W varies inversely as the square of t. If W = 12 when t = 2. Find t when W = 27. Options : 27t = 24 27t2=48 108 = 12t2 12t=54

The problem states that "W varies inversely as the square of t". This can be written as W = k/(t^2), where k is the constant of variation.

Step 1: Find the constant of variation (k) We know that W = 12 when t = 2. Substituting these values into the equation gives us 12 = k/(2^2) or 12 = k/4. Solving for k gives us k = 12 * 4 = 48.

Step 2: Find t when W = 27 Now we know that W = 48/(t^2). We can substitute W = 27 into the equation and solve for t. This gives us 27 = 48/(t^2).

Step 3: Solve for t To solve for t, we first multiply both sides of the equation by t^2 to get 27t^2 = 48. Then, we divide both sides by 27 to isolate t^2, giving us t^2 = 48/27.

Finally, we take the square root of both sides to solve for t. This gives us t = sqrt(48/27) = 1.33 (rounded to two decimal places).

So, the correct option is none of the given options.