One end of a steel rod of radius 10.0 {mm} and length 50.0 {cm} is clamped on a horizontal table. The other end of the rod is pulled with a force of magnitude 10.0 \times \pi {kN}. This force is uniform across the flat surface of the rod and is perpendicular to it. The change in the length of the rod due to this applied force is (Use Young's modulus =2.0 \times 10^{11} {N} \/ {m}^2 )

TS EAMCET 20-July-2022 Shift 1 Solved Paper

Section: Physics

Question : 93 of 160

Marks: +1, -0

One end of a steel rod of radius 10.0mm and length 50.0cm is clamped on a horizontal table. The other end of the rod is pulled with a force of magnitude 10.0×πkN. This force is uniform across the flat surface of the rod and is perpendicular to it. The change in the length of the rod due to this applied force is
(Use Young's modulus =2.0×1011N∕m2 )

0.25mm
0.75mm
0.50mm
1.0mm

Validate

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