Let's analyze each statement:
A. The terminal velocity $ V $ is given by:
$$ V = \frac{2}{9} \frac{r^2 (\rho_s - \rho_l)g}{\eta} $$
where $ r $ is the radius of the ball, $ \rho_s $ is the density of the ball, $ \rho_l $ is the density of the liquid, $ g $ is the acceleration due to gravity, and $ \eta $ is the viscosity of the liquid.
Since $ V \propto r^2 $, the graph between terminal velocity $ V $ and $ r $ (or $ R $) will be a parabola.
So, statement A is correct.
B. The terminal velocity depends on the radius (or diameter) of the ball. Different diameter balls will have different terminal velocities for a given liquid.
So, statement B is incorrect.
C. Measurement of terminal velocity is dependent on the temperature. The viscosity of the liquid is temperature-dependent. As temperature increases, the viscosity of most liquids decreases, affecting the terminal velocity.
So, statement C is correct.
D. This experiment can be utilized to assess the density of a given liquid. By measuring the terminal velocity of a ball with known density and radius, and knowing the viscosity, we can solve for the density of the liquid in the equation for terminal velocity.
So, statement D is correct.
E. If balls are dropped with some initial speed, the value of $ \eta $ will not change. The viscosity is a property of the liquid and does not depend on the initial speed of the ball. Although the ball takes a longer/shorter amount of time depending on whether it is thrown or released, this does not affect the viscosity.
So, statement E is incorrect.
Conclusion:
The correct statements are A, C, and D.
Final Answer:
The final answer is $ (2)\ \text{A, C and D only} $.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)