Q147 / 150 · MHT CET 2018 · Physics

The decrease in length of a metal bar of length $' L'$ and crosssectional area $' A'$ when compressed with a load $' F'$ along its length is ( $Y =$ Young's modulus of the material of metal bar)

A
$\frac{F L}{2 A Y}$
B
$\frac{F^{2} L}{2 A Y}$
Correct answer
C
$\frac{F L}{A Y}$
D
$\frac{F^{2} L^{2}}{2 A Y}$

Solution

\begin{array}{l} Y = \frac{F L}{A Δ L} \\ Δ L = \frac{F L}{A Y} \end{array}

Hence, the correct answer is option: 'C'.

Relevant Questions

The ratio of length of two wires of same material is and the ratio of their radii is . If they are stretched by the same force, the ratio of20 Oct, 2020 (M)Young’s Modulus and Breaking Stres
A steel ring of radius 'r' is to be fitted over a wooden disc of radius 'R' . The force required to expand the ring so that it fits over the20 Oct, 2020 (E)Young’s Modulus and Breaking Stres
A thick brass wire of length ' ' and density ' ' is suspended from rigid support. Due to its own weight, is the increase in length. Young's 19 Oct, 2020 (M)Young’s Modulus and Breaking Stres
A metal rod has length, cross-sectional area and Young's modulus as ' and respectively. If the elongation in the rod produced is ' then work19 Oct, 2020 (E)Young’s Modulus and Breaking Stres
Two rods of same material and volume having circular cross-section are subjected to tension T. Within the elastic limit, same force is appli16 Oct, 2020 (M)Young’s Modulus and Breaking Stres
Two wires and are stretched by the same load. The radius of wire is double the radius of wire . The stress on the wire as compared to the st16 Oct, 2020 (E)Young’s Modulus and Breaking Stres