A wire of length '2m' is clamped horizontally between two fixed supports. A mass m = 5 kg is hung from the middle of the wire. The vertical depression in the wire in equilibrium is: (Young's modulus of wire = 2.4 x 10^9 N/m2, cross-sectional area = 1 cm2)
Correct answer: A. 4.68 cm
- A. 4.68 cm
- B. 1.52 cm
- C. 1.12 cm
- D. 0.58 cm
Explanation
The vertical depression in the wire can be calculated using the formula for the depression of a wire with a central load: δ = (W * L3) / (48 * Y * I), where W is the weight of the mass, L is the length of the wire, Y is Young's modulus, and I is the moment of inertia (I = A2 / 4π for a circular cross-section). Using the given parameters, the correct depression is found to be 4.68 cm.equation 2T sin θ = mg⇒ 2(YA/a) x sin θ. sin θ = mg⇒ 2YA/a x. x²/a² = mg⇒ x = [a³mg / 2YA](1/3) = [1m × 5kg × 10m/s² / 2 × (2.4 × 10⁹ N/m²) × 10⁻⁴ m²](1/3) = 4.68 cmThe other options are incorrect as they do not align with this calculation using the provided values.
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