NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.1
The NCERT Solutions for Class 10 Maths Chapter 13 Surface area of volumes are undoubtedly an essential study materials for the students studying in CBSE Class 10 NCERT Solutions provided here along with the downnloads PDF can help the students prepare effectively for their first term exams The chapter is a continuation of what was taught in chapter probability in Class 9 and further explains the different concept related to it.
Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:-
Volume of cube = 64 cm3
A3 = 643
A = 4 cm
Surface area of the cuboid = 2(lb + bh + hl)
= 2(8 x 4 + 4 x 4 + 4 x 8 )
= 2(32 + 16 + 32)
= 160 cm2
Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm Find the inner surface area of the vessel.
Solution:-
Total inner surface area of vessel = Inner surface area of hemisphere + Inner surface area of cylinder = 2∏r2 + 2∏rh
= 2x (7)2 + 2∏ x 7 x 6
= 98∏ + 84∏
= 182 x 22/7
= 572 cm2
Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:-
Surface area of the hemisphere = 2∏r2
= 2 x 22/7 x 3.5 x 3.5
= 2 x 22 x 35 x 35/7 x 10 x 10
= 77 cm2
Slant height of the conical portion = √(12)2 + (3.5)2
= √144 + 12.25
= √156.25
= 12.5 cm
Curved surface area of the colonial portion = ∏rl = ∏ x 3.5 x 12.5
= 22/7 x 35/10 x 125/10 cm2
= 11 x 25/2 cm2
= 275/2 cm2
Total surface of the toy = Surface area of the hemisphere
+ Surface area of the conical portion
= 77 + 275/2
= 214.5 cm2
Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution:-
Surface area of the solid = Curved surface area of a cube + Curved surface area of a hemisphere – Area of the base of hemisphere
= 6a2 + 2∏r2 – ∏r2
= [6(7)2 + 2∏(3.5)2 – ∏(3,5)2]
= [294 + 77/7] cm2
= 332.5 cm2
Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:-
Inner surface area of the hemisphere = 2∏r2
= 2∏(1/2)2
= 2∏l2/4
= ∏l2/2
Area of circular portion of the hemisphere = ∏r2
= ∏(l/2)2
= ∏l2/4
Remaining surface area of the cubical box = ( Surface area of the cubical box + Inner surface area of the hemisphere – Area of the circular region)
= 6l2 + ∏l2/4 – ∏l2/4
= 6l2 + ∏l2/2 – ∏l2/4
= l2/4 (24 + ∏)
Q7. A tent is in the shape of a cylindrical surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively. And the slant height of the top is 2.8 m find the area of the canvas used for making the tent Also , find the cost of the canvas of the tent at the rate of Rs 500 per m2
Solution:-
Curved surface area of the cylindrical part = 2∏rh
= 2 x 22/7 x 2 x 2.1 m2
= 26.4 m2
Curved surface area of the top = ∏rl
= 22/7 x 2 x 2.8
= 17.6 m2
Total area of canvas = 26.4 + 17.6 = 44 m2
Cost of canvas = Rs 500/m2
Total Cost = 44 x 500 = Rs 22,000
Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm , a conical cavity of the same height and same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:-
Total surface area of the remaining solid = Outer curved surface area of cylinder + Area of bottom of cylinder + Inner curved surface area of conical cavity
= 2∏rh + ∏r2 + ∏rl
= ∏r(2h + r + l)
= 22/7 x 0.7[2 x 2.4 + 0.7 + √(0.7)2 + (2.4)2]
= 18 cm2.
Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder If the height ofd cylinder is 10 cm , and its base of radius 3.5 cm find the total surface area of the article.
Solution:-
Curved surface area of the cylinder = 2∏rh
= 2 x 22 x 35 x 10/7 x 10
= 220 cm2
Inner surface area of a hemispherical cavity = 2∏r2
= 2 x 22/7 x 35 x 35/10 x 10
= 77 cm2
Inner surface area of both hemispherical cavity = 77 cm2 + 77 cm2
= 154 cm2
Total surface area of the solid = Curved surface area of the solid + Inner surface area of both hemispherical ends
= 220 cm2 + 154 cm2
= 374 cm2