the given molten mass of metal is to be cast into the shape of uniform radius the M.I. of the castled mass about the symmetric axis will be maximum in case of the :
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From given,
The volume of half-cylinder is,
V=
2
1
πr
2
h
⇒h=
πr
2
2V
total surface area = area of rectangular base + area of 2 semi circular ends + area of curved surface
S=h×2r+2×
2
1
πr
2
+
2
1
×2πrh
∴S=2rh+πr
2
+πrh
upon substituting the value of "h", we get,
S=2r(
πr
2
2V
)+πr
2
+πr(
πr
2
2V
)
differentiating the above eqaution and equating it to zero, we get,
dr
dS
=0
⇒2πr−
r
2
1
(
π
4V
+2V)=0
upon solving the above equation, we get,
r
3
=
π
2
2V
+
π
V
again differentiating the equation, we get,
dr
2
d
2
S
=2π+
r
3
1
(
π
4V
+2V)>0 minima
resubstituting the value of V, we get, r as,
r=
2π
h
(2+π)
h
2r
=
π+2
π
Hence the ratio of length of cylinder to the diameter of its semi circular ends.
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