Answer:
e) None of these is true
Explanation:
Given that
Temperature = 0 K
We know that relationship between kelvin and Farenheit scale

Now by putting the values


So F= - 459.67°F
So we can say that 0 K is equal to - 459.67°F.
So the our option e is correct.
Answer:
The magnitude of minimum potential difference is 1800 V
Explanation:
Given:
Electric field 
Gap between electrodes
m
For finding the minimum potential difference,


V
Therefore, the magnitude of minimum potential difference is 1800 V
Answer:
i. 6.923 V
ii. The e.m.f. = 22.5 V
Explanation:
i. The given parameters are;
Length of potentiometer = 1 m
The resistance of the potentiometer = 10 Ω
The e. m. f. of the attached cell = 9 V
The current, I flowing in the circuit = e. m. f/(Total resistance)
The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A
The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire
The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V
ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

Where:
E = e.m.f. of the balance point cell
= Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω
= Resistance of the cell in the circuit = 3 Ω
V = e.m.f. attached cell = 9 V

E = 7.5*3 = 22.5 V
The e.m.f. = 22.5 V
Answer:
k_max = 31.82 w/mk
k_min = 17.70 w/mk
Explanation:
a) the maximum thermal conductivity is given as

where k_m is thermal conductvitiy of metal
k_p is thermal conductvitiy of carbide
v_m = proportion of metal in the cement = 0.17
v_p = proportion of carbide in the cement = 0.83
= 66*0.17 + 28*0.83
k_max = 31.82 w/mk
b) the minimum thermal conductivity is given as

= \frac{28+66}{28*0.17 +66*0.83}
k_min = 17.70 w/mk