RD Sharma solution class 8 chapter 22 Mensuration III(Volume and Surface Area of a Right Circular Cylinder) Exercise 22.1

Exercise 22.1

Page-22.10

Question 1:

Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.

Answer 1:

$$\begin{gathered} \mathrm{Let} r\mathit{ }\mathrm{and} h\mathit{ }\mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{and} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylinder}. \\ \mathrm{Given}: \\ \mathrm{r}=\frac{7}{2} \mathrm{cm} \\ \mathrm{h}=60 \mathrm{cm} \\ \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder} =2\mathrm{\pi }\times \mathrm{r}\times \mathrm{h} \\ =2\times \frac{22}{7}\times \frac{7}{2}\times 60 \\ =22\times 60=1320 {\mathrm{cm}}^2 \\ \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi }\times \mathrm{r}\times (\mathrm{r}+\mathrm{h}) \\ =2\times \frac{22}{7}\times \frac{7}{2}\times (\frac{7}{2}+60)=22\times \frac{127}{2}=11\times 127=1397{\mathrm{cm}}^2 \end{gathered}$$

Question 2:

The curved surface area of a cylindrical road is 132 cm². Find its length if the radius is 0.35 cm.

Answer 2:

$$\begin{gathered} \mathrm{Consider} h \mathrm{to} \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{rod}. \\ \mathrm{Given}: \\ \mathrm{Radius}, r=0.35 \mathrm{cm} \\ \mathrm{Curved} \mathrm{surface} \mathrm{area}=132 {\mathrm{cm}}^2 \\ \mathrm{We} \mathrm{know}: \\ \mathrm{Curved} \mathrm{surface} \mathrm{area}=2\times \mathrm{\pi }\times \mathrm{r}\times \mathrm{h} \\ 132=2\times \frac{22}{7}\times 0.35\times \mathrm{h} \\ h=\frac{132\times 7}{2\times 22\times 0.35} \\ h=60 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{rod} \mathrm{is} 60 \mathrm{cm}. \end{gathered}$$

Question 3:

The area of the base of a right circular cylinder is 616 cm² and its height is 2.5 cm. Find the curved surface area of the cylinder.

Answer 3:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} a \mathrm{right} \mathrm{circular} \mathrm{cylinder}= 616 {\mathrm{cm}}^2 \\ \mathrm{Height}= 2.5 cm \\ \mathrm{Let} r \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{a} \mathrm{right} \mathrm{circular} \mathrm{cylinder}. \\ {\mathrm{\pi r}}^2 =616 \\ \Rightarrow {\mathrm{r}}^2=616\times \frac{7}{22} \\ \Rightarrow {\mathrm{r}}^2=196 \\ \Rightarrow \mathrm{r}=14 \mathrm{cm} \\ \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{right} \mathrm{circular} \mathrm{cylinder}=2\mathrm{\pi rh}=2\times \frac{22}{7}\times 14\times 2.5= 220 {\mathrm{cm}}^2 \end{gathered}$$

Question 4:

The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.

Answer 4:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Height}, h=15 \mathrm{c}\mathrm{m} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{cylinder}=88 {\mathrm{cm}}^2 \\ \mathrm{Let}\mathit{ }r\mathit{ }\mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{cylinder}. \\ \mathrm{The} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi r} \\ 88=2\times \frac{22}{7}\times \mathrm{r} \\ \mathrm{r}=\frac{88\times 7}{2\times 22}=14 \mathrm{cm} \\ \mathrm{Curved} \mathrm{surface} \mathrm{area} =2\times \mathrm{\pi }\times \mathrm{r}\times \mathrm{h} \\ =2\times \frac{22}{7}\times 14\times 15 \\ =1320 {\mathrm{cm}}^{\mathit{2}} \\ \mathrm{Total} \mathrm{surface} \mathrm{area} =2\times \mathrm{\pi }\times \mathrm{r}\times (\mathrm{r}+\mathrm{h}) \\ =2\times \frac{22}{7}\times 14\times (14+15) \\ =2552 \mathrm{c}{\mathrm{m}}^{\mathit{2}} \end{gathered}$$

Question 5:

A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find The total surface area of the solid thus generated.

Answer 5:

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{rectangular} \mathrm{strip} \mathrm{of} 25 cm\; \times \; 7\; cm \mathrm{is} \mathrm{rotated} \mathrm{about} \mathrm{the} \mathrm{longer} side, \mathrm{we} \mathrm{have}: \\ \mathrm{Height}, h=25 \mathrm{cm} \\ \mathrm{Radius}, r=7\mathrm{cm} \\ \therefore \mathrm{Total} \mathrm{surface} \mathrm{area} =\mathrm{2\pi r(r+h)\; =\; 2\pi (7)(25+7)} \\ =\; 14\pi (32) \\ =\; 448\pi \mathrm{cm}\mathit{²} \\ =448\times \frac{22}{7}\mathrm{cm}\mathit{²} \\ =1408 \mathrm{cm}\mathit{²} \end{gathered}$$

Question 6:

A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a cylinder. Find the total surface area of the cylinder thus generated.

Answer 6:

$$\begin{gathered} \mathrm{The} \mathrm{rectangular} \mathrm{sheet} \mathrm{of} \mathrm{paper} 44 \mathrm{cm}\times 20 \mathrm{cm} \mathrm{is} \mathrm{rolled} \mathrm{along} \mathrm{its} \mathrm{length} \mathrm{to} \mathrm{form} \mathrm{a} \mathrm{cylinder}. \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{is} 20 \mathrm{cm} \mathrm{and} \mathrm{circumference} \mathrm{is} 44 \mathrm{cm}. \\ \mathrm{We} \mathrm{have}: \\ \mathrm{Height}, h=20 \mathrm{cm} \\ \mathrm{Circumference}= 2\mathrm{\pi }r=44 \mathrm{cm} \\ \therefore \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{is} S=2\mathrm{\pi }rh \\ =44\times 20 \mathrm{cm}² \\ =880 \mathrm{cm}² \end{gathered}$$

Question 7:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{radii} \mathrm{of} \mathrm{two} \mathrm{cylinders} \mathrm{be} 2r \mathrm{and} 3r, \mathrm{respectively}, \mathrm{and} \mathrm{their} \mathrm{heights} be 5h \mathrm{and} 3h, \mathrm{respectively}. \\ \mathrm{Let} {\mathrm{S}}_1 \mathrm{and} {\mathrm{S}}_2 \mathrm{be} \mathrm{the} \mathrm{curved} \mathrm{surface} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{cylinder}. \\ {\mathrm{S}}_1= Curved \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{of} \mathrm{height} 5h \mathrm{and} \mathrm{radius} 2r \\ {\mathrm{S}}_2= Curved \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{of} \mathrm{height} 3h \mathrm{and} \mathrm{radius} 3r \\ \therefore {\mathrm{S}}_1:{\mathrm{S}}_2=2\times \mathrm{\pi }\times r\times h : 2\times \mathrm{\pi }\times r\times h \\ =\frac{2\times \mathrm{\pi }\times 2r\times 5h}{ 2\times \mathrm{\pi }\times 3r\times 3h} \\ =10 : 9 \end{gathered}$$

Question 8:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Prove that its height and radius are equal.

Answer 8:

$$\begin{gathered} \mathrm{Let}\mathit{ }{S}_1 \mathrm{and} S_{\mathit{2}} \mathrm{be} \mathrm{the} \mathrm{curved} \mathrm{surface} \mathrm{area} \mathrm{and} \mathrm{total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{cylinder}, \mathrm{respectively}. \\ \mathrm{Then}, S_1=2\mathrm{\pi }rh , {\mathrm{S}}_2=2\mathrm{\pi }r\left(r\mathit{+}h\right) \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}: \\ \mathit{ }{S}_1:S_2=1:2 \\ 2\pi rh : 2\pi r\left(r+h\right) =1 : 2 \\ h : \left(r+h\right) = 1 : 2 \\ \frac{h}{r+h}=\frac{1}{2} \\ 2h=r+h \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }h\mathit{=}r \\ \mathrm{Therefore}, \mathrm{the} \mathrm{height} \mathrm{and} \mathrm{the} \mathrm{radiu}s \mathrm{are} equal. \end{gathered}$$

Page-22.11

Question 9:

The curved surface area of a cylinder is 1320 cm² and its base has diameter 21 cm. Find the height of the cylinder.

Answer 9:

$$\begin{gathered} \mathrm{Let} h \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylinder}. \\ \mathrm{Given}: \\ \mathrm{Curved} \mathrm{surface} \mathrm{area}, S=1320 {\mathrm{cm}}^2 \\ \mathrm{Diameter}, d=21 \mathrm{cm} \\ \mathrm{Radius},\mathit{ }r=10.5 \\ \mathrm{S}=2\mathrm{\pi }rh \\ 1320=2\mathrm{\pi }\times 10.5\times h \\ \mathit{ }\mathit{ }h=\frac{1320}{2\mathrm{\pi }\times 10.5} \\ \mathit{ }\mathit{ }h=20 \mathrm{cm} \end{gathered}$$

Question 10:

The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.

Answer 10:

$$\begin{gathered} \mathrm{Let}\mathit{ }r \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{cylinder}. \\ \mathrm{Height},\mathit{ }h= 10.5 \mathrm{cm} \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{curved} \mathrm{surface}, S_1=2\mathrm{\pi }rh \\ S\mathrm{um} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{its} \mathrm{two} \mathrm{circular} \mathrm{faces}, S_2=2\mathrm{\pi }{r}^2 \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ 3S_2=2S_1 \\ 3\times 2{\mathrm{\pi r}}^2=2\times 2\mathrm{\pi }rh \\ 6r=4h \\ 3r=2h \\ \mathit{ }\mathit{ }r=\frac{2}{3}\times 10.5 \mathrm{cm} \\ =7 \mathrm{cm} \end{gathered}$$

Question 11:

Find the cost of plastering the inner surface of a well at Rs 9.50 per m², if it is 21 m deep and diameter of its top is 6 m.

Answer 11:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Height}, h=21 \mathrm{m} \\ \mathrm{Diameter}, d= 6 \mathrm{m} \\ \mathrm{Radius}, r= 3 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{surface} \mathrm{of} \mathrm{the} \mathrm{well}, S=2\mathrm{\pi }rh \\ =2\mathrm{\pi }\times 3\times 21 {\mathrm{m}}^2 \\ = 2\times \frac{22}{7}\times 3\times 21 {\mathrm{m}}^2 \\ =396 {\mathrm{m}}^2 \\ \mathrm{According} \mathrm{to} \mathrm{question}, \mathrm{the} \mathrm{cost} \mathrm{per} {\mathrm{m}}^2 \mathrm{is} \mathrm{Rs} 9.50. \\ \therefore \mathrm{Inner} \mathrm{surface} \mathrm{cost} \mathrm{is} \mathrm{Rs} 396\times 9.50=\mathrm{Rs} 3762 \end{gathered}$$

Question 12:

A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.

Answer 12:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Diameter}, d=20 \mathrm{cm} \\ \mathrm{Radius}, r= 10 \mathrm{cm} \\ \mathrm{Height},\mathit{ }h=14 \mathrm{cm} \\ \mathrm{Area} \mathrm{inside} \mathrm{the} \mathrm{cylindrical} \mathrm{vessel} \mathrm{that} \mathrm{is} \mathrm{to} \mathrm{be} \mathrm{tin}-\mathrm{plated}=S \\ S=2\mathrm{\pi }rh+\mathrm{\pi }{r}^2 \\ =2\mathrm{\pi }\times 10\times 14+\pi \times {10}^2 \\ =280\mathrm{\pi }+100\mathrm{\pi } \\ =380\times \frac{22}{7} {\mathrm{cm}}^2 \\ =\frac{8360}{7} {\mathrm{cm}}^2 \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ \mathrm{Cost} \mathrm{per} 100 {\mathrm{cm}}^2 = 50 \mathrm{paise} \\ \mathrm{Cost} \mathrm{per} {\mathrm{cm}}^2 = \mathrm{Rs} 0.005 \\ \mathrm{Cost} \mathrm{of} \mathrm{tin}-\mathrm{plating} \mathrm{the} \mathrm{area} \mathrm{inside} \mathrm{the} \mathrm{cylindrical} \mathrm{vessel}= Rs 0.005\times \frac{8360}{7} \\ =\mathrm{Rs} \frac{41.8}{7}=\mathrm{Rs} 5.97 \end{gathered}$$

Question 13:

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.

Answer 13:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Inner} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{well} =3.5 \mathrm{m} \\ \therefore \mathrm{Inner} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{well}, r=1.75 \mathrm{m} \\ \mathrm{Depth} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{well}, h=10 \mathrm{m} \\ \mathrm{Inner} \mathrm{curved} \mathrm{surface} \mathrm{area}, S=2\mathrm{\pi }rh \\ \mathrm{S}=2\mathrm{\pi }\times 1.75\times 10 {\mathrm{m}}^2=2\times \frac{22}{7}\times 1.75\times 10 {\mathrm{m}}^2 \\ =110 {\mathrm{m}}^2 \\ \mathrm{Cost} \mathrm{of} \mathrm{plastering} 1{\mathrm{m}}^2 \mathrm{area} =\mathrm{Rs} 4 \\ \mathrm{Cost} \mathrm{of} \mathrm{plastering} 110 {\mathrm{m}}^2 \mathrm{area} =\mathrm{Rs} \left(110\times 4\right)=\mathrm{Rs} 440 \end{gathered}$$

Question 14:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground?

Answer 14:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{roller} =84 \mathrm{cm} \\ \therefore \mathrm{Radius}, r=\frac{\mathrm{Diameter}}{2}=42 \mathrm{cm} \\ \mathrm{In} 1 \mathrm{revolution}, \mathrm{it} \mathrm{covers} \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{its} \mathrm{lateral} \mathrm{surface} \mathrm{area}. \\ \mathrm{Roller} \mathrm{is} \mathrm{a} \mathrm{cylinder} \mathrm{of} \mathrm{height}, h= 120 \mathrm{cm} \\ \mathrm{Radius} = 42 \mathrm{cm} \\ \mathrm{Lateral} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi }rh \\ =2\times \frac{22}{7}\times 42\times 120 \\ =31680 {\mathrm{cm}}^2 \\ \mathrm{It} \mathrm{takes} 500 \mathrm{complete} \mathrm{revolutions} \mathrm{to} \mathrm{level} \mathrm{a} \mathrm{playground}. \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{field} =31680\times 500=15840000 {\mathrm{cm}}^2 \left({\mathrm{1\; cm}}^2=\frac{1}{10000}{\mathrm{m}}^2\right) \\ \therefore {\mathrm{15840000cm}}^2=1584 {\mathrm{m}}^2. \\ \mathrm{Thus}, \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{field} \mathrm{in} {\mathrm{m}}^2 \mathrm{is} 1584 {\mathrm{m}}^2. \end{gathered}$$

Question 15:

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

Answer 15:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{pillars}=0.5 \mathrm{m} \\ \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{pillars}, r=0.25 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{pillars}, h=4 \mathrm{m} \\ \mathrm{Number} \mathrm{of} \mathrm{pillars} =21 \\ \mathrm{Rate} \mathrm{of} \mathrm{cleaning} = \mathrm{Rs} 2.50 \mathrm{per} \mathrm{square} \mathrm{metre} \\ \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{pillar}=2\mathrm{\pi }rh \\ =2\times \frac{22}{7}\times 0.25\times 4 \\ =2\times \frac{22}{7} \\ =\frac{44}{7} {\mathrm{m}}^2 \\ \therefore \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{pillar}=\frac{44}{7} {\mathrm{m}}^2 \\ \mathrm{Cost} \mathrm{of} \mathrm{cleaning} 21 \mathrm{pillars} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{Rs} 2.50 \mathrm{per} {\mathrm{m}}^2= \mathrm{Rs} 2.5\times 21\times \frac{44}{7} \\ =7.5\times 44 \\ \therefore \mathrm{Cost} \mathrm{of} \mathrm{cleaning} 21 \mathrm{pillars} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{Rs} 2.50 \mathrm{per} {\mathrm{m}}^2= \mathrm{Rs} 330 \end{gathered}$$

Question 16:

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

Answer 16:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=4620 {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{ring}= 115.5 {\mathrm{cm}}^2 \\ \mathrm{Height}, h=7 \mathrm{cm} \\ \mathrm{Let} R \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{ring} \mathrm{and} r \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{ring}. \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{ring} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ 115.5=\mathrm{\pi }\left(R^2-\mathit{ }{r}^2\right) \\ {\mathrm{R}}^2- {\mathrm{r}}^2=115.5\times \frac{7}{22} \\ \mathit{(}R\mathit{+}r\mathit{)}\mathit{(}R\mathit{-}r\mathit{)}=36.75 ........... \left(\mathrm{i}\right) \\ \mathrm{Total} \mathrm{surface} \mathrm{area} = \mathrm{Inner} \mathrm{curved} \mathrm{surface} \mathrm{area} + \mathrm{Outer} \mathrm{curved} \mathrm{surface} \mathrm{area} + \mathrm{Area} \mathrm{of} \mathrm{bottom} \mathrm{and} \mathrm{top} \mathrm{rings} \\ 4620=2\mathrm{\pi }rh+2\mathrm{\pi }Rh+2\times 115.5 \\ 2\mathrm{\pi }h(R+r)=4620-231 \\ R+r=\frac{4389\times 7}{2\times 22\times 7} \\ R+r=\frac{399}{4} ........... \left(\mathrm{ii}\right) \\ \mathrm{Substituting} \mathrm{the} \mathrm{value} \mathrm{of} R+r \mathrm{from} \mathrm{the} \mathrm{equation} \left(\mathrm{ii}\right) \mathrm{in} \left(\mathrm{i}\right): \\ \frac{399}{4}\mathit{(}R\mathit{-}r\mathit{)}=36.75 \\ (R-r)=36.75\times \frac{4}{399}=0.368 \mathrm{cm} \\ \therefore \mathrm{Thickness} \mathrm{of} \mathrm{the} \mathrm{cylinder}= (R-r)=0.368 \mathrm{cm} \end{gathered}$$

Question 17:

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m², find the circumference of its base.

Answer 17:

$$\begin{gathered} \mathrm{Let} r \mathrm{and} h \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{solid} \mathrm{cylinder}. \\ \mathrm{Given}: \\ r+h=37 \mathrm{m} \\ \mathrm{Total} \mathrm{surface} \mathrm{area}, S=2\mathrm{\pi }r\left(r+h\right) \\ 1628 =2\mathrm{\pi }\times r\times 37 \\ \mathrm{r}=\frac{1628}{2\mathrm{\pi }\times 37} \\ =\frac{1628}{232.477} \\ =7 \mathrm{m} \\ \mathrm{Circumference} \mathrm{of} \mathrm{its} \mathrm{base}, S_{\mathit{1}}=2\mathrm{\pi r} \\ =\left(2\times \frac{22}{7}\times 7\right) \mathrm{m} \\ =44 \mathrm{m} \end{gathered}$$

Question 18:

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

Answer 18:

$$\begin{gathered} \mathrm{Let} S_1 \mathrm{and} S_2 \mathrm{be} \mathrm{the} \mathrm{total} \mathrm{surface} \mathrm{area} \mathrm{and} \mathrm{curved} \mathrm{surface} \mathrm{area}, \mathrm{respectively}. \\ \mathrm{Given}: \\ \mathrm{Height},\mathit{ }h=7.5 \mathrm{cm} \\ \mathrm{Radius},\mathit{ }r= 3.5 \mathrm{cm} \\ {\mathrm{S}}_1=2\pi r\left(r+h\right) \\ {\mathrm{S}}_2=2\mathrm{\pi }rh \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}: \\ \frac{S_1}{S_2}=\frac{2\mathrm{\pi }r\left(r\mathit{+}h\right)}{2\mathrm{\pi }rh} \\ \frac{S_1}{S_2}=\frac{\mathrm{r}+\mathrm{h}}{\mathrm{h}} \\ \frac{S_1}{S_2}=\frac{3.5+7.5}{7.5} \\ \frac{S_1}{S_2}=\frac{11}{7.5}=\frac{110}{75}=\frac{22}{15} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{ratio} \mathrm{is} 22:15. \end{gathered}$$

Question 19:

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 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm².

Answer 19:

$$\begin{gathered} \mathrm{Let} r \mathrm{cm} \mathrm{and}\mathit{ }h\mathit{ }\mathrm{cm} \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{vessel}. \\ \mathrm{Given}: \\ \mathrm{Radius}, r= 70 \mathrm{cm} \\ \mathrm{Height}, h=1.4 \mathrm{m}=140 \mathrm{cm} \\ \mathrm{Rate} \mathrm{of} \mathrm{tin}-\mathrm{plating} = \mathrm{Rs} 3.50 \mathrm{per} 1000 \mathrm{square} \mathrm{centimetre} \\ \mathrm{Cost} \mathrm{of} \mathrm{tin}-\mathrm{plating} \mathrm{the} \mathrm{cylindrical} \mathrm{vessel} \mathrm{on} \mathrm{both} \mathrm{the} \mathrm{surfaces} (\mathrm{inner} \mathrm{and} \mathrm{outer}): \\ \mathrm{Total} \mathrm{suface} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{vessel}= \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{and} \mathrm{the} \mathrm{outer} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{base} + \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{and} \mathrm{the} \mathrm{outer} \mathrm{curved} \mathrm{surface} \\ =2\left(\mathrm{\pi }{r}^2+2\mathrm{\pi }rh\right) \\ =2\mathrm{\pi }r\left(\mathrm{r}+2h\right) \\ =2\times \frac{22}{7}\times 70\times \left(70+2\times 140\right) \\ =44\times 10\times 350 \\ =154000 {\mathrm{cm}}^2 \\ \mathrm{Cost} \mathrm{of} \mathrm{painting} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{Rs} 3.50 \mathrm{per} 1000 {\mathrm{cm}}^2=154000\times \frac{3.50}{1000}=\mathrm{Rs} 539 \\ \mathrm{Therefore}, \mathrm{cost} \mathrm{of} \mathrm{painting} \mathrm{is} \mathrm{Rs} 539. \end{gathered}$$