Sankalp NEET Full Test-2 Question-26 Solution

Question 26: 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 :

(1) 8 U

(2) 16 U

(3) 2 U

(4) 4 U

Answer: Option (2)

Explanation:

The potential energy stored in a stretched spring is given by

U=\frac{1}{2}kx^{2}

where k is the spring constant and x is the extension of the spring.

Thus, the potential energy of a spring is proportional to the square of the extension.

U \propto x^{2}

Let the initial extension be

x_{1}=2\,\mathrm{cm}

and the corresponding potential energy be

U_{1}=U

The new extension is

x_{2}=8\,\mathrm{cm}

The ratio of potential energies is

\frac{U_{2}}{U_{1}}=\left(\frac{x_{2}}{x_{1}}\right)^{2}

Substituting values,

\frac{U_{2}}{U}=\left(\frac{8}{2}\right)^{2}=4^{2}=16 U_{2}=16U

Hence, the potential energy stored in the spring when stretched by 8 cm is 16U.

Therefore, the correct answer is Option (2).