Question

block of 7cm×5cm×5cm floats on water with 1cm×5cm×5cm× out of water. Relative density of material of block is?​

Answer 1

Answer:

6×5×5×5 g

Explanation:

Density of water =1 g/cc (P

w

)

Density of steed =7 g/cc (P

s

)

Buoyant force:

B=P

w

V

g

Apparent wt

W=mg−B=[(P

s

−P

w

)V

g

]f

W=[(7−1)×5×5×5 g]f

w=[6×5×5×5 g]f, f symbolyses force (unit)

Answer 2

Given:

Dimensions of the box $=7cm$ × $5cm$ × $5cm$

Dimension of the box out of the water $=1cm$ × $5cm$ × $5cm$

To find:

Relative density of the material of the block.

Explanation:

Archimedes principle states that the when an object having a finite mass is immersed in a fluid, the result is that the fluid is displaced from the container holding it and the weight of this displaced fluid is exactly equal to the weight of the object.

If, the mass of the object is $M_{o}$ and that of the water displaced is $M_{w}$ hence, according to Archimedes principle, on immersing the object in the fluid,

                                     $M_{o}.g =M_{w}.g$

Where g is the acceleration due to gravity.

We know, the density of any object is given by the ratio of mass of the object and the relative volume of it.

$d=\frac{M}{V}$

$M=d$ × $V$

Hence, at equilibrium, the mass of the object of density $d_{1}$ and volume $V_{1}$ immersed in the fluid of density $d_{2}$ and volume of the fluid as $V_{2}$. Therefore, the relation will be,

$d_{1} .V_{1} =d_{2} .V_{2}$

Solution:

According to the question, we have been given that an object of dimensions $7cm$ × $5cm$ × $5cm$ is immersed in the water but only some part of the box is immersed in the water remaining part of the box lies outside the water meniscus whose dimensions are $1cm$ × $5cm$ × $5cm$.

Step 1

Volume of the box $V_{b} =7cm$ × $5cm$ × $5cm$  

$V_{b} =175cm^{3}$

$V_{b} =175$ × $10^{-6}m^{3}$

Hence,

The volume of the box that lies inside the water level is

$V=$ ( $7cm$ × $5cm$ × $5cm$ ) - ( $1cm$ × $5cm$ × $5cm$ )

$V=175cm^{3}-25cm^{3}$

$V=150cm^{3}$

$V=150$ × $10^{-6}m^{3}$

Hence, the volume of the box inside the water is 150 × 10⁻⁶m³ and also, most importantly this will be the exact volume of the water that will be displaced according to the Archimedes principle.

Step 2

Therefore,

The weight of the box = weight of water displaced.

Let, the density of the box be $d_{b}$ and, we know, the density of the water is $d_{w} =997kg/m^{3}$.

$d_{b}$ × $V_{b}$ × $g$ $=d_{w}$ × $V_{}$ × $g$

$d_{b}$ × $175$ × $10^{-6}$ × $9.8=997$ × $150$ × $10^{-6}$ × $9.8$

$d_{b}$ × $175=997$ × $150$

$d_{b} =\frac{149550}{175}$

$d_{b}=854kg/m^{3}$

Final answer:

Hence, the density of the block is 854 kg/m³.