NLM O16

Step 1: Analyze Equilibrium in Figure-II

In Figure-II, the three cords meet at a point in equilibrium. The angles between the cords are:

• Angle between A and B: $90^\circ$
• Angle between A and C: $135^\circ$
• Angle between B and C: $360^\circ – (90^\circ + 135^\circ) = 135^\circ$

Using Lami’s Theorem for the tensions $T_A, T_B, T_C$: $$ \frac{T_A}{\sin 135^\circ} = \frac{T_B}{\sin 135^\circ} = \frac{T_C}{\sin 90^\circ} $$ Since $\sin 135^\circ = \frac{1}{\sqrt{2}}$ and $\sin 90^\circ = 1$: $$ \frac{T_A}{1/\sqrt{2}} = \frac{T_B}{1/\sqrt{2}} = \frac{T_C}{1} $$ $$ T_A \sqrt{2} = T_B \sqrt{2} = T_C $$ From this, we derive the critical force relationships: $$ T_A = T_B \quad \text{and} \quad T_C = \sqrt{2} T_A $$ Therefore, $T_C > T_A$.

Step 2: Check Validity of Statements

The tension in a cord is given by $T = k(l – l_0)$. We need to satisfy $T_A = T_B$ and $T_C > T_A$.

Let’s evaluate Option (a): $l_A = l_B < l_C$ and $k_A = k_B < k_C$.

• Since $l_A = l_B$ and $k_A = k_B$, the tensions $T_A$ and $T_B$ would be equal ($T_A = T_B$), which satisfies the first condition.
• Since $l_C > l_A$, the extension in C is greater than in A ($x_C > x_A$).
• Since $k_C > k_A$, the stiffness of C is greater than A.
• The tension $T_C = k_C x_C$. Since both factors $k_C$ and $x_C$ are greater than their counterparts in A, the product $T_C$ will definitely be greater than $T_A$.

This satisfies the physical requirement $T_C > T_A$. Therefore, the conditions in option (a) are valid for the equilibrium shown.