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A constant force F is applied to one end (P) of a rod of mass m and length l on a surface. A transverse pulse is created at P and travels to Q (the other end). Match the cases below with the correct travel times. List-I (cases): (I) No friction, m=4 kg, l=20 m, F=5 N (II) Friction coefficient 1/20, m=9 kg, l=32 m, F=8 N (III) Friction coefficient given, m=7 kg, l=14 m, F=2 N (IV) Friction coefficient 1/10, m=10 kg, l=20 m, F=2 N List-II (travel times): (P) 12 s (Q) 20 s (R) 8 s (S) 14 s (T) 7 s

1. (I)-(P), (II)-(Q), (III)-(R), (IV)-(S)
2. (I)-(R), (II)-(S), (III)-(P), (IV)-(T)
3. (I)-(T), (II)-(P), (III)-(Q), (IV)-(R)
4. (I)-(S), (II)-(R), (III)-(T), (IV)-(Q)

Correct answer: (I)-(R), (II)-(S), (III)-(P), (IV)-(T)

Solution

Travel time = 2*sqrt(ml/F). Case I: 2*sqrt(4*20/5)=8s. Case II: 2*sqrt(9*32/8)=12s. Case III: 2*sqrt(7*14/2)=14s. Case IV: 2*sqrt(10*20/2)=20s. Matching: I->R, II->P, III->S, IV->Q.

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