Q109 / 200 · NEET 2023 (All India) · Physics

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:

Correct answer
A
$16 U$
B
$2 U$
C
$4 U$
D
$8 U$

Solution

Potential energy stored in spring is given by,
$U = \frac{1}{2} K x^{2} .$
First case:
$U = \frac{1}{2} K {(2 \times 10^{- 2})}^{2} . . . E q (1)$
Second case:
$U^{'} = \frac{1}{2} K {(8 \times 10^{- 2})}^{2} . . . E q (2)$
From both equations, we get
$\frac{U}{U^{'}} = \frac{1}{16} \Rightarrow U^{'} = 16 U$

Hence, the correct answer is option: 'A'.

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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:

Solution

Relevant Questions

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: