A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to understand the motion of a small steel ball dropping into a viscous fluid like glycerine.
When the steel ball is dropped, it initially accelerates due to gravity. However, as it moves through the glycerine, it experiences an upward viscous drag force, which increases with velocity. Eventually, the ball reaches a constant velocity known as terminal velocity, where the net force on the ball becomes zero.
The forces acting on the ball are:
- Gravitational force \(mg\) (downward)
- Buoyant force \(F_B\) (upward, due to glycerine)
- Viscous drag force \(F_d\) (upward, increases with velocity)
As these forces balance, the ball reaches its terminal velocity \(v_t\).
The velocity-time graph for this scenario will show:
- Initially, an accelerating slope as the ball speeds up due to gravity.
- Then, it will transition to a horizontal line once terminal velocity is reached, indicating constant velocity.
The correct graph representation is an initial steep slope that gradually levels off to a constant value, which matches with the provided image:
Hence, the correct answer is the option represented by the image above, where the velocity initially increases and then becomes constant after reaching terminal velocity.
Approach Solution -2
The forces acting on the ball as it moves through glycerine include the gravitational force (\(mg\)) and the viscous drag force (\(F_v\)). The motion is described by:
\[ mg - F_v - F_b = ma \]
Where:
\[ F_b = \left( \frac{4}{3} \pi r^3 \right) g (\rho_L), \quad F_v = 6 \pi \eta r v \]
Equating forces and simplifying:
\[ \left( \frac{4}{3} \pi r^3 \right) g (\rho - \rho_L) = m \frac{dv}{dt} \]
Let:
\[ K_1 = \frac{4}{3} \pi r^3 (\rho - \rho_L) g, \quad K_2 = \frac{6 \pi \eta r}{m} \]
The differential equation becomes:
\[ \frac{dv}{dt} = K_1 - K_2 v \]
Integrating:
\[ \int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt \]
Solution yields:
\[ v = \frac{K_1}{K_2} \left( 1 - e^{-K_2 t} \right) \]
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