Question

please help maths legend solve 2 questions please

Answer 1

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▶ Question ( 7 ) - The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m^2.



▶ Explanation :-


It can be observed that a roller is cylindrical


Height ( h ) of the cylindrical roller = Length of roller = 120 cm


Diameter of the cylindrical roller = 84 cm


Radius ( r ) of circular end of roller = 84 / 2 = 42 cm


CSA of roller = 2πrh


= 2 × 22 / 7 × 42 × 120


= 2 × 22 × 6 × 120


= 31680 cm^2


Area of field = 500 × CSA of roller


= ( 500 × 31680 ) cm^2


= 15840000 cm^2


= 1584 m^2




▶ Question ( 8 ) - The inner diameter of a circular well is 3.5m. It is 10m deep.

Find : i)its inner curved surface area

ii)The cost of plastering this curved surface at the rate of Rs40 per m^2.



▶ Explanation :-


i) Inner diameter of a circular well = 3.5 m


Inner Radius ( r ) of a circular well = 3.5 / 2 = 1.75 m


Depth ( h ) of circular well = 10 m


Inner curved surface area = 2πrh


= 2 × 22 / 7 × 1.75 × 10


= 44 × 0.25 × 10


= 110 m^2


ii) Cost of plastering 1 m^2 area = Rs. 40


Cost of plastering 110 m^2 area = 110 × 40


= Rs. 4400


Therefore, Cost of plastering the CSA of this wall is Rs. 4400




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Answer 2

7) Diameter of roller = 84 cm

Radius of roller = 84/2 = 42 cm

Length (Height) = 120 cm

Curved Surface Area = 2πrh = 2×22/7×42×120 = 31680 cm² .

So , area covered in 1 revolution = 31680 cm²

Area covered in 500 revolutions = 31680×500 = 15840000 cm² .

8) Inner diameter = 3.5 m

Inner radius = 3.5/2 m

Height (Depth) = 10 m

(I) Inner Curved Surface Area = 2πrh = 2×22/7×3.5/2×10 = 110m².


(ii) Cost of plastering 1 m² = ₹ 40

Cost of plastering 110 m² = ₹40×110 = ₹4400