1. Weight variation with distance from Earth's centre

   The gravitational force, and hence the weight, varies inversely as the square of the distance from the centre of the Earth:

   $W \propto \frac{1}{r^2}$

   On the surface of the Earth:

   $W_0 \propto \frac{1}{R^2}$

2. Given distance from the centre

   The object is at a distance

   $r = \frac{5R}{4}$

   from the centre of the Earth.

   Therefore its new weight is:

   $W = W_0\left(\frac{R}{r}\right)^2$

   Substituting $r = \frac{5R}{4}$:

   $W = W_0\left(\frac{R}{5R/4}\right)^2$

   $W = W_0\left(\frac{4}{5}\right)^2$

   $W = W_0\cdot \frac{16}{25}$

3. Decrease in weight

   The decrease is:

   $\Delta W = W_0 - W = W_0 - \frac{16}{25}W_0$

   $\Delta W = \frac{9}{25}W_0$

4. Percentage decrease

   $\text{Percentage decrease} = \frac{\Delta W}{W_0}\times 100$

   $= \frac{9}{25}\times 100 = 36\%$

5. Option check

   • A: $36\%$ ✅
   • B: $50\%$ ❌
   • C: $64\%$ ❌
   • D: $25\%$ ❌

Therefore, the correct answer is A: $36\%$.