Question:
At \( t = 0 \), a particle of mass \( m \) having velocity \( v_0 \) starts moving through a liquid kept in a horizontal tube and experiences a drag force \( F_d = - k \frac{dv}{dt} \). It covers a distance \( L \) before coming to rest. If the times taken to cover the distances \( L/2 \) and \( L \) are \( t_2 \) and \( t_4 \), respectively, then the ratio \( t_2/t_4 \) (ignoring gravity) is:
At \( t = 0 \), a particle of mass \( m \) having velocity \( v_0 \) starts moving through a liquid kept in a horizontal tube and experiences a drag force \( F_d = - k \frac{dv}{dt} \). It covers a distance \( L \) before coming to rest. If the times taken to cover the distances \( L/2 \) and \( L \) are \( t_2 \) and \( t_4 \), respectively, then the ratio \( t_2/t_4 \) (ignoring gravity) is:
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When dealing with motion under drag force, the relationship between time and distance is logarithmic. Use this relationship to calculate the times for different distances.
Updated On: Nov 18, 2025
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Solution and Explanation
Step 1: Analyze the motion under drag force.
The drag force is proportional to the velocity, which means the motion follows a logarithmic decay. Using the equation for motion under a drag force: \[ F_d = - k \frac{dv}{dt} \] Step 2: Calculate the times taken to cover \( L/2 \) and \( L \).
The time to cover half the distance \( L/2 \) is related to the time to cover the full distance \( L \) by the kinematic equation. From this, we can derive the ratio \( t_2/t_4 \).
Step 3: Conclusion.
The ratio \( t_2/t_4 \) is 2.
The drag force is proportional to the velocity, which means the motion follows a logarithmic decay. Using the equation for motion under a drag force: \[ F_d = - k \frac{dv}{dt} \] Step 2: Calculate the times taken to cover \( L/2 \) and \( L \).
The time to cover half the distance \( L/2 \) is related to the time to cover the full distance \( L \) by the kinematic equation. From this, we can derive the ratio \( t_2/t_4 \).
Step 3: Conclusion.
The ratio \( t_2/t_4 \) is 2.
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