Two strings (A, B) having linear densities $\mu_{A}$ = 2 × 10⁻⁴ kg·m⁻¹ and $\mu_{B}$ = 4 × 10⁻⁴ kg·m⁻¹ and lengths L₍ᴀ₎ = 2.5 m and L₍ʙ₎ = 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 :
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Question

Two strings (A, B) having linear densities $\mu_{A}$ = 2 × 10⁻⁴ kg·m⁻¹ and $\mu_{B}$ = 4 × 10⁻⁴ kg·m⁻¹ and lengths L₍ᴀ₎ = 2.5 m and L₍ʙ₎ = 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 :

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G1 Physics · Accepted Answer

Wave speed on a stretched string is v = √[ T ─── μ ]. For string A, v₍ᴀ₎ = √[ 500 ────────── 2 × 10⁻⁴ ] = 5√10 × 10² m·s⁻¹. For string B, v₍ʙ₎ = √[ 500 ────────── 4 × 10⁻⁴ ] = 5√5 × 10² m·s⁻¹. The travel times are t₁ = L₍ᴀ₎ ─── v₍ᴀ₎, t₂ = L₍ʙ₎ ─── v₍ʙ₎. Therefore, t₁ ─── t₂ = L₍ᴀ₎ v₍ʙ₎ ─────── L₍ʙ₎ v₍ᴀ₎ . Substitute the given values: t₁ ─── t₂ = 2.5 × 5√5 ──────────── 1.5 × 5√10 = 5 ─── 3 × 1 ─── √2 ≈ 1.18. Correct option: D.