What is the cost price of steel coating of a
cylindrical iron rod at the rate of Rs.3.5per
100cm2
? If r = 18cm and h=54cm
Pls answer fast iwill mark as brainliest
Answers
Solution :-
Let us assume that, height of cylindrical rod is h cm.
so,
→ Volume of iron ball = 4 * volume of each cylindrical rod .
→ (4/3) * π * (radius)³ = 4 * π * (radius)² * height .
→ (4/3) * π * (36/2)³ = 4 * π * (12/2)² * h
4π will be cancel,
→ 18 * 18 * 18 = 3 * 6 * 6 * h
→ 18 * 18 = 6h
→ h = 54 cm. (Ans.a)
then,
→ CSA of 1 cylindrical rod = 2 * π * r * h = 2 * 3.14 * 6 * 54 = 2034.72 cm².
therefore,
→ 100 cm² cost = Rs.3.5
→ 1 cm² cost = (3.5/100)
→ 2034.72 cm² cost = (3.5/100) * 2034.72 = Rs.71.21 (Ans.)
Let us assume that, n number of small ball formed by blacksmith.
so,
→ Volume of iron ball = n * volume of each small ball .
→ (4/3) * π * (R)³ = n * (4/3) * π (r)³
(4/3) and π will be cancel from both sides,
→ (R)³ = n * (r)³
→ (36/2)³ = n * (9)³
→ (18)³ = n * (9)³
→ 18 * 18 * 18 = n * 9 * 9 * 9
→ n = 2 * 2 * 2
→ n = 8 (Ans.c)
Hence, the number of small ball formed by blacksmith are 8 .
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