One end of the steel rod is clamped to the roof and the other end is attached to mass of 1000 ~kg as shown in the figure. The length of the rod is 50 ~cm and its cross-sectional area is 1000 ~mm ^2 . The change in the length of the rod due to the weight of the mass is (Young's modulus of steel =2 10¹¹ Nm ⁻² and acceleration due to gravity =10 ~ms ⁻² )

Step-by-step solution

As, young's modulus of a rod under tension, Y= F / A l / l = F l A l We have, change in length l= F . l A . Y Here, force on rod, F=1000 10=10,000 ~N Length of rod, l=50 ~cm =50 10⁻² ~m Area of rod, A=1000 ~mm ^2=1000 10⁻⁶ ~m ^2 Substituting these values in eq. (i), we get hange of length of rod, aligned l & = 10000 50 10⁻² 1000 10⁻⁶ 2 10¹¹ & =25 10⁻⁶ ~m & =0.025 ~mm aligned