AP EAMCET20225 Jul 2022Morning ShiftPhysicsMechanical Properties of SolidsActual
Two wires A and B of same cross-section are connected end to end. When same tension is created in both wires, the elognation in B wire is twice the elongation in A wire. If L_A and L_B are the initial lengths of the wires A and B respectively, then (Young's modulus of material of wire A=2 10¹¹ Nm ⁻² and Young's modulus of material of wire B=1.1 10¹¹ Nm ⁻² ).
Options
- AL_A L_B = 10 11
- BL_A L_B = 4 5
- CL_A L_B = 9 11
- DL_A L_B = 3 7
Correct answer
A. L_A L_B = 10 11
Step-by-step solution
The given situation is shown below. Here, elongation in wire B is twice of elongation in wire A on the application of same tension T . i.e L_B=2 L_A Young's modulus of wire A and B are given as Y_A= T L_A A L_A and Y_B= T L_B A L_B Y_A Y_B = L_A L_B L_B L_A 2 10¹¹ 1.1 10¹¹ = L_A L_B 2 L_A L_A 2 11 = L_A L_B 2 L_A L_B = 1 11 = 10 11