Question

The volume of a metal sphere is increased by 1% of its original volume. When it is heated from 320k to 522k. Calculate the cubical expansion of the metal

Answer 1

THE ANSWER IS THE FOLLOW..............

γ = Coefficient of Volumetric expansion

γ = Coefficient of Volumetric expansionV = initial Volume before expansion

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 V

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 deg

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given as

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔT

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γ

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶Coefficient of linear expansion is given as

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶Coefficient of linear expansion is given asα = γ/3

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶Coefficient of linear expansion is given asα = γ/3α = (62.5 x 10⁻⁶)/3

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶Coefficient of linear expansion is given asα = γ/3α = (62.5 x 10⁻⁶)/3α = 20.83 x 10⁻⁶

γ = Coefficient of Volumetric expansionV = initial Volume before expansionΔV = expansion of Volume = 0.15% of V = 0.0015 VΔT = change in temperature = 24 degchange in Volume is given asΔV = V γ ΔTinserting the values0.0015 V = V (24) γγ = 62.5 x 10⁻⁶Coefficient of linear expansion is given asα = γ/3α = (62.5 x 10⁻⁶)/3α = 20.83 x 10⁻⁶Read more on Brainly.in - https://brainly.in/question/5180850#readmore

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Answer 2

The cubical expansion of the metal is $4.95\times10^{-5}\ K^{-1}$

Explanation:

Given that,

Initial temperature = 320 K

Final temperature = 522 K

The volume of a metal sphere is increased by 1% of its original volume.

We need to calculate the change the volume

Using formula of change in volume

$\Delta V=\dfrac{1}{100}V_{0}$

Where, V₀ = original volume

$\Delta V=0.10 V_{0}$

We need to calculate the cubical expansion of the metal

Using formula of change in volume

$\Delta V=V_{0}\times\gamma\times\Delta T$

$\gamma=\dfrac{\Delta V}{V_{0}\Delta T}$

Put the value into the formula

$\gamma=\dfrac{0.10 V_{0}}{V_{0}\times(522-320)}$

$\gamma=\dfrac{1}{202}\times10^{-2}$

$\gamma=0.0000495\ K^{-1}$

$\gamma=4.95\times10^{-5}\ K^{-1}$

Hence, The cubical expansion of the metal is $4.95\times10^{-5}\ K^{-1}$

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Topic : cubical expansion

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