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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Solids are compared through crystal structure, amorphous arrangement, elasticity, plastic deformation, stress, strain and Young’s modulus. Electrical behavior depends on available charge carriers and band structure, so conductors, insulators and semiconductors must be distinguished from mechanical properties such as strength, stiffness and brittleness.
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