RD Sharma 2020 solution class 9 chapter 19 Surface Area and Volume of a Right Circular Cylinder FBQS

FBQS

Page-19.30

Question 1:

In a cylinder, if radius is doubled and height is halved, curved surface area will be _________.

Answer 1:

Let r and h be the radius and height of the cylinder, respectively.

∴ Curved surface of the original cylinder, C₁ = 2$\mathrm{\pi }$rh 

If the radius is doubled and height is halved, then

Radius of the new cylinder, R = 2r

Height of the new cylinder, H = $\frac{h}{2}$

∴ Curved surface of the new cylinder, C₂ = $2\mathrm{\pi }RH=2\mathrm{\pi }\times 2r\times \frac{h}{2}$ = 2$\mathrm{\pi }$rh

Thus, the curved surface area of the new cylinder is same as the curved surface area of the original cylinder.

In a cylinder, if radius is doubled and height is halved, curved surface area will be __same__.

Question 2:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is _________.

Answer 2:

Let r₁ be the radius and h₁ be the height of first cylinder & r₂ be the radius and h₂ be the height of second cylinder.

It is given that,

$\frac{r_1}{r_2}=\frac{2}{3}$  and $\frac{h_1}{h_2}=\frac{5}{3}$       .....(1)

Now,

$\frac{\mathrm{Volume} \mathrm{of} \mathrm{first} \mathrm{cylinder}}{\mathrm{Volume} \mathrm{of} \mathrm{second} \mathrm{cylinder}}$

$=\frac{\mathrm{\pi }{r}_1^2{h}_1}{\mathrm{\pi }{r}_2^2{h}_2}$

$={\left(\frac{r_1}{r_2}\right)}^2\times \frac{h_1}{h_2}$

$={\left(\frac{2}{3}\right)}^2\times \frac{5}{3}$             [Using (1)]

$=\frac{4}{9}\times \frac{5}{3}$

$=\frac{20}{27}$

∴ Volume of first cylinder : Volume of second cylinder = 20 : 27

Thus, the ratio of volumes of two cylinders is 20 : 27.

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is ___20 : 27___.

Question 3:

If the radius of a right circular cylinder is doubled and its curved surface area is not changed, then its height must be _________.

Answer 3:

Let r be the radius and h be the height of the cylinder.

∴ Curved surface of the cylinder, C₁ = 2$\mathrm{\pi }$rh

Let the height of the new cylinder be H.

Radius of the new cylinder, R = 2r        (Given)

∴ Curved surface of the new cylinder, C₂ = 2$\mathrm{\pi }$RH = 2$\mathrm{\pi }$(2r)H

It is given that, the curved surface areas of the new cylinder is not changed.

∴ C₂ = C₁

⇒ 2$\mathrm{\pi }$(2r)H = 2$\mathrm{\pi }$rh

⇒ $H=\frac{h}{2}$

Thus, the height of the new cylinder is half of the height of given cylinder.

If the radius of a right circular cylinder is doubled and its curved surface area is not changed, then its height must be __halved__.

Question 4:

The difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is _________.

Answer 4:

Let r and h be the radius and height of the cylinder, respectively.

r = 10 cm    (Given)

Total surface area of the cylinder = 2$\mathrm{\pi }$r(r + h)

Curved surface area of the cylinder = 2$\mathrm{\pi }$rh

∴ Difference between the total surface area and curved surface area of the cylinder

= Total surface area of the cylinder − Curved surface area of the cylinder

= 2$\mathrm{\pi }$r(r + h) − 2$\mathrm{\pi }$rh

= 2$\mathrm{\pi }$r² + 2$\mathrm{\pi }$rh − 2$\mathrm{\pi }$rh

= 2$\mathrm{\pi }$× (10 cm)²           (r = 10 cm)

= 2$\mathrm{\pi }$× 100 cm²

= 200$\mathrm{\pi }$ cm²

Thus, the difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is 200$\mathrm{\pi }$ cm².

The difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is ___200$\mathrm{\pi }$ cm²___.

Question 5:

The cost of painting the total outside surface of a closed cylinder at ₹ 3 per cm² is ₹ 2772. If the height of the cylinder is 2 times the radius, then its volume is __________.

Answer 5:

Let r cm and h cm be the radius and height of the closed cylinder, respectively.

h = 2r    (Given)

Total surface area of the cylinder

$=2\mathrm{\pi }r\left(r+h\right)$

$=2\mathrm{\pi }r\left(r+2r\right)$           (h = 2r)

$=2\mathrm{\pi }r\times 3r$

$=6\mathrm{\pi }{r}^2{\mathrm{cm}}^2$

Rate of painting the surface of closed cylinder = ₹ 3 per cm²

∴ Total cost of painting the total outside surface of closed cylinder = ₹ 3 per cm² × $6\mathrm{\pi }{r}^2 {\mathrm{cm}}^2$ = ₹ $18\mathrm{\pi }{r}^2$

⇒ ₹ $18\mathrm{\pi }{r}^2$ = ₹ 2772

$\Rightarrow 18\times \frac{22}{7}\times r^2=2772$

$\Rightarrow r^2=\frac{2772\times 7}{22\times 18}=49$

$\Rightarrow r=\sqrt{49}=7 \mathrm{cm}$

So, h = 2r = 2 × 7 cm = 14 cm

∴ Volume of the cylinder

$=\mathrm{\pi }{r}^{2}h$

$=\frac{22}{7}\times {\left(7\right)}^2\times 14$

$=2156 {\mathrm{cm}}^3$

Thus, the volume of cylinder is 2156 cm³.

The cost of painting the total outside surface of a closed cylinder at ₹ 3 per cm² is ₹ 2772. If the height of the cylinder is 2 times the radius, then its volume is ___2156 cm³__.

Question 6:

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

Answer 6:

Let r and h be the radius and height of the cylinder, respectively.

It is given that, the ratio between the radius of base and the height of a cylinder is 2 : 3.

So, r = 2x and h = 3x, where x is constant

Volume of the cylinder = 1617 cm³

$\Rightarrow \mathrm{\pi }{r}^{2}h=1617$

$\Rightarrow \frac{22}{7}\times {\left(2x\right)}^2\times 3x=1617$

$\Rightarrow \frac{22}{7}\times 4x^2\times 3x=1617$

$\Rightarrow x^3=\frac{1617\times 7}{12\times 22}={\left(\frac{7}{2}\right)}^3$

$\Rightarrow x=\frac{7}{2}\mathrm{cm}$

$\therefore r=2x=2\times \frac{7}{2}=7\mathrm{cm}$ and $h=3x=3\times \frac{7}{2}=\frac{21}{2} \mathrm{cm}$

Total surface area of the cylinder

$=2\mathrm{\pi }r\left(r+h\right)$

$=2\times \frac{22}{7}\times 7\times \left(7+\frac{21}{2}\right)$

$=44\times 17.5$

$=770 {\mathrm{cm}}^2$

Thus, the total surface area of the cylinder is 770 cm².

The ratio between the radius of base and the height of a cylinder is 2 : 3. If its volume is 1617 cm³, the total surface area of the cylinder is ___770 cm²___.

Question 7:

The ratio of the total surface area to the lateral surface area of a cylinder of base radius r and height h is __________.

Answer 7:

Let r and h be the radius and height of the cylinder, respectively.

Total surface of the cylinder = 2$\mathrm{\pi }$r(r + h)

Lateral surface area of the cylinder = 2$\mathrm{\pi }$rh

$\therefore \frac{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}}{\mathrm{Lateral} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}}$

$=\frac{2\mathrm{\pi }r\left(r+h\right)}{2\mathrm{\pi }rh}$

$=\frac{r+h}{h}$

So,

Total surface area of the cylinder : Lateral surface area of the cylinder = r + h : h

Thus, the ratio of the total surface area to the lateral surface area of a cylinder of base radius r and height h is r + h : h.

The ratio of the total surface area to the lateral surface area of a cylinder of base radius r and height h is ___r + h : h___.

Question 8:

The radius of a wire is decreased to one-third. If volume remains the same, the length will be increased __________.

Answer 8:

Let r units and l units be the radius and length of the wire, respectively.

∴ Original volume of the wire = $\mathrm{\pi }$r²l cu. units

When the radius of the wire is decreased to one-third, then the new radius of the wire, R = $\frac{r}{3}$ units.

Suppose the new length of the wire is L units.

∴ New volume of the wire = $\mathrm{\pi }{\left(\frac{r}{3}\right)}^{2}L$ cu. units

It is given that, the volume of the wire remains the same.

$\therefore \mathrm{\pi }{\left(\frac{r}{3}\right)}^{2}L=\mathrm{\pi }{r}^{2}l$

$\Rightarrow \frac{r^2}{9}\times L=r^2\times l$

$\Rightarrow L=9l$

Thus, the length of the wire will be increased 9 times.

The radius of a wire is decreased to one-third. If volume remains the same, the length will be increased ___9 times___.

Question 9:

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

Answer 9:

Let r₁ be the radius and h₁ be the height of first cylinder & r₂ be the radius and h₂ be the height of second cylinder.

$\therefore \frac{r_1}{r_2}=\frac{2}{3}$ and $\frac{h_1}{h_2}=\frac{5}{3}$          .....(1)              (Given)

Now,

$\frac{\mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{first} \mathrm{cylinder}}{\mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{second} \mathrm{cylinder}}$

$=\frac{2\mathrm{\pi }{r}_1{h}_1}{2\mathrm{\pi }{r}_2{h}_2}$

$=\frac{r_1}{r_2}\times \frac{h_1}{h_2}$

$=\frac{2}{3}\times \frac{5}{3}$           [Using (1)]

$=\frac{10}{9}$

∴ Curved surface area of first cylinder : Curved surface area of second cylinder = 10 : 9

Thus, the ratio of the curved surface areas of the two cylinders is 10 : 9.

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface areas is ___10 : 9___.

Question 10:

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. The volume of the cylinder is __________.

Answer 10:

It is given that, a rectangular tin sheet is 12 cm long and 5 cm broad. When it is rolled along its length to form a cylinder by making the opposite edges just to touch each other, then the circumference of the base of the cylinder so formed is same as the length of the rectangular sheet and the height of the cylinder is same as the breadth of the rectangular sheet.

Let r and h be the radius and height of the cylinder, respectively.

Height of the cylinder, h = 5 cm       .....(1)

Also,

Circumference of base of cylinder = Length of the rectangular tin sheet = 12 cm

⇒ 2$\mathrm{\pi }$r = 12 cm

$\Rightarrow r=\frac{6}{\mathrm{\pi }}$               .....(2)

∴ Volume of the cylinder

$=\mathrm{\pi }{r}^{2}h$

$=\mathrm{\pi }\times {\left(\frac{6}{\mathrm{\pi }} \mathrm{cm}\right)}^2\times 5 \mathrm{cm}$           [Using (1) and (2)]

$=\frac{180}{\mathrm{\pi }} {\mathrm{cm}}^3$

Thus, the volume of the cylinder so formed is $\frac{180}{\mathrm{\pi }} {\mathrm{cm}}^3$.

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. The volume of the cylinder is $\underline{ \frac{180}{\mathrm{\pi }} {\mathrm{cm}}^3 }$.

Question 11:

If the area of the curved surface and the circumference of the base of a right circular cylinder are 4400 cm² and 110 cm respectively, then the volume of the cylinder is _________.

Answer 11:

Let r and h be the radius and height of the cylinder, respectively.

Curved surface area of the cylinder = 4400 cm²        (Given)

∴ 2$\mathrm{\pi }$rh = 4400 cm²         .....(1)

Also, circumference of the base = 110 cm                  (Given)

∴ 2$\mathrm{\pi }$r = 110 cm             .....(2)

$\Rightarrow 2\times \frac{22}{7}\times r=110 \mathrm{cm}$

$\Rightarrow r=\frac{110\times 7}{44}=\frac{35}{2}\mathrm{cm}$

Now, dividing (1) by (2), we get

$\frac{2\mathrm{\pi }rh}{2\mathrm{\pi }r}=\frac{4400 {\mathrm{cm}}^2}{110 \mathrm{cm}}$

$\Rightarrow h=40 \mathrm{cm}$

∴ Volume of the cylinder

$=\mathrm{\pi }{r}^{2}h$

$=\frac{22}{7}\times {\left(\frac{35}{2} \mathrm{cm}\right)}^2\times 40 \mathrm{cm}$

$=\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\times 40 {\mathrm{cm}}^3$

$=38500 {\mathrm{cm}}^3$

Thus, the volume of the cylinder is 38500 cm³.

If the area of the curved surface and the circumference of the base of a right circular cylinder are 4400 cm² and 110 cm respectively, then the volume of the cylinder is ___38500 cm³___.

Question 12:

The altitude of a right circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface increases is _________.

Answer 12:

Let r and h be the radius and height of the cylinder, respectively.

∴ Area of the base of the cylinder = $\mathrm{\pi }$r²   

Lateral surface area of the original cylinder = 2$\mathrm{\pi }$rh       .....(1)

Now,

Height of the new cylinder, H = 6h      (Given)        .....(2)

Let the radius of the new cylinder be R.

Area of the base of the new cylinder = $\frac{\mathrm{\pi }{r}^2}{9}$        (Given)

$\therefore \mathrm{\pi }{R}^2=\frac{\mathrm{\pi }{r}^2}{9}$

$\Rightarrow R^2=\frac{r^2}{9}$

$\Rightarrow R=\sqrt{\frac{r^2}{9}}=\frac{r}{3}$            .....(3)

Now,

Lateral surface area of the new cylinder

$=2\mathrm{\pi }RH$

$=2\mathrm{\pi }\times \frac{r}{3}\times 6h$             [From (1) and (2)]

$=4\mathrm{\pi }rh$

$=2\times 2\mathrm{\pi }rh$

= 2 × Lateral surface area of the original cylinder         [From (1)]

Thus, the lateral surface area of the new cylinder increases 2 times.

The altitude of a right circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface increases is ___2___.

Question 13:

Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is _________.

Answer 13:

The length and breadth of each rectangular sheet of paper is 30 cm and 18 cm, respectively.

When the sheet of paper is rolled along its length to form a cylinder, then the circumference of the base of the cylinder so formed is same as the length of the rectangular sheet and the height of the cylinder is same as the breadth of the rectangular sheet.

Let r be the radius and h be the height of the cylinder formed in this case.

∴ h = 18 cm

Circumference of base = 30 cm

⇒ 2$\mathrm{\pi }$r = 30 cm

$\Rightarrow r=\frac{15}{\mathrm{\pi }}$ cm

∴ Volume of the cylinder formed in this case, V₁ = $\mathrm{\pi }{r}^{2}h=\mathrm{\pi }\times {\left(\frac{15}{\mathrm{\pi }} \mathrm{cm}\right)}^2\times 18 \mathrm{cm}$         .....(1)

When the sheet of paper is rolled along its breadth to form a cylinder, then the circumference of the base of the cylinder so formed is same as the breadth of the rectangular sheet and the height of the cylinder is same as the length of the rectangular sheet.

Let R be the radius and H be the height of the cylinder formed in this case.

∴ H = 30 cm

Circumference of base = 18 cm

⇒ 2$\mathrm{\pi }$R = 18 cm

$\Rightarrow R=\frac{9}{\mathrm{\pi }}$ cm

∴ Volume of the cylinder formed in this case, V₂ = $\mathrm{\pi }{R}^{2}H=\mathrm{\pi }\times {\left(\frac{9}{\mathrm{\pi }} \mathrm{cm}\right)}^2\times 30 \mathrm{cm}$         .....(2)

Now,

$\frac{V_1}{V_2}=\frac{\mathrm{\pi }\times {\left({\displaystyle \frac{15}{\mathrm{\pi }}}\mathrm{cm}\right)}^2\times 18 \mathrm{cm}}{\mathrm{\pi }\times {\left({\displaystyle \frac{9}{\mathrm{\pi }}} \mathrm{cm}\right)}^2\times 30}=\frac{5}{3}$

⇒ V₁ : V₂ = 5 : 3

Thus, the ratio of the volumes of the two cylinders so formed is 5 : 3.

Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is __5 : 3__.

Question 14:

If the sum of the radius of the base and the height of a solid cylinder is 37 m and the total surface area of the cylinder is 1628 m², then its volume is ___________.

Answer 14:

Let r and h be the radius and height of the cylinder, respectively.

Now,

r + h = 37 m     (Given)        .....(1)

Total surface of the cylinder = 1628 m²         (Given)

$\therefore 2\mathrm{\pi }r\left(r+h\right)=1628 {\mathrm{cm}}^2$

$\Rightarrow 2\times \frac{22}{7}\times r\times 37 \mathrm{cm}=1628 {\mathrm{cm}}^2$         [Using (1)]

$\Rightarrow r=\frac{1628\times 7}{44\times 37}=7 \mathrm{cm}$         .....(2)

Substituting the value of r in (1), we get

7 cm + h = 37 cm

⇒ h = 37 − 7 = 30 cm        .....(3)

∴ Volume of the solid cylinder

$=\mathrm{\pi }{r}^{2}h$

$=\frac{22}{7}\times {\left(7 \mathrm{cm}\right)}^2\times 30 \mathrm{cm}$            [From (2) and (3)]

$=4620 {\mathrm{cm}}^3$

Thus, the volume of the solid cylinder is 4620 cm³.

If the sum of the radius of the base and the height of a solid cylinder is 37 m and the total surface area of the cylinder is 1628 m², then its volume is ___4620 cm³___.

Question 15:

A largest possible right-circular cylinder is cut from a wooden cube of edge 7 cm. The volume of the wood left over after cutting the cylinder is __________.

Answer 15:

The largest possible right circular cylinder that can be cut off from a wooden cube of given edge is such that the diameter of the cylinder is same as the edge of the cube and height of the cylinder is same as the edge of the cube.

Let r be the radius and h be the height of the largest right circular cylinder cut from a wooden cube of edge 7 cm.

∴ Diameter of the cylinder = Edge of the cube = 7 cm

⇒ 2r = 7 cm

⇒ r = $\frac{7}{2}$ cm

Height of the cylinder = Edge of the cube = 7 cm

∴ h = 7 cm

Now,

Volume of the wood left over after cutting the cylinder

= Volume of the cube − Volume of the cylinder

= (Edge)³ − $\mathrm{\pi }$r²h

$={\left(7 \mathrm{cm}\right)}^3-\frac{22}{7}\times {\left(\frac{7}{2} \mathrm{cm}\right)}^2\times 7 \mathrm{cm}$

= 343 cm³ − 269.5 cm³

= 73.5 cm³

Thus, the volume of the wood left over after cutting the cylinder is 73.5 cm³.

A largest possible right-circular cylinder is cut from a wooden cube of edge 7 cm. The volume of the wood left over after cutting the cylinder is ___73.5 cm³___.

Question 16:

Four times the sum of the areas of the two circular faces of a cylinder is equal to the twice its curved surface area. The height of the cylinder is ________ the radius.

Answer 16:

Let r be the radius and h be the height of the cylinder.

It is given that,

4 × Sum of areas of the two circular faces of the cylinder = 2 × Curved surface area of the cylinder

$\Rightarrow 4\times \left(\mathrm{\pi }{r}^2+\mathrm{\pi }{r}^2\right)=2\times 2\mathrm{\pi }rh$

$\Rightarrow 8\mathrm{\pi }{r}^2=4\mathrm{\pi }rh$

$\Rightarrow 2r=h$

Or h = 2r

Thus, the height of the cylinder is twice the radius of the cylinder.

Four times the sum of the areas of the two circular faces of a cylinder is equal to the twice its curved surface area. The height of the cylinder is __2 times__ the radius.