(A) For wire $A$: length $L_A = L$, radius $R_A = R$, area $A_A = \pi R^2$.For wire $B$: length $L_B = 3L$, radius $R_B = 2R$, area $A_B = \pi (2R)^2

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(A) For wire $A$: length $L_A = L$, radius $R_A = R$, area $A_A = \pi R^2$.For wire $B$: length $L_B = 3L$, radius $R_B = 2R$, area $A_B = \pi (2R)^2 = 4\pi R^2$.When the wires are joined in series and pulled by a force $F$, the tension in both wires is the same.The elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$.Given that the elongation in both wires is equal, $\Delta L_A = \Delta L_B$.$\frac{F L_A}{A_A Y_A} = \frac{F L_B}{A_B Y_B}$Substituting the values:$\frac{F L}{(\pi R^2) Y_A} = \frac{F (3L)}{(4\pi R^2) Y_B}$$\frac{1}{Y_A} = \frac{3}{4 Y_B}$Rearranging for the ratio:$\frac{Y_B}{Y_A} = \frac{3}{4}$.

Class 11 Physics Mechanical Properties of Solids Young’s Modulus

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Two metal wires $A$ and $B$ have lengths $L$ and $3L$ respectively. The radii of the cross-sectional circular areas of wires $A$ and $B$ are $R$ and $2R$, respectively. These wires are joined end-to-end along their axis. When one end of the combined system is fixed and the other end is pulled with a constant force $F$, the elongation in both wires is equal. If $Y_A$ and $Y_B$ are the Young's moduli of wires $A$ and $B$, respectively, then the ratio $Y_B / Y_A$ is:

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A
$3/4$
B
$4/3$
C
$2/3$
D
$3/2$

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Mechanical Properties of Solids

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Young’s Modulus

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$A$ steel wire and a copper wire are joined end to end having equal cross-sections. The elongation of the two wires is found to be equal under tension. What is the ratio of the length of the steel wire to the length of the copper wire? (Young's modulus of steel $= 2.0 \times 10^{11} \ N \ m^{-2}$ and Young's modulus of copper $= 1.1 \times 10^{11} \ N \ m^{-2}$)

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Write the equation of bending in a rod. Write the unit of bending and its dimensional formula.

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When the temperature of a wire of length $L_0$ is increased by $T$,what is the energy density? The volume expansion coefficient of the wire is $\gamma$ and the Young's modulus is $Y$.

$A$ steel rod has a radius of $10 \,mm$ and a length of $1 \,m$. $A$ $80 \,kN$ force stretches it along its length. If the Young's modulus of the rod is $2 \times 10^{11} \,N/m^2$, then the change in length is

$A$ pendulum made of a uniform wire of cross-sectional area $A$ has a time period $T$. When an additional mass $M$ is added to its bob,the time period changes to $T_M$. If the Young's modulus of the material of the wire is $Y$,then $\frac{1}{Y}$ is equal to ($g$ = gravitational acceleration).

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