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प्रश्न
From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find :
- the surface area of remaining solid,
- the volume of remaining solid,
- the weight of the material drilled out if it weighs 7 gm per cm3.
उत्तर
i. Total surface area of cuboid = 2(lb + bh + lh)
= 2(42 × 30 + 30 × 20 + 20 × 42)
= 2(1260 + 600 + 840)
= 2 × 2700
= 5400 cm2
Diameter of the cone = 14 cm
Radius of the cone = `14/2` = 7 cm
Area of circular base = πr2
= `22/7 xx 7 xx 7`
= 154 cm2
Area of curved surface area of cone = πrl
= `22/7 xx 7 xx sqrt(7^2 + 24^2)`
= `22sqrt(49 + 576)`
= 22 × 25
= 550 cm2
Surface area of remaining part = 5400 + 550 – 154 = 5796 cm2
ii. Dimensions of rectangular solids = (42 × 30 × 20) cm
Volume = (42 × 30 × 20) = 25200 cm3
Radius of conical cavity (r) = 7 cm
Height (h) = 24 cm
Volume of cone = `1/3pir^2h`
= `1/3 xx 22/7 xx 7 xx 7 xx 24`
= 1232 cm3
Volume of remaining solid = (25200 – 1232) = 23968 cm3
iii. Weight of material drilled out
= 1232 × 7 g
= 8624 g
= 8.624 kg
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