First slide

Pressure in a fluid-mechanical properties of fluids

Question

A cubical block of wood of side a and density $ρ$ floats in water of density 2 $ρ$ . The lower surface of the cube just touches the free end of a massless spring of force constant
k fixed at the bottom of the vessel. The weight W put over the block so that it is completely immersed in water without wetting the weight is

Moderate

A

$a (a^{2} ρ g + k)$
B

$a (a ρ g + 2 k)$
C

$a (\frac{a ρ g}{2} + 2 k)$
D

$a (a^{2} ρ g + \frac{k}{2})$

Solution

Since density of block is 1/2(density of water), 50% of its volume is immersed in water.
In the second case, half of the volume of the block is further immersed in water.
Therefore, W=Extra upthrust + Spring force
$= B u o y a n t f o r c e + S p r i n g f o r c e = V_{i} \times ρ_{l} \times g + k x = a \times a \times \frac{a}{2} \times 2 ρ \times g + k (\frac{a}{2}) = a (a^{2} ρ g + \frac{k}{2})$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App