The ratio of radii of two wires of same material is $2: 1$. If these wires are stretched by equal force, the ratio of stresses produced in them is $.............$
Easy
Download our app for free and get started
We know,
$\text { Stress }=\frac{\text { Force }}{\text { Area }}$
So, Stress $\times$ Area $=$ Force
$S \times A=F$
$\left\{\begin{array}{l}S=\text { Stress } \\ F=\text { Force } \\ A=\text { Area } \\ r=\text { radius }\end{array}\right\}$
$\because ($Since$)$ Force applied on the wires is equal we can relate two conditions as
$S_1 A_1=S_2 A_2$
$\frac{S_1}{S_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}$
$\frac{S_1}{S_2}=\frac{r^2}{(2 r)^2}=\frac{r^2}{4 r^2}=\frac{1}{4}$
$\left\{\begin{array}{l}\text { Where } \\ S_1-\text { Stress in } 1^{\text {st }} \text { wire } \\ A_1-\text { Area of } 1^{\text {st }} \text { wire } \\ r_1-\text { Radius of } 1^{\text {st }} \text { wire } \\ S_2-\text { Stress in } 2^{\text {nd }} \text { wire } \\ A_2-\text { Area of } 2^{\text {nd }} \text { wire } \\ r_2-\text { Radius of } 2^{\text {nd }} \text { wire }\end{array}\right\}$
$\text { Stress }=\frac{\text { Force }}{\text { Area }}$
So, Stress $\times$ Area $=$ Force
$S \times A=F$
$\left\{\begin{array}{l}S=\text { Stress } \\ F=\text { Force } \\ A=\text { Area } \\ r=\text { radius }\end{array}\right\}$
$\because ($Since$)$ Force applied on the wires is equal we can relate two conditions as
$S_1 A_1=S_2 A_2$
$\frac{S_1}{S_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}$
$\frac{S_1}{S_2}=\frac{r^2}{(2 r)^2}=\frac{r^2}{4 r^2}=\frac{1}{4}$
$\left\{\begin{array}{l}\text { Where } \\ S_1-\text { Stress in } 1^{\text {st }} \text { wire } \\ A_1-\text { Area of } 1^{\text {st }} \text { wire } \\ r_1-\text { Radius of } 1^{\text {st }} \text { wire } \\ S_2-\text { Stress in } 2^{\text {nd }} \text { wire } \\ A_2-\text { Area of } 2^{\text {nd }} \text { wire } \\ r_2-\text { Radius of } 2^{\text {nd }} \text { wire }\end{array}\right\}$
Download our appand 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
- 1The relation between $\gamma ,\,\eta $ and $K$ for a elastic material isView Solution
- 2If the force constant of a wire is $K,$ the work done in increasing the length of the wire by $l$ isView Solution
- 3A solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. A massless piston of area $a$ floats on the surface of the liquid, covering entire crosssection of cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left( {\frac{{dr}}{r}} \right)$ isView Solution
- 4The area of cross-section of a wire of length $1.1$ metre is $1$ $mm^2$. It is loaded with $1 \,kg.$ If Young's modulus of copper is $1.1 \times {10^{11}}\,N/{m^2}$, then the increase in length will be ......... $mm$ (If $g = 10\,m/{s^2})$View Solution
- 5View SolutionThe 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
- 6The ratio of two specific heats of gas ${C_p}/{C_v}$ for argon is $1.6$ and for hydrogen is $1.4$. Adiabatic elasticity of argon at pressure $P$ is $E.$ Adiabatic elasticity of hydrogen will also be equal to $E$ at the pressureView Solution
- 7A rod of length $l$ and radius $r$ is joined to a rod of length $l/2$ and radius $r/2$ of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $\theta°$, the twist angle at the joint will beView Solution
- 8If the compressibility of water is $\sigma $ per unit atmospheric pressure, then the decrease in volume $V$ due to $P$ atmospheric pressure will beView Solution
- 9The area of cross-section of a railway track is $0.01\, {m}^{2}$. The temperature variation is $10^{\circ} {C}$. Coefficient of linear expansion of material of track is $10^{-5} /{ }^{\circ} {C}$. The energy stored per meter in the track is ...... ${J} / {m} .$View Solution
(Young's modulus of material of track is $10^{11} \,{Nm}^{-2}$ ))
- 10An iron rod of length $2m$ and cross section area of $50\,m{m^2}$, stretched by $0.5\, mm$, when a mass of $250\, kg$ is hung from its lower end. Young's modulus of the iron rod isView Solution