Question

A massless rod of length L is hinged at one end and is held horizontal by two identical vertical wires of length $y_0$ and area of cross section A. Wires are tied at distance $\frac{L}{4}$ and $\frac{3L}{4}$ from hinged end. Young's modulus, tension and extension in left wire is $Y_1,T_1$ and $\Delta l$, and that in right wire is $Y_2,T_2$ and $\Delta l_2$. A force F is applied at free end of rod as shown in figure, then $T_1=x_{1}F,T_2=x_{2}F$, $\Delta l_1=x_3\frac{y_{0}F}{AY},\Delta l_2=x_4\frac{y_{0}F}{AY}$.
If $Y_1=Y$ and $Y_2=2Y$ match correct list.

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(I)}& x_1& \text{(P)}& \frac{4}{19}\\ \text{(II)}& x_2& \text{(Q)}& \frac{12}{19}\\ \text{(III)}& x_3& \text{(R)}& \frac{24}{19}\\ \text{(IV)}& x_4& \text{(S)}& \frac{8}{17}\\ & & \text{(T)}& \frac{12}{17}\\ & & \text{(U)}& \frac{16}{17}\end{array}$

• A. I $\to$ P, II $\to$ Q, III $\to$ P, IV $\to$ R
• B. I $\to$ R, II $\to$ S, III $\to$ V, IV $\to$ P
• C. I $\to$ R, II $\to$ U, III $\to$ T, IV $\to$ V
• D. I $\to$ P, II $\to$ R, III $\to$ P, IV $\to$ Q

Answer 1

The correct option is D I $\to$ P, II $\to$ R, III $\to$ P, IV $\to$ Q

I $\to$ P, II $\to$ R, III $\to$ P, IV $\to$ Q

Balancing the torque at the hinge results in,
$T_1\frac{L}{4}+T_2\frac{3L}{4}=FL$
$\Rightarrow \frac{T_1}{4}+\frac{3T_2}{4}=F---\left(1\right)$
$T_1=Y_{1}A\frac{\Delta l_1}{l_1}$ and
$T_2=Y_{2}A\frac{\Delta l_2}{l_2}$
Because of the hinge, the extensions in the strings are related by the equation,
$\Delta l_2=3\Delta l_1$
and given that
$l_1=l_2$
$Y_1=Y$ and $Y_2=2Y$
This implies $T_2=6T_1$
Substituting these in (1) results in,
$\frac{Y_1{A}_1\Delta l_1}{l}19=4F$
$\Delta l_1=\frac{4Fl}{19Y_1{A}_1}=\frac{4F{y}_0}{19YA}$
$\Delta l_2=3\Delta l_1=\frac{12F{y}_0}{19YA}$

Also, $T_2=6T_1$
Substituting these in (1) gives,
$T_1=\frac{4F}{19}$ and
$T_2=\frac{24F}{19}$