Solved papers for JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)
done JEE Main Online Paper (Held on 9 April 2013) Total Questions - 1
question_answer1) Two springs of force constants 300 N/m (Spring A) and 400 N/m (Spring B) are joined together in compressed by 8.75 cm. The ratio of energy stored in A and B is\[\frac{{{E}_{A}}}{{{E}_{B}}}\]. The \[{}^{{{E}_{A}}}/{}_{{{E}_{B}}}\]is equal to:
JEE Main Online Paper (Held On 09 April 2013)