5 Marks Questions - Maths STD 9 Questions - Vidyadip

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17 questions · timed · auto-graded

Question 15 Marks

If 1cm³ of case iron weighs 21g, find the weight of a cast iron pipe of length 1m with a bore of 3cm in which the thickness of the metal is 1cm.

Answer

Internal radius $=\Big(\frac{3}{2}\Big)\text{cm}=1.5\text{cm}$
And, external radius $= (1.5 + 1)\text{cm} = 2.5\text{cm}$
Volume of cast iron $=\big[\pi\times(2.5)^2\times100-\pi\times(1.5)^2\times100\big]\text{cm}^3$
$=\pi\times100\times\big[(2.5)^2-(1.5)^2\big]\text{cm}^3$
$=\frac{22}{7}\times100\times\big[6.25-2.25\big]\text{cm}^3$
$=\Big(\frac{22}{7}\times100\times4\Big)\text{cm}^3$
Weight $=\Big(\frac{22}{7}\times100\times4\times\frac{21}{1000}\Big)\text{kg}$
$\big[\therefore1\text{kg}=1000\text{g}\big]$
$=26.4\text{kg}$
$\therefore$ the weight of the iron pipe n= 26.4kg.

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Question 25 Marks

The diameter of a cylinder is 28cm and its height is 40cm. Find the curved surface, total surface area and the volume of the cylinder.

Answer

Here, diameter = 28cm
Radius $=\Big(\frac{28}{2}\Big)\text{cm}=14\text{cm}$ and,
height = 40cm
$\therefore$ Curved surface area $=(2\pi\text{rh})$
$=\Big(2\times\frac{22}{7}\times14\times40\Big)\text{cm}^2$
$=3520\text{cm}^2$
$\therefore$Total surface area $=(2\pi\text{rh}+2\pi\text{r}^2)$
$=\Big(2\times\frac{22}{7}\times14\times4+2\times\frac{22}{7}\times14^2\Big)$
$=(3520+1232)=4752\text{cm}^2$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times14^2\times40\Big)\text{cm}^3$
$=\Big(\frac{22}{7}\times14\times14\times40\Big)\text{cm}^3$
$=24640\text{cm}^3.$

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Question 35 Marks

A heap of wheat is in the form of a cone of diameter 9m and height 3.5m. Find its volume. How much canvas cloth is required to just cover the heap? $\big(\text{Use}\ \pi = 3.14\big).$

Answer

Radius of a conical heap, r = 4.5m
Height of a conical tent, h = 3.5
Volume of a conical heap $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times3.14\times4.5\times4.5\times3.5\Big)\text{m}^3$
$=\big(3.14\times1.5\times4.5\times3.5\big)\text{m}^3$
$=74.1825\text{m}^3$
$\therefore$ Slant height of a conical tent,
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{4.5^2+3.5^2}$
$\text{l}=\sqrt{\Big(\frac{9}{2}\Big)+\Big(\frac{7}{2}\Big)^2}$
$\text{l}=\sqrt{\frac{81}{4}+\frac{49}{4}}$
$\text{l}=\frac{\sqrt{130}}{2}\text{m}$
Curved surface area of a conical tent $=\pi\text{rl}$
$=\Big(3.14\times\frac{9}{2}\times\frac{\sqrt{130}}{2}\Big)\text{m}^2$
$=80.54\text{m}^2$
Hence, 54m² canvas is required to just cover the heap.

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Question 45 Marks

A cloth having an area of 165m² is shaped into the form of a conical tent of radius 5m. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

How many students can sit in the tent if a student, on an average, occupies $\frac{5}{7}\text{m}^2$ on the ground?
Find the volume of the cone.

Answer

Area of the floor of the tent $=\pi\text{r}^2$

$=\Big(\frac{22}{7}\times5\times5\Big)\text{m}^2$

$=\frac{550}{7}\text{m}^2$

Area required by 1 students $=\frac{5}{7}\text{m}^2$

$\therefore$ Required number of students $=\frac{\frac{550}{7}}{\frac{5}{7}}=110$

Curved surface area of the tent = Area of the cloth = 165m²

$\Rightarrow\pi\text{rl}=165$

$\Rightarrow\frac{22}{7}\times5\times\text{l}=165$

$\Rightarrow\text{l}=\frac{165\times7}{22\times5}=\frac{21}{2}\text{m}$

Now,

$\text{l}^2=\text{r}^2+\text{h}^2$

$\Rightarrow\Big(\frac{21}{2}\Big)^2=5^2+\text{h}^2$

$\Rightarrow\text{h}^2=\frac{441}{4}-25=\frac{341}{4}=85.25$

$\Rightarrow\text{h}=9.23\text{m}$

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

$=\Big(\frac{1}{3}\times\frac{22}{7}\times5\times5\times9.23\Big)\text{m}^3$

$=241.7\text{m}^3$

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Question 55 Marks

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius and height of each has the ratio 3 : 4.

Answer

Let the curved surface areas of cylinder and cone be 8x and 5x.
Then $2\pi\text{rh}=8\text{x}\ ...(\text{i})$
And, $\pi\text{r}\sqrt{\text{h}^2+\text{r}^2}=5\text{x}\ ...(\text{ii})$
Squaring both sides of equation (i), we have
$(2\pi\text{rh})^2=(8\text{x})^2$
$4\pi^2\text{r}^2\text{h}^2=64\text{x}^2\ ...(\text{iii})$
From (ii) we have,
$\pi\text{r}\sqrt{\text{h}^2+\text{r}^2}=5\text{x}$
Squaring both sides,
$\Rightarrow\pi^2\text{r}^2(\text{h}^2+\text{r}^2)=25\text{x}^2\ ...(\text{iv})$
$\Rightarrow\frac{4\pi^2\text{r}^2\text{h}^2}{\pi^2\text{r}^2(\text{h}^2+\text{r}^2)}=\frac{64}{25}$ [Dividing (iii) by (iv)]
$\Rightarrow\frac{\text{h}^2}{(\text{h}^2+\text{r}^2)}=\frac{16}{25}$
$\Rightarrow9\text{h}^2=16\text{r}^2$
$\Rightarrow\frac{\text{r}^2}{\text{h}^2}=\frac{9}{16}$
$\Rightarrow\frac{\text{r}}{\text{h}}=\frac{3}{4}$
$\therefore$ Thus ratio of radius and height = 3 : 4

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Question 65 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 1cm³ of iron weighs 7.5g. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Here, height (h) of cylinder = 2.8m = 280cm and diameter = 20cm.
$\Rightarrow\text{radius}=\Big(\frac{20}{2}\Big)=10\text{cm}$
height (H) of the cone = 42cm
$\therefore$ Volume pf the pillar $=\Big(\pi\text{r}^2+\frac{1}{3}\pi\text{r}^2\text{H}\Big)\text{cm}^3$
$=\pi\text{r}^2\Big(\text{h}+\frac{1}{3}\text{H}\Big)\text{cm}^3$
$=\frac{22}{7}\times10\times10\Big(280+\frac{1}{3}\times42\Big)\text{cm}^3$
$=\frac{2200}{7}\times\big[280+14\big]$
$=92400\text{cm}^3$
$\therefore$ Weight of pillar $=\Big(\frac{92400\times7.5}{1000}\Big)\text{kg}=693\text{kg}$

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Question 75 Marks

The total surface area of a cylinder is 462cm². Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer

Curved surface area $=\frac{1}{3}\times(\text{total}\ \text{surface}\ \text{area})$
$=\Big(\frac{1}{3}\times462\Big)\text{cm}^2=154\text{cm}^ 2$
(Total surface area) - (Curved surface area)
= (462 - 154)cm² = 308cm²
$\Rightarrow2\pi\text{r}^2=308$
$\Rightarrow2\times\frac{22}{7}\times\text{r}^2=308$
$\Rightarrow\text{r}^2=\frac{308\times7}{44}=49$
$\Rightarrow\text{r}=\sqrt{49}=7\text{cm}$
Now, curved surface area $=2\pi\text{rh}=154\text{cm}^2$
$=2\times\frac{22}{7}\times7\times\text{h}=154\text{cm}^2$
$=\text{h}=\frac{154}{44}=3.5\text{cm}$
Now, r = 7cm and h = 3.5cm
Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times7\times7\times3.5\Big)\text{cm}^3$
$=539\text{cm}^3$
$\therefore$ The volume of the cylinder = 539cm³.

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Question 85 Marks

The total surface area of a solid cylinder is 231cm² and its curved surface area is $\frac{2}{3}$ of the total surface area. Find the volume of the cylinder.

Answer

Curved surface area $=\frac{2}{3}\times(\text{total}\ \text{surface}\ \text{area})$
$=\Big(\frac{2}{3}\times231\Big)\text{cm}^2=154\text{cm}^ 2$
(Total surface area) - (Curved surface area)
= (231 - 154)cm² = 77cm²
$\Rightarrow2\pi\text{r}^2=77\text{cm}^2$
$\Rightarrow2\times\frac{22}{7}\times\text{r}^2=77$
$\Rightarrow\text{r}^2=\frac{77\times7}{44}=\frac{49}{4}$
$\Rightarrow\text{r}=\sqrt{\frac{49}{4}}=\frac{7}{2}\text{cm}$
Now, $=2\pi\text{rh}=154\text{cm}^2$
$=2\times\frac{22}{7}\times7\times\text{h}=154\text{cm}^2$
$=\text{h}=\frac{154}{22}=7\text{cm}$
Now,$\text{r} =\frac{7}{2}\text{cm}$ and h = 3.5cm
Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times7\Big)\text{cm}^3$
$=269.5\text{cm}^3$
Volume of the cylinder = 269.5cm³.

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Question 95 Marks

How many cubic centimetres of iron are there in an open box whose external dimension are 36cm, 25cm and 16.5cm, the iron being 1.5cm thick throughout? If 1cm³ of iron weighs 15g, find the weight of the empty box in kilograms.

Answer

The external dimensions of the box are 36cm, 25cm and 16.5cm.
Thickness of the iron = 1.5cm
$\therefore$ Inner length of the box = 36 - 1.5 -1.5 = 33 cm
Inner breadth of the box = 25 - 1.5 - 1.5 = 22cm
Inner height of the box = 16.5 − 1.5 = 15cm
Now,
Volume of iron in the open box
= Volume of the outer box - Volume of the inner box
= 36 × 25 × 16.5 - 33 × 22 × 15
= 14850 - 10890
= 3960cm³
It is given that 1cm³ of iron weights 15g.
$\therefore$ Weight of the empty box = 3960 × 15 = 59400g $=\frac{59400}{1000}=59.4\text{kg}$ (1kg = 1000g)

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Question 105 Marks

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616cm².

Answer

Curved surface area $=2\pi\text{rh}$
Total surface area $=2\pi\text{r}(\text{h}+\text{r})$
Since they are in the ratio of 1 : 2
$\therefore\frac{2\pi\text{rh}}{2\pi\text{r}(\text{h}+\text{r})}=\frac{1}{2}$
$\Rightarrow\frac{\text{h}}{\text{h}+\text{r}}=\frac{1}{2}$
$\Rightarrow2\text{h}=\text{h}+\text{r}$
$\Rightarrow2\text{h}-\text{h}=\text{r}$
$\Rightarrow\text{h}=\text{r}$
$2\pi\text{r}(\text{h}+\text{r})=616\text{cm}^2$
$\Rightarrow4\pi\text{r}^2=616\text{cm}^2$ [Putting = r]
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=616$
$\Rightarrow\text{r}^2=\frac{616\times7}{88}=49$
$\Rightarrow\text{r}=\sqrt{49}=7\text{cm}$
Then, r = 7cm and h = 7cm
$\therefore\text{Volume}=\big(\pi\text{r}^2\text{h}\big)$
$=\Big(\frac{22}{7}\times7\times7\times7\Big)\text{cm}^3=1078\text{cm}^3$
$\therefore$ The volume of the cylinder = 1078cm³.

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Question 115 Marks

A juiceseller has a large cylindrical vessel of base radius 15cm filled up to a height of 32cm with orange juice. The juice is filled in small cylindrical glasses of radius 3cm up to a height of 8cm, and sold for ₹ 15 each. How much money does he received by selling the juice completely?

Answer

Radius (r) of cylindrical vassel = 15cm
Height (h) of cylindrical vessel = 32m
Volume of cylindrical vessel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times15\times15\times32\Big)\text{cm}^3$
$=\frac{158400}{7}\text{cm}^3$
Radius of small cylindrical glass = 3cm
Height of a small cylindrical glass glass = 8cm
Volume of each small cylindrical glass $=\Big(\frac{22}{7}\times3\times3\times8\Big)\text{cm}^3$
$=\frac{1584}{7}\text{cm}^3$
Number of small glasses filled $=\frac{\text{Volume}\ \text{of}\ \text{cylindrical}\ \text{vessel}}{\text{Volume}\ \text{of}\ \text{each}\ \text{glass}}$
$=\frac{\frac{158400}{7}}{\frac{1584}{7}}$
$=100$
Cost of 1 glass = ₹ 15
⇒ Cost of 100 glasses = ₹ (15 × 100) = ₹ 1500

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Question 125 Marks

The difference between inside and outside surfaces of a cylindrical tube 14cm long, is 88cm². If the volume of the tube is 176cm³, find the inner and outer radii of the tube.

Answer

Let R cm and r cm be the outer and inner of the cylindrical tube.
We have, length of tube = h = 14cm
Now,
Outside surface area - inner surface area = 88cm²
$\Rightarrow2\pi\text{Rh}-2\pi\text{rh}=88$
$\Rightarrow2\pi(\text{R}-\text{r})\text{h}=88$
$\Rightarrow2\times\frac{22}{7}\times(\text{R}-\text{r})\times14=88$
$\Rightarrow2\times22\times(\text{R}-\text{r})\times2=88$
$\Rightarrow\text{R}-\text{r}=\frac{88}{2\times88\times2}=1\ ...(\text{i})$
It is given that the volume of the tube = 176cm³
⇒ External volume - internal volume = 176cm³
$\Rightarrow\pi\text{R}^2\text{h}-\pi\text{r}^2\text{h}=176$
$\Rightarrow\pi(\text{R}^2-\text{r}^2)\text{h}=176$
$\Rightarrow\frac{22}{7}\times(\text{R}-\text{r})(\text{R}+\text{r})\times14=176$
$\Rightarrow22\times1\times(\text{R}+\text{r})\times2=176$ [Using (i)]
$\Rightarrow\text{R}+\text{r}=\frac{176}{22\times2}$
$\Rightarrow\text{R}+\text{r}=4\ ...(\text{ii})$
Adding (i) and (ii), we get
2R = 5
⇒ R = 2.5cm
⇒ 2.5 - r = 1
⇒ r = 1.5cm
Thus, the inner and outer radii of the tube are 1.5cm and 2.5cm respectively.

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Question 135 Marks

A rectangular sheet of paper 30cm × 18cm can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders, thus formed.

Answer

When the sheet is folded along, it forms a cylinder of height, h₁ = 18cm and perimeter of base equal to 30cm.
Let r₁ be the radius of the bese and V₁ be is volume.
Then,
$2\pi\text{r}_1=30$
$\Rightarrow\text{r}_1=\frac{30}{2\pi}=\frac{15}{\pi}$
Now,
$\text{V}_1=\pi\text{r}_1^2\text{h}_1$
$=\pi\times\Big(\frac{15}{\pi}\Big)^2\times18=\Big(\frac{225}{\pi}\times18\Big)\text{cm}^3$
Again, when the sheet is folded along its breadth, it forms a cylinder of height, h₂ = 30cm and perimeter of base equal to 18cm.
Let, r₂ be the base and V₂ be is volume.
Then,
$\Rightarrow2\pi\text{r}_2=18$
$\Rightarrow\text{r}_2=\frac{18}{2\pi}=\frac{9}{\pi}$
Now,
$\text{V}_2=\pi\text{r}_2^2\text{h}_2$
$=\pi\Big(\frac{9}{\pi}\Big)^2\times30=\Big(\frac{81}{\pi}\times30\Big)\text{cm}^3$
$\therefore\frac{\text{V}_1}{\text{V}_2}=\frac{\Big(\frac{225}{\pi}\times18\Big)\text{cm}^3}{\Big(\frac{81}{\pi}\times30\Big)\text{cm}^3}=\frac{225\times18}{81\times30}=\frac{5}{3}$
$\Rightarrow\text{V}_1:\text{V}_2=5:3$

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Question 145 Marks

A well with inside diameter 10m 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 the well = 5m
Depth of the well = 8.4m
Volume of the earth dug out = Volume of well
$=\Big(\frac{22}{7}\times5\times5\times8.4\Big)\text{m}^3$
$=660\text{m}^2$
Width of the embankment = 7.5m
External radius of the embankment, R = (5 + 7.5)m = 12.5m
Internal radius of the embankment $=\pi(\text{R}^2-\text{r}^2)$
$=\Big[\frac{22}{7}(12.5^2-5^2)\Big]\text{m}^2$
$=\Big[\frac{22}{7}\times(156.25-25)\Big]\text{m}^2$
$=\Big[\frac{22}{7}\times131.25\Big]\text{m}^2$
$=412.5\text{m}^2$
Volume of the embankment = Volume of the earth dug out = 660m²
Height of the embankment $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{embankment}}{\text{Area}\ \text{of}\ \text{the}\ \text{embankment}}$
$=\frac{660\text{m}^3}{412.5\text{m}^2}$
$=1.6\text{m}$

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Question 155 Marks

The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617cm³, find the total surface area of the cylinder.

Answer

Let the radius (r) = 2x cm and height (h) = 3x cm
Then, Volume of cylinder $=(\pi\text{r}^2\text{h})$
Volume $=\Big[\frac{22}{7}\times(2\text{x})^2\times3\text{x}\Big]$
Volume $=\Big[\frac{22}{7}\times4\text{x}^2\times3\text{x}\Big]$
Volume $=\frac{22}{7}\times12\text{x}^3$
$\Rightarrow1617=\frac{22}{7}\times12\text{x}^3$
$\big[\therefore$ Volume given = 1617cm³$\big]$
$\Rightarrow12\text{x}^3=\frac{1617\times7}{22}$
$\Rightarrow \text{x}^3=\frac{1617\times7}{22\times12}=\Big(\frac{7}{2}\Big)^3$
$\Rightarrow\text{x}=\frac{7}{2}$
$\therefore$ radius $=2\text{x}=2\times\frac{7}{2}=7\text{cm}$
and height $=3\text{x}=3\times\frac{7}{2}=\frac{21}{2}\text{cm}$
Total surface area $=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\Big(\frac{21}{2}+7\Big)\text{cm}^2$
$=44\times\Big(\frac{21+14}{2}\Big)\text{cm}^2$
$=(22\times35)\text{cm}^2=770\text{cm}^2$

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Question 165 Marks

Find the volume, curved surface area and the total surface area of a cone having base radius 35cm and height is 12cm. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Radius of a cone, r = 35cm
Height of a cone, h = 12cm
Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times35\times35\times12\Big)\text{cm}^3$
$=15400\text{cm}^3$
Slant height of a cone, $\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{35^2+12^2}$
$\text{l}=\sqrt{1369}$
$\text{l}=37\text{cm}$
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times35\times37\Big)\text{cm}^2$
$=4070\text{cm}^2$
Total surface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times35(37+35)\Big]\text{cm}^2$
$=\big[22\times5\times72\big]\text{cm}^2$
$=7920\text{cm}^2$

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Question 175 Marks

A soft drink is available in two packs:

A tin can with a rectangular base of length 5cm, breadth 4cm and height 15cm.
A plastic cylinder with circular base of diameter 7cm and height 10cm.

Which container has greater capacity and by how much?

Answer

For a tin of rectangular base,

Length = 5cm

Breadth = 4cm

Height = 15cm

$\therefore$ Volume of a tin can = Length × Breadth × Height

= (5 × 4 × 15)cm³

= 300cm³

For a cylinder with circular base,

Diameter = 7

⇒ Radius $=\text{r}=\frac{7}{2}\text{cm}$

Height = h = 10cm

$\therefore$ Volume of cylinder $=\pi\text{r}^2\text{h}$

$=\Big(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times10\Big)\text{cm}^3$

$=385\text{cm}^3$

⇒ Volume of plastic cylinder is greater than volume of a tin can.

Difference in volume = (385 - 300) = 85cm³

Thus, a plastic cylinder has more capacity that a tin can by 85cm³.

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5 Marks Questions - Maths STD 9 Questions - Vidyadip