The base unit of time in the metric and SI system is the second.
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
101397.16 pa
Explanation:
The pressure recorded will be equal to pgh
Where p = density of mercury = 13.6x10^3 kg/m^ 3
g = acceleration due to gravity 9.81 m/s^2
h = height of mercury in the column = 760 mm = 760x10^-3 m
Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa
