A) \[300gm\]
B) \[\frac{1525}{4}gm\]
C) \[\frac{1555}{4}gm\]
D) \[\frac{3125}{8}gm\]
Correct Answer: D
Solution :
| Let the amount of salt present in tank is ?m? gm |
| \[\frac{dm}{dt}=10-{{\left( \frac{m}{100+t} \right)}^{4}}\]\[\Rightarrow \frac{dm}{dt}+\frac{4m}{100+t}=10\] \[\Rightarrow {{\left( 100+t \right)}^{4}}m=\frac{10{{(100+t)}^{5}}}{5}+c\] \[\Rightarrow m{{\left( 100+t \right)}^{4}}=2{{\left( 100+t \right)}^{5}}+c\]At \[t=0,{{m}_{0}}=50gm\] |
| \[50{{\left( 100 \right)}^{4}}=2{{\left( 100 \right)}^{5}}+c\] |
| \[c=-150\times {{100}^{4}}\] |
| \[\therefore {{m}_{t}}=\frac{2{{\left( 100+t \right)}^{5}}}{{{\left( 100+t \right)}^{4}}}-\frac{150\times {{100}^{4}}}{{{\left( 100+t \right)}^{4}}}\] |
| \[\therefore {{m}_{100}}=2\left( 100+100 \right)-\frac{150\times {{100}^{4}}}{{{\left( 200 \right)}^{4}}}\] |
| \[{{m}_{100}}=400-\frac{150}{16}\] |
| \[{{m}_{100}}=400-\frac{75}{8}=\frac{3200-75}{8}=\frac{3125}{8}\] |
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