Suppose now you drop the beaker in the previous question from some height. Will the
range of liquid from orifice B still be more? Justify your answer.
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Answer:
Range =2
(H−h)h
So, R
max
=H
According to question,
R=
2
H
=2
(H−h)h
⇒
4
H
=
(H−h)h
⇒
16
H
2
=(H−h)h
⇒
16
H
2
−Hh+h
2
=0
⇒h=
2
1±
1−4×1×
16
1
H=
2
1±
2
3
H=(
2
1
−
4
3
)H=0.067H
and x+2h=H
⇒x=H−2h=0.866H
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