3 Marks Question - MATHS STD 9 Questions - Vidyadip

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22 questions · timed · auto-graded

Question 13 Marks

The diameter of a copper sphere is 18cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108m, find its diameter.

Answer

Given that diameter of a copper sphere = 18cm
Radius of the sphere = 9cm
Length of the wire = 108m = 10800cm
Volume of cylinder = Volume of sphere
$\pi^2_1\text{h}=\frac{4}{3}\pi\text{r}^3_2$
$\text{r}^2_1\times10800=\frac{4}{3}\times9\times9\times9$
$\text{r}^2_1=0.009$
$\text{r}_1=0.3\text{cm}$
Therefore Diameter $=2\times0.3=0.6\text{cm}$

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Question 23 Marks

The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer

Let the diameter of the earth be d
Then,
Diameter of moon will be $\frac{\text{d}}{4}$
Radius of earth $=\frac{\text{d}}{2}$
Radius of moon $=\frac{\frac{\text{d}}{2}}{4}=\frac{\text{d}}{8}$
Surface area of moon $=\frac{4\pi\big(\frac{\text{d}}{8}\big)^2}{4\pi\big(\frac{\text{d}}{2}\big)^2}$
$=\frac{4}{64}=\frac{1}{16}$
Thus the required ratio of the surface areas is $\frac{1}{16}$

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Question 33 Marks

A hemispherical bowl made of brass has inner diameter $10.5\ cm.$ Find the cost of tin plating it on the inside at the rate of $Rs. 4$ per $100\ cm^2$

Answer

Inner diameter of hemispherical bowl $= 10.5\ cm$
Radius $=\frac{10.5}{2}=5.25\text{ cm}$
Surface area of hemispherical bowl $=2\pi\text{r}^2$
$= 2 \times 3.14 \times (5.25)^2$
$= 173.25\ cm^2$
Cost of tin plating $100\ cm^2$ area $= Rs. 4$
area $=\text{Rs}.\ \frac{4\times173.25}{100}=\text{Rs}. 6.93$
Cost of tin plating $173.25\ cm^2$
Thus, the cost of tin plating the inner side of hemispherical bowl is $Rs. 6.93$

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Question 43 Marks

A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is $17.6\ cm,$ find the cost of painting it, given the cost of painting is $Rs. 5$ per $100\ cm^2$

Answer

Given that only the rounded surface of the dome to be painted,
we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done.
Now, circumference of the dome $= 17.6\ cm$
Therefore $2\pi\text{r}=17.6$
$2\times\frac{22}{7}\times\text{r}=17.6\text{m}$
So, the radius of the dome $2\pi\text{r}^2$
$=2\times\frac{22}{7}\times2.8\times2.8$
$=49.28\text{ m}^2$
Cost of painting $100\ cm^2$ is $Rs. 5$
So, the cost of painting $1m^2 = Rs. 500$
Therefore the cost of painting the whole dome $= Rs. 500 \times 49.28$
$= Rs. 24640$

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Question 53 Marks

A cylindrical tub of radius 12cm contains water to a depth of 20cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75cm. What is the radius of the ball?

Answer

Radius of cylindrical tub = 12cm
Depth = 20cm
Let r be the radius of the ball
Then
Volume of the ball = Volume of water raised
$\frac{4}{3}\pi\text{r}^3=\pi\text{r}^2\text{h}$
$\text{r}^3=\frac{3.14\times(12)^2\times6.75\times3}{4}$
$\text{r}^3=729$
$\text{r}=\sqrt[3]{729}$
$\text{r}=9\text{cm}$
Therefore radius of the ball = 9cm

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Question 63 Marks

A shopkeeper has one laddoo of radius 5cm. With the same material, how many laddoos of radius 2.5cm can be made?

Answer

Volume of laddoo having radius = 5cm
i.e Volume $(\text{V}_1) =\frac{4}{3\pi\text{r}^3}$
$(\text{V}_1) =\frac{4}{3}\times\frac{22}{7}\times(2.5)^3$
$(\text{V}_1) =\frac{1375}{21}\text{cm}^3$
Therefore number of laddoos $=\frac{\text{V}_1}{\text{V}_1}=\frac{11000}{1375}=8$

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Question 73 Marks

A capsule of medicine is in the shape of a sphere of diameter $3.5 \ mm.$ How much medicine $(mm^3)$ is needed to fill this capsule?

Answer

Given that
Diameter of capsule $= 3.5 \ mm$
Radius $\frac{3.5}{2}=1.75\ mm$
Volume of spherical sphere $=\frac{4}{3\pi\text{r}^3}$
$=\frac{4}{3}\times\frac{22}{7}\times(1.75)^3$
$= 22.458 \ mm^3$
Therefore $22.46\ mm^3$ of medicine is required.

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Question 83 Marks

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

Answer

Given
Radius of the hemisphere = Volume of cone
$\frac{2}{3}\pi\text{r}^3_1=\frac{1}{3}\pi\text{r}^2_2\text{h}$
$\frac{2}{3}\times7^3=\frac{1}{3}\text{r}^2_2\times49$
$\text{r}^2_2=\frac{2\times7\times7\times7\times3}{3\times49}$
$\text{r}^2_2=2058147$
$\text{r}_2=3.47\text{cm}$
Therefore radius of the base = 3.74cm

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Question 93 Marks

A cylinder of same height and radius is placed on top of a hemisphere. Find the curved surface area of the shape if the length of the shape is 7cm.

Answer

Given length of the shape = 7cm

But length = r + r
2r = 7cm
r = 3.5cm
Also; h = r
Total surface area of shape $=2\pi\text{rh}+2\pi\text{rh}^2$
$=2\pi\text{rr}+2\pi\text{r}^2$
$=2\pi\text{rr}+2\pi\text{r}^2$
$=4\times\frac{22}{7}\times(3.5)^2=154\text{cm}^2$

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Question 103 Marks

A cube of side $4\ cm$ contains a sphere touching its side. Find the volume of the gap in between.

Answer

It is given that
Cube side $= 4\ cm$
Volume of cube $= (4\ cm)^3 = 64\ cm^3$
Diameter of the sphere $=$ Length of the side of the cube $= 4\ cm$
Therefore radius of the sphere $= 2\ cm$
Volume of the sphere $=\frac{4}{3\pi\text{r}^3}=\frac{4}{3}\times\frac{22}{7}\times(2)^3=33.52\ cm^3$
Volume of gap $=$ Volume of cube $-$ Volume of sphere
$= 64\ cm^3 - 33.52\ cm^3 = 30.48\ cm^3$

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Question 113 Marks

The dome of a building is in the form of a hemisphere. Its radius is $63\ dm.$ Find the cost of painting it at the rate of Rs. $2$ per $sq \ m.$

Answer

Dome radius $= 63dm = 6.3m$
Inner surface area of dome $=2\pi\text{r}^2$
$= 2 \times 3.14 \times (6.3)^2$
$= 249.48m^2$
Now, cost of $1m^2 = Rs. 2$
Therefore cost of $249.48m^2 = Rs. (249.48 \times 2) = Rs. 498.96$

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Question 123 Marks

A hemispherical tank is made up of an iron sheet $1\ cm$ thick. If the inner radius is $1m,$ then find the volume of the iron used to make the tank.

Answer

Given that,
Inner radius of the hemispherical tank $= 1m = r_1$
Thickness of the hemispherical tank $= 1\ cm = 0.01m$
Outer radius of hemispherical tank $= (1 + 0.01) = 1.01m = r_2$
Volume of iron used to make the tank
$=\frac{2}{3}\pi(\text{r}^3_2-\text{r}^3_1)$
$=\frac{2}{3}\times\frac{22}{7}\big[(1.01)^3-1^3\big]$
$=\frac{44}{21}\big[(1.0303)-1\big]\text{ m}^3$
$= 0.06348\ m^3$

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Question 133 Marks

A cylindrical tub of radius 16cm contains water to a depth of 30cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9cm. What is the radius of the ball?

Answer

Let r be the radius of the iron ball
Radius of the cylinder = 16cm
Then,
Volume of iron ball = Volume of water raised in the hub
$\frac{4}{3}\pi\text{r}^3=\pi\text{r}^2\text{h}$
$\frac{4}{3}\text{r}^3=(16)^3\times9$
$\text{r}^3=\frac{27\times16\times16}{4}$
$\text{r}^3=1728$
$\text{r}=12\text{cm}$
Therefore radius of the ball = 12cm.

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Question 143 Marks

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be $1.4m$ and its length is $8m,$ find the cost of painting it on the outside at the rate of $Rs. 10$ per $m^2.$

Answer

Diameter of a cylinder $= 1.4m$
Therefore radius of cylinder $= 1.42 = 0.7m$
Height of cylinder $= 8m$
Therefore surface area of tank $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\times\frac{22}{7}\times0.7\times8+2\times\frac{22}{7}\times(0.7)^2$
$=\frac{176}{5}+\frac{77}{25}$
$=38.28\text{ m}^2$
Now cost of $1\ m^2 = Rs. 10$
Therefore cost of $38.28\ m^2$
$= Rs. 382.80$

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Question 153 Marks

A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.

Answer

Let $r$ be the common radius
Height of the cone $=$ height of the cylinder $= 2r$
Let
$v_1 =$ Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$v_1 =$ Volume of cylinder $=\pi\text{r}^2\text{h}=\pi\text{r}^2\times2\text{r}$
$v_1 =$ Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^3$
Now
$v_{1}: v_{2}: v_{3}\frac{4}{3}\pi\text{r}^3:2\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3$
$= 4 : 6 : 2$
$ = 2 : 3 : 1$

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Question 163 Marks

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44cm, each bullet being 4cm in diameter?

Answer

In the given problem, we have a lead cube which is remolded into small spherical bullets
Here, edge of the cube (s) = 44cm
Diameter of the small spherical bullets (d) = 4cm
Now, let us take the number of small bullets be x
So, the total volume of x spherical bullets is equal to the volume of the lead cube.
Therefore, we get,
Volume of the x bullets = volume of the cube
$\text{x}\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}}{2}\Big)^3=\text{s}^3$
$\text{x}\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)\Big(\frac{4}{2}\Big)^3=(44)^3$
$\text{x}\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)(2)^3=85184$
$\text{x}=\frac{(85184)(3)(7)}{(22)(4)(8)}$
$\text{x}=2541$
Therefore, 2541 small bullets can be made from the given lead cube.

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Question 173 Marks

The diameter of a sphere is $6\ cm.$ It is melted and drawn into a wire of diameter $0.2\ cm.$ Find the length of the wire.

Answer

Given that
Diameter of sphere $= 6\ cm$
Radius of sphere $=\frac{\text{d}}{2}=\frac{6}{2}=3\text{ cm}=\text{r}_1$
Diameter of the wire $= 0.2\ cm$
Radius of the wire $= 0.1\ cm = r_2$
Volume of sphere $=$ Volume of wire
$\frac{4}{3}\pi\text{r}^3_1$
$=\pi\text{r}^2_2\text{h}$
$=\frac{4}{3}\times3\times3\times3$
$=0.1\times0.1\times\text{h}$
$\text{h}=\frac{4\times3\times3}{0.1\times0.1}$
$\text{h}=3600\text{ cm}$
$\text{h}=36\text{m}$
Therefore length of wire $= 36m$

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Question 183 Marks

The radius of the internal and external surfaces of a hollow spherical shell are $3\ cm$ and $5\ cm$ respectively. If it is melted and recast into a solid cylinder of height $2\frac{2}{3}\ cm.$ Find the diameter of the cylinder.

Answer

Given that,
Internal radius of the sphere $= 3\ cm = r_1$
External radius of the sphere $= 5\ cm = r_2$
Height of the cylinder $=\frac{8}{3}\text{cm}=\text{h}$
Volume of the spherical shell $=$ Volume of cylinder
$\frac{4}{3}\pi(\text{r}^3_2-\text{r}^3_1)=\pi\text{r}^2_3\text{h}$
$\frac{4}{3}(5^3-3^3)=\frac{8}{3}\text{r}^2_3$
$\text{r}^2_3=\frac{4\times98\times3}{3\times8}$
$\text{r}_3=\sqrt{49}$
$\text{r}_3=7\text{ cm}$
Therefore diameter of the cylinder $= 2($radius$) = 14\ cm$

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Question 193 Marks

If a sphere of radius $2r$ has the same volume as that of a cone with circular base of radius $r$, then find the height of the cone.

Answer

In the given problem, we are given a cone and a sphere which have equal volumes. The dimensions of the two are;
Radius of the cone $(r_c) = r$
Radius of the sphere $(r_s) = 2r$
Now, let the height of the cone $= h$
Here, Volume of the sphere $=$ volume of the cone
$\Big(\frac{4}{3}\Big)\pi\text{r}^3_3=\Big(\frac{1}{3}\Big)\pi\text{r}^2_\text{c}\text{h}$
$\Big(\frac{4}{3}\Big)\pi(2\text{r})^3=\Big(\frac{1}{3}\Big)\pi(\text{r})^2\text{h}$
$\Big(\frac{4}{3}\Big)(8\text{r}^3)=\Big(\frac{1}{3}\Big)\text{r}^2\text{h}$
Further, solving for $h$
$\text{h}=\frac{(4)(8\text{r}^3)}{(\text{r}^2)}$
$\text{h}=32\text{r}$
Therefore, the height of the cone is $32r$.

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Question 203 Marks

The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?

Answer

Diameter of moon $=\frac{1}{4}\text{th}$ diameter of earth
Let the diameter of earth be d, so radius $=\frac{\text{d}}{4}$
Radius $=\frac{\frac{\text{d}}{2}}{4}=\frac{\text{d}}{8}$
Volume of moon $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\Big(\frac{\text{d}}{8}\Big)^3=\frac{4}{3}\times\frac{1}{512}$
Volume of earth $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\Big(\frac{\text{d}}{2}\Big)^3=\frac{4}{3}\times\frac{1}{8}\pi\text{d}^3$
$\frac{\text{Volume of moon}}{\text{Volume of earth}}=\frac{\frac{4}{3}\times\frac{1}{512}\pi\text{d}^3}{\frac{4}{3}\times\frac{1}{8}\pi\text{d}^3}$
$=\frac{1}{64}$
Thus the volume of the moon is $\frac{1}{64}$ of volume of earth.

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Question 213 Marks

Find the volume of a sphere whose surface area is $154\ cm^2.$

Answer

In the given problem,
we have to find the volume of a sphere whose surface area is given.
So, let us first find the radius of the given sphere.
Surface area of the sphere $(S) = 154\ cm^2$
Let the radius of the sphere be rcm
Now, we know that surface area of the sphere $=4\pi\text{r}^2$
So,
$154=4\Big(\frac{22}{7}\Big)(\text{r})^2$
$\text{r}^2=\frac{(154)(7)}{(4)(22)}$
$\text{r}^2=12.25$
Further, solving for $r$
$\text{r}=\sqrt{12.25}$
$\text{r}=3.5$
Therefore, the radius of the given sphere is $3.5\ cm.$
Now, the volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
$=\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)(3.5)^3$
$=179.66\text{ cm}^3$
Therefore, the volume of the given sphere is $179.66\ cm^3.$

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Question 223 Marks

If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.

Answer

In the given problem, we are given a sphere inscribed in a cube.
So, here we need to find the ratio between the volume of a cube and volume of sphere.
This means that the diameter of the sphere will be equal to the side of the cube.
Let us take the diameter as $d.$
Here,
Volume of a cube $(V_1) = S^3 = d^3$
Volume of a sphere $(V_2) =\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}}{2}\Big)^3$
$=\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}^3}{8}\Big)$
$=\frac{\pi\text{d}^3}{6}$
Now, the ratio of the volume of sphere to the volume of the cube $=\frac{\text{V}_1}{\text{V}_3}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{\text{d}^3}{\Big(\frac{\pi\text{d}^3}{6}\Big)}$
$=\frac{6}{\pi}$
So, the ratio of the volume of cube to the volume of the sphere is $6:\pi.$

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3 Marks Question - MATHS STD 9 Questions - Vidyadip