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Question

A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg m3 is supported by a vertical spring and is half dipped in water as shown in figure. (a) Find the elongation of the spring in equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant = 500 N m1

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Solution

Given, d = 10 cm

r = 5 cm

h = 20 cm

ρb=8000kg/m3 = 8 gm/cc

k= 500 N/m = 500×103dynecm

(a) Here, F+U =mg [where F = kx]

kx+Vρwg = mg

500×103×(x)+(πr2)×(h2)×1×1000

= πr2×h×ρb×1000

500×103×(x)

= πr2×h×1000(ρ12)

= π×(5)2×20×1000(ρb12)

50x=π×25×2×(ρb12)

x= π(80.5)

or x = π×7.5 = 23.5 cm

ma=kX+πr2×ρw×g(k+πr2×ρw×g)X

ω2(X)=k+πr2×ρw×gm× (X)

[because a = w2 X in SHM]

T=2πmk+πr2×g

= 2ππ×25×20×8500+103+π×25×1×1000

= 0.935 s.


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