The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm², is 2 × 10¹¹ N / m². The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 1/4 of its value on the earth. The elongation of wire is (Take g on the earth =10 m / s²) :

Solution

Physics sir · 3 steps

Take the weight with the planet's g, put the area in m², then use Δ L = FL/AY.

1. Load on the wire. On this planet g' = 1/4 × 10 = 2.5 m s⁻². The tension in the wire is the block's weight there: F = 4 × 2.5 = 10 N.
2. Area in SI. Since 1 mm = 10⁻³ m, we get 1 mm² = 10⁻⁶ m². So A = 3 × 10⁻⁶ m².

   Δ L = FL/AY = (10 × 6)/((3 × 10⁻⁶) × (2 × 10¹¹)) = 60/(6 × 10⁵) = 10⁻⁴ m

3. So. Δ L = 10⁻⁴ m = 0.1 mm, which is option (c).

The idea to keep. Elongation is Δ L = FL/AY, and F is the weight in the local gravity. Put every quantity in SI first. Remember that 1 mm² = 10⁻⁶ m².

Answer from the NTA final answer key. Solution by PrepWiser.