The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied
Length $100 \,cm,$ Diameter $1 \,mm$
Length $200\, cm,$ Diameter $2\, mm$
Length $300$ $cm$, Diameter $3 \,mm$
Length $50$ $cm,$ Diameter $0.5$ $mm$
One end of a metal wire is fixed to a ceiling and a load of $2 \mathrm{~kg}$ hangs from the other end. A similar wire is attached to the bottom of the load and another load of $1 \mathrm{~kg}$ hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be____________.
[Area of cross section of wire $=0.005 \mathrm{~cm}^2$, $\mathrm{Y}=2 \times 10^{11}\ \mathrm{Nm}^{-2}$ and $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$
A steel uniform rod of length $2L$ cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young’s modulus for steel, find the extension in the length of the rod.
A steel wire of diameter $0.5 mm$ and Young's modulus $2 \times 10^{11} N m ^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 m$. A vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0 mm$, is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 kg$, the vernier scale division which coincides with a main scale division is. . . . Take $g =10 m s ^{-2}$ and $\pi=3.2$.
The length of wire becomes $l_1$ and $l_2$ when $100\,N$ and $120\,N$ tensions are applied respectively. If $10l_2=11l_1$, the natural length of wire will be $\frac{1}{x} l_1$. Here the value of $x$ is ........
Two wire $A$ and $B$ are stretched by same force. If, for $A$ and $B, Y_A: Y_B=1: 2, r_A: r_B=3: 1$ and $L_A: L_B=4: 1$, then ratio of their extension $\left(\frac{\Delta L_A}{\Delta L_B}\right)$ will be .............