2 Mark Question - Maths STD 9 Questions - Vidyadip

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

A cylinder of radius 12cm contains water to a depth of 20cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75cm. Find the radius of the ball. $\Big(\text{Use }\pi=\frac{22}{7}\Big).$

Answer

Given that: Radius of the cylinder = 12cm = $r_1$
Volume of water raised = Volume of the sphere$=\pi\text{r}^2_1\text{h}=\frac{4}{3}\pi\text{r}^3_2$
$=12\times12\times6.75=\frac{4}{3}\text{r}^3_2$
$=\text{r}^3_2=\frac{12\times12\times6.75\times3}{4}$
$=\text{r}^3_2=729$
$=\text{r}_2=9\text{cm}$
Radius of the sphere is 9cm

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Question 22 Marks

A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Answer

Given that Volume of the cone = Volume of the hemisphere$\frac{1}{3\pi\text{r}^2\text{h}}=\frac{2}{3\pi\text{r}^3}$
$\text{r}^2\text{h}=2\text{r}^3$
$\text{h}=2\text{r}$
$\frac{\text{h}}{\text{r}}=\frac{1}{1}\times2=2$
Therefore Ratio of their heights = 2 : 1

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Question 32 Marks

A hemispherical tank has the inner radius of 2.8m. Find its capacity in liters.

Answer

Radius of the tank = 2.8m Therefore Capacity $=\frac{2}{3\pi\text{r}^3}$$=\frac{2}{3}\times\frac{22}{7}\times(2.8)^3=45.994\text{m}^2$
$1\text{m}^3=1000\mid$
Therefore capacity in litres = 45994 litres

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Question 42 Marks

A metallic sphere of radius 10.5cm is melted and thus recast into small cones, each of radius 3.5cm and height 3cm. Find how many cones are obtained.

Answer

Given that Metallic sphere of radius = 10.5cm Cone radius = 3.5cm Height of radius = 3cm Let the number of cones obtained be x$\text{v}_\text{s}=\text{x}\times\text{v}\text{ cone}$
$\frac{4}{3}\pi\text{r}^3=\text{x}\times\frac{1}{3}\pi\text{r}^2\text{h}$
$\text{x}=\frac{4\times10.5\times10.5\times10.5}{3.5\times3.5\times3}$
$\text{x}=126$
Therefore number of cones = 126

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Question 52 Marks

A cylindrical jar of radius 6cm contains oil. Iron spheres each of radius 1.5cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters?

Answer

Given that, Radius of the cylinder jar = 6cm = $r_1$
Level to be rised = 2cm
adius of each iron sphere = 1.5cm = $r_2$
Numberofsphere $=\frac{\text{Volume of cylinder}}{\text{Volume of sphere}}$
$=\frac{\pi\text{r}^2_1\text{h}}{\frac{4}{3}\pi\text{r}^3_2}$
$=\frac{\text{r}^2_1\text{h}}{\frac{4}{3}\text{r}^3_2}$
$=\frac{6\times6\times2}{43\times1.5\times1.5\times1.5}$
Number of sphere = 16

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Question 62 Marks

How many bullets can be made out of a cube of lead, whose edge measures $22cm$, each bullet being $2cm$ in diameter?

Answer

Cube edge = $22cm$
Therefore volume of the cube = $(22)^3$ = $10648cm^3$ And,
Volume of each bullet $=\frac{4}{3\pi\text{r}^3}$
$=\frac{4}{3}\times\frac{22}{7}\times(1)^3$
$=\frac{4}{3}\times\frac{22}{7}$
$=\frac{88}{21}\text{cm}^3$
Number of bullets $=\frac{\text{Volume of cube}}{\text{Volume of bullet}}=\frac{\frac{10648}{88}}{21}=2541$

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Question 72 Marks

A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Answer

Given that A cone and a hemisphere have equal bases and volumes $v_{\text {cone }}=v_{\text {hemisphere }}^{3 \pi \pi^{2^2 h}}=\frac{2}{9 \pi^{r^3}}$
$\text{r}^2\text{h}=2\text{r}^3$
$\text{h}=2\text{r}$
$\text{hr}=\frac{2}{1}$
$\text{h}:\text{r}=2:1$
Therefore the ratio is $2 : 1$

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Question 82 Marks

Assuming the earth to be a sphere of radius 6370km, how many square kilometers is the area of the land if three-fourths of the earth’s surface is covered by water?

Answer

$\frac{3}{4^\text{th}}$ of earth surface is covered by water.Therefore $\frac{1}{4^\text{th}}$ of earth surface is covered by land
Therefore Surface area covered by land $=\frac{1}{4}\times4\pi\text{r}^2$
$=\frac{1}{4}\times4\times\frac{22}{7}\times(6370)^2$
$=127527400\text{km}^2$

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Question 92 Marks

A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6cm and 4cm. Find the height of water in the cylinder.

Answer

It is given that Volume of water in hemispherical bowl = Volume of cylinder$\frac{2}{3}\pi\text{r}^3_1=\pi\text{r}^2_2\text{h}$
$\frac{2}{3}\pi(6)^3=\pi(4)^2\text{h}$
$\text{h}=\frac{2}{3}\times\frac{6\times6\times6}{4\times4}$
$\text{h}=9\text{cm}$

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Question 102 Marks

Find the surface area of a sphere of radius $14cm.$

Answer

In the given problem, we have to find the surface area of a sphere of a given radii.
Radius of the sphere (r) = $14cm$
So, surface area pf the sphere
$=4\pi\text{r}^2$$=4\Big(\frac{22}{7}\Big)(14)^2$
$=2464\text{cm}^2$
Therefore, the surface area of the given sphere of radius $14cm$ is $2464cm^2.$

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Question 112 Marks

Find the total surface area of a hemisphere of radius 10cm.

Answer

In the given problem, we have to find the total surface area of a hemisphere of a given radii. Radius of the hemisphere (r) = 10cm So, total surface area of the hemisphere $=3\pi\text{r}^2$$=3\Big(\frac{22}{7}\Big)(10)^2$
$=942.86\text{cm}^2$ Therefore, the total surface area of the given hemisphere of radius 10cm is 942.86cm.

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Question 122 Marks

A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio $1 : 2 : 3.$

Answer

Given that, A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight.
We know that $v_{\text {cone }} v _{\text {hemisphere }} v _{\text {Cylinder }} \frac{1}{3} \pi r ^2 h: \frac{2}{3} \pi r ^3: \pi r ^2 h$
Multiplying by $3 \pi r ^2 h: 2 \pi r ^3: 3 \pi r ^2 h$
$\pi r ^3: 2 \pi r ^3: 3 \pi r ^3\left(\therefore r = h\right.$ and $\left.r ^2 h= r ^3\right)$
$1: 2: 3$ Therefore the ratio is $1: 2: 3$.

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Question 132 Marks

Find the total surface area of a hemisphere and a solid hemisphere each of radius 10cm $(\pi=3.14)$

Answer

The surface area of the hemisphere $=2\pi\text{r}^2$$=2\times3.14\times(10)^2$
$=628\text{cm}^2$
The surface area of solid hemisphere $=2\pi\text{r}^2$$=3\times3.14\times(10)^2$
$=942\text{cm}^2$

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Question 142 Marks

A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5cm and 7cm respectively.
Find the height to which the water will rise in the cylinder.

Answer

Given that Volume of water in the hemispherical bowl= Volume of water in the cylinder
Let h be the height to which water rises in the cylinder Inner radii of the bowl
$= r_1 = 3.5cm$
Inner radii of the bowl
$= r_2 = 7cm$
$\frac{2}{3}\pi\text{r}^3_1=\pi\text{r}^2_2\text{h}$
$\text{h}=\frac{2\text{r}^3_1}{3\text{r}^2_2}$
$\text{h}=\frac{2(3.5)^3}{3(7^2)}$
$\text{h}=\frac{7}{12}\text{cm}$

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Question 152 Marks

Find the radius of a sphere whose surface area is $154 cm^2$

Answer

In the given problem, we have to find the radius of a sphere whose surface area is given.
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 3.5cm.

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Question 162 Marks

The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7m. Find the area available to the motorcyclist for riding.

Answer

In the given problem, the area available for the motorcyclist for riding will be equal to the surface area of the hollow sphere.
So here, we have to find the surface area of a hollow sphere of a given diameter.
Diameter of the sphere (d) = 7m
So, surface area of the sphere
$=4\pi\Big(\frac{\text{d}}{2}\Big)^2$
$=4\Big(\frac{22}{7}\Big)\Big(\frac{7}{2}\Big)^2$
$=4\Big(\frac{22}{7}\Big)(3.5)^2$
$=154\text{m}^2$
Therefore, the area available for the motorcyclist for riding is $154m^2$

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Question 172 Marks

hemispherical bowl is made of steel 0.25cm thick. The inside radius of the bowl is 5cm. Find the volume of steel used in making the bowl.

Answer

Inner radius = 5cm Outer radius = 5 + 0.25 = 5.25 Volume of steel used = Outer volume-Inner volume$=\frac{2}{3}\times\pi\times\big((5.25)^3-(5)^3\big)$
$=\frac{2}{3}\times\frac{22}{7}\times\big((5.25)^3-(5)^3\big)$
$=41.282\text{cm}^3$

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Question 182 Marks

The surface area of a sphere is $5544cm^2$, find its diameter.

Answer

Surface area of a sphere is $5544 cm 24 \pi r ^2=5544$
$4 \times 3.14 \times(r)^2=5544$
$r^2=\frac{5544 \times 7}{88}$
$r=\sqrt{21 cm \times 21 cm}$
$r=21 cm$
Diameter $=2($ radius $)=2(21)=42 cm$

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Question 192 Marks

The largest sphere is carved out of a cube of side 10.5cm. Find the volume of the sphere.

Answer

Side of cube = 10.5cm Volume of sphere = v Diameter of the largest sphere = 10.5cm 2r = 10.5 r = 5.25cm Volume of sphere $=\frac{4}{3\pi\text{r}^3=\frac{4}{3}}\times\frac{22}{7}\times5.25\times5.25\times5.25$$\text{v}=\frac{11\times441}{8}\text{cm}^3$
$\text{v}=606.375\text{cm}^3$

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Question 202 Marks

If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Answer

Let $v_1$ and $v_2$ be the volumes of the first and second sphere respectively
Radius of the first sphere = $r$
Radius of the second sphere = $2r$
Therefore, $\frac{\text{Volume of frist sphere}}{\text{Volume of second sphere}}=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi(2\text{r})^3}$
$=\frac{1}{8}$

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Question 212 Marks

A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4cm. Calculate the radius of the base of the cylinder.

Answer

Given that Height of the cylinder $=\frac{2}{3\text{ diameter}}$ We know that Diameter = 2(radius)$\text{h}=\frac{2}{3}\times2\text{r}=\frac{4}{3\text{r}}$
Volume of the cylinder = Volume of the sphere$\pi\text{r}^2\text{h}=\frac{4}{3\pi\text{r}^3}$
$\pi\times\text{r}^2\times(\frac{4}{3\text{r}})=\frac{4}{3\pi(4)^3}$
$(\text{r})^3=(\text{4)}^3$
$\text{r}=4\text{cm}$

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2 Mark Question - Maths STD 9 Questions - Vidyadip