Triple point of water on absolute scale A, \(T_1\) = 200 A
Triple point of water on absolute scale B, \(T_2\) = 350 B
Triple point of water on Kelvin scale, \(T_K\) = 273.15 K
The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.
\(T_1 = T_K\)
200 A = 273.15 K
∴ A = \(\frac{273.15}{200}\)
The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
\(T_2 = T_K\)
350 B = 273.15
∴ B = \(\frac{273.15}{350}\)
\(T_A\) is triple point of water on scale A.
\(T_B\) is triple point of water on scale B.
∴ \(\frac{273.15}{200}\) x \(T_A\) = \(\frac{273.15}{350}\) x \(T_B\)
\(T_A = \frac{200}{350} T_B\)
Therefore, the ratio \(T_A : T_B\) is given as 4 : 7.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = \(R_0\) [1 + α (T – \(T_0\))]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |