A cylindrical block of wood $(d e n s i t y = 650 k g m^{- 3})$ , of base area $30 c m^{2}$ and height $54 c m$ , floats in a liquid of density $900 k g$ $m^{- 3}$ . 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)

52 c m

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26 c m

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39 c m

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65 c m

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Solution

The correct option is C $39 c m$
As block is floating it's weight should be equal to buoyancy force
$ρ_{w o o d} V_{c y l i n d e r} g = ρ_{l i q u i d} V_{d i s p l a c e d} g$
$V_{d i s p l a c e d} = \frac{ρ_{w o o d} V_{c y l i n d e r}}{ρ_{l i q u i d}}$
$V_{d i s p l a c e d} = \frac{ρ_{w o o d} V_{c y l i n d e r}}{ρ_{l i q u i d}}$ ...(i)
After displacing by small distance x, the net force on cylinder will be
$F = B u o y a n c y - w = ρ_{l i q u i d} (V_{d i s p l a c e d} + A_{c y l i n d e r} Δ x) g - ρ_{w o o d} V_{c y l i n d e r} g$
$F = ρ_{l i q u i d} V_{d i s p l a c e d} g + ρ_{l i q u i d} A_{c y l i n d e r} Δ x g - ρ_{w o o d} V_{w o o d} g$ the net force on cylinder becomes
from equation (i) $ρ_{w o o d} V_{c y l i n d e r} g = ρ_{l i q u i d} V_{d i s p l a c e d} g$
$F = ρ_{l i q u i d} A_{c y l i n d e r} Δ x g$
$m a = ρ_{l i q u i d} A_{c y l i n d e r} Δ x g$
$a = \frac{ρ_{l i q u i d} A_{c y l i n d e r}}{m} Δ x$
$a = ω^{2} Δ x$
$ω^{2} = \frac{ρ_{l i q u i d} A_{c y l i n d e r}}{ρ_{c y l i n d e r} V_{c y l i n d e r}}$
$ω^{2} = \frac{ρ_{l i q u i d} A_{c y l i n d e r}}{ρ_{w o o d} A_{c y l i n d e r} h_{c y l i n d e r}}$
$ω^{2} = \frac{ρ_{l i q u i d}}{ρ_{w o o d} h_{c y l i n d e r}}$
$ω^{2} = \frac{900}{650 \times .54}$
This should be equal to angular frequency of simple pendulum
$ω = \sqrt{\frac{g}{l}}$
$\sqrt{\frac{900}{650 \times .54}} = \sqrt{\frac{g}{l}}$
$l = g \frac{650 \times .54}{900}$
$l = .06 \times 65$
$= 39 c m$

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