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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > Two opposite forces F1 = 120 N and F2 = 80 N act on an elastic plank of modulus of elasticity Y = 2 x 10^11 N/m^2 and length l=1m placed over a smooth horizontal surface. The cross-sectional area of the plank is S = 0.5 m^2. The change in length of the plank is n ×10^−11 m. Find the value of n .

Two opposite forces F1 = 120 N and F2 = 80 N act on an elastic plank of modulus of elasticity Y = 2 x 10^11 N/m^2 and length l=1m placed over a smooth horizontal surface. The cross-sectional area of the plank is S = 0.5 m^2. The change in length of the plank is n ×10^−11 m. Find the value of n .

A body of mass M is attached to lower end of a metal wire mark out the correct statements. whose upper end is fixed. The elongation of wire is l. In this situation