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5.Work, Energy, Power and Collision
easy
If a long spring is stretched by $0.02\, m$, its potential energy is $U$. If the spring is stretched by $0.1\, m$ then its potential energy will be
A
$\frac{U}{5}$
B
$U$
C
$5U$
D
$25U$
(AIPMT-2003)
Solution
(d)$U \propto {x^2}$
==> $\frac{{{U_2}}}{{{U_1}}} = {\left( {\frac{{{x_2}}}{{{x_1}}}} \right)^2} = {\left( {\frac{{0.1}}{{0.02}}} \right)^2} = 25$
${U_2} = 25U$
Standard 11
Physics
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