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 = 2πrh 

If the radius is doubled and height is halved, then

Radius of the new cylinder, R = 2r

Height of the new cylinder, Hh2

∴ Curved surface of the new cylinder, C2 = 2πRH=2π×2r×h2 = 2πrh

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,

r1r2=23  and h1h2=53       .....(1)

Now,

Volume of first cylinderVolume of second cylinder

=πr12h1πr22h2

=r1r22×h1h2

=232×53             [Using (1)]

=49×53

=2027

∴ 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 = 2πrh

Let the height of the new cylinder be H.

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

∴ Curved surface of the new cylinder, C2 = 2πRH = 2π(2r)H

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

C2 = C1

⇒ 2π(2r)H = 2πrh

H=h2

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 = 2πr(r + h)

Curved surface area of the cylinder = 2πrh

∴ 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

= 2πr(r + h) − 2πrh

= 2πr2 + 2πrh − 2πrh

= 2π× (10 cm)2           (r = 10 cm)

= 2π× 100 cm2

= 200π cm2

Thus, the difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is 200π cm2.

The difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is ___200π 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

=2πrr+h

=2πrr+2r           (h = 2r)

=2πr×3r

=6πr2cm2

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 × 6πr2 cm2 = ₹ 18πr2

⇒ ₹ 18πr2 = ₹ 2772

18×227×r2=2772

r2=2772×722×18=49

r=49=7 cm

So, h = 2r = 2 × 7 cm = 14 cm

∴ Volume of the cylinder

=πr2h

=227×72×14

=2156 cm3

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

πr2h=1617

227×2x2×3x=1617

227×4x2×3x=1617

x3=1617×712×22=723

x=72cm

r=2x=2×72=7cm and h=3x=3×72=212 cm

Total surface area of the cylinder

=2πrr+h

=2×227×7×7+212

=44×17.5

=770 cm2

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 = 2πr(r + h)

Lateral surface area of the cylinder = 2πrh

Total surface area of the cylinderLateral surface area of the cylinder

=2πrr+h2πrh

=r+hh

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 = πr2l cu. units

When the radius of the wire is decreased to one-third, then the new radius of the wire, R = r3 units.

Suppose the new length of the wire is L units.

∴ New volume of the wire = πr32L cu. units

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

πr32L=πr2l

r29×L=r2×l

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.

r1r2=23 and h1h2=53          .....(1)              (Given)

Now,

Curved surface area of first cylinderCurved surface area of second cylinder

=2πr1h12πr2h2

=r1r2×h1h2

=23×53           [Using (1)]

=109

∴ 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

⇒ 2πr = 12 cm

r=6π               .....(2)

∴ Volume of the cylinder

=πr2h

=π×6π cm2×5 cm           [Using (1) and (2)]

=180π cm3

Thus, the volume of the cylinder so formed is 180π cm3.

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      180π cm3     .

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)

∴ 2πrh = 4400 cm2         .....(1)

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

∴ 2πr = 110 cm             .....(2)

2×227×r=110 cm

r=110×744=352cm

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

2πrh2πr=4400 cm2110 cm

h=40 cm

∴ Volume of the cylinder

=πr2h

=227×352 cm2×40 cm

=227×352×352×40 cm3

=38500 cm3

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 = πr2   

Lateral surface area of the original cylinder = 2πrh       .....(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 = πr29        (Given)

πR2=πr29

R2=r29

R=r29=r3            .....(3)

Now,

Lateral surface area of the new cylinder

=2πRH

=2π×r3×6h             [From (1) and (2)]

=4πrh

=2×2πrh

= 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

⇒ 2πr = 30 cm

r=15π cm

∴ Volume of the cylinder formed in this case, V1 = πr2h=π×15π cm2×18 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

⇒ 2πR = 18 cm

R=9π cm

∴ Volume of the cylinder formed in this case, V2 = πR2H=π×9π cm2×30 cm         .....(2)

Now,

V1V2=π×15πcm2×18 cmπ×9π cm2×30=53

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)

2πrr+h=1628 cm2

2×227×r×37 cm=1628 cm2         [Using (1)]

r=1628×744×37=7 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

=πr2h

=227×7 cm2×30 cm            [From (2) and (3)]

=4620 cm3

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 = 72 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 − πr2h

=7 cm3-227×72 cm2×7 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

4×πr2+πr2=2×2πrh

8πr2=4πrh

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.

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