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Resistive force proportional to object velocity
At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as
`R = -bv`
Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity.
Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here.
Thus `mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v`
Solving the equation
`v = (mg)/(b) (1- e^(-bt//m))`
where e=2.71 is the base of the natural logarithm
The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed `v_(1)` and then on it continues to move with zero acceleration
`mg - b_(T) =0`
`rArr m_(T) = (mg)/(b)`
Hence `v = v_(T) (1-e^((vt)/(m)))`
In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table.
`{:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):}`
If buoyant force were also taken into account then value of terminal speed would have

Options

A. decreased
B. increased
C. remained same
D. increased or decreased depending upon density of liquid.

Answer

A

Explanation

By taking into account buoyant force, effective constant force in downward direction would decrease. Initially it was mg then it becomes `(mg - F_(b))`
Thus `g_("eff")` will have value less than g so
`V_(T)(=(mg_("eff"))/(b))` will increase.