Question

A wire of length '2m' is clamped horizontally between two fixed support. A mass m = 5 kg is hanged from middle of wire. The vertical depression in wire in equilibrium is (young modulus of wire = 2.4 × 10⁹ N/m², cross-sectional area = 1 cm²) –

• A. 4.68 cm
• B. 1.52 cm
• C. 1.12 cm
• D. 0.58 cm

Answer 1

Answer: (A)

4.68 cm

Answer 2

equation

2T sin q = mg

Ž $2.\left( \frac{YA}{a} \right)$x sin q. sin q = mg

Ž $\frac{2YA}{a}$x. $\frac{x^{2}}{a^{2}}$= mg

Ž x = $\left\{ \frac{a^{3}mg}{2YA} \right\}^{1/3}$

= $\left\{ \frac{1m \times 5kg \times 10m/s^{2}}{2 \times (2.4 \times 10^{9}N/m^{2}) \times 10^{–4}m^{2}} \right\}^{1/3}$ = 4.68 cm