RD Chapter 16 Surface Areas and Volumes Ex 16.2 Solutions
Question - 11 : - A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of tent and the cost of the canvas at Rs 100 per square meter.
Answer - 11 : -
Given,
Diameter of thecylinder = 20 m
So, its radius of thecylinder (R) = 10 m
Height of the cylinder(h1) = 2.5 m
Radius of the cone =Radius of the cylinder (r) = 15 m
Height of the Cone (h2)= 7.5 m
Let us consider L asthe slant height of the Cone, then we know that
L2 = r2 +h22
L2 =152 + 7.52
L2 =225 + 56.25
L2 =281.25
L = 12.5 m
Now,
Volume of the cylinder= πR2h1 = V1
V1 =π(10)2 2.5
V1 =250 πm3
Volume of the Cone= 1/3 × 22/7 × r2 × h22 = V2
V2 = 1/3× 22/7 × 102 × 7.5
V2 =250π m3
So, the total capacityof the tent = volume of the cylinder + volume of the cone = V1 +V2
V = 250 π + 250 π
V = 500π m3
Hence, the totalcapacity of the tent is 500π m3
And, the total area ofcanvas required for the tent is S = 2 πRh1 + πrL
S = 2(π)(10)(2.5) +π(10)(12.5)
S = 550 m2
Therefore, the totalcost for canvas is (100) (550) = Rs. 55000
Answer - 12 : -
Given,
Diameter of thehemisphere = 2 m
So, the radius of thehemisphere (r) = 1 m
Height of the cylinder(h1) = 2 m
And, the volume of theCylinder = πr2h1 = V1
V1 =π(1)22
V1 = 22/7× 2 = 44/7 m3
As at each of the endsof the cylinder, hemispheres are attached.
So, totally there are2 hemispheres.
Then the volume of twohemispheres = 2 × 2/3 × 22/7 × r3 = V2
V2 = 2× 2/3 × 22/7 × 13
V2 = 22/7× 4/3 = 88/21 m3
Thus,
The volume of theboiler (V) = volume of the cylindrical portion + volume of the two hemispheres
V = V1 +V2
V= 44/7 + 88/21
V = 220/21 m3
Therefore, the volumeof the boiler 220/21 m3
Answer - 13 : -
Given,
Diameter of thehemisphere = 3.5 m
So, the radius of thehemisphere (r) = 1.75 m
Height of the cylinder(h) = 14/3 m
We know that, volumeof the Cylinder = πr2 h1 = V1
V1 =π(1.75)2 x 14/3 m3
The volume of thehemispherical bottom = 2 × 2/3 × 22/7 × r3 = V2
V2 = 2/3× 22/7 × 1.753 m3
Therefore,
The total volume ofthe vessel (V) = volume of the cylinder + volume of the hemisphere
V = V1 +V2
V = π(1.75)2 x14/3 + 2/3 × 22/7 × 1.753
V = π(1.75)2 (14/3+ 2/3 x 1.75)
V = 56.15 m2
Hence, the volume ofthe vessel = V = 56.15 m3
Now,
Internal surface areaof solid (S) = Surface area of the cylinder + Surface area of the hemisphere
S = 2 πr h1 +2 πr2
S = 2 π(1.75)(143) + 2π(1.75)2
S = 70.51 m3
Therefore, theinternal surface area of the solid is 70.51 m3
Question - 14 : - A solid is composed of a cylinder with hemispherical ends. If the complete length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Rs.10 per dm2.
Answer - 14 : -
Given,
Radius of thehemispherical end (r) = 7 cm
Height of the solid =(h + 2r) = 104 cm
⇒ h + 2r = 104
⇒ h = 104 − (2 × 7)
So, h = 90 cm
We know that,
The curved surfacearea of the cylinder (S1) = 2 πr h
S1= 2π(7)(90)
S1 =3960 cm2
Next,
Curved surface area ofthe two hemisphere (S2) = 2 (2πr2)
S2 = 2x 2π(7)2
S2 =616 cm2
Therefore,
The total curvedsurface area of the solid (S) = Curved surface area of the cylinder + Curvedsurface area of the two hemispheres
S = S1 +S2
S = 3960 + 616
S = 4576 cm2 =45.76 dm2
Given that the cost ofpolishing the 1 dm2 surface of the solid is Rs. 10
So, the cost ofpolishing the 45.76 dm2 surface of the solid = Rs (10 x 45.76)= Rs. 457.6
Therefore,
The cost of polishingthe whole surface of the solid is Rs. 457.60
Question - 15 : - A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16cm and height of 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic cms of the cork dust will be required?
Answer - 15 : -
Given,
Depth of the cylindricalvessel = Height of the cylindrical vessel = h = 42 cm (common for both)
Inner diameter of thecylindrical vessel = 14 cm
So, the inner radiusof the cylindrical vessel = r1 = 14/2 = 7 cm
Outer diameter of thecylindrical vessel = 16 cm
So, the outer radiusof the cylindrical vessel = r2 = 16/2 = 8 cm
Now,
The volume of thecylindrical vessel
V = π(r22 –r12 ) x h
= π(82 –72 ) x 42
= 22/7 x 15 x 42
V = 1980 cm3
Therefore, the totalspace between the two vessels is 1980 cm3, which is also the amountof cork dust required.
Question - 16 : - A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making roller is 9 cm. Find the mass of the road roller, if 1 cm3 of the iron has 7.8 gm mass.
Answer - 16 : -
Given,
Height/length of thecylindrical road roller = h = 1 m = 100 cm
Internal Diameter ofthe cylindrical road roller = 54 cm
So, the internalradius of the cylindrical road roller = 27 cm = r
Also given, thethickness of the road roller (T) = 9 cm
Let us assume that theouter radii of the cylindrical road roller be R.
T = R – r
9 = R – 27
R = 27 + 9
R = 36 cm
Now,
The volume of the ironsheet (V) = π × (R2 − r2) × h
V = π × (362 −272) × 100
V = 1780.38 cm3
Hence, the volume ofthe iron sheet = 1780.38 cm3
It’s given that, massof 1 cm3 of the iron sheet = 7.8 gm
So, the mass of1780.38 cm3 of the iron sheet = 1388696.4gm = 1388.7 kg
Therefore, the mass ofthe road roller is 1388.7 kg
Question - 17 : - A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13cm. Find the inner surface area of the vessel.
Answer - 17 : -
Given,
Diameter of thehemisphere = 14 cm
So, the radius of thehemisphere = 7 cm
Total height of thevessel = 13 cm = h + r
Now,
Inner surface area ofthe vessel = 2 π r (h + r)
= 2 (22/7)(7)(13)
= 572 cm2
Therefore, the innersurface area of the vessel is 572 cm2
Question - 18 : - A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer - 18 : -
Given,
Radius of the conicalportion of the toy = 3.5 cm = r
Total height of thetoy = 15.5 cm = H
If H is the length ofthe conical portion
Then,
Length of the cone (h)= H – r = 15.5 – 3.5 = 12 cm
Now, we know that
The curved surfacearea of the cone (S1) = πrL, where L is the slant height of thecone.
L2 = r2 +h2
L2 =3.52 + 122
L2 =12.25 + 144 = 156.25
L = 12.5
So,
S1 = π(3.5)(12.5)
S1 =137.5 cm2
Next, the curved surfacearea of the hemisphere (S2) = 2πr2
S2 =2π (3.5)2
S2 =77 cm2
Therefore,
The total surface areaof the toy (S) = Curved surface area of the cone + curved surface area of thehemisphere
S = S1 +S2
S = 137.5 + 77
S = 214.5 cm2
Hence, the totalsurface area of the children’s toy is 214.5 cm2