A circular disc of diameter d is immersed vertically in a liquid of density P. The top most point of the disc just touches the liquid surface. What is the depth of Centre of pressure ?

\frac{3}{8} D

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\frac{5}{4} D

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\frac{5}{8} D

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\frac{3}{4} D

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Solution

The correct option is C $\frac{5}{8} D$
$C e n t r e o f p r e s s u r e = ¯ h + \frac{I_{g}}{A ¯ h}$

$Y_{p} = \frac{D}{2} + \frac{\frac{π D^{4}}{64}}{\frac{π D^{2}}{4} \times \frac{D}{2}} = \frac{5}{8} D$

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Complete the following sentences :
(a) Pressure at a depth h in a liquid of density $ρ$ is _____ .
(b) Pressure is ___ in all directions about a point in a liquid.
(c) Pressure at all points at the same depth is ___ .
(d) Pressure at a point inside a liquid is ___ to its depth.
(e) Pressure of a liquid at a given depth is ___ to the density of liquid.

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(1) A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will be in vertically ………….. direction.

(2) When an object sinks into the liquid, the density of object is ………….. than density of the liquid.

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A circular disc of diameter d is immersed vertically in a liquid of density P. The top most point of the disc just touches the liquid surface. What is the depth of Centre of pressure ?

The correct option is C 58DCentre of pressure=¯h+IgA¯h Yp=D2+πD464πD24×D2=58D

The correct option is C $\frac{5}{8} D$
$C e n t r e o f p r e s s u r e = ¯ h + \frac{I_{g}}{A ¯ h}$

$Y_{p} = \frac{D}{2} + \frac{\frac{π D^{4}}{64}}{\frac{π D^{2}}{4} \times \frac{D}{2}} = \frac{5}{8} D$