A rod of length $L$ at room temperature and uniform area of cross section $A$, is made of a metal having coefficient of linear expansion $\alpha {/^o}C$. It is observed that an external compressive force $F$, is applied on each of its ends, prevents any change in the length of the rod, when it temperature rises by $\Delta \,TK$. Young’s modulus, $Y$, for this metal is
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Wires $A$ and $B$ are connected with blocks $P$ and $Q$ as shown. The ratio of lengths, radii and Young's modulus of wires $A$ and $B$ are $r, 2r$ and $3r$ respectively ($r$ is a constant). Find the mass of block $P$ if ratio of increase in their corresponding lengths is $1/6r^2$. The mass of block $Q$ is $3M$.
A cubical solid aluminium (bulk modulus $=-V \frac{ dP }{ dV }=70 GPa$ ) block has an edge length of $1 m$ on the surface of the earth. It is kept on the floor of a $5 km$ deep ocean. Taking the average density of water and the acceleration due to gravity to be $10^3 kg m ^{-3}$ and $10 ms ^{-2}$, respectively, the change in the edge length of the block in $mm$ is . . . . .
The mass and length of a wire are $M$ and $L$ respectively. The density of the material of the wire is $d$. On applying the force $F$ on the wire, the increase in length is $l$, then the Young's modulus of the material of the wire will be
A steel wire of lm long and $1\,m{m^2}$ cross section area is hang from rigid end. When weight of $1\,kg$ is hung from it then change in length will be given ..... $mm$ $(Y = 2 \times {10^{11}}N/{m^2})$
A material has Poisson's ratio $0.50.$ If a uniform rod of it suffers a longitudinal strain of $2 \times {10^{ - 3}}$, then the percentage change in volume is
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