When a load of 80 ~N is suspended from a string, its length is 101 ~mm . If a load of 100 ~N is suspended, its length is 102 ~mm . If a load of 160 ~N is suspended from it, then length of the string is (Assume the area of crosssection unchanged)

Step-by-step solution

aligned & Natural length (l)= l₁ T₂-l₂ T₁ T₂-T₁ & = 101 100-102 80 (100-80) aligned aligned & = 101 100-102 80 20 & = 101 100 20 - 102 80 20 & =505-408=97 ~cm & Again T₁ T₂ = 4 l₃-97 & 80 160 = 4 l₃-97 or, l₃-97=8 & l₃=105 ~mm or 10.5 ~cm aligned