Two wires of equal length and cross-sectional area are suspended as shown in figure. Their Young's modullare `Y_1` and`Y_2` respectively. The equivalent Young's modulil will bebr>

Options

A. `Y_1 + Y_2`
B. `(Y_1 + Y_2)/2`
C. `(Y_1 Y_2)/(Y_1 + Y_2)`
D. `sqrt(Y_1 Y_2)`

Answer

B

Explanation

Given that the wires have equal length and cross section We know that
`F = K DeltaX `
Where K is spring constant
Let `K_1 and K_2` be the spring constants of the two wires.
`therefore K = F/(DeltaL) = (YA)/L`
therefore equivalent spring constant `(K_(eq))`
` K_(eq) = K_1 + K_2`
`(Y_(eq) xx 2A)/(L) = (Y_1 A)/(L_1) + (Y_2 A)/(L_2)`
Here `L_1` and `L_2` are equal
`therefore Y_(eq) (2A)/L = A/L (Y_1 + Y_2)`
`implies Y_(eq) = (Y_1 + Y_2)/2`