Surface Areas and Volumes: 9 Math
NCERT Exemplar Exercise 13.1: 9th math
Write the correct answer in each of the following
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (1) The radius of a sphere is 2r, then its volume will be
(A) 4/3 ℼ r3
(B) 4 ℼ r3
(C) `(8\ pi\ r^3)/3`
(D) 32/3 ℼ r3
Answer (D) 32/3 ℼ r3
Explanation
Given, Radius of a sphere = 2r
Thus, volume of the given sphere = ?
Here, we know that Volume of a sphere = 4/3 ℼ r3
Thus, volume of the given sphere
= 4/3 × ℼ × 2r × 2r × 2r
= 4/3 × ℼ × 8 r3
= 32/3 ℼ r3
Thus, volume of the given sphere = (D) 32/3 ℼ r3 Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (2) The total surface area of a cube is 96 cm2. The volume of the cube is:
(A) 8 cm3.
(B) 512 cm3
(C) 64 3
(D) 27 cm 3
Answer (C) 64 cm2
Explanation
Given, the total Surface area of a cube = 96 cm2
Thus, volume of the cube = ?
Now, we know that, the Total Surface Area of a Cube = 6 a2
Therefore, the total surface area of the given cube
96 cm2 = 6 × (a)2
⇒ a2 = 96/6 cm2
⇒ a2 = 16
⇒ a = √ 16 cm2
⇒ a = 4 cm
Now, we know that volume of a cube = a3
= 4 cm × 4 cm × 4 cm
= 64 cm3
Thus, volume of the given cube = (C) 64 cm2 Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (3) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is:
(A) 4.2 cm
(B) 2.1 cm
(C) 2.4 cm
(D) 1.6 cm
Answer (B) 2.1 cm
Explanation
Given, the height of a cone = 8.4 cm
And, the radius of the base of the cone = 2.1 cm
Thus, radius of the sphere made after melting the given cone = ?
Now, we know that volume of a cone
= 1/3 ℼ r2 h
Now, the volume of the given cone,
= 1/3 × 22/7 × 2.1 cm × 2.1 cm × 8.4 cm
= 22 × 0.7 cm × 0.3 cm × 8.4 cm
= 38.808 cm3
Thus, volume of the given cone = 38.808 cm3
Now, as per question the given cone is melted and transformed into a sphere
Now, we know that, Volume of a Sphere = 4/3 ℼ r3
Thus, volume of the given sphere
= 38.808 cm3 = 4/3 × 22/7 × r3
`=>r^3= (38.808 cm^3xx3xx7)/(4xx22)`
⇒ r3 = 9.261 cm3
`=>r = root3(9.261 cm^3)`
⇒ r = 2.1 cm
Thus, radius of the given sphere = 2.1 cm Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (4) In a cylinder, radius is doubled and height is halved, curved surface area will be
(A) halved
(B) doubled
(C) same
(D) four times
Answer (C) same
Explanation
Given, radius of a cylinder is doubled and height is halved,
Then area of the given cylinder = ?
Let radius of the cylinder = r
Now, as per question, radius is doubled, i.e. radius = 2r
And, let height = h
Now, as per question, height is halved, i.e. height =h/2
Now, we know hta, Curved Surface Area of a Cylinder
= 2 ℼ r h
Now, after doubled of radius and halved of height, the curved surface area of the given cylinder
= 2 × ℼ × 2r × `h/2`
= 2 ℼ rh
Thus, area of the given cylinder remains the same even after doubling of radius and halved of height.
Thus, Answer = (C) Same
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (5) The total surface area of a cone whose radius is `r/2` and slant height `2l` is
(A) 2 ℼ r (`l` + r)
(B) `pi\ r\ (l + r/4)`
(C) ℼ r (`l` + r)
(D) 2 ℼ r `l`
Answer (B) `pi\r(l + r/4)`
Explanation
Given, Radius of a cone = r/2
And, slant height of the cone = 2`l`
Thus, total surface area of the given cone = ?
Now, we know that Total Surface Area of a Cone
= ℼ r (`l` + r)
Thus, total surface area of the given cone
Thus, total surface area of the given cone `=pi\ r(l+r/4)`
Thus, option (B) `=pi\ r(l+r/4)` is correct option
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