An object of mass 1 kg is launched with an initial speed of vo into a large tank
filled with a viscous liquid. The liquid exerts a resistive force (drag) of the form
D = $\alpha$v on any object that is moving inside it, where v is the instantaneous speed
of the object and $\alpha$ = 1 kg/s. If the effect of gravity is ignored, the time taken by
the object to slow down to the speed vo/2 is ________ s (rounded off to 2 decimal
places).
Assume that the tank is sufficiently large for the above deceleration to happen
inside the tank.
Official Answer Key:
0.69
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.50)**
**1. Intuitive Physical Setup & Given Parameters**:
- Mass of the object: $m = 1\text{ kg}$
- Initial launch speed: $v(0) = v_0$
- Resistive fluid drag force: $D = \alpha v$, where $\alpha = 1\text{ kg/s}$
- Gravity is neglected.
- Target speed: $v(t) = \frac{v_0}{2}$
We need to determine the time $t$ taken to decelerate to half the initial speed.
**2. Step-by-Step Mathematical Derivation**:
1. **Equation of Motion**:
By Newton's second law:
m \frac{dv}{dt} = -D = -\alpha v
2. **Separation of Variables**:
\frac{dv}{v} = -\frac{\alpha}{m} dt
3. **Integrate from $t = 0$ ($v = v_0$) to $t$ ($v = v_0/2$)**:
\int_{v_0}^{v_0/2} \frac{dv}{v} = -\frac{\alpha}{m} \int_0^t dt
\ln\left(\frac{v_0/2}{v_0}\right) = -\frac{\alpha}{m} t
\ln\left(\frac{1}{2}\right) = -\frac{\alpha}{m} t \implies -\ln(2) = -\frac{\alpha}{m} t
t = \frac{m}{\alpha} \ln(2)
4. **Substitute Numerical Values**:
- $m = 1\text{ kg}$
- $\alpha = 1\text{ kg/s} $t = \frac{1}{1} \ln(2) = \ln(2) \approx 0.69315\text{ s}
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $\ln(2) \approx 0.693\text{ s}$.
- **Rounding Requirement**: Two decimal places $\implies \mathbf{0.69}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.69** (Acceptable Range: **0.68 to 0.70**)
- **Exam Takeaway**: Linear drag force $F = -\alpha v$ produces pure exponential velocity decay $v(t) = v_0 e^{-(\alpha/m)t}$. The half-life time is always $t_{1/2} = \frac{m}{\alpha}\ln 2$.