A blacksmith have big iron ball of diameter
36cm. He first formed four identical
cylindrical rod out of this ball for some
construction work. When he finished his
work with the cylindrical rod, he again
mixed all the iron rod and formed some
identical small balls, each of radius 9cm.
If the diameter of each cylinder
formed by him is 12cm, then height
of each of them is-
(a) 54cm
(b) 36cm
(c) 45cm
(d) 72cm
The number of small balls formed by the
blacksmith?
(a) 6
(b) 10
(c) 8
(d) 16
What is the cost price of steel coating of a
cylindrical iron rod at the rate of Rs.3.5per
100cm2
?
(a) Rs.79.20
(b) Rs.89.20
(c) Rs.85.56
(d) Rs.71.28
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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