Physics · 2013
JEE Main · 9 April 2013, Shift 1 · Q3
Two springs of force constants 300 N/m (Spring A) and 400 N/m (Spring B) are joined together in series. The combination is compressed by 8.75 cm. The…
Two springs of force constants $\displaystyle 300$ N/m (Spring A) and $\displaystyle 400$ N/m (Spring B) are joined together in series. The combination is compressed by $\displaystyle 8.75$ cm. The ratio of energy stored in $\displaystyle A$ and $\displaystyle B$ is $\displaystyle \frac{E_A}{E_B}$. Then $\displaystyle \mathrm{E}_{\mathrm{A}} / \mathrm{E}_{\mathrm{B}}$ is equal to :
Official answer
(1)
$\displaystyle \frac{4}{3}$
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JEE Main 2013 Physics question, with the answer from NTA’s final answer key. Where our answers come from.