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Two spheres of radii in the ratio $1:2$ and densities in the ratio $2:1$ and of the same specific heat, are heated to the same temperature and left in the same surroundings. Their rate of falling temperature will be in the ratio:
A) 2:1
B) 1:1
C) 1:2
D) 1:4
Answer
119.1k+ views
Hint: First of all write the given quantities. Use the formula for rate of cooling i.e., $\dfrac{{{{d\theta }}}}{{{{dt}}}}{{ = }}\dfrac{{{1}}}{{{{ms}}}}\dfrac{{{{dQ}}}}{{{{dt}}}}{{ = }}\dfrac{{{{AK}}}}{{{{ms}}}}$. Now, find the value of mass per unit area and find out the rate of cooling is proportional to which quantity. Finally, rate of cooling is dependent on the radius and density of the sphere by deriving the relation, rate of cooling is proportional to $\dfrac{{{{{r}}_{{2}}}{{ }}{{{\rho }}_{{2}}}}}{{{{{r}}_{{1}}}{{ }}{{{\rho }}_{{1}}}}}$. Then substitute the values and find out the required ratio.
Complete step by step solution:
Given: The ratio of the radius of two spheres is $1:2$
The ratio of the densities of two spheres is $2:1$
Specific heat of both the spheres are same
Both the spheres are heated to same temperature and then both are left in surroundings
To find: The ratio of the rate of cooling for two spheres
Formula for rate of cooling is given by,
$\Rightarrow \dfrac{{{{d\theta }}}}{{{{dt}}}}{{ = }}\dfrac{{{1}}}{{{{ms}}}}\dfrac{{{{dQ}}}}{{{{dt}}}}{{ = }}\dfrac{{{{AK}}}}{{{{ms}}}}$
Formula for mass is given by,
${{mass = volume \times density}}$
Thus, the mass of the sphere is $\dfrac{4}{3}{{\pi }}{{{r}}^3}{{ \times \rho }}$.
Mass per unit area is $\dfrac{{\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}{{ \times \rho }}}}{{{{4\pi }}{{{r}}^{{2}}}}}{{ = }}\dfrac{{{1}}}{{{3}}}{{r\rho }}$.
So, the rate of cooling is proportional to $\dfrac{{{1}}}{{{{r\rho }}}}$.
Thus, the ratio of cooling for two sphere is $\dfrac{{{{{r}}_{{2}}}{{ }}{{{\rho }}_{{2}}}}}{{{{{r}}_{{1}}}{{ }}{{{\rho }}_{{1}}}}}$.
On substituting the values, we get
Hence, the ratio of cooling for two sphere is $\dfrac{{{{2 \times 1 }}}}{{{{1 \times 2 }}}}{{ = }}\dfrac{{{1}}}{{{1}}}$.
Thus, the ratio of cooling for two spheres is $1:1$.
Therefore, option (B) is the correct choice.
Note: Newton's Law of Cooling states that the rate of change of the temperature of an object is directly proportional to the difference between its own temperature and the temperature of its surroundings. The drawback of Newton’s law of cooling is that the temperature of surroundings should be constant during the cooling of the body.
Complete step by step solution:
Given: The ratio of the radius of two spheres is $1:2$
The ratio of the densities of two spheres is $2:1$
Specific heat of both the spheres are same
Both the spheres are heated to same temperature and then both are left in surroundings
To find: The ratio of the rate of cooling for two spheres
Formula for rate of cooling is given by,
$\Rightarrow \dfrac{{{{d\theta }}}}{{{{dt}}}}{{ = }}\dfrac{{{1}}}{{{{ms}}}}\dfrac{{{{dQ}}}}{{{{dt}}}}{{ = }}\dfrac{{{{AK}}}}{{{{ms}}}}$
Formula for mass is given by,
${{mass = volume \times density}}$
Thus, the mass of the sphere is $\dfrac{4}{3}{{\pi }}{{{r}}^3}{{ \times \rho }}$.
Mass per unit area is $\dfrac{{\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}{{ \times \rho }}}}{{{{4\pi }}{{{r}}^{{2}}}}}{{ = }}\dfrac{{{1}}}{{{3}}}{{r\rho }}$.
So, the rate of cooling is proportional to $\dfrac{{{1}}}{{{{r\rho }}}}$.
Thus, the ratio of cooling for two sphere is $\dfrac{{{{{r}}_{{2}}}{{ }}{{{\rho }}_{{2}}}}}{{{{{r}}_{{1}}}{{ }}{{{\rho }}_{{1}}}}}$.
On substituting the values, we get
Hence, the ratio of cooling for two sphere is $\dfrac{{{{2 \times 1 }}}}{{{{1 \times 2 }}}}{{ = }}\dfrac{{{1}}}{{{1}}}$.
Thus, the ratio of cooling for two spheres is $1:1$.
Therefore, option (B) is the correct choice.
Note: Newton's Law of Cooling states that the rate of change of the temperature of an object is directly proportional to the difference between its own temperature and the temperature of its surroundings. The drawback of Newton’s law of cooling is that the temperature of surroundings should be constant during the cooling of the body.
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