Physics · Mechanical Properties Of Fluids · NEET
It is a curve, not a straight slanted line. A straight slanted line would mean acceleration stays the same (like free fall). But in a viscous liquid the drag force grows as speed grows, so acceleration keeps dropping. The curve starts steep, then bends and flattens into a horizontal line at the terminal velocity.
The graph flattens because the ball stops speeding up. At the start, weight is bigger than drag, so it accelerates. As speed rises, the viscous drag (which depends on speed) rises too. When drag plus buoyancy equal the weight, the net force is zero, so acceleration is zero and speed stays constant. A constant speed on a v-vs-t graph is a horizontal line.
No. Acceleration zero means the speed is not changing, not that speed is zero. The ball keeps moving down at a steady, maximum speed. This is a common trap: zero acceleration and zero velocity are different things. On the graph the line is flat but sits at a positive height above the time axis.
In theory it approaches terminal velocity like an asymptote and gets extremely close but never mathematically touches it. For NEET you treat the flat part of the curve as the terminal velocity. The important point is the SHAPE: concave curve rising and then leveling off to a horizontal asymptote.
In near-free fall (ignoring air drag) the speed increases steadily, so the v-vs-t graph is a straight slanted line through the origin with slope g. In a viscous liquid, drag caps the speed, so the graph is a bent curve that flattens. The key difference: free fall = straight line, viscous fall = curve that saturates.
A spherical ball is dropped in a long column of a highly viscous liquid. Which curve in the graph shown represents the speed of the ball (v) as a function of time (t)?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Terminal velocity is the maximum, constant speed reached by a body falling through a fluid, when the downward weight is exactly balanced by upward drag plus buoyancy, so the net force and acceleration become zero.
It is a concave curve: it starts at the origin, rises steeply, then bends and flattens into a horizontal straight line (a horizontal asymptote) at the terminal velocity value.
The slope equals the acceleration. As speed increases, viscous drag increases, so net force decreases, which means acceleration (the slope) decreases. Finally the slope becomes zero at terminal velocity.
Yes. Since acceleration starts large and falls to zero, the a-vs-t graph starts high and decays toward zero, while the v-vs-t graph starts at zero and rises to a constant. They are complementary.
Yes. Terminal velocity is proportional to r squared and to (density of ball minus density of liquid). A larger radius or a denser ball gives a higher flat line on the v-vs-t graph, but the SHAPE (rise then flatten) stays the same.
