The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line
Easy
Download our app for free and get startedPlay store
(a) $l = \frac{{FL}}{{AY}}$ $\Rightarrow$ $l \propto \frac{1}{{{r^2}}}$ $(Y, L$ and $F$ are constant$)$

i.e. for the same load, thickest wire will show minimum elongation. So graph $D$ represent the thickest wire.

art

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Similar Questions

  • 1
    The elongation of a wire on the surface of the earth is $10^{-4} \; m$. The same wire of same dimensions is elongated by $6 \times 10^{-5} \; m$ on another planet. The acceleration due to gravity on the planet will be $\dots \; ms ^{-2}$. (Take acceleration due to gravity on the surface of earth $=10 \; m / s ^{-2}$ )
    View Solution
  • 2
    A wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gain by the wire is $...... J ($take $g=10 \,m / s ^2 )$
    View Solution
  • 3
    The length of an elastic string is a metre when the longitudinal tension is $4\, N$ and $b$ metre when the longitudinal tension is $5\, N$. The length of the string in metre when the longitudinal tension is $9\, N$ is
    View Solution
  • 4
    The density and breaking stress of a wire are $6 \times$ $10^4 \mathrm{~kg} / \mathrm{m}^3$ and $1.2 \times 10^8 \mathrm{~N} / \mathrm{m}^2$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is $\frac{1^{\text {rd }}}{3}$ of the value on the surface of earth. The maximum length of the wire with breaking is ............ $\mathrm{m}$ (take, $\mathrm{g}=$ $\left.10 \mathrm{~m} / \mathrm{s}^2\right)$
    View Solution
  • 5
    $Assertion :$ Solids are least compressible and gases are most compressible.
    $Reason :$ solids have definite shape and volume but gases do not have either definite shape or definite volume.
    View Solution
  • 6
    The diagram shows stress v/s strain curve for the materials $A$ and $B$. From the curves we infer that
    View Solution
  • 7
    A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
    View Solution
  • 8
    A wire of cross sectional area $A$, modulus of elasticity $2 \times 10^{11} \mathrm{Nm}^{-2}$ and length $2 \mathrm{~m}$ is stretched between two vertical rigid supports. When a mass of $2 \mathrm{~kg}$ is suspended at the middle it sags lower from its original position making angle $\theta=\frac{1}{100}$ radian on the points of support. The value of $A$ is. . . . . .  $\times 10^{-4} \mathrm{~m}^2$ (consider $\mathrm{x}<\mathrm{L}$ ).

    (given: $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

    View Solution
  • 9
    The work done in stretching an elastic wire per unit volume is 
    View Solution
  • 10
    The bulk modulus of rubber is $9.1\times 10^8\,N/m^2$. To ......... $m$ depth a rubber ball be taken in a lake so that its volume is decreased by $0.1\,\%$ .
    View Solution