1. Use Stokes’ law for terminal velocity

For a small sphere of radius $R$ falling through a viscous liquid,

$6\pi \eta R v_t = \text{effective weight} = \frac{4}{3}\pi R^3 (\rho_s-\rho_l)g$

So,

$v_t = \frac{2R^2(\rho_s-\rho_l)g}{9\eta}$

Thus,

$v_t \propto R^2$

2. Check statement A

A says: Graph between terminal velocity $V$ and $R$ will be a parabola.

Since

$V = kR^2$

this is the equation of a parabola in the $V$-$R$ plane.

✅ A is correct.

3. Check statement B

B says: The terminal velocities of different diameter balls are constant for a given liquid.

This is false because terminal velocity depends on radius as

$v_t \propto R^2$

So different diameter balls will have different terminal velocities.

❌ B is incorrect.

4. Check statement C

C says: Measurement of terminal velocity is dependent on the temperature.

Yes, because viscosity $\eta$ changes with temperature, and

$v_t = \frac{2R^2(\rho_s-\rho_l)g}{9\eta}$

Hence terminal velocity depends on temperature through viscosity.

✅ C is correct.

5. Check statement D

D says: This experiment can be utilized to assess the density of a given liquid.

From

$v_t = \frac{2R^2(\rho_s-\rho_l)g}{9\eta}$

if $\eta$, $R$, $\rho_s$, and $v_t$ are known, one can determine $\rho_l$.

So the setup can indeed be used to estimate density of the liquid.

✅ D is correct.

6. Check statement E

E says: If balls are dropped with some initial speed, the value of $\eta$ will change.

This is false. Viscosity is a property of the liquid and does not change because the ball is given an initial speed. The ball may take some time to reach terminal velocity, but the measured viscosity of the liquid remains the same if terminal velocity is correctly used.

❌ E is incorrect.

7. Correct combination

Correct statements are:

$A,\ C,\ D$

So the correct option is:

$\boxed{\text{C: A, C and D only}}$

8. Comparison with stored correct answer

Stored correct answer = C

Derived answer = C

They match.