 RS Aggarwal Solutions Class 10 Ex 19D

RS Aggarwal Class 10 Ex 19D Chapter 19

QUESTION -1: In a corner of a rectangular field with dimensions 35 m x 22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

Solution:

We have,

Length of the field, = l = 35 m,

Width of the field, b = 22 m,

Depth of the well, H = 8 m and

Radius of the well, R = 14/2 7 m,

Let the rise in the level of the field be h.

Now,

Volume of the earth on remaining part of the field = Volume of the earth dug out

=> Area of the remaining field x h = Volume of the well

=> (Area of the field – Area of base of the well) x h = $\pi R^{2}H$

$\Rightarrow \left ( lb-\pi R^{2} \right )\times h=\pi R^{2}H$

=> (35 x 22 – 22/7 x 7 x 7) x h = 22/7 x 7 x 7 x 8

=> (770 – 154) x h = 1232

=> 616 x h = 1232

=> h = 1232/616

$Therefore, h=2m$

So, the rise in the level of the field is 2m.

QUESTION -2: A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 18 cm and diameter 49 cm to cover its whole surface. Find the length and the volume of the wire. If the density of copper be 8.8 g per cu-cm, find the weight of the wire.

Solution:

We have,

Diameter of a copper wire, d = 6 mm = 0.6 cm,

Radius of the copper wire, r = 0.6/2 = 0.3 cm,

Length of the cylinder, H = 18 cm

Radius of the cylinder, R = 49/2 cm

The number of rotations of the wire on the cylinder = $\frac{length of the cylinder,H }{diameter of the copper wire, d} = \frac{18}{0.6}$ = 30

The circumference of the base of the cylinder = $2\pi R = 2 \times \frac{22}{7} \times \frac{49}{2} = 154 \; cm$

So, the length of the wire, h = 30 × 154 = 4620 cm = 46.2 m

Now, the volume of the wire = πr2h

= 22/7 × 0.3 × 0.3× 4620 = 1306.8 cm3

Also, the weight of the wire = Volume of the wire × Density of the wire

= 1306.8 × 8.8 = 11499.84 g = 11.5 kg

QUESTION -3: A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (choose value of π as found appropriate)

Solution: We have,

$In \Delta ABC, \angle B = 90^{\circ}, AB = l_{1} = 15 \; cm \; and \; BC = l_{2} = 20 \; cm$

Let OD = OB = r, AO = h1 and CO = h2

Using Pythagoras theorem,

$AC = \sqrt{AB^{2} + BC^{2}}$

$= \sqrt{15^{2} + 20^{2}}$

$= \sqrt{225 + 400}$

$= \sqrt{625}$

h = 25 cm

As, $ar\left ( \Delta ABC \right ) = \frac {1} {2} \times AC \times BO = \frac {1} {2} \times AB \times BC$

$\Rightarrow AC \times BO = AB \times BC$

$\Rightarrow 25r = 15 \times 20$

$\Rightarrow r = \frac{15 \times 20 }{25}$

r = 12 cm

Now,

Volume of the double cone so formed = volume of cone 1 + volume of cone 2

$= \frac {1} {3} \pi r^ {2} h_ {1} + \frac {1} {3} \pi r^ {2} h_ {2}$

$= \frac {1} {3} \pi r^ {2}\left (h_ {1} + h_ {2} \right)$

$= \frac {1} {3} \pi r^{2}h$

$= \frac {1} {3} \times 3.14 \times 12 \times 12 \times 25$

= 3768 cm3

Also,

Surface area of the solid so formed = CAS of cone 1 + CSA of cone 2

= $= \pi r l_{1} + \pi r l_{2}$

= $= \pi r \left ( l_{1} + l_{2} \right )$

= $\frac{22} {7} \times 12 \times \left (15 + 20 \right)$

= $\frac{22}{7} \times 12 \times 35$ = 1320 cm2

EXERCISE 19C

Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass.

Solution:

We have,

Height of the frustum, h = 14 cm,

Base radii, $R=\frac{16}{2}=8cm$ and $r=\frac{12}{2}=6cm$

The capacity of the glass = Volume of the frustrum

= $\frac{1}{3}\Pi h(R^{2}+r^{2}+rR)$

= $\frac{1}{3}\times \frac{22}{7}\times 14\times (8^{2}+6^{2}+8\times 6)$

= $\frac{1}{3}\times 22\times 2\times (64+36+48)$

= $\frac{44}{3}\times 148$

= $\frac{6512}{3}cm^{3}$

$\approx 2170.67cm^{3}$

So, the capacity of the glass is 2170.67 cm3.

Question 2: The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area. [Use $\Pi =3.14$]

Solution:

We have, Height, h = 8 cm,

Base radii, R = 18 cm and r = 12 cm

Also, the slant length, l = $\sqrt{(R-r)^{2}+h^{2}}$

= $\sqrt{(18-12)^{2}+8^{2}}$

= $\sqrt{6^{2}+8^{2}}$

= $\sqrt{36+64}$

= $\sqrt{100}$

= 10 cm

Now,

Total surface area of the solid frustum = $\Pi (R+r)l+\Pi R^{2}+\Pi r^{2}$

= 3.14 x (18 + 12) x 10 + 3.14 x 182 + 3.14 x 122

= 3.14 x 30 x 10 + 3.14 x 324 + 3.14 x 144

= 3.14 x (300 + 324 + 144)

= 3.14 x 768

= 2411.52 cm2

So, the total surface area of the solid frustum is 2411.52 cm2.

Question 3: A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find

(i) the volume of water which can completely fill the bucket;

(ii) the area of the metal sheet used to make the bucket.

Solution:

We have,

Height, h = 24 cm,

Upper base radius, R = 14 cm and lower base radius, r = 7 cm

Also, the slant height, l = $\sqrt{(R-r)^{2}+h^{2}}$

= $\sqrt{(14-7)^{2}+24^{2}}$

= $\sqrt{7^{2}+24^{2}}$

= $\sqrt{49+576}$

= $\sqrt{625}$

= 25 cm

(i) Volume of the bucket = $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times \frac{22}{7}\times 24\times (14^{2}+7^{2}+14\times 7)$

= $\frac{22}{7}\times 8\times (196+49+98)$

= $\frac{176}{7}\times 343$

= 8624 cm3

So, the volume of water which can completely fill the bucket is 8624 cm3.

(ii) Surface area of the bucket = $\Pi (R+r)l+\Pi r^{2}$

= $\frac{22}{7}\times (14+7)\times 25+\frac{22}{7}\times 7\times 7$

= $\frac{22}{7}\times 21\times 25+22\times 7$

= $22\times 3\times 25+22\times 7$

= 1650 + 154

= 1804 cm2

So, the area of the metal sheet used to make the bucket is 1804 cm2.

Question 4: A container, open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs. 21 per liter.

Solution:

We have,

Height, h = 24 cm,

Upper radius, R = 20 cm and

Lower radius, r = 8 cm

Now,

Volume of the container = $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times \frac{22}{7}\times 24\times (20^{2}+8^{2}+20\times 8)$

= $\frac{176}{7}\times (400+64+160)$

= $\frac{176}{7}\times 624$

= $\frac{109824}{7}$ cm3

= $\frac{109.824}{7}$ L (As, 1000 cm3 = 1 L)

So, the cost of milk in the container = $\frac{109.824}{7}\times 21$

= 329.472

$\approx$ 329.47

Hence, the cost of milk which can completely fill the container is 329.47.

Question 5: A container, open at the top and made up of metal sheet, is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the cost of metal sheet used to make the container, if it costs Rs. 10 per 100 cm2.

Solution:

We have,

Height, h = 16 cm,

Radius of the upper end, R = $\frac{40}{2}=20cm$ and

Radius of the lower end, r = $\frac{16}{2}=8cm$

Also, the slant height, l = $\sqrt{(R-r)^{2}+h^{2}}$

= $\sqrt{(20-8)^{2}+16^{2}}$

= $\sqrt{12^{2}+16^{2}}$

= $\sqrt{144+256}$

= $\sqrt{400}$

= 20 cm

Now,

Total surface area of the container = $\Pi (R+r)l+\Pi r^{2}$

= $\frac{22}{7}\times (20+8)\times 20+\frac{22}{7}\times 8\times 8$

= $\frac{22}{7}\times 28\times 20+\frac{22}{7}\times 64$

= $\frac{22}{7}\times 560+\frac{22}{7}\times 64$

= $\frac{22}{7}\times (560+64)$

= $\frac{22}{7}\times 624$

= $\frac{13728}{7}cm^{2}$

So, the cost of metal sheet used = $\frac{13728}{7}\times \frac{10}{100}$

= $\frac{13728}{70}$

$\approx$ 196.11

Hence, the cost of metal sheet used to make the container is Rs. 196.11.

Question 6: The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and the total surface are. [Take $\Pi =\frac{22}{7}$]

Solution:

Greater radius = R = 33 cm

Smaller radius = r = 27 cm

Slant height = l = 10 cm

Using the formula for height of a frustum:

Height = h = $\sqrt{l^{2}-(R-r)^{2}}$

= $\sqrt{10^{2}-(33-27)^{2}}$

= $\sqrt{10^{2}-(6)^{2}}$

= $\sqrt{100-36}$

= $\sqrt{64}$ = 8 cm

Capacity of the frustum

= $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times \frac{22}{7}\times 8(33^{2}+27^{2}+33\times 27)$

= $\frac{22\times 8}{3\times 7}\times 2709$ = 22704 cm3

Surface area of the frustum

= $\Pi R^{2}+\Pi r^{2}+\Pi l(R+r)$

= $\Pi \left [ R^{2}+r^{2}+l(R+r) \right ]$

= $\frac{22}{7}\left [ 33^{2}+27^{2}+10(33+27) \right ]$

= $\frac{22}{7}\left [ 1089+729+10(60) \right ]$

= $\frac{22\times 2418}{7}$ = 7599.43cm2

Question 7: A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. Find how many litres of water the bucket can hold. [Take $\Pi =\frac{22}{7}$]

Solution:

Greater diameter of the frustum = 56 cm

Greater radius of the frustum = R = 28 cm

Smaller diameter of the frustum = 42 cm

Radius of the smaller end of the frustum = r = 21 cm

Height of the frustum = h = 15 cm

Capacity of the frustum

= $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times \frac{22}{7}\times 15(28^{2}+21^{2}+28\times 21)$

= $\frac{22\times 5}{7}\times 1813$

= 28490 cm3 = 28.49 litres

Question 8: A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of metal sheet used is Rs. 15 per 100 cm2. [Use $\Pi =3.14$]

Solution:

Greater radius of the frustum = R = 20 cm

Smaller radius of the frustum = r = 8 cm

Height of the frustum = h = 16 cm

Slant height, l, of the frustum

= $\sqrt{h^{2}+(R-r)^{2}}$

= $\sqrt{16^{2}+(20-8)^{2}}$

= $\sqrt{16^{2}+(12)^{2}}$

= $\sqrt{256+144}$

= $\sqrt{400}$ = 20 cm

Surface area of the frustum

= $\Pi r^{2}+\Pi l(R+r)$

= $\Pi \left [ r^{2}+l(R+r) \right ]$

= $\frac{22}{7}\left [ 8^{2}+20(20+8) \right ]$

= $\frac{22}{7}\left [ 64+20(20+8) \right ]$

= $\frac{22\times 624}{7}$ = 1961.14 cm2

100 cm2 of metal sheet costs Rs. 15.

So, total cost of the metal sheet used for fabrication

= $\frac{1961.14}{100}\times 15$

= Rs. 294.171

Question 9: A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs. 20 per litre and the cost of metal sheet used if it costs Rs. 10 per 100 cm2. [Use $\Pi =3.14$]

Solution:

Greater diameter of the bucket = 30 cm

Radius of the bigger end of the bucket = R = 15 cm

Diameter of the smaller end of the bucket = 10 cm

Radius of the smaller end of the bucket = r = 5 cm

Height of the bucket = 24 cm

Slant height, l

= $\sqrt{h^{2}+(R-r)^{2}}$

= $\sqrt{24^{2}+(15-5)^{2}}$

= $\sqrt{24^{2}+(10)^{2}}$

= $\sqrt{576+100}$

= $\sqrt{676}$ = 26 cm

Capacity of the frustum

= $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times 3.14\times 24(15^{2}+5^{2}+15\times 5)$

= $3.14\times 8\times 325$

= 8164 cm3 = 8.164 litres

A litre of milk cost Rs. 20.

So, total cost of filling the bucket with milk = 8.164 x 20 = Rs. 163.28

Surface area of the bucket

= $\Pi r^{2}+\Pi l(R+r)$

= $\Pi \left [ 5^{2}+26(15+5) \right ]$

= $3.14\left [ 25+26(20) \right ]$

= $3.14\left [ 25+520 \right ]$

= 1711.3 cm2

Cost of 100 cm2 of metal sheet is Rs. 10.

So, cost of metal used for making the bucket = $\frac{1711.3}{100}\times 10$

= Rs. 171.13

Question 10: A container in the shape of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm, is completely filled with oil. If each cm3 of oil has mass 1.2 g, then find the cost of oil in the container if it costs Rs. 40 per kg.

Solution:

We have,

Height, h = 14 cm,

Radius of upper end, R = $\frac{35}{2}=17.5cm$ and

Radius of lower end, r = $\frac{30}{2}=15cm$

Now,

Volume of the container = $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$

= $\frac{1}{3}\times \frac{22}{7}\times 14\times (17.5^{2}+15^{2}+17.5\times 15)$

= $\frac{44}{3}\times (306.25+225+262.5)$

= $\frac{44}{3}\times 793.75$

= $\frac{34925}{3}$ cm3

So, the mass of the oil that is completely filled in the container = $\frac{34925}{3}\times 1.2$

= 13970 kg

= 13.97 kg

Therefore, The cost of the oil in the container = 40 x 13.97 = Rs. 558.80

Question 11: A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket.

Solution:

We have,

Radius of upper end, R = 28 cm and

Radius of lower end, r = 21 cm

Let the height of the bucket be h.

Now,

Volume of water the bucket can hold = 28.49 L

$\Rightarrow$ Volume of bucket = 28490 cm3 (As, 1 L = 1000 cm3)

$\Rightarrow$ $\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)$ = 28490

$\Rightarrow$ $\frac{1}{3}\times \frac{22}{7}\times h\times (28^{2}+21^{2}+28\times 21)$ = 28490

$\Rightarrow$ $\frac{22h}{21}\times 49\times (16+9+12)$ = 28490

$\Rightarrow$ $\frac{22h}{3}\times 7\times 37$ = 28490

$\Rightarrow$ h = $\frac{28490\times 3}{22\times 7\times 37}$

Therefore, h = 15 cm

So, the height of the bucket is 15 cm.

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The class size and frequency of modal class is

Warranty period(years)011223344556Number of components61016251620