Surface Areas and Volumes: 9 Math
NCERT Exercise 13.4 part-2: 9th math
Surface Areas And Volumes Class nine Math NCERT Exercise 13.4 Question (5) A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.
Solution
Given, inner diameter of the hemispherical bowl = 10.5 cm
&therefor; inner radius (r) = 10.5/2 = 5.25 cm
And, Rate of painting of the bowl = Rs 16 per 100 cm2
Thus, rate of painting of bowl for 1 cm2 = Rs 16/100
= Rs 0.16
Now, we know that the Curved Surface Area of a Hemisphere = 2ℼr2
Thus, curved surface area of given hemispherical bowl
= 2 × `22/7` × 5.25 cm × 5.25 cm
= 2 × 22 × 0.75 cm × 5.25 cm
= 44 × 3.9375 cm2
= 173.25 cm2
Thus, curved surface area of given hemispherical bowl = 173.25 cm2
Now, as given in the question, for 1cm2 cost of painting = Rs 0.16
Therefore, for 173.25 cm2, the cost of painting = Rs 0.16 × 173.25
= Rs 27.72
Thus, cost of painting of the given hemispherical bowl = Rs 27.72 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.4 Question (6) Find the radius of a sphere whose surface area is 154cm2.
Solution
Given, surface area of sphere= 154 cm2
Therefore, radius (r) = ?
Now, we know that, Surface Area of a Sphere = 4 ℿ r2
Thus, Surface Area of Given Sphere
⇒ 154 = 4 × `22/7` × r2
⇒ 154 × 7 = 4 × 22 × r2
⇒ 154 × 7 = 88 × r2
`=>r^2 = (154xx7)/(22xx4)`
`=>r=sqrt((7xx7)/4)`
⇒ r=7/2 cm
⇒ r = 3.5 cm
Thus, radius of the given sphere = 3.5 cm Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.4 Question (7) The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution
Given, Diameter of the moon = 1/4 of the diameter of the earth
Thus, ratio of their curved surface area = ?
Let, radius of moon = r
Therefore, diameter of moon = 2r
Now, according to question, diameter of earth = 4 × diameter of moon
Thus, diameter of earth
= 4 ×2 r = 8 r
Thus, radius of earth = 8r/2 = 4r
Now, we know that Curved Surface Area of a Sphere
= 4 ℼ r2
Thus, Curved Surface Area of Moon (CSAmoon)
= 4 ℿ r2
And, Curved Surface Area of Earth (CSAearth)
= 4 ℿ (4r)2
= 4 ℿ 16 r2
Thus, ratio of curved surface area of moon and curved surface area of earth
= CSAmoon : CSAearth
= 4 ℿ r2 : 4 ℿ 16 r2
`=(4\ pi\ r^2)/(4\ pi\ 16\ r^2)`
= 1/16 = 1:16
Thus, radio of curved surface area of moon and curved surface area of earth = 1 : 16 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.4 Question (8) A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution
Given, Thickness of hemispherical bowl = 0.25 cm
And, Inner radius of hemispherical bowl = 5 cm
Therefore, Outer Curved Surface Area of given bowl = ?
Now, since inner radius of bowl = 5 cm, and thickness = 0.25 cm
Thus, outer radius of given hemispherical bowl = Inner radius + thickness
= 5 cm + 0.25 cm = 5.25 cm
Now, we know that, Curved Surface Area of a Hemishpere
= 2 ℼ r2
Therefore, Outer Curved Surface Area of given hemispherical bowl
= 2 × `22/7` × 5.25 cm × 5.25 cm
= 2 × 22 × 0.75 cm × 5.25 cm
= 44 × 3.9375 cm2
= 173.25 cm2Answer
Thus, outer curved surface area of given hemispherical bowl = 173.25 cm2 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.4 Question (9) A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) Surface area of the sphere,
(ii) Curved surface area of the cylinder,
(ii) Ratio of the areas obtained in (i) and (ii)
Solution
Given, radius of sphere = r
Therefore, radius of the cylinder = r [Since cylinder just encloses the sphere]
Therefore, Height of cylinder = 2r (diameter of sphere)
(i) Surface Area of the Sphere
We know that, Surface area of a sphere
= 4 ℿ r2
Thus, Surface area of given sphere
= 4 ℼ r2 Answer
(ii) Curved Surface Area of the Cylinder
We know that, Curved Surface Area of a Right Circular Cylinder
= 2 ℿ r h
Thus, Curved Surface Area of given Right Circular Cylinder which encloses the given sphere
= 2 × ℿ r × 2 r
[Because both radius and height of the right circular cylinder which encloses the given sphere = r]
= 2 ℿ 2× r2
= 4 ℿ r2
Thus, curved surface area of given right circular cylinder = 4 ℼ r2 Answer
(iii) Ratio of the areas obtained in (i) and (ii)
Here curved surface area of sphere = 4 ℼ r2
[As calculated in section (i)]
And, curved surface area of right circular cylinder
= 4 ℼ r2
[As calculated in section (ii)]
Thus, ratio of curved surface area of given sphere and curved surface area of given right circular cylinder
= 4 ℼ r2 : 4 ℼ r2
`=(4\ pi\ r^2)/(4\ pi\ r^2)`
= 1 : 1
Thus, Ratio of the areas obtained in (i) and (ii) = 1 : 1 Answer
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