A wire PQ has length 4.8 m and mass 0.06 kg. Another wire QR has length 2.56 m and mass 0.2 kg. Both wires have same radii and are joined as a single wire. This wire is under tension of 80 N. A wave pulse of amplitude 3.5 cm is sent along the wire PQ from end P. the time taken by the wave pulse to travel along the wire from point P to R is? (a) 0.1 s (b) 0.12 s (c) 0.14 s (d) 0.16 s

Answer: (c). PQ: μ₁ = 0.06/4.8 = 0.0125 kg/m, v₁ = √(80/0.0125) = 80 m/s, t₁ = 4.8/80 = 0.06 s. QR: μ₂ = 0.2/2.56 = 0.078 kg/m, v₂ = √(80/0.078125) = 32 m/s, t₂ = 2.56/32 = 0.08 s. t = 0.06 + 0.08 = 0.14 s So the answer is (c) 0.14 s.

A wire PQ has length 4.8 m and mass 0.06 kg. Another wire QR has… | MHT CET 2023 (14 May, Shift 2)

Q19MHT CET 2023 · 14 MAY · SHIFT 2Difficult

Speed of Waves

A wire $PQ$ has length $4.8\text{ m}$ and mass $0.06\text{ kg}$ . Another wire QR has length $2.56\text{ m}$ and mass $0.2\text{ kg}$ . Both wires have same radii and are joined as a single wire. This wire is under tension of $80\text{ N}$ . A wave pulse of amplitude $3.5\text{ cm}$ is sent along the wire $PQ$ from end $P$ . the time taken by the wave pulse to travel along the wire from point P to R is?

Show answer and solution

Answer: (c)

PQ: $\mu_1 = 0.06/4.8 = 0.0125$ kg/m, $v_1 = \sqrt{80/0.0125} = 80$ m/s, $t_1 = 4.8/80 = 0.06$ s.
QR: $\mu_2 = 0.2/2.56 = 0.078$ kg/m, $v_2 = \sqrt{80/0.078125} = 32$ m/s, $t_2 = 2.56/32 = 0.08$ s.
$t = 0.06 + 0.08 = 0.14$ s
So the answer is (c) 0.14 s.

Report a mistake

Speed of Waves question from MHT CET 2023 (14 May, Shift 2)

More questions on Speed of Waves