
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density $\rho $ remains uniform throughout the volume. The rate of fractional change in density ($\dfrac{1}{\rho }\dfrac{{d\rho }}{{dt}}$ ) is constant. The velocity v of any point on the surface of the expanding sphere is proportional to:
1) R
2) R$^3$
3) R$^{⅔}$
4) 1/R
Answer
565.8k+ views
Hint:In the given question we are provided with sphere whose volume is taken as;
$\dfrac{{4\pi {R^3}}}{3}$and mass of the sphere is taken as m.
Density is the ratio of mass upon volume, we have mass and volume of sphere now we can easily calculate the change in density and if change in density takes place change in volume also be calculated as the sphere is expanding according to the question.
$D = \dfrac{{mass}}{{volume}}$
Complete step by step solution:
We are being given that change in density is constant:
$\dfrac{1}{\rho }\dfrac{{d\rho }}{{dt}}$= constant ......................1
According to the definition of density we have density is equal to the ratio of mass upon volume.
$\rho = \dfrac{m}{v}$
Volume of sphere is: $\dfrac{{4\pi {R^3}}}{3}$ and mass of the sphere is m.
Change in density will bring change in mass and volume, thus we can write:
$ \Rightarrow \dfrac{{4\pi {R^3}}}{{3m}}\dfrac{d}{{dt}}\dfrac{m}{{4\pi {R^3}}}$= constant (we have substituted the value mass and volume in expression 1)
$ \Rightarrow {R^3}\dfrac{d}{{dt}}{R^{ - 3}}$= constant (cancelling the constant terms which were common to both the numerator and denominator).
$ \Rightarrow - 3{R^3}\left( {{R^{ - 4}}} \right)\dfrac{{dR}}{{dt}}$= constant
$ \Rightarrow \dfrac{{ - 3}}{R}\dfrac{{dR}}{{dt}}$= constant
$ \Rightarrow \dfrac{{dR}}{{dt}} \propto R$
Thus change in volume is proportional to R.
Option 1 is correct.
Note:If we want to check whether the object is dense or not we can perform an experiment: let the object to fall in a bucket of water if the object will be dense then it will sink at the bottom of the bucket and if the object will be not dense it will float at the surface of the water. We can take the example of balloon and iron nail for this experiment to learn about density of the object.
$\dfrac{{4\pi {R^3}}}{3}$and mass of the sphere is taken as m.
Density is the ratio of mass upon volume, we have mass and volume of sphere now we can easily calculate the change in density and if change in density takes place change in volume also be calculated as the sphere is expanding according to the question.
$D = \dfrac{{mass}}{{volume}}$
Complete step by step solution:
We are being given that change in density is constant:
$\dfrac{1}{\rho }\dfrac{{d\rho }}{{dt}}$= constant ......................1
According to the definition of density we have density is equal to the ratio of mass upon volume.
$\rho = \dfrac{m}{v}$
Volume of sphere is: $\dfrac{{4\pi {R^3}}}{3}$ and mass of the sphere is m.
Change in density will bring change in mass and volume, thus we can write:
$ \Rightarrow \dfrac{{4\pi {R^3}}}{{3m}}\dfrac{d}{{dt}}\dfrac{m}{{4\pi {R^3}}}$= constant (we have substituted the value mass and volume in expression 1)
$ \Rightarrow {R^3}\dfrac{d}{{dt}}{R^{ - 3}}$= constant (cancelling the constant terms which were common to both the numerator and denominator).
$ \Rightarrow - 3{R^3}\left( {{R^{ - 4}}} \right)\dfrac{{dR}}{{dt}}$= constant
$ \Rightarrow \dfrac{{ - 3}}{R}\dfrac{{dR}}{{dt}}$= constant
$ \Rightarrow \dfrac{{dR}}{{dt}} \propto R$
Thus change in volume is proportional to R.
Option 1 is correct.
Note:If we want to check whether the object is dense or not we can perform an experiment: let the object to fall in a bucket of water if the object will be dense then it will sink at the bottom of the bucket and if the object will be not dense it will float at the surface of the water. We can take the example of balloon and iron nail for this experiment to learn about density of the object.
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