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Q.

A string of uniform cross-sectional area is suspended from one of its ends. A transverse pulse is produced at its lower end and the pulse starts propagating up the string. Then the acceleration of the pulse is (Take g = 10 m/s2)

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a

5 m/s2

b

10 m/s2

c

zero

d

8 m/s2

answer is A.

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Detailed Solution

At a distance x from the lower end, tension in the string Tx=μ.x.g where μ is the mass per unit length of the string. ∴ Vx=Txμ=gx∴ Acceleration a=Vx.dVxdx=gx12gx=g2=5 m/s2
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