Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is $2 \,cm$, then how much is the elongation in steel and copper wire respectively? Given, $Y_{\text {steel }}=20 \times 10^{11} \,dyne / cm ^2$, $Y_{\text {copper }}=12 \times 10^{11} \,dyne / cm ^2$
$1.25 \,cm ; 0.75 \,cm$
$0.75 \,cm ; 1.25 \,cm$
$1,15 \,cm , 0.85 \,cm$
$0.85 \,cm ; 1.15 \,cm$
Two wires are made of the same material and have the same volume. However wire $1$ has crosssectional area $A$ and wire $2$ has cross-section area $3A$. If the length of wire $1$ increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire $2$ by the same amount?
The length of wire, when $M_1$ is hung from it, is $I_1$ and is $I_2$ with both $M_1$ and $M_2$ hanging. The natural length of wire is ........
A uniform heavy rod of mass $20\,kg$. Cross sectional area $0.4\,m ^{2}$ and length $20\,m$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x \times 10^{-9} m$. The value of $x$ is
(Given. Young's modulus $Y =2 \times 10^{11} Nm ^{-2}$ અને $\left.g=10\, ms ^{-2}\right)$
On all the six surfaces of a unit cube, equal tensile force of $F$ is applied. The increase in length of each side will be ($Y =$ Young's modulus, $\sigma $= Poission's ratio)
The mean distance between the atoms of iron is $3 \times {10^{ - 10}}m$ and interatomic force constant for iron is $7\,N\,/m$The Young’s modulus of elasticity for iron is