2 Marks Questions - Maths STD 9 Questions - Vidyadip

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Question 12 Marks

The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20cm and height 10m. How much concreate mixture would be required to build 14 such pillars?

Answer

Radius (r) of pillar = 20cm $=\frac{20}{100}\text{m}$
Height (h) of pillar = 10m
$\therefore$ Volume of 1 pillar $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{20}{100}\times\frac{20}{100}\times10\Big)\text{m}^3$
$=\frac{44}{35}\text{m}^3$
⇒ Volume of concreate mixture in 14 pillars $=14\times\frac{44}{35}=17.6\text{m}^3$

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Question 22 Marks

The outer diameter of a spherical shell is 12cm and its inner diameter is 8cm. Find the volume of metal contained in the shell. Also, find its outer surface area. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$

Answer

Outer radius of the spherical shell = 6cm
Inner radius of the spherical shell = 4cm
Volume of metal contained in the shell $=\frac{4}{3}\times\frac{22}{7}(6^3-4^3)$
$=\frac{88}{21}\times(216-64)$
$=\frac{88}{21}\times152$
$=636.95\text{cm}^3$
$\therefore$ Outer surface area $=4\times\frac{22}{7}\times6\times6$
$=452.57\text{cm}^2$

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Question 32 Marks

The surface area of sphere is $(576\pi)\text{cm}^2.$ Find its volume.$\big(\text{Take}\ \pi=\frac{22}{7}\big).$

Answer

Surface area of the sphere $=(576\pi)\text{cm}^2$
Suppose that r cm is the radius of the sphere.
Then $4\pi\text{r}^2=576\pi$
$\Rightarrow\text{r}^2=\frac{576}{4}=144$
$\Rightarrow\text{r}=12\text{cm}$
$\therefore$ Volume of the sphere $=\frac{4}{3}\times\pi\times12\times12\times12\text{cm}^3$
$=2304\text{cm}^3$

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Question 42 Marks

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

Answer

Volume of the solid metallic cuboid = 9m × 8m × 2m = 144m3
Volume of each solid cube = (Edge)³ = (2)³ = 8m³
$\therefore$ Number of cubes formed $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{solid}\ \text{metallic}\ \text{cuboid}}{\text{Volume}\ \text{of}\ \text{each}\ \text{solid}\ \text{cube}}$
$=\frac{144}{8}=18$
Thus, the number of cubes so formed is 18.

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Question 52 Marks

Find the total surface area of a cone, if its slant height is 21m and diameter of its base is 24m. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Radius of a cone, r = 12cm
Slant height of a cone, l = 21cm
Totalsurface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times12(21+12)\Big]\text{m}^2$
$=\Big(\frac{22}{7}\times12\times33\Big)\text{m}^2$
$=1244.57\text{m}^2$

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Question 62 Marks

A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates, each measuring 1.5m × 1.25m × 0.5m, that can be stored in the godown.

Answer

Volume of the godown = 40m × 25m × 15m = 15000m³
Volume of each wooden create = 1.5m × 1.25m × 0.5m = 0.9375m³
$\therefore$ Maximum number of wooden creates that can be stored in the godown
$=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{godown}}{\text{Volume}\ \text{of}\ \text{each}\ \text{wooden}\ \text{crate}}$
$=\frac{15000}{0.9375}$
$=16000$

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Question 72 Marks

A cylindeical tub of radius 12cm contains water to a depth of 20cm. A sphrical iron ball is dropped into the tub and thus the leval of water is raised by 6.75cm. What is the radius of the ball.

Answer

Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub = 12cm
Depth of the tub = 20cm
Now, volume of the ball = volume of water raised in the cylinder
$\Rightarrow\frac{4}{3}\pi\text{r}^3=\pi\times12^2\times6.75$
$\Rightarrow\text{r}^3=\frac{144\times6.75\times3}{4}$
$=36\times6.5\times3=729$
$\Rightarrow\text{r}=9\text{cm}$
$\therefore$ The radius of the ball is 9cm.

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Question 82 Marks

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

Answer

Let their heights be h and 3h
And, their radii be 3r and r.
Then, $\text{V}_1=\frac{1}{3} \pi(3\text{r})^2\times\text{h}$
And, $\text{V}_2=\frac{1}{3} \pi\text{r}^2\times3\text{h}$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\frac{\frac{1}{3}\pi(3\text{r})^2\times\text{h}}{\frac{1}{3}\pi\text{r}^2\times3\text{h}}=\frac{3}{1}$
$\therefore\text{V}_1:\text{V}_2=3:1$

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Question 92 Marks

The surface areas of two spheres are in the 1 : 4. Find the ratio of their volumes.

Answer

Suppose that the radii of the spheres are r and R.
We have:
$\frac{4\pi\text{r}^2}{4\pi\text{R}^2}=\frac{1}{4}$
$\Rightarrow\frac{\text{r}}{\text{R}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Now, ratio of the volumes $=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}=\Big(\frac{\text{r}}{\text{R}}\Big)^3$
$=\Big(\frac{1}{2}\Big)^3=\frac{1}{8}$
$\therefore$ The ratio of the volumes of the sphere is 1 : 8.

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Question 102 Marks

A hemispherical bowl of internal radius 9cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3cm and height 4cm How many bottles required to empty the bowl?

Answer

Internal radius of the hemispherical bowl = 9cm
Radius of a cylindrical shaped bottle = 1.5cm
Height of a bottle = 4cm
Number of bottles required to empty the bowl $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{hemispherical}\ \text{bowl}}{\text{Volume}\ \text{of}\ \text{a}\ \text{cylindrical}\ \text{shaped}\ \text{bottle}}$
$=\frac{\frac{2}{3}\pi\times9^3}{\pi\times1.5^2\times4}$
$=\frac{2\times9\times9\times9}{3\times1.5\times1.5\times4}$
$=54$
$\therefore$ 54 bottles are required to empty the bowl.

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Question 112 Marks

A matchbox measures 4cm × 2.5cm × 1.5cm. What is the volume of a packet containing 12 such matchboxes?

Answer

Volume of each matchbox = 4 × 2.5 × 1.5 = 15cm³
$\therefore$ Volume of 12 matchbox = 12 × 15 = 180cm³
Thus, the volume of a packet containing 12 such matchboxes is 180cm³.

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Question 122 Marks

A patient in a hospital is given soup in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer

Radius (r) of cylindrical bowl $=\Big(\frac{7}{2}\Big)\text{cm}=3.5\text{cm}$
Height (h) up to which the bowl is filled with soup = 4cm
Volume of soup in 1bowl = pr²h
$=\Big(\frac{22}{7}\times(3.5)^2\times4\Big)\text{cm}^3$
$=154\text{cm}^3$
Hence, volume of soup in 250 bowls = (250 × 154)cm³ = 38500cm³ = 38.5 litres
Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

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Question 132 Marks

The volume of a cuboid is 1536m³. Its length is 16m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

Answer

Length of the cuboid = 16m
Suposs that the breadth and height of the cuboid are 3x m and 2x m, respectively.
Then 1536 = 16 × 3x × 2x
⇒ 1536 = 16 × 6x²
$\Rightarrow\text{x}^2=\frac{1536}{96}=16$
$\Rightarrow\text{x}=\sqrt{16}=4$
$\therefore$ The breadth and height of the cuboid are 12m and 8m, respectively.

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Question 142 Marks

Find the curved surface area of a cone with base radius 5.25cm and slant height 10cm. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Radius of a cone, r = 5.25cm
Slant height of a cone, l = 10cm
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times5.25\times10\Big)\text{cm}^2$
$=165\text{cm}^2$

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Question 152 Marks

A right circular cone is 3.6cm high and the radius of its base is 1.6cm. It is melted and recast into a right circular cone having base radius 1.2cm. Find its height.

Answer

Here, height of cone = 3.6cm and radius = 1.6cm
After melting, its radius = 1.2cm
Volume of original cone = Volume of cone after melting
$\therefore\frac{1}{3}\pi\times1.6\times1.6\times3.6$
$=\frac{1}{3}\pi\times1.2\times1.2\times\text{h}$
$\Rightarrow\text{h}=\frac{\frac{1}{3}\pi\times1.6\times1.6\times3.6}{\frac{1}{3}\pi\times1.2\times1.2}=6.4\text{cm}$
$\therefore$ height of new cone = 6.4cm.

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Question 162 Marks

Find the volume, the lateral surface area, the totle surface area and the diagonal of a cube, each of whose edges measures 9m. $\big(\text{Take}\sqrt{3}=1.73\big)$

Answer

Here, a = 9m
Volume of the cube = a³ = 9³m³ = 729m³
Lateral surface area of the cube = 4a² = 4 × 9²m² = 4 × 81m² = 324m²
Total surface area of the cube = 6a²6 × 9²m² = 6 × 81m² = 486m²
$\therefore$ Diagonal of the cube $=\sqrt{3}\text{a}=\sqrt{3}\times9=15.57\text{m}$

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Question 172 Marks

How many persons can be accommodated in a dining hall of dimension (20m × 16m × 4.5m), assuming that each person requires 5 cubic metres of air?

Answer

Volume of the dining hall = (20 × 16 × 4.5) m³
= 1440m³
Volume of air required by each person = 5m³
$\therefore$ Capacity of the dining hall $=\frac{\text{Volume}\ \text{of}\ \text{dining}\ \text{hall}}{\text{Volume}\ \text{of}\ \text{air}\ \text{required}\ \text{by}\ \text{each}\ \text{each}\ \text{person}}$
$=\frac{1440}{5}=288\ \text{persons}$

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Question 182 Marks

A conical pit of diameter 3.5m is 12m deep. What is its capacity in kilolitres? $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Hint: 1m³ = 1 kilolitre

Answer

Radius of a conical pit, $\text{r}=\frac{3.5}{2}\text{m}$
Depth of a conical pit, h = 12m
Volume of the conical pit $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times\frac{3.5}{2}\times\frac{3.5}{2}\times12\Big)\text{m}^3$
$=38.5\text{m}^3$
$=38.5\ \text{kiloletre}$

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Question 192 Marks

In a shower, 5cm of rain falls. Find the volume of water that falls on 2 hectares of ground.

Answer

Volume of the water that falls on the ground = area of ground × depth

= 20000 × 0.05m³

= 1000m³

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Question 202 Marks

The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.

Answer

Suppose that the radii are r and 2r.
Now, ratio of the surface areas $=\frac{4\pi\text{r}^2}{4\pi(2\text{r})^2}=\frac{\text{r}^2}{4\text{r}^2}=\frac{1}{4}$
$=1:4$
$\therefore$ The ratio of their surface areas is 1 : 4.

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Question 212 Marks

Three cubes of metal with edges 3cm, 4cm and 5cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

Answer

Three cubes of metal with edges 3cm, 4cm and 5cm are melted to form a single cube.
$\therefore$ Volume of the new cube = sum of the volumes the old cubes
= (3³ + 4³ + 5³)cm³
= (27 + 64 + 125)cm³
= 216cm³
Suppose the edge of the new cube = x cm
Then we have:
Then 216 = x³
$\Rightarrow\text{x}=\sqrt[3]{216}=6$
$\therefore$ Lateral surface area of the new cube = 4x²cm² = 4 × 6²cm² = 144cm²

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Question 222 Marks

Find the length of the longest pole that can be put in a room of dimension (10m × 10m × 5m).

Answer

Length of the longest pole = length of the diagonal of the room
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}\text{m}$
$=\sqrt{10^2+10^2+5^2}\text{m}$
$=\sqrt{100+100+25}$
$=\sqrt{225}=15\text{m}$

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Question 232 Marks

A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? (Given, 1m³ = 1000 litres.)

Answer

Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135m³
$\therefore$ Volume of water in litres = 135 × 1000 = 135000L (1m³ = 1000L)
Thus, the water tank can hold 135000L of water.

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Question 242 Marks

The volume of a cube is 512cm³. Find its surface area.

Answer

Suppose that the side of the given cube is x cm.

Volume of the cube = 512cm³

Then 512 = x³

$\Rightarrow\text{x}=\sqrt[3]{512}=8$

i.e., the side of the cube is 8cm.

$\therefore$ Surface area of the cube = 6x²cm²= 6 × 8²cm² = 384cm²

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2 Marks Questions - Maths STD 9 Questions - Vidyadip