91Ó°ÊÓ

First, we need to find the area of each plate to determine the ratio of their areas. The area of a circle is given by the formula \(A = \pi r^2\), where A is the area and r is the radius of the circle. For the small plate with radius of 1m: \(A_1 = \pi (1)^2 = \pi\) For the large plate with radius of 2m: \(A_2 = \pi (2)^2 = 4\pi\) Step 2: Find the ratio of areas of the two plates

Now, let's find the ratio between the areas of the small plate to the large plate: \(\frac{A_1}{A_2} = \frac{\pi}{4\pi}\) Simplifying this expression gives: \(\frac{A_1}{A_2} = \frac{1}{4}\) Step 3: Determine the ratio of thrusts on the two plates

Since both plates are submerged at the same depth and in the same liquid, the pressure \(P\) acting on both plates will be the same. The thrust (force) on each plate can be calculated as the product of the pressure and the area of the plate: Thrust on small plate = \(P \times A_1\) Thrust on large plate = \(P \times A_2\) Now, we'll find the ratio of thrusts on the small plate to the large plate: \(\frac{Thrust_{small}}{Thrust_{large}} = \frac{P \times A_1}{P \times A_2} = \frac{A_1}{A_2}\) From Step 2, we already know that the ratio \(\frac{A_1}{A_2} = \frac{1}{4}\). Therefore, the ratio of thrusts on the small plate to the large plate is also \(\frac{1}{4}\). Comparing this to the given answer choices, we can see that the correct answer is: (C) 1:4