E5. A glass flask whose volume is exactly 1000 Cm² at 0⁰C is filled levelfull of mercury at this temp. when the flask and mercury are heated to 100 ⁰C 15 - 2 Cm² of mercury overflow.
The coefficiant of cubical expansion of Hg is 1.82 X 10^-4 Compute the coefficiant of linear expansion of glass.
Solution:- Hg overflow = 15 - 2 cm²
1000 (1 + ) - 1000(1 + _g ) = 15 - 2
_g = _e -
_g = 0.00003 / ⁰C
_g = = 1 X 10^-5 / ⁰C.

E6. A pendulam clock loses 12 Sec. a day if the temperature is 40 ⁰C and fast by 4 Sec. a day if the temperature is 20 ⁰C . Find the temperature at which the clock will show correct time and the coefficiant of linear expansion of the metal of the pendulum shaft.
Solution:- Time lost per day
= 1/2 (40 - T) X 86400 = 12
Time gained per day
= 1/2 (T - 20) X 86400 = 4
= 1.85 X 10^-5
T = 25 ⁰C.
Medium:-
M1. A versel is filled completely with 500 gm. of water and 1000 gm. of mercury. When 21200 cal. of heat is given to it , water of mass 3.52 g overflows. Calculate the coefficiant of volume expansion of mercury. The expansion of the versel way be neglected. Coefficiant of volume expansion of water = 1.5 x 10 ^-4 / ⁰C, density of mercury = 13.6 g/cc, density water = 1 gm / cc and specific heat of mercury = 0.03 cal/ g/⁰C.
Solution:- Water overflow = final volume of water + final volume of Hg - final volume of versel.

= 40 ⁰C
_e = 1.84 X 10 ^-4 (⁰C) ^-1

M2. a metallic slab weighs 50g in the air. If it is immersed in a liquid at a temperature 25 ⁰C . It weighs 45g. When the temperature of the liquid is raised to 100 ⁰C it weighs 45 - 1 gm. Calculate the coefficient of cubical expansion of liquid assuming the linear coefficient of the metal to be 12 X 10 ^-6 C ^-1.

Solution:- V₁₀₀ = V ₂₅ ( 1 + (t ₂ - t ₁))
= 3
V₂₅ - d₂₅ = Loss of wt at 25 ⁰C
= 50 - 45 = 5 g
V₁₀₀ . d₁₀₀ = Loss of wt at 100 ⁰C
= 50 - 45 - 1 = 4.9 g

_e = 3 X 10 ^-4 / ⁰C.

M2. A clock with a brass pendulum shaft keeps correct time at a certain temperature.
(a) How closely must be temperature controlled if the clock is not to be gain or lose more than 1 sec. a day ? Does the answer depend on the period of the pendulum.
(b) Will an increase of temperature cause the clock to gain or lose ? (_brass = 2 X 10 ^-5 / ⁰ C)

Solution:- No. of seconds lost or gained per day
= X 86400