Answer:
a. the density will not change
b. D' = 0.125 D
So, the density will change by a factor of 0.125
Explanation:
a.
Density is the material property and the value of density is constant for all solid materials. So, when the dimensions of the a solid are increased, while the material is same, then the material must be added to the object for increasing its dimensions. So, with the increase in the volume, the mass of the object also increases. And as a result the density of the object remains constant.
Since, here the material remains the same.
Therefore, the density will not change
b.
Density = mass/Volume
D = m/V ------------ equation (1)
Now,
V = LWH ----------- equation (2)
Now, if each dimension increases by a factor of 2, the volume becomes:
V' = (2L)(2W)(2H)
V' = 8 LWH
using equation (2)
V' = 8 V
So, for constant mass, density becomes:
D' = m/V'
D' = m/8V
using equation (1)
D' = D/8
D' = 0.125 D
So, the density will change by a factor of 0.125
(a) The density of the object will be reduced by one-eighth times on increasing the dimension by a factor of two and keeping the material constant.
(b) With constant mass and increasing the dimensions by a factor of 2, the density will reduce by the factor of 1/8.
The degree measure of the compactness of an object is known as the density of an object. Mathematically, it is expressed as the ratio of the mass of the object to its volume.
(a)
Let us consider an object of dimension 'a'. As per the above definition, the density of an object is,
[tex]\rho = \dfrac{m}{V}[/tex]
Here,
m is the mass of the object.
V is the volume of the object. And its value is, [tex]V = a^{3}[/tex].
So the density becomes,
[tex]\rho = \dfrac{m}{a^{3}}[/tex] ..............................................................(1)
Now, if the dimensions of the object is increased by a factor of 2. Then, new density becomes,
[tex]\rho' =\dfrac{m}{(2a) \times (2a) \times (2a)}\\\\\\ \rho' =\dfrac{m}{8a^{3}}\\\\[/tex]
Substitute the value of equation (1) in the above expression as,
[tex]\rho' =\dfrac{1}{8} \times \dfrac{m}{a^{3}}\\\\\rho' =\dfrac{1}{8} \times \rho[/tex]
Thus, we can conclude that the density of the object will be reduced by one-eighth times on increasing the dimension by a factor of two and keeping the material constant.
(b)
The density of the material is equal to the ratio of the mass of the object (m) and the volume of the object (V). So, if the mass remains unchanged and the dimensions increased by the factor of 2, then the result will remain same as the above part.
Thus, we can conclude that with constant mass and increasing the dimensions by a factor of 2, the density will reduce by the factor of 1/8.
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