Two strings (A,B) having linear densities μ_A=2× 10⁻⁴ kg/m and, μ_B=4× 10⁻⁴ kg/m and lengths L_A=2.5 m and L_B=1.5 m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t₁ and t₂, respectively, to reach the joint. The ratio t₁/t₂ is:

Solution

Physics sir · 4 steps

Same tension in both strings, so only length and linear density set each travel time.

1. Speed on a string. A pulse on a string moves at v = √(T/μ). The tension is 500 N in both strings, but μ differs, so the speeds differ.
2. Time to the joint. Each pulse covers its own string, so t = L/v = L√(μ/T). In the ratio, the common T cancels.

   t₁/t₂ = L_A/L_B√(μ_A/μ_B) = 2.5/1.5√((2× 10⁻⁴)/(4× 10⁻⁴)) = 5/3×1/√2

3. Check with speeds. v_A = √(500/(2× 10⁻⁴)) ≈ 1581 m s⁻¹ and v_B ≈ 1118 m s⁻¹. So t₁ ≈ 1.58 ms and t₂ ≈ 1.34 ms.
4. So. t₁/t₂ = 1.667 × 0.707 ≈ 1.18, which is option (b).

The idea to keep. For pulses on strings under the same tension, t = L√(μ/T). Take the ratio and let T cancel. A heavier string is slower, so its time per metre is longer.

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