EXERCISE 13.2
Q 1, Ex 13.2, Page No 216, Surface Areas & Volumes, Class 9th Mathematics
Page No 216:
Question 1:
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. Assume π $=\frac{22}{7}$Answer:
Height (h) of cylinder = 14 cmLet the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
⇒ 2πrh = 88 cm2 (r is the radius of the base of the cylinder)
⇒ πdh = 88 cm2 (d = 2r)
⇒$\frac{22}{7} \times d \times 14 \mathrm{cm}=88 \mathrm{cm}^{2}$
⇒ d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Q 2, Ex 13.2, Page No 216, Surface Areas & Volumes, CBSE Class 9th
Question 2:
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Height (h) of cylindrical tank = 1 mBase radius (r) of cylindrical tank=\left(\frac{140}{2}\right)
=70cm
=0.7m
Area of sheet required=Total surface area of tank=2πr(r+h)
$=\left[2 \times \frac{22}{7} \times 0.7(0.7+1)\right] \mathrm{m}^{2}$
$=(4.4 \times 1.7) \mathrm{m}^{2}$
=7.48 m2
Therefore, it will require 7.48 m2 area of sheet.
Q 3, Ex. 13.2 Page No 216 - Surface Areas & Volumes - NCERT Class 9th Maths
Question 3:
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area. $\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
Inner radius $\left(r_{1}\right)$ of cylindrical pipe $=\left(\frac{4}{2}\right) \mathrm{cm}=2 \mathrm{cm}$Outer radius $\left(r_{2}\right)$of cylindrical pipe $=\left(\frac{4.4}{2}\right) \mathrm{cm}$
=2.2cm
Height (h) of cylindrical pipe = Length of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe$=2 \pi r_{1} h$
$=\left(2 \times \frac{22}{7} \times 2 \times 77\right) \mathrm{cm}^{2}$
$=968 \mathrm{cm}^{2}$
(ii) CSA of outer surface of pipe $=2 \pi r_{2} h$
$=\left(2 \times \frac{22}{7} \times 2.2 \times 77\right) \mathrm{cm}^{2}$
$=(22 \times 22 \times 2.2) \mathrm{cm}^{2}$
$=1064.8 \mathrm{cm}^{2}$
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
$=2 \pi r_{1} h+2 \pi r_{2} h+2 \pi\left(r_{2}^{2}-r_{1}^{2}\right)$
$=\left[968+1064.8+2 \pi\left\{(2.2)^{2}-(2)^{2}\right\}\right] \mathrm{cm}^{2}$
$=\left(2032.8+2 \times \frac{22}{7} \times 0.84\right) \mathrm{cm}^{2}$
$=(2032.8+5.28) \mathrm{cm}^{2}$
$=2038.08 \mathrm{cm}^{2}$
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
Q 4, Ex 13.2, Page No 217 - Surface Areas & Volumes - NCERT Class 9th Maths
Page No 217:
Question 4:
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
It can be observed that a roller is cylindrical.Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller $=\left(\frac{84}{2}\right) \mathrm{cm}$
=42cm
CSA of roller = 2πrh
$=\left(2 \times \frac{22}{7} \times 42 \times 120\right) \mathrm{cm}^{2}$
$=31680 \mathrm{cm}^{2}$
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 15840000 cm2
= 1584 m2
Q 5, Ex 13.2, Page No 217, Surface Areas & Volumes, Class 9th Mathematics
Question 5:
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m2. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Height (h) cylindrical pillar = 3.5 mRadius (r) of the circular end of pillar $=\frac{50}{2}$
=25cm
= 0.25 m
CSA of pillar = 2πrh
$=\left(2 \times \frac{22}{7} \times 0.25 \times 3.5\right) \mathrm{m}^{2}$
$=(44 \times 0.125) \mathrm{m}^{2}$
$=5.5 \mathrm{m}^{2}$
Cost of painting 1 m2 area = Rs 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs 68.75
Therefore, the cost of painting the CSA of the pillar is Rs 68.75.
Q 6, Ex 13.2, Page No 217, Surface Areas & Volumes, Class 9th Mathematics
Question 6:
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Let the height of the circular cylinder be h.Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
2πrh = 4.4 m2
$\left(2 \times \frac{22}{7} \times 0.7 \times h\right) \mathrm{m}=4.4 \mathrm{m}^{2}$
h = 1 m
Therefore, the height of the cylinder is 1 m
Q 7, Ex 13.2, Page No 217, Surface Areas & Volumes, 9th R D Sharma Solutions
Question 7:
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find(i) Its inner curved surface area,
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m2. $\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
Inner radius (r) of circular well$=\left(\frac{3.5}{2}\right) \mathrm{m}$=1.75m
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
$=\left(2 \times \frac{22}{7} \times 1.75 \times 10\right) \mathrm{m}^{2}$
= (44 × 0.25 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
Cost of plastering 1 m2 area = Rs 40
Cost of plastering 110 m2 area = Rs (110 × 40)
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
Q 8, Ex 13.2, Page No 217, Surface Areas & Volumes, Class 9th
Question 8:
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 mRadius (r) of circular end of pipe$=\frac{5}{2}$ = 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2πrh
$=\left(2 \times \frac{22}{7} \times 0.025 \times 28\right) \mathrm{m}^{2}$
= 4.4 m2
The area of the radiating surface of the system is 4.4 m2
Q 9, Ex 13.2, Page No 217, Surface Areas & Volumes, CBSE Class 9th Mathematics
Question 9:
Find(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if $\frac{1}{12}$of the steel actually used was wasted in making the tank. $\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
Height (h) of cylindrical tank = 4.5 mRadius (r) of the circular end of cylindrical tank $=\left(\frac{4.2}{2}\right) \mathrm{m}$
=2.1m
(i) Lateral or curved surface area of tank = 2πrh
$=\left(2 \times \frac{22}{7} \times 2.1 \times 4.5\right) \mathrm{m}^{2}$
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2πr (r + h)
$=\left[2 \times \frac{22}{7} \times 2.1 \times(2.1+4.5)\right] \mathrm{m}^{2}$
= (44 × 0.3 × 6.6) m2
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
$\therefore A\left(1-\frac{1}{12}\right)=87.12 \mathrm{m}^{2}$
$\Rightarrow A=\left(\frac{12}{11} \times 87.12\right) \mathrm{m}^{2}$
$\Rightarrow \mathrm{A}=95.04 \mathrm{m}^{2}$
Therefore, 95.04 m2 steel was used in actual while making such a tank.
Q 10, Ex. 13.2, Page No 217, Surface Areas & Volumes, Class 9th Maths
Question 10:
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. $\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius (r) of the circular end of the frame of lampshade $=\left(\frac{20}{2}\right) \mathrm{cm}$
=10cm
Cloth required for covering the lampshade = 2πrh
$=\left(2 \times \frac{22}{7} \times 10 \times 35\right) \mathrm{cm}^{2}$
= 2200 cm2
Hence, for covering the lampshade, 2200 cm2 cloth will be required.
Q 11, Ex 13.2, Page No 217, Surface Areas & Volumes - NCERT Solutions Class 9
Question 11:
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of the circular end of cylindrical penholder = 3 cmHeight (h) of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
= 2πrh + πr2
$=\left[2 \times \frac{22}{7} \times 3 \times 10.5+\frac{22}{7} \times(3)^{2}\right] \mathrm{cm}^{2}$
$=\left(132 \times 1.5+\frac{198}{7}\right) \mathrm{cm}^{2}$
$=\left(198+\frac{198}{7}\right) \mathrm{cm}^{2}$
$=\frac{1584}{7} \mathrm{cm}^{2}$
Area of cardboard sheet used by 1 competitor $=\frac{1584}{7} \mathrm{cm}^{2}$
Area of cardboard sheet used by 35 competitors
$=\left(\frac{1584}{7} \times 35\right) \mathrm{cm}^{2}$
= 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be bought.
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