a tank has a volume of 2000m³ it is filled with water in half an hour from a well 50 meters below the tank level calculate the power of the engine used to pump the water if 40% of the power is wasted what would be the actual power density of water is 1000kg m-³
Answers
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given that,
Rs. 25,000 invested for 2 years at compound interest, if the rates for the successive years be 4 and 5 per cent per year.
So, we have
Principal, P = Rs 25000
Rate of interest, r = 4 % per annum compounded annually.
Time, n = 1 year
Rate of interest, R = 5 % per annum compounded annually
Time, m = 1 year
We know,
Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years and R % per annum compounded annually for next m years is given by
\begin{gathered}\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} {\bigg[1 + \dfrac{R}{100} \bigg]}^{m} \: }} \\ \end{gathered}
Amount=P[1+
100
r
]
n
[1+
100
R
]
m
So, on substituting the values, we get
\begin{gathered}\rm \: Amount = 25000 {\bigg[1 + \dfrac{4}{100} \bigg]}^{1} {\bigg[1 + \dfrac{5}{100} \bigg]}^{1} \\ \end{gathered}
Amount=25000[1+
100
4
]
1
[1+
100
5
]
1
\begin{gathered}\rm \: Amount = 25000 {\bigg[1 + \dfrac{1}{25} \bigg]} {\bigg[1 + \dfrac{1}{20} \bigg]} \\ \end{gathered}
Amount=25000[1+
25
1
][1+
20
1
]
\begin{gathered}\rm \: Amount = 25000 {\bigg[ \dfrac{25 + 1}{25} \bigg]} {\bigg[ \dfrac{20 + 1}{20} \bigg]} \\ \end{gathered}
Amount=25000[
25
25+1
][
20
20+1
]
\begin{gathered}\rm \: Amount = 25000 {\bigg[ \dfrac{26}{25} \bigg]} {\bigg[ \dfrac{21}{20} \bigg]} \\ \end{gathered}
Amount=25000[
25
26
][
20
21
]
\begin{gathered}\rm\implies \:Amount = Rs \: 27300 \\ \end{gathered}
⟹Amount=Rs27300
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Additional Information :-
1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: }} \\ \end{gathered}
Amount=P[1+
100
r
]
n
2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: }} \\ \end{gathered}
Amount=P[1+
200
r
]
2n
3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: }} \\ \end{gathered}
Amount=P[1+
400
r
]
4n
4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: }} \\ \end{gathered}
Amount=P[1+
1200
r
]
12n
Answer:
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Explanation:
Here, Volume of water =30 m
3
t=15 min=15×60 s=900 s;Height,h=40 m
Efficiency, η=30%
Density of water =10
3
kg m
−3
∴ Mass of water pumped =Volume×Density
=(30 m
3
)(10
3
kg m
−3
)=3×10
4
kg
P
output
=
t
W
=
t
mgh
=
900 s
(3×10
4
kg)(10 ms
−2
)(40 m)
=
3
4
×10
4
W
Efficiency, η=
P
input
P
output
P
input
=
η
P
output
=
3×
100
30
4×10
4
=
9
4
×10
5
=44.4×10
3
W=44.4 kW.