Two wires of equal lengths are made of the sa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The formula for the extension in a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,a

Vedclass · Accepted Answer

One-fourth for wire $A$ as for wire $B$

Vedclass · Answer

(D) The formula for the extension in a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the wires are made of the same material,$Y$ is the same. Given that $F$ and $L$ are also identical,we have $l \propto \frac{1}{A}$.Since $A = \pi r^2$,where $r$ is the radius,we get $l \propto \frac{1}{r^2}$.Given that the diameter of wire $A$ is twice that of wire $B$,the radius $r_A = 2r_B$.Therefore,$\frac{l_A}{l_B} = \left( \frac{r_B}{r_A} \right)^2 = \left( \frac{r_B}{2r_B} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.This implies $l_A = \frac{1}{4} l_B$,which means the increase in length for wire $A$ is one-fourth that of wire $B$.

Two wires of equal lengths are made of the same material. Wire $A$ has a diameter that is twice as that of wire $B$. If identical weights are suspended from the ends of these wires,the increase in length is

Similar Questions

For a solid rod,the Young's modulus of elasticity is $3.2 \times 10^{11} \, N m^{-2}$ and density is $8 \times 10^3 \, kg m^{-3}$. The velocity of longitudinal wave in the rod will be $......... \times 10^{3} \, m s^{-1}$.

$A$ wire is suspended from a rigid support and a weight $W$ is attached to its free end,causing an extension of $1.0 \, mm$. If the same wire is passed over a pulley and a weight $W$ is attached to each of its two ends,what will be the extension in the wire in $mm$?

$A$ metallic rod breaks when the strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^9 \,N/m^2$. The area of cross-section required to support a load of $10^4 \,N$ is:

$A$ cylindrical rod made of aluminum has a length of $1 \,m$ and a diameter of $10 \,cm$. The rod is subjected to a tensile force of $100 \,kN$. Calculate the elongation in the rod. (Young's modulus of aluminum $= 70 \,GPa$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module