A physicist took a hollow metallic sphere of radius 4 cm at a temperature of 27-C- He heated metallic sphere up to 127-C and found that diameter of the sphere increased to 8-2 cm- then what is the coefficient of real expansion-

Given initial radius r_1= 4 cm at temperature t_1=27^∘C ,then T_1=t_1+27 K T_1=273+27 T_1=300 K Final diameter is d_2=2 r_2 ⇒ ...

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Byju's Answer

Standard XII

Physics

Area and Volume Expansion

A physicist t...

Question

A physicist took a hollow metallic sphere of radius 4 cm at a temperature of $27^{\circ} C$ . He heated metallic sphere up to $127^{\circ} C$ and found that diameter of the sphere increased to 8.2 cm, then what is the coefficient of real expansion.

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Solution

Given initial radius $r_{1} = 4 c m$ at temperature
$t_{1} = 27^{\circ} C$ ,
then $T_{1} = t_{1} + 27 K$
$T_{1} = 273 + 27$
$T_{1} = 300 K$
Final diameter is $d_{2} = 2 r_{2}$
$\Rightarrow r_{2} = \frac{d_{2}}{2}$
Given $d_{2} = 8.2 c m$
then $r_{2} = 4 c m$
And final temp $T_{2} = t_{2} + 273$
$= 127 + 273 = 400 K$
Now coefficient of a real expansion $β$
$\frac{A_{2} - A_{1}}{A_{1} (T_{2} - T_{1})} = \frac{[(4.1)^{2} (4)^{2}] π}{16 π (400 - 300)}$
$= \frac{0.81}{16 π (100)}$
$0.05 \times 10^{- 2} = 5 \times 10^{- 4} K^{- 1}$

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