Surface Areas & Volume
Mathematics Class Tenth
RD Sharma exercise 16.1 q21 to 25
Question (21) 500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is 0.04 m3 ?
Solution:
[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (21) (a) Find the total volume of water displace by multiplying number of person and average displacement. (b) Using given dimension of rectangular tank and volume of water find the rise of level of water in tank.]
Given, Total number of person to dip in water in tank = 500
Average displacement of water by a person = 0.04 m3
Length of tank = 80 m
Width of tank = 50 m
Therefore, rise in level of water in tank, i.e. height of water in tank = ?
Calculation of volume of water
Total displacement of water = total number of person × average displacement
= 500 × 0.04 m3
= 20 m3
The volume of displacement of water = volume of water in tank
Calculation of rise in level of water in tank
Let the height of water rise = h
We know that, volume of water in rectangular tank = Length × Width × Height
Thus, volume of water in tank = 80 m × 50 m × h
⇒ 20 m3 = 4000 m2 × h
Thus, `h = (20\ m^3)/(4000\ m^2)`
⇒ h = 0.005 m
= 0.005 × 100 cm = 0.5 cm
Thus, rise in the level of water in the tank = 0.5 cm Answer
Question (22) A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the soil by two centimeters?
Solution:
[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (22) (a) Find the volume of raised level of oil using height and radius of cylindrical jar. (b) This volume of raised level of oil is equal to the sum of the volume of total spheres. (c) Calculate the volume of one sphere. (d) Calculate the number of spheres required after dividing sum total volume of balls by volume of one sphere.]
Given, Radius of iron sphere = 1.5
Radius of cylindrical jar = 6 cm
Level of oil to be raised, i.e. height of level of oil to be raised = 2 cm
Thus, number of sphere required to rise the oil upto given level = ?
Calculation of volume of raised oil level
We know that, volume of cylinder `=pi\ r^2\ h`
Thus, volume of raised oil in cylinder `=pi\ (6\ cm)^2\ 2\ cm`
Or, Volume of raised oil in cylinder `=pi\ 72\ cm^3`
This volume of raised oil is due to spheres immersed in oil. This means the volume of raised oil in cylinder = Sum of volume of total spheres
Thus, volume of total spheres = Volume of raised oil in cylinder `=pi\ 72\ cm^3`
[Here, `pi` is left as it is for ease of calculation. However volume can be calculated by taking the value of `pi=22/7` also which will not affect the result.]
Calculation of volume of one sphere
We know that, volume of a sphere `=4/3\ pi\ r^3`
Thus volume of one given sphere `=4/3\ pi\ (1.5)^3`
`=4/3\ pi\ 3.375\ cm^3`
`=4xx1.125\ pi\ cm^3`
Or, volume of one sphere `=4.5\ pi\ cm^3`
Calculation of number of sphere required
Now, since the volume of water raised is due to spheres immersed.
Thus, number of sphere = Total volume of oil raised, i.e. volume of total spheres/Volume of one sphere
`=( pi\ 72\ cm^3)/( 4.5\ pi\ cm^3)`
= 16
Thus, number of spheres required to raise the oil upto given height = 16 Answer
Question (23) A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Solution:
[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (23) (a) Find the volume of material in hollow sphere. (b) Since sphere is melted into cone, thus, volume of material of sphere will be equal to the volume of cone. (c) Now, using calculated volume of cone, find the height and slant height of cone using given radius.]
Given, Internal radius of hollow sphere (r) = 2 cm
And, external radius of hollow sphere (R) = 4 cm
And, radius of base of cone made by the material of sphere = 4 cm
Thus, height of cone and slant height of cone = ?
Calculation of volume of material of hollow sphere
We know that, volume of material of a hollow sphere `=4/3\ pi(R^3–r^2)`
Thus, volume of material of given hollow sphere `=4/3\ pi\ ((4\ cm)^3–(2\ cm)^3)`
`=4/3\ pi\ (64\ –\ 8)\ cm^3`
`=4/3\ pi\ xx 56\ cm^3`
Thus, volume of material of hollow sphere `= 4/3\ pi\ xx 56\ cm^3`
[Here, `pi` is left as it is for ease of calculation. However volume can be calculated by taking the value of `pi=22/7` also which will not affect the result.]
Calculation of height of given cone
Since hollow sphere is melted into cone, thus volume of cone = volume of material of hollow sphere
Thus, volume of cone `= 4/3\ pi\ xx 56\ cm^3`
Now, we know that, volume of a cone `=1/3\ pi\ r^2\ h`
Thus, volume of given cone `=1/3\ pi\ (4\ cm)^2\ h`
`=>4/3\ pi\ xx 56\ cm^3` `= 1/3\ pi\ 16\ cm^2\ h`
`:. h = (4xx3xx\ pi\ 56\ cm^3)/(3xxpi\ 16\ cm^2)`
Or, h = 14 cm
Thus, height of cone = 14 cm
Calculation of slant height of cone
Here, we have, height (h) 14 cm
And, radius of cone (r) = 4 cm
Now, since height, radius and slant height of a cone makes a right angle triangle,
Thus, according to Pythagoras theorem
`l^2 = h^2 + r^2`
`=> l^2 = (14\ cm)^2 + (4\ cm)^2`
`=> l^2 = 196\ cm^2 + 16\ cm^2`
`=>l^2 = 212\ cm^2`
`=> l = sqrt(212\ cm^2)`
`=> l = 14.56\ cm`
Thus, slant height of given cone = 14.56 cm
Thus, height of cone = 14 cm and slant height of cone = 14.56 cm Answer
Question (24) The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm2 of the surface is 10 paise. Find the total cost of paint the vessel all over.
Solution:
[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (24) (a) Find the external and internal surface area of hemisphere. (b) Find the surface area of rim of hemisphere (c) Add all, i.e. internal, external and surface area of rim. This will give the total surface area of hollow hemisphere. (c) Calculate the cost of paint of total surface area of hollow hemisphere by multiplying the rate of paint with total surface area.]
Given, external diameter of hollow hemispherical vessel = 25.2 cm
Thus, external radius of hollow hemispherical vessel (R) = 25.2/2 = 12.6 cm
And, internal diameter of hollow hemispherical vessel = 21 cm
Thus, internal radius (r) = 21/2 = 10.5 cm
Cost of paint for 1 cm2 = 10 paise
Thus, total cost of painting of vessel all over = ?
Calculation of total surface are of hollow hemispherical vessel
We know that, surface area of a hemisphere `=2\ pi\ r^2`
Thus, total surface area of given hollow hemispherical vessel
= External surface area + Internal surface area + surface area of upper rim
`=2\ pi\ R^2 + 2\ pi\ r^2 + pi(R^2–r^2)`
`=pi[2(12.6\ cm)^2\ +\ 2(10.5\ cm)^2` `+\ {(12.6\ cm)^2\ –\ (10.5\ cm)^2}]`
`=22/7[(2xx158.76)\ +\ (2xx110.25)` `+\ (158.76\ –\ 110.25) ] cm^2`
`=22/7(317.52+220.5+48.51)\ cm^2`
`=22/7xx586.53\ cm^2`
= 1843.38 cm2
Thus, total surface area of given hollow hemispherical vessel = 1843.38 cm2
Calculation of cost of painting
Now, ∵ cost of painting for 1 cm2 = 10 paise
∴ cost of painting for 1843.38 cm2
= 1843.38 cm2 × 10 paise
= 184.34 Rupees
Thus, cost of painting of given hemispherical vessel = Rs 184.34 Answer
Question (25) A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball of radius 9 cm is dropped into the tub and thus the level of water is raised by h cm. What is the value of h?
Solution:
[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (25) (a) Find the volume of spherical ball. (b) The volume of spherical ball is equal to the volume of water raised to given height h. (c) Using volume of water raised and radius of cylindrical tub, find the value of h.]
Given, Radius of spherical ball = 9 cm
Radius of cylindrical tub = 12 cm
And depth of water in tube = 20 cm.
Raised height of water after ball dropped into tub = h
Thus, value of h = ?
Calculation of volume of spherical ball
We know that volume of sphere `=4/3\ pi\ r^3`
Thus, volume of given spherical ball `=4/3\ pi\ (9\ cm)^3`
`=4xx3xx9xx9xxpi\ cm^3`
⇒ Volume of spherical ball `=972\ pi\ cm^3`
[Here, `pi` is left as it is for ease of calculation. However volume can be calculated by taking the value of `pi=22/7` also which will not affect the result.]
Since, volume of water in cylindrical tub is raised because of dropping of spherical ball, thus volume of water raised = volume of spherical ball.
Calculation of raised height (h) of water in cylindrical tub
We know that, volume of a cylinder `=pi\ r^2\ h`
Thus, volume of raised water in cylinder `=pi\ (12\ cm)^2\ h`
`=> 972\ pi\ cm^3 = pi\ 144\ cm^2xxh`
`=> h = (972\ pi\ cm^3)/(144\ pi\ cm^2)`
Thus, h = 6.75 cm
Thus, value of required, height, h = 6.75 cm Answer
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