Question

Two springs A and B (kA = 2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is
(a) E/2
(b) 2E
(c) E
(d) E/4

Answer 1

Answer ⇒ Option (b). 2E

Explanation/Reason ⇒ 
As per as the given conditions, KA = 2 KB. 
Therefore, For the constant amount of the force on both the springs,
  $m_{B} = 2 m_{A}$

Thus, energy stored in the springs A = E  ------(i) 
Also, E = F × $m_{A}$ 

where, F is the constant force. 
Energy stored in the B = F × $m_{B}$
 Energy stored in B = 2E  ------(ii)

From the Equations (i) and (ii), Energy stored in B is twice that of A i.e., 2E.

Hence, option (b) is correct. 

Hope it helps.

Answer 2

Answer:

the correct ans is B

Explanation:

hope it helps