On applying a stress of $20 \times {10^8}N/{m^2}$ the length of a perfectly elastic wire is doubled. Its Young’s modulus will be
A$40 \times {10^8}N/{m^2}$
B$20 \times {10^8}N/{m^2}$
C$10 \times {10^8}N/{m^2}$
D$5 \times {10^8}N/{m^2}$
Medium
Download our app for free and get started
B$20 \times {10^8}N/{m^2}$
b (b) Young’s modules =$\frac{{{\rm{stress}}}}{{{\rm{strain}}}}$
As the length of wire get doubled therefore strain $= 1$
$Y =$ strain $= 20 \times 10^8 N/ m^2$
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.*
A substance breaks down by a stress of $10^6 N/m^2$. If the density of the material of the wire is $3×10^3 kg/m^3$, then the length of the wire of the substance which will break under its own weight when suspended vertically, is ......... $m$
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 ........
The Young's modulus of the material of a wire is $6 \times {10^{12}}\,N/{m^2}$ and there is no transverse strain in it, then its modulus of rigidity will be
The bulk modulus of rubber is $9.1 \times 10^8\,N/m^2$. To what depth a rubber ball be taken in a lake so that its volume is decreased by $0.1\%$ ? ....... $m$
$A$ rod of length $1000\, mm$ and coefficient of linear expansion $a = 10^{-4}$ per degree is placed symmetrically between fixed walls separated by $1001\, mm$. The Young’s modulus of the rod is $10^{11} N/m^2$. If the temperature is increased by $20^o C$, then the stress developed in the rod is ........... $MPa$
A 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, $\left( {\frac{{dr}}{r}} \right)$ is