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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

 
Answers (1)
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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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