Question

Page 8 (i) Assume that the storage tank is completely sealed and is to be filled with diesel from an opening at the top. Find the capacity, in m', of the tank, inclusive of the conical roof.​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: a \: storage \: tank \: constructed \: by \: a \: farmer \: is \: made \: up \: of \: two \: geometrical \: shapes \: namely \: cylinder(lower \: part) \: and \: cone(upper \: part) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (given)$

$so \: thus \: then \: radius \: of \: both \: cylindrical \: and \: conical \: shape \: would \: be \: equal \\ so \: let \: the \: radius \: of \: entire \: storage \: tank \: be \: r \: (inclusive \: of \: conical \: part \: and \: cylindrical) \\ here \: diameter(d) = 8.cm \\ thus \: radius(r) = 4.cm$

$moreover \\ for \: a \: cylindrical \: part \\ vertical \: height \: ie \: perpendicular \: height(h) = 3.5 \: cm \\ \\ for \: a \: conical \: part \: \\ height(H) = 1.2 \: cm$

$so \: thus \: combined \: volume(inclusive \: volume) \: of \: storage \: tank = volume \: of \: cylindrical \: part + volume \: of \: conical \: part$

$so \: we \: know \: that \\ volume \: of \: conical \: part = 1/3 \times \pi \: r.square \: H \\ volume \: of \: cylindrical \: part = \pi \: r.squre \: h$

$thus \: then \\ volume \: of \: storage \: tank = 1/3 \times \pi \: r.square \: H + \pi \: r.square \: h \\ = \pi \: r.square(1/3.H + h) \\ = 3.14 \times (4).square (1/3 \times 1.2 + 3.5) \\ = 3.14 \times 16(0.4 + 3.5) \\ = 50.24 \times 3.9 \\ = 195.936 \\ \\ so \: volume \: of \: storage \: tank = 195.936 \: cubic.cm$

$but \: as \: we \: know \: that \\ 1l = 1000 \: cubic.cm \\ so \: 195.936 \: cubic.cm = 195.936 \times 1000 \: l \\ = 195936 \: l$

$so \: thus \: capacity \: of \: given \: storage \: tank \: is \: about \: 195936 \: l$