Question

A small steel ball falls through a syrup at a constant speed of $10cm{s}^{-1}$. If the steel ball is pulled upwards with a force equal to twice its effective weight, neglecting buoyant force, how fast will it move upwards?

• A. $10cm{s}^{-1}$
• B. $20cm{s}^{-1}$
• C. $5cm{s}^{-1}$
• D. $zerocm{s}^{-1}$

Answer 1

The correct option is A $10cm{s}^{-1}$

As the ball falls under it's own weight with a constant speed, the syrup is able to generate enough drag to balance the ball's weight and restrict the ball at 10cm/s
Hence, when the ball is pulled upwards with twice the weight, it's weight gets cancelled out and the remaining weight-worth of force is cancelled out by drag of syrup equivalent to velocity of 10cm/s