9th Maths 13.4
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.1
NCERT Class 9th solution of Exercise 13.2
NCERT Class 9th solution of Exercise 13.3
NCERT Class 9th solution of Exercise 13.5
NCERT Class 9th solution of Exercise 13.6
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 13.8
Exercise 13.4
Assume π=227, unless stated otherwise.
Q1. Find the surface area of a sphere of radius:
i) 10.5cm ii) 5.6cm iii) 14cm
Sol. :
i) Surface Area of a Sphere =4πr2
=4×227×(10.5)2
=1386cm2
Answer:
Surface Area of a Sphere 1386cm2.
ii) Surface Area of a Sphere =4πr2
=4×227×(5.6)2
=394.24cm2
Answer:
Surface Area of a Sphere 394.24cm2.
iii) Surface Area of a Sphere =4πr2
=4×227×(14)2
=2464cm2
Answer:
Surface Area of a Sphere 2464cm2.
Q2. Find the surface area of a sphere of diameter:
i) 14cm ii) 21cm iii) 3.5m
Sol. ;
i)
Given:
Radius of Sphere is r=142=7
Solve:
Surface Area of a Sphere =4πr2
=4×227×(7)2
=616cm2
Answer:
Surface Area of a Sphere 616cm2.
ii)
Given:
Radius of Sphere is r=212
Solve:
Surface Area of a Sphere =4πr2
=4×227×(212)2
=1386cm2
Answer:
Surface Area of a Sphere 1386cm2.
iii)
Given:
Radius of Sphere is r=3.52
Solve:
Surface Area of a Sphere =4πr2
=4×227×(3.57)2
=38.5cm2
Answer:
Surface Area of a Sphere 38.5cm2.
Q3. Find the total surface area of a hemisphere of radius 10cm. (Use π=3.14)
Sol. :
Given:
Radius of a hemisphere is r=10cm.
To Find:
Total Surface Area of a Hemisphere.
Solve:
Total Surface Area of a Hemisphere =3πr2
=3×3.14×(10)2
=942cm2
Answer:
Total Surface Area of a Hemisphere 942cm2.
Q4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Sol. :
Given:
Radius of Spherical balloon r1=7cm,r2=14cm
To Find:
Ratio of Surface Areas.
Solve:
Ratio of Surface Areas =4π(r1)24π(r2)2
=(7)2(14)2=(714)2=(12)2=14
Answer:
Ratio of Surface Areas of the balloon in the two cases is 1:4.
Q5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of ₹16per100m2.
Sol. :
Given:
Radius of Hemisphere r=10.52=5.25cm
Rate ₹16 per 100m2
To Find:
Cost of tin-plating.
Solve:
Curved Surface Area of Hemisphere =2πr2
=2×227×(5.25)2
=173.25cm2
Cost of tin-plating =Area×Rate
=₹173.25×16100
=₹27.72
Answer:
Cost of tin-plating ₹27.72.
Q6. Find the radius of a sphere whose surface area is 154cm2.
Sol. :
Given:
Surface Area of Sphere 154cm2.
To Find:
Radius
Solve:
Surface Area of Sphere =4πr2
154=4×227×r2
r2=154×722
r2=7×7
r=7cm
Answer:
Radius of Sphere 7cm.
Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol. :
Given:
Radius of earth is re=D2
Radius of moon is rm=D2×4=D8
To Find:
Ratio of Surface Area.
Solve:
Ratio of Surface Area =4π(rm)24π(re)2
=r2mr2e
=(D8)2(D2)2
=16414=(164)×(41)=464=116
Answer:
Ratio of their Surface Areas 1:16.
Q8. A hemispherical bowl is made of steel, 0.25cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl.
Sol. :
Given:
Inner Radius r=5cm.
Thickness of steel =0.25cm
Outer Radius R=5+0.25=5.25cm
To Find:
Outer Curved Surface Area of Bowl.
Solve:
Curved Surface Area of Hemisphere = Outer Curved Surface Area of Bowl
=2πR2
=2×227×(5.25)2
=173.25cm2
Answer:
Outer Curved Surface Area of Bowl 173.25cm2.
Q9. 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,
iii) ratio of the areas obtained in i) and ii).
Sol. :
i) Surface Area of the Sphere =4πr2
ii)
Radius=r, and Height
Curved Surface Area of the Cylinder =2pirh=2pi r times 2r
=4pir^2
Answer:
Curved Surface Area of the Cylinder =4pir^2.
iii)
Ratio of the Areas =(4pir^2)/(4pir^2)=1/1
Answer:
Ratio of the Areas 1:1.
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