First slide

Thermal expansion

Question

Volume of mercury in the bulb of a thermometer is $10^{-6} m^{3}$ , area of cross section of capillary tube is $2 \times 10^{-7} m^{2}$ . If the temperature is raised by $100^{\circ} C$ , the length of mercury is ( $γ_{Hg} =18 \times {10^{- 5}}^{\circ} C^{-1}$ )

Moderate

A

9 cm
B

9 mm
C

18 cm
D

5 cm

Solution

$\begin{array}{rcl} v o l u m e o f m e r c u r y i n t h e b u l b o f t h e r m o m e t e r i s \\ {V=10}^{-6} m^{3} : \\ a r e a o f c r o s s s e c t i o n o f c a p i l l a r t y t u b e i s A=2 \times 10^{-7} m^{2} \\ c h a n g e i n t e m p e r a t u r e i s {ΔT=100}^{o} C \\ l e n g t h o f m e r c u r y i n t h e t u b e i s l & = & ? \\ r e a l e x p a n s i o n o f m e r c u r y i s γ_{reav} & = & γ_{r} =18 \times {10^{-5}}^{\circ} c^{- 1} \\ γ_{r} = \frac{ΔV}{V.Δt} \\ γ_{r} = \frac{A×l}{V.Δt} \\ l= \frac{γ_{r} .V.Δt}{A} \\ s u b s t i t u t e t h e g i v e n v a l u e s \\ l & = & \frac{18 \times 10^{- 5} \times 10^{- 6} \times 100}{2 \times 10^{- 7}} \\ l {=9×10}^{-2} m \\ L=9 cm \end{array}$

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