First slide

Thermal expansion

Question

Two liter glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains same. The volume of the mercury in side the flask is
$(α f o r g l a s s = 9 \times 10^{- 6} /^{0} C, γ_{H g} = 1.8 \times 10^{- 4} /^{0} C)$ $x \times 10^{- 6} m^{3}$ where x=

Moderate

A

3
B

30
C

300
D

3000

Solution

$a t a l l t e m p e r a t u r e s t h e v o l u m e o f t h e a i r i n s i d e t h e f l a s k r e m a i n s s a m e V_{C} γ_{C} = V_{L} γ_{L}; h e r e V i s v o l u m e o f c o n t a i n e r a n d v o l u m e o f l i q u i d; γ c o e f f i c i e n t o f v o l u m e e x p a n s i o n o f c o n t a i n e r a n d l i q u i d V_{L} = \frac{V_{C} γ_{C}}{γ_{L}} g i v e n V_{C} = 2 l i t e r; γ_{C} = 3 α = 3 x 9 x 10^{- 6} / {}^{0}C; γ_{L} = 1.8 x 10^{- 4} / {}^{0}C, s u b s t i t u t e V_{L} = \frac{2 l i t e r x 3 x 9 x 10^{- 6}}{1.8 x 10^{- 4}} \sin c e 1 l i t e r = 10^{- 3} m^{3} V_{L} = 300 x 10^{- 6} m^{3} = v o l u m e o f m e r c u r y i n t h e f l a s k$

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