S. Bitter Problem of day-to-day life can be solved by sweet words. Which lesson teaches us this message. (a) In astrologer's day (b) The Cop and the anthem (c) Into the wild (d) On saving please
Answers
Answer:
Answer:-
Internal \: diameter \: of \: the \: pipe = 2m.Internaldiameterofthepipe=2m.
So, its \: radius = 1cm = \frac{1}{100} mSo,itsradius=1cm=
100
1
m
\begin{gathered}Water \: that \: flows \: out \: through \: the \\ pipe \: in \: 6ms {}^{ - 1} \\ so \: volume \: of \: water \: that \: flows \: out \: \\ through \: the \: pipe \: in \: \\ 1sec \: = \pi \times \frac{1 {}^{2} }{100} \times 6m {}^{3} \end{gathered}
Waterthatflowsoutthroughthe
pipein6ms
−1
sovolumeofwaterthatflowsout
throughthepipein
1sec=π×
100
1
2
×6m
3
\begin{gathered}∴ In \: 30 \: minutes,volume \: of \: water \: \\ flow = \pi \frac{1}{100 \times 100} \times 6 \times 30 \times 60m {}^{3} \end{gathered}
∴In30minutes,volumeofwater
flow=π
100×100
1
×6×30×60m
3
This must be equal to the volume of water that rises in the cylindrical tank after 30 minutes and height up to which it rises say h.
Radius \: of \: tank = 60cm = \frac{60}{100} mRadiusoftank=60cm=
100
60
m
Volume \: = \pi( \frac{60}{100} ) {}^{2} hVolume=π(
100
60
)
2
h
\begin{gathered}=\pi( \frac{60}{100} ) {}^{2} h = \\ \pi\times \frac{1}{100 \times 100} \times 6\times 30\times 60\end{gathered}
=π(
100
60
)
2
h=
π×
100×100
1
×6×30×60
= > \frac{60 \times 60}{100 \times 100} h = \frac{60 \times 30 \times 60}{100 \times 100}=>
100×100
60×60
h=
100×100
60×30×60
= > h = \frac{3 \times 36}{36} = 3m=>h=
36
3×36
=3m
\fbox\pink{So,required height will be 3m}
So,required height will be 3m
Explanation:
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