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Application of Derivatives: Water Flowing Container Rate Proportional Square Amount Wate

JEE Maths question with a full step-by-step solution.

Question
Water is flowing into a container at a rate proportional to the square of the amount of water present inside the container. If at t=0t = 0 the volume of water inside the tank is V0V_0, and it takes 66 hours for the volume to become 2V02V_0, then the time (in hours) taken for the volume to become 4V04V_0 is equal to
Solution
Answer: 9
Step 1: Turn the sentence into a differential equation. "Rate of flow is proportional to the square of the amount present" means
dVdt=kV2,\frac{dV}{dt} = kV^{2} ,
 for some constant k>0k > 0. Step 2: Separate the variables.
dVV2=k dt.\frac{dV}{V^{2}} = k\,dt .
 Step 3: Integrate both sides.
∫V−2 dV=∫k dt  ⟹  −1V=kt+C.\int V^{-2}\,dV = \int k\,dt \implies -\frac{1}{V} = kt + C .
 Step 4: Fix CC from the initial condition V=V0V = V_0 at t=0t = 0.
−1V0=C.-\frac{1}{V_0} = C .
 Step 5: Substitute back and tidy into a convenient form.
−1V=kt−1V0  ⟹  1V=1V0−kt....(i)-\frac{1}{V} = kt - \frac{1}{V_0} \implies \frac{1}{V} = \frac{1}{V_0} - kt . \qquad \text{...(i)}
 Step 6: Find kk using "V=2V0V = 2V_0 when t=6t = 6".
12V0=1V0−6k  ⟹  6k=1V0−12V0=12V0,\frac{1}{2V_0} = \frac{1}{V_0} - 6k \implies 6k = \frac{1}{V_0} - \frac{1}{2V_0} = \frac{1}{2V_0} ,
k=112V0.k = \frac{1}{12V_0} .
 Step 7: Now set V=4V0V = 4V_0 in (i) and solve for tt.
14V0=1V0−kt  ⟹  kt=1V0−14V0=34V0.\frac{1}{4V_0} = \frac{1}{V_0} - kt \implies kt = \frac{1}{V_0} - \frac{1}{4V_0} = \frac{3}{4V_0} .
 Step 8: Substitute the value of kk.
t=34V0⋅1k=34V0×12V0=9.t = \frac{3}{4V_0} \cdot \frac{1}{k} = \frac{3}{4V_0} \times 12V_0 = 9 .
 Step 9: Sanity check against the given data. Putting t=6t = 6 back gives 1V=1V0−612V0=12V0\dfrac{1}{V} = \dfrac{1}{V_0} - \dfrac{6}{12V_0} = \dfrac{1}{2V_0}, so V=2V0V = 2V_0 . Answer: 99.
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