Question

$\huge\bold\red{QUESTION}$
Two spheres made of same substance have diameters in the ratio 1 : 2. Their thermal capacities are in the ratio of ___
(a) 1 : 2
(b) 1 : 8
(c) 1 : 4
(d) 2 : 1

Answer 1

$\\\;\underbrace{\underline{\sf{Question's\;\;Analysis\;:-}}}$

Here the Concept of Thermal Capacity has been used. We see we are given the ratio of the diameter of the spheres which are made of same material. First we can assume a constant to attain radii. Then we can find density since we are not given the mass. After finding all these, we can find the ratio of their Thermal Capacities.

Let's do it !!

________________________________

★ Formula Used :-

$\\\;\boxed{\sf{Density,\;\rho\;=\;\bf{\dfrac{Mass}{Volume}}}}$

$\\\;\boxed{\sf{Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\;\times\;\pi r^{3}}}}$

$\\\;\boxed{\sf{Thermal\;Capacity,\;C\;=\;\bf{Mass\;\times\;Specific\;Heat}}}$

________________________________

★ Solution :-

» Ratio of diameters of spheres = 1 : 2

» Both spheres are made of same substance.

• Let the constant by which the ratio of diameter should be multiplied to attain original value be 'x'.

• Also since, both spheres are made of same material then their Specific Heat will be same.

• Diameter of Sphere 1, d₁ = 1x

• Diameter of Sphere 2, d₂ = 2x

• Radius of Sphere 1, r₁ = ½(d) = 0.5 x

• Radius of Sphere 2, r₂ = ½(d) = 1 x

• Specific heat of both Sphere = S

• Mass of first sphere = m₁

• Mass of second sphere = m₂

• Thermal Capacity of Sphere 1 = C₁

• Thermal Capacity of Sphere 2 = C₂

• Density of both Spheres = ρ

Mass of the both spheres are different and their values too. But their density can be same since they are made of same substance and same figure.

________________________________

~ For the mass of both Spheres ::

We know that,

$\\\;\;\sf{:\rightarrow\;\;Density,\;\rho\;=\;\bf{\dfrac{Mass}{Volume}}}$

And volume of spheres is given as,

$\\\;\;\sf{:\rightarrow\;\;Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\;\times\;\pi r^{3}}}$

Now combining these both equations, we get,

$\\\;\;\sf{:\Rightarrow\;\;Density,\;\rho\;=\;\bf{\dfrac{Mass}{\bigg(\dfrac{4}{3}\;\times\;\pi r^{3}\bigg)}}}$

→ Mass of First Sphere, m₁ :

$\\\;\;\sf{:\Rightarrow\;\;\rho\;=\;\bf{\dfrac{m_{1}}{\bigg(\dfrac{4}{3}\;\times\;\pi r_{\tt{1}} ^{3}\bigg)}}}$

$\\\;\;\underline{\underline{\bf{:\Rightarrow\;\;m_{1}\;=\;\bf{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{1}} ^{3}\;\times\;\rho\bigg)}}}}$

→ Mass of Second Sphere, m₂ :

$\\\;\;\sf{:\Rightarrow\;\;\rho\;=\;\bf{\dfrac{m_{2}}{\bigg(\dfrac{4}{3}\;\times\;\pi r_{\tt{2}} ^{3}\bigg)}}}$

$\\\;\;\underline{\underline{\bf{:\Rightarrow\;\;m_{2}\;=\;\bf{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{2}} ^{3}\;\times\;\rho\bigg)}}}}$

________________________________

~ For the Ratio of Thermal Capacities ::

We know that, Thermal Capacity is given as,

$\\\;\;\sf{:\mapsto\;\;Thermal\;Capacity,\;C\;=\;\bf{Mass\;\times\;Specific\;Heat}}$

Also, ratio of both Thermal Capacities is given as,

$\\\;\;\sf{:\mapsto\;\;Ratio\;of\;Thermal\;Capacities\;=\;\bf{\dfrac{C_{1}}{C_{2}}}}$

Now combining these both equations, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{Thermal\;Capacity\;of\;Sphere\;1}{Thermal\;Capacity\;of\;Sphere\;2}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{Mass\;1\;\times\;Specific\;Heat}{Mass\;2\;\times\;Specific\;Heat}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{m_{1}\;\times\;S}{m_{2}\;\times\;S}}}$

Now by applying the masses which we got above in this equation, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{1}} ^{3}\;\times\;\rho\;\times\;S\bigg)}{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{2}} ^{3}\;\times\;\rho\;\times\;S\bigg)}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{4\;\times\;\pi\;\times\;r_{\tt{1}} ^{3}\;\times\;\rho\;\times\;S\;\times\;3}{4\;\times\;\pi\;\times\;r_{\tt{2}} ^{3}\;\times\;\rho\;\times\;S\;\times\;3}}}$

Cancelling 4, 3, ρ, S and π , we get

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{r_{\tt{1}} ^{3}}{r_{\tt{2}} ^{3}}}}$

Now by applying the value of radius of both spheres, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{(0.5)^{3}}{(1)^{3}}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{0.125}{1}}}$

Since, one term here is in decimals which cannot be done into ratio. So first we need to convert this decimal to a nearest unit digit that is 1.

Multiplying the term by 8, we get

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{0.125}{1}\;\times\;\dfrac{8}{8}}}$

$\\\;\;\bf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{1}{8}}}$

$\\\;\;\bf{:\Longrightarrow\;\;C_{1}\;:\;C_{2}\;=\;\bf{1\;:\;8}}$

So, option b.) 1 : 8 is correct.

$\\\;\underline{\boxed{\tt{Thus,\;\;ratio\;\;of\;\;Thermal\;\;Capacities\;\;of\;\;Spheres\;\;=\;\bf{1\;:\;8}}}}$

________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;Specific\;Heat\;=\;\dfrac{Heat}{Mass\;\times\;Temperature}}$

Answer 2

★ Solution :-

» Ratio of diameters of spheres = 1 : 2

» Both spheres are made of same substance.

• Let the constant by which the ratio of diameter should be multiplied to attain original value be 'x'.

• Also since, both spheres are made of same material then their Specific Heat will be same.

Diameter of Sphere 1, d₁ = 1x

Diameter of Sphere 2, d₂ = 2x

Radius of Sphere 1, r₁ = ½(d) = 0.5 x

Radius of Sphere 2, r₂ = ½(d) = 1 x

Specific heat of both Sphere = S

Mass of first sphere = m₁

Mass of second sphere = m₂

Thermal Capacity of Sphere 1 = C₁

Thermal Capacity of Sphere 2 = C₂

Density of both Spheres = ρ

Mass of the both spheres are different and their values too. But their density can be same since they are made of same substance and same figure.

________________________________

~ For the mass of both Spheres ::

We know that,

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;Density,\;\rho\;=\;\bf{\dfrac{Mass}{Volume}}}\end{gathered}

:→Density,ρ=

Volume

Mass

And volume of spheres is given as,

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\;\times\;\pi r^{3}}}\end{gathered}

:→VolumeofSphere=

3

4

×πr

3

Now combining these both equations, we get,

\begin{gathered}\\\;\;\sf{:\Rightarrow\;\;Density,\;\rho\;=\;\bf{\dfrac{Mass}{\bigg(\dfrac{4}{3}\;\times\;\pi r^{3}\bigg)}}}\end{gathered}

:⇒Density,ρ=

(

3

4

×πr

3

)

Mass

→ Mass of First Sphere, m₁ :

\begin{gathered}\\\;\;\sf{:\Rightarrow\;\;\rho\;=\;\bf{\dfrac{m_{1}}{\bigg(\dfrac{4}{3}\;\times\;\pi r_{\tt{1}} ^{3}\bigg)}}}\end{gathered}

:⇒ρ=

(

3

4

×πr

1

3

)

m

1

\begin{gathered}\\\;\;\underline{\underline{\bf{:\Rightarrow\;\;m_{1}\;=\;\bf{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{1}} ^{3}\;\times\;\rho\bigg)}}}}\end{gathered}

:⇒m

1

=(

3

4

πr

1

3

×ρ)

→ Mass of Second Sphere, m₂ :

\begin{gathered}\\\;\;\sf{:\Rightarrow\;\;\rho\;=\;\bf{\dfrac{m_{2}}{\bigg(\dfrac{4}{3}\;\times\;\pi r_{\tt{2}} ^{3}\bigg)}}}\end{gathered}

:⇒ρ=

(

3

4

×πr

2

3

)

m

2

\begin{gathered}\\\;\;\underline{\underline{\bf{:\Rightarrow\;\;m_{2}\;=\;\bf{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{2}} ^{3}\;\times\;\rho\bigg)}}}}\end{gathered}

:⇒m

2

=(

3

4

πr

2

3

×ρ)

________________________________

~ For the Ratio of Thermal Capacities ::

We know that, Thermal Capacity is given as,

\begin{gathered}\\\;\;\sf{:\mapsto\;\;Thermal\;Capacity,\;C\;=\;\bf{Mass\;\times\;Specific\;Heat}}\end{gathered}

:↦ThermalCapacity,C=Mass×SpecificHeat

Also, ratio of both Thermal Capacities is given as,

\begin{gathered}\\\;\;\sf{:\mapsto\;\;Ratio\;of\;Thermal\;Capacities\;=\;\bf{\dfrac{C_{1}}{C_{2}}}}\end{gathered}

:↦RatioofThermalCapacities=

C

2

C

1

Now combining these both equations, we get,

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{Thermal\;Capacity\;of\;Sphere\;1}{Thermal\;Capacity\;of\;Sphere\;2}}}\end{gathered}

:⟹

C

2

C

1

=

ThermalCapacityofSphere2

ThermalCapacityofSphere1

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{Mass\;1\;\times\;Specific\;Heat}{Mass\;2\;\times\;Specific\;Heat}}}\end{gathered}

:⟹

C

2

C

1

=

Mass2×SpecificHeat

Mass1×SpecificHeat

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{m_{1}\;\times\;S}{m_{2}\;\times\;S}}}\end{gathered}

:⟹

C

2

C

1

=

m

2

×S

m

1

×S

Now by applying the masses which we got above in this equation, we get,

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{1}} ^{3}\;\times\;\rho\;\times\;S\bigg)}{\bigg(\dfrac{4}{3}\:\pi\:r_{\tt{2}} ^{3}\;\times\;\rho\;\times\;S\bigg)}}}\end{gathered}

:⟹

C

2

C

1

=

(

3

4

πr

2

3

×ρ×S)

(

3

4

πr

1

3

×ρ×S)

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{4\;\times\;\pi\;\times\;r_{\tt{1}} ^{3}\;\times\;\rho\;\times\;S\;\times\;3}{4\;\times\;\pi\;\times\;r_{\tt{2}} ^{3}\;\times\;\rho\;\times\;S\;\times\;3}}}\end{gathered}

:⟹

C

2

C

1

=

4×π×r

2

3

×ρ×S×3

4×π×r

1

3

×ρ×S×3

Cancelling 4, 3, ρ, S and π , we get

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{r_{\tt{1}} ^{3}}{r_{\tt{2}} ^{3}}}}\end{gathered}

:⟹

C

2

C

1

=

r

2

3

r

1

3

Now by applying the value of radius of both spheres, we get,

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{(0.5)^{3}}{(1)^{3}}}}\end{gathered}

:⟹

C

2

C

1

=

(1)

3

(0.5)

3

\begin{gathered}\\\;\;\sf{:\Longrightarrow\;\;\dfrac{C_{1}}{C_{2}}\;=\;\bf{\dfrac{0.125}{1}}}\end{gathered}

:⟹

C

2

C

1

=

1

0.125

Since, one term here is in decimals which cannot be done into ratio. So first we need to convert this decimal to a nearest unit digit that is 1.

Multiplying the term by 8, we get

:⟹

C

2

C

1

=

1

0.125

×

8

8

:⟹

C

2

C

1

=

8

1

:⟹C

1

:C

2

=1:8

So, option b.) 1 : 8 is correct.

Thus,ratioofThermalCapacitiesofSpheres=1:8