## RD Sharma Solutions Class 9 Chapter 19 Exercise 19 1

Q1. Curved surface area of a right circular cylinder is \(4.4m^{2}\)

Solution:

Given that

Radius of the base of the cylinder (r) =0.7m

Curved surface area of cylinder (C.S.A) = \( 4.4m^{2}\)

Let the height of the cylinder be h

The curved surface area of a cylinder is given by: \(2\pi rh\)

\(2\pi rh\)

2*3.14*0.7*h=\(4.4m^{2}\)

h=\(4.4\div 2*3.14*0.7\)

=1m

Therefore the height of the cylinder is 1m.

Q2. 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. (Take \(\pi =3.14\)

Solution:

Given that

Height of cylinder=length of cylindrical pipe

=28m

Radius(r) of circular end of pipe=5/2=2.5cm=0.025 m

Curved surface area of cylindrical pipe=\(2\pi rh\)

Therefore the area of radiating surface of the system is \( 4.4m^{2}\)

Q3. 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 \(m^{2}\)

Solution:

Given that

Height of cylindrical pillar=3.5m

Radius(r) of circular end of pillar=50/2=25cm=0.25m

Curved surface area of cylindrical pillar=\(2\pi\)

=\(5.5m^{2}\)

The cost of whitewashing Â 1m^{2} is Rs 12.50

Cost of whitewashing \(5.5m^{2}\)

Thus the cost of whitewashing the pillar is Rs68.75

Q4. It is required to make a closed cylindrical tank of height Â 1m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same? (Take \(\pi =3.14\)

Solution:

Height of the cylindrical tank(h)=1 m

Base radius of cylindricall tank(r)= \(\frac{140}{2}\)

Area of sheet required =total surface area of tank= \(2\pi rh\)

=2*3.14*0.7(0.7+1)

=4.4*1.7 =\(7.48m^{2}\)

Therefore it will require \(7.48m^{2}\)

Q5.A solid cylinder has a total surface area of 462\(cm^{2}\)

Solution:

Given that

Curved or lateral surface area=\(\frac{1}{3}\)

\(2\pi rh=\frac{1}{3}(2\pi rh+2\pi r^{2})\)

\(4\pi rh=2\pi r^{2}\)

2h=r

Total surface area =462\(cm^{2}\)

Curved surface area=\(\frac{1}{3}*462\)

\(2\pi rh=154\)

\(2*3.14*2*h^{2}\)

\(h^{2}\)

h=\(\frac{49}{4}\)

= \(\frac{7}{2}\)

NowÂ r=2h

Therefore r=2*\(\frac{7}{2}\)

The height and the radius of the cylinder is \(\frac{7}{2}\)

Q6. The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7cm. Find the thickness of the cylinder.

Solution:

Let the inner radius of the hollow cylinder be \(r_{1}\)

The outer radius of the hollow cylinder be \(r_{2}\)

Then,

\(2\pi r_{1}*h+2\pi r_{2}*h+2\pi r_{2}^{2}-2\pi r_{1}^{2}\)

\(\pi r_{1}^{2}-\pi r_{2}^{2}=115.5\)

Now solving eq(a)

\(2\pi r_{1}h+2\pi r_{2}h+2\pi r_{2}^{2}-2\pi r_{1}^{2}\)

=> \(2\pi h(r_{1}+r_{2})+2\pi(r_{2}^{2}-r_{1}^{2})\)

=> \(2\pi h(r_{1}+r_{2})+2(\pi r_{2}^{2}-\pi r_{1}^{2})\)

Now putting the value of (b) in (a) we get

=> \(2\pi h(r_{1}+r_{2})+231=4620\)

=> \(2\pi *7(r_{1}+r_{2})\)

=> \(\pi (r_{1}+r_{2})\)

Now solving eq(b)

\(\pi r_{1}^{2}-\pi r_{2}^{2}=115.5\)

=> \(\pi r_{2}^{2}-\pi r_{1}^{2}=115.5\)

=> \(\pi (r_{2}+r_{1})(r_{2}-r_{1})\)

Dividing equation (d) by (c) we get

\(\frac{\pi r_{1}+r_{2}(r_{2}-r_{1)})}{\pi (r_{1}+r_{2})}\)

=> \(r_{2}-r_{1}\)

Q7. Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5m and 3.5 m respectively.

Solution:

Given that,

Radius of the cylinder(r)=3.5m

Height of the cylinder(h)=7.5m

Total Surface Area of cylinder (T.S.A)

= \(2\pi r(r+h))\)

Curved surface area of a cylinder(C.S.A)

= \(2\pi rh\)

Now ,

\(\frac{T.S.A}{C.S.A}\)

= \(\frac{h+r}{h}\)

Putting the values in eq(1)

= \(\frac{7.5+3.5}{7.5}\)

= \(\frac{11}{7.5}\)

=\(\frac{11*10}{7.5}\)

= \(\frac{22}{15}\)

= 22:15

Therefore the ratio is 22:15.

Q8.The total surface of a hollow metal cylinder, open at both ends of an external radius of 8cm and heightÂ 10cm is 338\(cm^{2}\)

Solution:

Given that

The external radius of the cylinder(R)=8cm

Height of the cylinder(h)=10 cm

The total surface area of the hollow cylinder(T.S.A) = \(338\pi cm^{2}\)

As we know that,

\(2\pi r*h+2\pi R*h+2\pi R*^{2}-2\pi r^{2}\)

=> \(h(r+R)+(R+r)(R-r)\)

=> \(10(8+r)+(8+r)(8-r)\)

=> \(80+10r+64-r^{2}\)

=> \(r^{2}-10r+25\)

=> \(r=5\)

\(R-r\)

Q9.A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000\(cm^{2}\)

Solution:

Given that

Radius of the vessel (r)=70cm

Height ofÂ the vessel (h)=1.4m=140cm

The area to be tin coated

=2*(\(2\pi rh+\pi r^{2}\)

=\(2\pi r(2h+r)\)

= 2*3.14*70[(2*140)+70]

=154000\(cm^{2}\)

Required cost=\(\frac{154000*3.5}{1000}\)

Q10. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:

(a)Inner curved surface area

(b)the cost of plastering this curved surface at the rate of Rs 40 per \(m^{2}\)

Solution:

The inner diameter of the well=3.5m

Inner radius=\(\frac{3.5}{2}\)

Height of the well=10m

(a) Inner curved surface area

=\(2\pi rh\)

=2*3.14*1.75*10

=110\(m^{2}\)

(b) Cost of painting 1\(m^{2}\)

=Rs 40

Â Cost of painting 110\(m^{2}\)

=Rs (50*110) =Rs 4400

Q11.Find the lateral surface area of a petrol storage tank is 4.2 m in diameter and 4.5 m high. How much steel was actually used, if 1/12^{th} of the steel actually used was wasted in making the closed tank?

Solution:

I t is given that

Diameter of cylinder=4.2m

Radius of cylinder=\(\frac{4.2}{2}\)

=2.1m

Height of cylinder=4.5m

Therefore,

Lateral or Curved surface area= \(2\pi rh\)

=2*3.14*2.1*4.5=59.4 \(m^{2}\)

Total surface area of tank = 2*Ï€*r(r+h)

=2*(\(\frac{22}{7}\)

= 87.12\(m^{2}\)

Let, A \(m^{2}\)

Area of iron present in cylinder=(\(y-\frac{A}{12}m^{2}\)

=\(\frac{11}{12}A\;m^{2}\)

Hence,

\(\frac{11}{12}A\)

=> A=\(\frac{12}{11}\)

=> A=(\(\frac{12}{11}*87.12)m^{2}\)

= \(95.04 m^{2}\)

Thus, \(m^{2}\)

Q12. The students of Â 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?Â (TakeÂ Â \(\pi =3.14\)

Solution:

It is given that

Radius of the circular part of the penholder (r) =3cm

The height of the penholder (h) =10.5 cm

Surface area of one penholder (S.A)

= Curved surface area of penholder +Area of theÂ circular base of penholder

= \(2\pi rh+\pi r^{2}\)

=(2*3.14*3*10.5)+3.14* \(3^{2}\)

=198+ \(\frac{198}{7}\)

= \(\frac{1584}{7}\)

The total area of cardboard sheet used by one competitor= \(\frac{1584}{7}\)

The total area of cardboard sheet used by 30 competitors= \(\frac{1584}{7}\)

=7920\(cm^{2}\)

Therefore, the school needs to buy 7920 \(cm^{2}\)

Q13. The diameter of roller 1.5 m long is 84cm.If it takes 100 revolutions to level a playground, Â find the cost of leveling this ground at the rate of 50 paise per square meter.

Solution:

Given that

Diameter of the roller (d)=85cm=0.84m

Length of the roller=1.5m

Radius of the roller(r) = \(\frac{d}{2}\)

= \(\frac{0.84}{2}\)

The total area of the playground covered by the roller in one revolution=covered area of the roller

Curved surface area of the roller= \(2\pi rh\)

=2*3.14*0.42*1.5

=0.12*22*1.5 \(m^{2}\)

Area of the playground=100*Area covered by roller in one revolution

= (100*0.12*22*1.5) \(m^{2}\)

=396 \(m^{2}\)

Now,

Cost of leveling 1 \(m^{2}\)

Cost of leveling 396 \(m^{2}\)

Hence, cost ofÂ leveling 396\(m^{2}\)

Q14. Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50m and height is 4m.What will be the cost of cleaning them at the rate of Rs 2.50 per square meter?

Solution:

Diameter of each pillar =0.5,

Radius of each pillar(r)= \(\frac{d}{2}\)

=\(\frac{0.5}{2}\)

Height of each pillar=4m

Lateral surface area of one pillar

=\(2\pi rh\)

Lateral surface area of 20 pillars=20*\(\frac{44}{7}m^{2}\)

Cost of cleaning one pillar =Rs2.50 per square meter

Cost of cleaning 20 pillars =Rs 2.50 *20*\(\frac{44}{7}m^{2}\)