The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be:

16U 2U 4U 8U

16U

$ U = \frac{1}{2} Kx^2$ $\text{ At x = 2 cm, } U = \frac{1}{2} K\times 2^2 = 2K$ $ \text{ At x = 8cm, } U'= \frac{1}{2} K 8^2 = 32 K = 16U$