 5$

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Question 123 Marks

The difference between outside and inside surface areas of cylindrical metallic pipe $14\ cm$ long is $44m^2$. If the pipe is made of $99cm^3$ of metal, find the outer and inner radii of the pipe.

Answer

Let inner radius of pipe be $r_1$
Radius of outer cylinder be $r_2$
Length of cylinder $(h) = 14cm.$
Surface area of hollow cylinder ($r_2 - r_1$) $=2\pi$
Given surface area of cylinder $= 44m^2$
$2\pi\text{Rh}-2\pi\text{r}\text{h}=44$
$2\pi\text{h}(\text{R}-\text{r})=44$
$2\times\frac{22}{7}\times14(\text{R}-\text{r})=44$
$\text{R}-\text{r}=\frac{44\times7}{2\times22\times14}$
$\text{R}-\text{r}=\frac{1}{2}\ ...(1)$
Now,
Volume of pipe $= 99cm^3$
$\pi\text{R}^2\text{h}-\pi_2\text{r}^2\text{h}=99$
$\pi\text{h}(\text{R}^2-\text{r}^2)=99$
$(\text{R}^2-\text{r}^2)=\frac{99}{\pi\times\text{h}}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99\times7}{22\times14}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99}{22\times2}$
Putting value of $\text{R}-\text{r}=\frac{1}{2}$
$(\text{R}+\text{r})\frac{1}{2}=\frac{99}{22\times2}$
$(\text{R}+\text{r})=\frac{9}{2}\ ...(2)$
Adding equation (1) and (2)
$\text{R}-\text{r}=\frac{1}{2}$
$\text{R}+\text{r}=\frac{9}{2}$
$2\text{R}=5$
$\text{R}=2.5$
And $\text{r}=\frac{9}{2}-\frac{5}{2}$
$\text{r}=2\text{cm}$

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Question 133 Marks

A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces?

Answer

Given that:
The base of the cone and hemisphere are equal.
Therefore,
Radius of the hemisphere = Height of the cone
Then, the salant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{\text{r}^2+\text{r}^2}$
$=\sqrt{2}\text{r}$
Curved surface area of hemisphere
$\text{S}_1=2\pi\text{r}^2$
And the curved surface area of the cone
$\text{S}_2=\pi\text{rl}$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}^2}{\pi\text{rl}}=\frac{2\pi\text{r}^2}{\pi\text{r}\sqrt{2}\text{r}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{\sqrt{2}}{1}$
Hence, the required ratio are $\sqrt{2}:1$

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Question 143 Marks

A solid metal sphere of 6cm diameter is melted and a circular sheet of thickness 1cm is prepared. Determine the diameter of the sheet.

Answer

Diameter of sphere = 6cm
Therefore,
Radius = 3cm.
Therefore,
Surface area of sphere
$=4\pi\text{r}^2$
$=4\times\pi\times(3)^2$
$=36\pi\ \text{cm}^2$
Area of the circular sheet $=\pi\text{r}^2$
Therefore,
Surface area of sphere = area of the circular sheet
$=\pi\text{r}^2=36\pi$
$\text{r}=6\text{cm}$
Therefore,
Diameter of the sheet = 2 × 6 = 12cm

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Question 153 Marks

If a cone and a sphere have equal radii and equal volumes. What is the ratio of the diameter of the sphere to the height of the cone?

Answer

Given that:
A cone and a sphere have equal radii and equal volume
Therefore,
Volume of cone = Volume of sphere
$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow r^2h = 4r^3$
$\Rightarrow h = 4r$
$\Rightarrow h = 2 (2r)$
$\Rightarrow h = 2d$
$\Rightarrow\frac{\text{h}}{\text{d}}=\frac{2}{1}$
$\Rightarrow d : h = 1 : 2$
Hence, the required eatio are $1 : 2$

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Question 163 Marks

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

Answer

Perimeter of the top of frustum $= 18cm$

$\therefore $ Radius ($r_1$) $=\frac{\text{C}}{2\pi}=\frac{18\times7}{2\times22}$
$=\frac{63}{22}\text{cm}$
and perimeter of the bottom $= 6cm$
$\therefore$ Radius ($r_2$) $ =\frac{\text{C}}{2\pi}=\frac{18\times7}{2\times22}=\frac{21}{22}\text{cm}$
and slant height $(l) = 4cm$
curved surface area $=\pi (\text{r}_1+\text{r}_2) \text{l} $
$=\frac{22}{7}\Big(\frac{63}{22}+\frac{21}{22}\Big)\times4$
$= \frac{22}{7}\times\frac{84}{22}\times4=48\text{cm}^2$

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Question 173 Marks

A hollow sphere of internal and external radii 2cm and 4cm respectively is melted into a cone of base radius 4cm. Find the height and slant height of the cone.

Answer

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is,
$\text{V}=\frac{4}{3}\pi\{(4)^3-(2)^3\}$
$=\frac{4}{3}\times\frac{22}{7}\times56$
$=\frac{32\times22}{3}$
The hollow sphere is melted to produce a right circular cone of base-radius 4cm. Let, the height and slant height of the cone be h cm and l cm respectively. Then, we have,
$\text{l}^2=(4)^2+\text{h}^2$
$\Rightarrow\text{l}^2=16+\text{h}^2$
The volume of the cone is,
$\text{V}_1=\frac{1}{3}\pi\text{r}^2_1\text{h}_1$
$=\frac{1}{3}\times\frac{22}{7}\times(4)^2\times\text{h}$
Since, the volume of the cone and hollow sphere are same, we have
$V_1 = V$
$\Rightarrow\frac{1}{3}\times\frac{22}{7}\times(4)^2\times\text{h}=\frac{32\times22}{3}$
$\Rightarrow\frac{1}{7}\times(4)^2\times\text{h}=32$
$\Rightarrow\text{h}=\frac{32\times7}{16}$
$\Rightarrow\ =14$
Then, we have
$\text{l}^2=16+(14)^2$
$\Rightarrow\ =212$
$\Rightarrow\text{l}=14.56$
Therefore, the height and the slant height of the cone are 14cm and 14.56cm respectively.

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Question 183 Marks

A sphere of diameter 5cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10cm. If the sphere is completely submerged, by how much will the level of water rise?

Answer

Radius of sphere $\text{r}=\frac{5}{2}\text{cm}$ radius of cylindrical vessel $\text{r}_1=\frac{10}{2}\text{cm}$
When the sphere is completely submerged into the vessel. The level of water will be raised let xbe height of level of raised water.

Therefore,
The volume of raised water in cylindrical vessel = volume of sphere
$\pi\times(5)^2\times\text{x}=\frac{4}{3}\pi\Big(\frac{5}{2}\Big)^3$
$25\text{x}=\frac{4\times125}{3\times8}$
$\text{x}=\frac{4\times125}{3\times8\times25}$
$=\frac{5}{6}\text{cm}$
$\text{x}=\frac{5}{6}\text{cm}$
Hence, the level of water rise $=\frac{5}{6}\text{cm}$

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Question 193 Marks

From a solid cube of side 7cm, a conical cavity of height 7cm and radius 3cm is hollowed out. Find the volume of the remaining solid.

Answer

Given that, side of a solid cube (a) = 7cm
Height of conical cavity i.e., cone, h = 7cm

Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.
Radius of conical cavity i. e., cone, r = 3cm
⇒ Diameter = 2 × r = 2 × 3 = 6cm
Since, the diameter is less than the side of a cube that means the base of a
conical cavity is not fit inhorizontal face of cube.
Now, volume of cube = (side)$^3 = a^3 = (7)^3 = 343cm^3$
and volume of conical cavity i.e., cone
$=\frac{1}{3}\pi\times\text{r}^2\times\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times3\times3\times7=66\text{cm}^3$
$\therefore$ Volume of remaining solid = Volume of cube - Volume of conical Cavity
$= 343 - 66 = 277cm^3$
Hence, the required volume of solid is $277cm^3$

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Question 203 Marks

If the areas of circular bases of a frustum of a cone are $4 cm^2$ and $9 cm^2$ respectively and the height of the frustum is 12 cm . What is the volume of the frustum?

Answer

Given that:
Area of circular ends of frustum are
$\pi\text{r}_1^2=4$
$\Rightarrow\text{r}_1^2=\frac{4}{\pi}$
and
$\pi\text{r}_2^2=9$
$\text{r}^2_2=\frac{9}{\pi}$
The height of frustum h = 12cm
Now, the Volume of frustum
$\text{v}=\frac{\text{h}}{3}(\text{r}_1^2+\text{r}_2^2+\text{r}_1\text{r}_2)$
$=\frac{12}{3}\pi\Big(\frac{4}{\pi}+\frac{9}{\pi}+\sqrt{\frac{4}{\pi}}\times\sqrt{\frac{9}{\pi}}\Big)$
$=4\pi\Big(\frac{13}{\pi}+\sqrt{\frac{36}{\pi^2}}\Big)$
$=4\pi\Big(\frac{13}{\pi}+\frac{6}{\pi}\Big)$
$=4\times19$
$=76\text{cm}^2$
Hence, the Volume of frustum is $76cm^3.$

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Question 213 Marks

A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone.

Answer

Let r be the radius of base and h be the height of cylinder and cone
Therefore,
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{1}{\frac{1}{3}}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{3}{1}$
$\Rightarrow\text{v}_1:\text{v}_2=3:1$
Hence, the required ratio are 3 : 1

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Question 223 Marks

A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.

Answer

Let r = 5x and h = 12x be the base radius and height of the cone and cylinder respectively.
Slant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(5\text{x})^2+(12\text{x}^2)^2}$
$=\sqrt{25\text{x}^2+144\text{x}^2}$
$\Rightarrow\text{l}=13\text{x}$
The total surface area of cylinder
$\text{S}_1=2\pi\text{r}(\text{h}+\text{r})$
The total surface area of cone
$\text{s}_2=2\pi\text{r}(\text{l}+\text{r})$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}(\text{h}+\text{r})}{\pi\text{r}(\text{l}+\text{r})}$
$=\frac{2(\text{h}+\text{r})}{(\text{l}+\text{r})}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2(12\text{x}+5\text{x})}{13\text{x}+5\text{x}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\times17\text{x}}{18\text{x}}$
$\Rightarrow\text{S}_1:\text{S}_2=17:9$
Hence, the required ratio are 17 : 9

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Question 233 Marks

A cylindrical tub of radius 5cm and length 9.8cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5cm and height of the cone outside the hemisphere is 5cm, find the volume of the water left in the tube.

Answer

Given radius of cylinderical tube (r) = 5cm.
Height of cylindrical tube (h) = 9.8cm
Volume of cylinder $=\pi\text{r}^2\text{h}$
$\text{V}_1=\pi(5)^2(9.8)=770\text{cm}^3$
Given radius of hemisphere (r) = 3.5cm
Height of cone (h) = 5cm
Volume of hemisphere $=\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\times\pi(3.5)^3=89.79\text{cm}^3$
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{\pi}{3}(3.5)^25=64.14\text{cm}^3$
Volume of cone + Volume of hemisphere $(V_2) = 39.79 + 64 = 154cm^3$

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Question 243 Marks

A solid iron pole having cylindrical portion 110cm high and of base diameter 12cm is surmounted by a cone 9cm high. Find the mass of the pole, given that the mass of $1cm^3$ of iron is 8gm.

Answer

Diamter of the base = 12cm
$\therefore$ Radius (r) $=\Big(\frac{12}{2}\Big)=6\text{cm}$
Height of the cylindrical portion ($h_1$) = 110cm
and height of conical portion ($h_2$) = 9cm

$\therefore$ Total volume of the pole
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2\Big)$
$=\frac{22}{7}\times(6)^2\Big(110+\frac{1}{3}\times9\Big)\text{cm}^2$
$=\frac{22}{7}+36(110+3)\text{cm}^3$
$=\frac{22}{7}\times36\times113\text{cm}^3$
$=\frac{89496}{7}\text{cm}^3$
Weight of $1cm^3$ of iron = 8gm
$\therefore$ Total weight $=\frac{89496}{7}\times8\text{gms}$
$=\frac{89496\times8}{7\times1000}\text{kg}=102.281\text{kg}$
$\therefore$ Mass of the pole = 102.281kg

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Question 253 Marks

50 circular plates each of diameter 14cm and thickness 0.5cm are placed one above the other to form a right circularcylinder. Find its total surface area.

Answer

Given that 50 circular plates each with diameter = 14cm
Radius of circular plates (r) = 7cm
Thickness of plates = 0.5
Since these plates are placed one above other so total thickness of plates 0.5 × 50 = 25cm
Total surface area of a cylinder $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{r}(\text{h}+\text{r})$
$= 2\times\frac{22}{7}\times7(25+7)$
$T.S.A = 1408cm^2$
$\therefore$ Total surface area of circular plates is $1408cm^2$

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Question 263 Marks

The vertical height of a conical tent is 42dm and the diameter of its base is 5.4m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm.

Answer

Radius of conicaltent, $\text{r}=\frac{5.4}{2}$
= 2.7m
= 27dm
Height of conical tent h = 42dm
The volume of conical tent
$=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times27\times27\times42$
$=22\times27\times27\times2$
$=32076\text{dm}^3$
Since, each person is to be allowed $29.16dm^3,$
Therefore,
$=\frac{\text{Volume of conical tent}}{\text{place to be allow to each person}}$
$=\frac{32076}{29.16}$
$=\frac{3207600}{2916}$
$\text{No. of persons}=1100$

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Question 273 Marks

If the total surface area of a solid hemisphere is $462cm^2$, find its volume $\Big(\text{use}\ \pi=\frac{22}{7}\Big)$

Answer

Let the radius of the hemisphere be r cm.
Total surface area of hemisphere = $462cm^2$
$\Rightarrow3\pi\text{r}^2=462$
$\Rightarrow3\times\frac{22}{7}\times(\text{r})^2=462$
$\Rightarrow r^2 = 49$
$\Rightarrow r = 7cm$
Now, the volume of hemisphere is given by
$\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\times\frac{22}{7}(7)^3$
$=\frac{2156}{3}$
$=718\frac{2}{8}\text{cm}^3$

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Question 283 Marks

Water flows through a cylindrical pipe, whose inner radius is 1cm, at the rate of 80cm/sec in an empty cylindrical tank, the radius of whose base is 40cm. What is the rise of water level in tank in half an hour?

Answer

Given, radius of tank, $r_1=40 cm$
Let height of water level in tank in half an hour $=h_1$
Also, given internal radius of cylindrical pipe, $r_1=1 cm$
and speed of water $=80 cm / s$ in 1 water flow $=80 cm$
$\therefore \ln 30(min)$ water flow $=80 \times 60 \times 30=144000 cm$
According to the question,
Volume of water in cylindrical tank = Volume of water flow the circular pipe in half an hour
$\Rightarrow\pi\text{r}^2_1\text{h}_1=\pi\text{r}^2_1\text{h}_2$
$\Rightarrow40\times40\times\text{h}_1=1\times144000$
$\therefore\text{h}_1=\Big(\frac{144000}{40\text{x}40}\Big)=90\text{cm}$
Hence, the level of water in cylindrical tank rises 90cm in half an hour.

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Question 293 Marks

An iron pillar consists of a cylindrical portion 2.8m high and 20cm in diameter and a cone 42cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5gm.

Answer

Volume of cylindrical portion
$=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\Big(\frac{20}{2}\Big)^2\times280$
$=88000\text{cm}^3$
Volume of conical portion
$\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times(10)^2\times42$
$=4400\text{cm}^3$
Total number
= 88000 + 4400
= 92400
So total height
= 92400 × 7.5
= 693000gm
= 693kg

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Question 303 Marks

A solid cone of base radius 10cm is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.

Answer

Radius of solid cone (r) = 10cm
Let total height = h
In $\triangle\text{AOB}$
C is mid point of AO and CD || OB

$\therefore\frac{\text{OB}}{\text{CD}}=\frac{\text{AO}}{\text{AC}}\Rightarrow\frac{10}{\text{CD}}=\frac{\text{h}}{\text{h}}$

$\Rightarrow\frac{10}{\text{CD}}=\frac{2}{1}\Rightarrow\text{CD}=\frac{10}{2}=5\text{cm}$
$\therefore\text{r}_2=5\text{cm}$
Volume of smaller cone
$=\frac{1}{3}\pi\text{r}^2_2\frac{\text{h}}{2}=\frac{1}{3}\pi\times5\times5\times\frac{\text{h}}{2}=\frac{25}{6}\pi\text{h}$
Volume of frustum $=\frac{1}{3}\pi\frac{\text{h}}{2}(\text{r}^2_1+\text{r}_1\text{r}_2+\text{r}_2^2)$
$=\frac{\text{h}\pi}{6}(10^2+10\times5+5^2)$
$\frac{\pi\text{h}}{6}(100+50+25)$
$=\frac{175}{6}\pi\text{h}$
Ratio between the upper part and lower part
$=\frac{25}{6}\pi\text{h}:\frac{175}{6}\pi\text{h}=1:7$

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Question 313 Marks

If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then write the ratio of their height.

Answer

Let $r_1 r_2$ be the radii and $h_1, h_2$ be the height of two cones
It is given that the ratio of the volimes of two cones.
$V_1: V_2 = 1 : 4$
and $2r_1 : 2r_2 = 4 : 5$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{5}$
the ratio of Volume of two cones.
$\Rightarrow\frac{\text{V}_1}{\text{v}_2}=\frac{\frac{1}{3}\pi\text{r}_1^2\text{h}_1}{\frac{1}{3}\pi\text{r}_2^2\text{h}_2}$
$\Rightarrow\frac{1}{4}=\Big(\frac{4}{5}\Big)^\times\frac{\text{h}_1}{\text{h}_2}$
$\Rightarrow\frac{1}{4}=\frac{16}{25}\times\frac{\text{h}_1}{\text{h}_2}$
$\Rightarrow\text{h}_1:\text{h}_2=25:64$
Hence, the required ratio are 25 : 64

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Question 323 Marks

A well with 10m inside diameter is dug 8.4m deep. Earth taken out of it is spread all around it to a width of 7.5m to form an embankment. Find the height of the embankment.

Answer

Radius of well
$\text{r}=\frac{10}{2}$
= 5m
Depth of well h = 8.4m
Clearly,
Volume of earth dugout
$=\pi(5)^2\times8.4$
$=\pi\times25\times8.4$
$=\frac{22\times25\times8.4}{7}\text{m}^3$

Let h' be the height of embankment
Clearly,
Embankment forms a cylindrical shell whose inner and outer radius are 5m and 12.5m respectively.
$\therefore$ Volume of the embankment
$=\pi\{(12.5)^2-(5)^2\}\times\text{h}'$
$=\pi\times17.5\times7.5\times\text{h}'\ \text{m}^3$
But, volume of earth dugout = volume of the embankment
$\frac{22\times25\times8.4}{7}=\frac{22}{7}\times17.5\times7.5\times\text{h}$
$\text{h}=\frac{258\times8.4}{17.5\times7.5}$
$\text{h}'=1.6\text{m}$

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Question 333 Marks

The radii of the base of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?

Answer

Let $r_1$ and $r_2$ be the radii of the base of a cylinder and a cone.The volume of cylinder $\text{V}_1=\pi\text{r}^2_1\text{h}_1...(1)$
The volume of cone $\text{V}_2=\pi\text{r}^2_2\text{h}_2...(2)$
Dividing (i) by (ii), the, we get
$\frac{\text{V}_1}{\text{V}_2}=\frac{\pi\text{r}^2_1\text{h}_1}{\frac{1}{3}\pi\text{r}^2_2\text{h}_2}$
$\frac{=3\times\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\times\Big(\frac{\text{h}_1}{\text{h}_2}\Big)}{\Big(\frac{\text{r}_1}{\text{r}_2}=\frac{3}{4},\frac{\text{h}_1}{\text{h}_2}=\frac{2}{3},\ \text{given}\Big)}$
$\frac{\text{V}_1}{\text{V}_2}=3\times\Big(\frac{3}{4}\Big)^2\times\frac{2}{3}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{9}{8}$
$\text{V}_1:\text{V}_2=9:8$

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Question 343 Marks

Water in a canal 1.5m wide and 6m deep is flowing with a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8cm of standing water is desired?

Answer

The canal is 1.5m wide and 6m deep. The water is flowing in the canal at 10km/hr. Hence, in 30 minutes, the length of the flowing standing water is
$=10\times\frac{30}{60}\text{km}$
= 5km
= 5000m
Therefore, the volume of the flowing water in 30 min is
$V_1 = 5000 \times 1.5 \times 6m^3$
Thus, the irrigated area in 30 min of 8cm = 0.08m standing water is
$=\frac{5000\times1.5\times6}{0.08}$
$= 562500m^2$

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Question 353 Marks

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24m. The height of the cylindrical portion is 11m while the vertex of the cone is 16m above the ground. Find the area of canvas required for the tent.

Answer

Given diameter of cylinder 24 m
Radius $( r )=\frac{24}{2}=12 m$
Given height of cylindrical part $\left(h_1\right)=11 m$
$\therefore$ Height of cone part $\left(h_2\right)=5 m$
Vertex of come above ground $=11+5=16 m$
Curved surface area of cone $\left( S _1\right)=\pi rl$
$=\frac{22}{7}\times12\times\text{l}$
Let l be slant height of cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2_2}$
$\Rightarrow\text{l}=\sqrt{12^2+5^2}=13\text{m}$
l = 13m
$\therefore$ Curved surface area of cone (5) $=\frac{22}{7}\times12\times13\text{m}^2$

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Question 363 Marks

A cylindrical container is filled with ice-cream, whose diameter is 12cm and height is 15cm. the whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.

Answer

Volume of cylindrical container
$=\pi\text{r}^2\text{h}$
$=\pi\times(6)^2\times15$
Amount of ice-cream distributed to 10 children $=\frac{\pi\times(6)^2\times15}{10}$
Therefore,
Height of conical portion = 2 × diameter of its bars
Let the diameter of bare = r
Height = 2r
Therefore,
Volume of the cones
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\text{h}+\frac{2}{3000}\pi\text{r}^3$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2(\text{h}+2\text{r})$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\Big(2\text{r}+2\times\frac{\text{r}}{2}\Big)$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\times3$
$\text{r}=\frac{\pi\text{r}^3}{4}$
Therefore,
Volume of the cones = amount distributed
$\frac{\pi\text{r}^3}{4}=\frac{\pi(6)^2\times15}{10}$
$\text{r}^3=\frac{4\times6\times6\times15}{10}=4\times6\times9$
$\text{r}=\sqrt[3]{6\times6\times6}=6\text{cm}$

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Question 373 Marks

The slant height of the frustum of a cone is 5cm. If the difference between the radii of its two circular ends is 4cm, write the height of the frustum.

Answer

Let $r_1$ and $r_2$ be the radius of frustum ends. $r_1-r_2=4 cm$
Slant height of the frustum cone
l = 5cm
$\Rightarrow\text{l}=\sqrt{\text{h}^2+(\text{r}_1-\text{r}_2)^2}$
$=5=\sqrt{\text{h}^2+4^2}$
Squaring both the sides.
$\Rightarrow 25 = h^2 + 16$
$\Rightarrow h^2 = 25 - 16$
$\Rightarrow h^2 = 9cm$
$\Rightarrow h = 3cm$
Hence, the height of frustum is 3cm

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Question 383 Marks

A container open at the top, is in the form of a frustum of a cone of height 24cm with radii of its lower and upper circular ends as 8cm and 20cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs. 21 per litre.

Answer

Upper radius (R) = 20cm
Lower radius (r) = 8cm
Height (h) = 24cm

$\therefore$ Volume of frustum
$=\frac{\pi}{3}(\text{R}^2+\text{Rr}+\text{r}^2)\text{h}$
$=\frac{22}{7\times3}[20^2+20\times8+8^2]\times24\text{cm}^3$
$\frac{22}{21}[400+160+64]\times24\text{cm}^3$
$=\frac{22}{21}\times624\times24\text{cm}^3=\frac{329472}{21}\text{cm}^3$
Volume of milk in it
$=\frac{329472}{21}\times\frac{1}{1000}\text{l}$
$\frac{329.472}{21}\text{l}$
Rate of milk = ₹ 21 per l
$\frac{329.472}{21}\times21$
= ₹ 329.47

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Question 393 Marks

From a solid cylinder of height 2.5cm and diameter 4.2cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.$\Big(\text{use}\ \pi=\frac{22}{7}\Big)$

Answer

Diameter of solid cylinder = 4.2cm
$\therefore$ Radius (r) $\Big(\frac{4.2}{2}\Big)=2.1\text{cm}$
Height (h) = 2.8cm

Slant height of cone
$=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(2.1)^2+(2.8)^2}=\sqrt{4.41+7.84}$
$=\sqrt{12.25}=3.5\text{cm}$
Total surface area of remaining solid = surface area of cylinder + surface area of cone
$=2\pi\text{rh}+\pi\text{r}^2+\pi\text{rl}$
$=\pi\text{r}(2\text{h}+\text{r}+\text{l})$
$=\frac{22}{7}\times2.1(2\times2.8+2.1+3.5)\text{cm}^2$
$=6.6(5.6+2.1+3.5)\text{cm}^2$
$=6.6(11.2)=73.92\text{cm}^2$

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Question 403 Marks

The rain water from a roof of dimensions 22m × 20m drains into a cylindrical vessel having diameter of base 2m and height 3.5m. If the rain water collected from the roof just fills the cylindrical vessel, then find the rain in cm.

Answer

Given, length of roof = 22m and breadth of roof = 20m
Let the rainfall be a cm.
Volume of water on the roof $=22\times20\times\frac{\text{a}}{100}=\frac{22\text{a}}{5}\text{m}^3$
Also, we have radius of base of the cylinderical vessel = 1m
and height of the cylindrical vessel = 3.5m
$\therefore$ Volume of water in the cylindrical vessel when it is just full
$=\Big(\frac{22}{7}\times1\times1\times\frac{7}{2}\Big)=11\text{m}^3$
Now, volume of water on the roof = volume of water in the vessel
$\Rightarrow\frac{22\text{a}}{5}=11$
$\therefore\text{a}=\frac{11\times5}{22}=2.5$
$\Big[\because$ volume of cylindricer = $\pi\times(\text{radius})^2\times\text{height}\Big]$
Hence, the rainfall is 2.5cm.

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Question 413 Marks

A hemisphere of lead of radius 7cm is cast into a right circular cone of height 49cm. Find the radius of the base.

Answer

Radius of hemisphere r = 7cm
The volume of hemisphere
$=\frac{2}{3}\pi\text{r}^2$
$=\frac{2}{3}\pi\times(7)^3$
$=\frac{2}{3}\pi\times343$
$=\frac{686}{3}\pi\ \text{cm}^3$
Since,
The hemisphere cast into the right circular cone.
The height of cone h = 49cm
Let x be the radius of cone.
Clearly,
Volume of cone = volume of hemisphere
$\frac{1}{3}\pi\times49=\frac{686}{3}\pi$
$\text{x}^2=\frac{686\times3}{49\times3}$
$=14$
$\text{x}^2=14$
$\text{x}=\sqrt{14}$
$\text{x}=3.74\text{cm}$
Thus, the radius of cone = 3.74cm

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Question 423 Marks

A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and contains $41\Big(\frac{19}{21}\Big)\text{m}^3$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Answer

Let total height of the building = Internal diameter of the dome = 2m

Radius of building (or dome) $=\Big(\frac{2\text{r}}{2}\Big)=\text{r}\ \text{m}$
Height of cylinder = 2r - r = r m
$\therefore$ Volume of the cylinder $=\pi\text{r}^2(\text{r})=\pi\text{r}^3\text{m}^3$
and volume of hemispherical dome cylinder $=\frac{2}{3}\pi\text{r}^3\text{m}^3$
$\therefore$ Total volume of the building = volume of the cylinder + volume of hemispherical dome.
$=\Big(\pi\text{r}^3+\frac{2}{3}\pi\text{r}^3\Big)\text{m}^3=\frac{5}{3}\pi\text{r}^3\text{m}^3$
According to the condition,
Volume of the building = volume of the air
$\Rightarrow\frac{5}{3}\pi\text{r}^3=41\frac{19}{21}\Rightarrow\frac{5}{3}\pi\text{r}^3=\frac{880}{21}$
$\Rightarrow\text{r}^3=\frac{880\times7\times3}{21\times22\times5}=\frac{40\times21}{21\times5}=8$
$\Rightarrow\text{r}^3=8\Rightarrow\text{r}=2$
$\therefore$ Height of the building = 2r = 2 × 2 = 4m.

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Question 433 Marks

A hemisphere of lead of radius 7cm is cast into a right circular cone of height 49cm. Find the radius of the base.

Answer

Radius of hemisphere of lead ($r_1$) = 7cm
$\therefore\frac{2}{3}\pi\times7\times7\times7=\frac{686\pi}{3}\text{cm}^3$
Now volume of right circular cone $=\frac{686\pi}{3}\pi\ \text{cm}^3$
Height (h) = 49cm
Let $r_2$ be the radius, then
Volume $=\frac{1}{3}\pi\text{r}_2^2\text{h}$
$\Rightarrow686\pi=\frac{1}{3}\pi\text{r}_2^2\times49\Rightarrow\text{r}_2^2=\frac{686\pi\times3}{1\times49\pi}$
$\Rightarrow\text{r}_2^2=42\Rightarrow\text{r}_2=\sqrt{42}$
$\therefore$ Radius $=\sqrt{42}=6.480\text{cm}$

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Question 443 Marks

A sphere of maximum volume is cut-out from a solid hemisphere of radius r, what is the ratio of the volume of the hemisphere to that of the cut-out sphere?

Answer

Given that:
A sphere of maximum volume is cut from a solid hemisphere of radius r.
Therefore radius of sphere $=\frac{\text{r}}{2}$
The volume of sphere $=\frac{4}{3}\pi\Big(\frac{\text{r}}{2}\Big)^3$
$\Rightarrow\text{v}_1=\frac{1}{6}\pi\text{r}^3...(1)$
The volume of hemisphere
$\Rightarrow\text{v}_2=\frac{2}{3}\pi\text{r}^3...(2)$
Dividing Eq. (1) by Eq. (2)
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{\frac{1}{6}\pi\text{r}^3}{\frac{2}{3}\pi\text{r}^3}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{1}{4}$
$\Rightarrow\text{v}_2:\text{v}_1=4:1$
Hence the required ratio are 4 : 1

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Question 453 Marks

A cylindrical vessel of diameter 14cm and height 42cm is fixed symmetrically inside a similar vessel of diameter 16cm and height 42cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required?

Answer

Diameter of inner cylinder = 14cm
$\therefore$ Radius (r) $=\Big(\frac{14}{2}\Big)=7\text{cm}$
Diameter of outer cylinder = 16cm
$\therefore$ Radius (R) $\Big(\frac{16}{2}\Big)=8\text{cm}$
Height (h) = 42cm

$\therefore$ space between the two cylinders
$=\pi\text{R}^2\text{h}-\pi\text{r}^2\text{h}$
$=\pi\text{h}(\text{R}^2-\text{r}^2)$
$=\frac{22}{7}\times42(8^2-7^2)\text{cm}^2$
$=22\times6\times(64-49)$
$=22\times6\times15\text{cm}^3$
$=1980\text{cm}^3$

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Question 463 Marks

A toy is in the form of a cone mounted on a hemisphere of radius 3.5cm. The total height of the toy is 15.5cm find the total surface area and volume of the toy.

Answer

Radius of the toy (r) = 3.5cm
Total height of the toy = 15.5cm
$\therefore$ Height of the conical part = 15.5 – 3.5 = 12cm
Slant height of the conical part (l)
$=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{(3.5)^2+(12)^2}$
$=\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{cm}$

Now total surface area of the toy = curved surface area of conical part + curved surface area of hemispherical part

$=\pi\text{rl}+2\pi\text{r}^2=\pi\text{r}(\text{l}+2\text{r})$

$=\frac{22}{7}\times3.5(12.5+2\times3.5)\text{cm}^2$

$=214.5\text{cm}^2$

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

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

$=\frac{1}{3}\times\frac{22}{7}(3.5)^2(12+2\times3.5)\text{cm}^3$

$=\frac{1}{3}\times\frac{22}{7}\times12.25(12+7)\text{cm}^3$

$=\frac{22}{3}\times1.75\times19\text{cm}^3$

$=\frac{731.5}{3}=243.83\text{cm}^3$

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Question 473 Marks

A cylindrical vessel with internal diameter 10cm and height 10.5cm is full of water. A solid cone of base diameter 7cm and height 6cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder (ii) left in the cylinder. $(\text{take}\ \pi=\frac{22}{7})$

Answer

Internal diameters of cylindrical vessel = 10cm
$\therefore$ Radius (r) $=\Big(\frac{10}{2}\Big)=5\text{cm}$
and height (h) = 10.5cm
$\therefore$ Volume of water filled in it
$=\pi\text{r}^2\text{h}=\pi\times(5)^2\times10.5\text{cm}^3$
$=\frac{22}{7}\times2.5\times\frac{105}{10}\text{cm}^3=825\text{cm}^3$
Diameter of the cone = 7cm
$\therefore$ Radius (r) $\frac{7}{2}\text{cm},$ Height ($h_1$) = 6cm
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{r}_1^2\text{h}_1$
$=\frac{1}{3}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times6\text{cm}^3=77\text{cm}^3$

$\therefore$ Water displaced out of the cylinder $= 77cm^3$
Water left in the vessal $= (825 - 77)cm^3= 748cm^3$

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Question 483 Marks

A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25cm and 20cm respectively. Find its height and area of tin used in its construction.

Answer

Since volume of frustum
$=\frac{\pi\text{h}}{3}(\text{R}^2+\text{Rr}+\text{r}^2)$
$=15250\text{cm}^3$
$\text{h}\times\frac{\pi}{3}\times\Bigg(\Big(\frac{25}{2}\Big)^2+(10)^2+\frac{25}{2}\times10\Bigg)$
$=15250$
$\text{h}\times\frac{\pi}{3}(156.25+22.5)$
$=15250$
$\text{h}=\frac{3\times7\times15250}{22\times381.25}$
$=38.18\text{cm}$
Area of the required
$=\pi(25.5+10)\sqrt{(12.5-10)+38.18}$
$=3017\text{cm}^2$

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Question 493 Marks

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?

Answer

Let $r_1$ and $r_2$ be the radii of two right circular cylinders and $h_1$ and $h_2$ be the heights.
Since,
Both the cylinder has the same volume.
Therefore,
$\pi\text{r}_1^2\text{h}_1=\pi\text{r}_2^2\text{h}_2$
$\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\frac{\text{h}_2}{\text{h}_1}$
$(\text{h}_1:\text{h}_2=1:2,\ \text{given})$
$\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{2}{1}\Big)$
$\text{r}_1:\text{r}_2=\sqrt{2}:1$

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Question 503 Marks

A toy is in the form of a cone of radius 3.5cm mounted on a hemisphere of same radius. The total height of the toy is 15.5cm. Find the total surface area of the toy.

Answer

Radius of cone (r) = 3.5cm
Total height of the toy = 15.5cm

Height of the conical part (h) = 15.5 – 3.5 = 12cm
$\therefore$ Slant height of the cone (l)
$=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{(3.5)^2+(12)^2}$
$=\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{cm}$
Now total srrface area of the toy = curved surface area of the conical part + curved surface area of hemispherical part.
$=\pi\text{rl}+2\pi\text{r}^2=\pi\text{r}(\text{l}+2\text{r})$
$=\frac{22}{7}\times3.5\ (12.5+2\times3.5)\text{cm}^2$
$=\frac{22}{7}\times\frac{7}{2}(12.5+7)\text{cm}^2$
$=11(19.5)=214.5\text{cm}^2$

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3 Marks Question - MATHS STD 10 Questions - Vidyadip