The Total solution for NCERT class 6-12
Answer - 1 : - The relation between kelvin scale and Celsius scale is T_K – 273.15 =T_C => T_C=T_K– 273.15
Answer - 2 : - As we know, triple point of water on absolute scale = 273.16 K, Size of one degree of kelvin scale on absolute scale A
Answer - 3 : - Here, R0 = 101.6 Ω; T0 = 273.16 K Case(i) R1= 165.5 Ω; T1 = 600.5 K, Case (ii) R2 =123.4 , T2 = ?Using the relation R = R0[1 + α (T – T0)]Case (i) 165.5 = 101.6 [1 + α (600.5 – 273.16)]
Answer - 4 : - (a) Triple point of water has a unique value i.e., 273.16 K. The melting point and boiling points of ice and water respectively do not have unique values and change with the change in pressure.(b) On Kelvin’s absolute scale, there is only one fixed point, namely, the triple-point of water and there is no other fixed point.(c) On Celsius scale 0 °C corresponds to the melting point of ice at normal pressure and the value of absolute temperature is 273.15 K. The temperature273.16 K corresponds to the triple point of water.(d)The Fahrenheit scale and Absolute scale are related as
Answer - 5 : -
Answer - 6 : -
T = 27 °C
At the temperature of 27 °C the length of the tape is 1m =100 cm
T1 = 45°C
At the temperature of 45 °C the length of the tape is 63 cm
Coefficient of linear expansion of steel = 1.20 × 10–5 K–1
Let L be the actual length of the steel rod and L’ is the lengthat 45 °C
L’ = L [1 + ∝(T1 – T)]
= 100 [1+ 1.20 × 10–5 (45-27)]
= 100[1+ 21.6 × 10–5]
= 100 + 2160 x 10–5
= 100 + 0.02160 = 100.02160
The actual length of the rod at 450 Cis
L2 =(100.02160 /100) x 63 = 63. 013608 cm
The length of the rod at 270 Cis 63.0 cm
Answer - 7 : -
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Answer - 10 : -