if circular metal sheet is 0.65 CM thick and of 50 cm in diameter is melted and recast into cylindrical bar with 8 cm diameter then length of bar will be
Answers
area = pi × r*2
= 625pi square cm
volume = 0.65 × 625pi cubic cm
after recasting the volume will remain same
volume of cylinder will be area × height
area = pi × r*2
= 16 pi
so height = 0.65 × 625 /16
= 25.4 CMS
Answer:
Answer:
Length of the cylindrical bar is 25.39 cm. (approx.)
Step-by-step explanation:
Thick / Height (h) of the circular metal sheet is 0.65 cm. = \frac{65}{100}
100
65
Radious (r) of the sheet = \frac{Diameter}{2}
2
Diameter
= \frac{50}{2}
2
50
= 25 cm.
Therefore, volume of the circular metal sheet = \pi r^{2}hπr
2
h
= \pi 25^{2}\frac{65}{100}π25
2
100
65
cm^{3}cm
3
Let, Height of the cylindrical bar is (H)
Radious (R) of the base of cylindrical bar = \frac{Diameter}{2}
2
Diameter
= \frac{8}{2}
2
8
=4 cm
Therefore, Volume of the cylindrical bar = \pi R^{2} HπR
2
H = \pi 4^{2}Hπ4
2
H cm^{3}cm
3
Since, circular metal sheet is melted and recast into cylindrical bar
So,
volume of the circular metal sheet = Volume of the cylindrical bar
⇒ \pi 25^{2}\frac{65}{100}π25
2
100
65
= \pi 4^{2} Hπ4
2
H
⇒H= \frac{25^{2}\frac{65}{100} }{4^{2} }H=
4
2
25
2
100
65
=25.390625 cm = 25.39 cm (approx.)