1. Check Assertion A

Pressure has dimension: $[P] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$

So, $[Pt] = [P][t] = \left(ML^{-1}T^{-2}\right)(T) = ML^{-1}T^{-1}$

Now, coefficient of viscosity $\eta$ has dimension:

Using Newton’s law of viscosity, $\text{viscous force} = \eta \times \text{area} \times \text{velocity gradient}$

Hence, $\eta = \frac{\text{Force}}{\text{Area} \times \text{velocity gradient}}$

Velocity gradient $= \dfrac{\text{velocity}}{\text{distance}}$, so $[\text{velocity gradient}] = \frac{LT^{-1}}{L} = T^{-1}$

Therefore, $[\eta] = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} = \frac{MLT^{-2}}{L^2T^{-1}} = ML^{-1}T^{-1}$

Thus, $[Pt] = [\eta]$

So Assertion A is true.

2. Check Reason R

Reason says: $\text{Coefficient of viscosity} = \frac{\text{Force}}{\text{Velocity gradient}}$

But the correct relation is: $\eta = \frac{\text{Force}}{\text{Area} \times \text{velocity gradient}}$

So the given reason is incorrect because the factor of area is missing.

Hence Reason R is false.

3. Conclusion

• Assertion A: True
• Reason R: False

Therefore, the correct option is: $\boxed{\text{C}}$

4. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, the answer agrees with the stored correct answer.