Inverse square variation with parameter change – UGC NET Question

The condition “x varies inversely as the square of y” means x = k/y² for some constant k. Using x = 9 when y = 2, we get 9 = k/4, so k = 36. For y = 3, we have x = 36/3² = 36/9 = 4. Therefore, the new value of x corresponding to y = 3 is 4.

Option A:
Option A, 3, would correspond to x = 36/y² giving 3 = 36/y², so y² = 12, which does not match y = 3. Hence, 3 cannot be the value of x when y is 3 under this variation law.

Option B:
Option B is correct because it uses the constant k = 36 derived from the first pair (x, y) = (9, 2) and substitutes y = 3 appropriately. The computation 36/9 is straightforward and yields 4, which fully respects the inverse-square relationship between x and y.

Option C:
Option C, 6, could result from treating x as inversely proportional to y instead of y², leading to x = 18/3. However, that would violate the stated condition involving the square of y, and it does not satisfy x = 9 when y = 2 if we consistently use the same rule.

Option D:
Option D, 8, might arise from doubling 4 under the mistaken belief that x halves when y increases from 2 to 3, but inverse-square variation is more rapid than simple inverse variation. The correct calculation yields 4, not 8, so this option is not consistent with the given dependency.