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# Need explanation for: A cylindrical block of wood (density=650 kg m-3), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m-3. The block is depressed slightly and then released. The time period of the resulting oscillations o

A cylindrical block of wood (density=650 kg m-3), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m-3. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :

• Option 1)

65 cm

• Option 2)

52 cm

• Option 3)

39 cm

• Option 4)

26 cm

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As we learned

Equation of S.H.M. -

$a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x$

$w= \sqrt{\frac{k}{m}}$

- wherein

$x= A\sin \left ( wt+\delta \right )$

Let Block is Hosting with h depth in water.

At equilibrium

$(A H\cdot P_{B})g=AhP_{1}g$

Let it is depressed by x

$\Rightarrow f_{net}-f_{up}-M block g$

$f_{net}=P_{l}g- AHP_{B}\cdot g\cdot A(h+x)$

$=-AxP_{l}g.$

$\Rightarrow A/H\cdot f_{Block}\frac{d^{2}x}{dt^{2}}=\not{Ax}P_{l}g$

$\Rightarrow \frac{a^{2}x}{dt^{2}}=-\left ( \frac{P_{l}g}{H\cdot P_{Block}} \right )\cdot x$

$w^{2}=\frac{P_{l}g}{HP_{B}}=\frac{g}{l}$ (for simple pendulum)

$\Rightarrow l=\frac{HP_{B}}{P_{l}}=\frac{650\times 54}{900}=39cm$

Option 1)

65 cm

Option 2)

52 cm

Option 3)

39 cm

Option 4)

26 cm

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