The thermo e.m.f. E in volts of a certain thermo couple is found to vary with temperature T of hot junction while coldjunction is kept at 0ºC

E = 40T – ^T²⁄₂₀

The neutral temperature of the couple is

100º C
200º C
400º C
800º C

Solution

The e.m.f. developed in a thermo-couple is given by

E = α T + ¹⁄₂ β T²

where T is the temperature of hot junction, cold junction being at 0ºC. The thermo electric power of the couple is

α + ^β⁄₂ T
α + β T
\(\frac{\alpha T^{2}}{2}+\frac{\beta T^{3}}{6}\)
α / 2β

Solution

Thermo-electric power S=^dE⁄_dT= α + β T

One junction of a certain thermocouple is at a fixed temperature Tr and the other junction is at a temperature T.The electormotive force for this is expressed by,

At temperature T= T₀/2, the thermo electric power is

k T₀/2
k T₀
\(\frac{KT_{0}^{2}}{2}\)
¹⁄₂K(T₀-T_i)²

Solution

The thermo e.m.f. of a thermocouple is given by E = 2164 t – 6.2 t². The neutral temperature and a temperature of inversion are

349, 174.5
174.5, 349
349, 698
698, 349

Solution

At neutral temperature, dE/dt = 0;

so 2164 – 6.2 × 2 × t_n = 0 or t_n= 174.5ºC.

At temperature of inversion, E = 0

⇒ 2164 ti– 6.2 \(t_{i}^{2}\)= 0 Hence ti= 349ºC.

A 4 μF conductor is charged 50 volts and then its plates are joined through a resistance of 1 k μ. The heat produced in the resistance is

0.16 J
1.28 J
0.64 J
0.32 J

Solution

The energy stored in the capacitor = ¹⁄₂ CV² ; This energy will be converted into heat in the resistor.

A fuse wire with aradius of 1 mm blows at 1.5 A. If the fuse wire of the same material should blow at 3.0 A, the radius of the wire must be

4^1/3 mm
√2 mm
0.5 mm
8.0 mm

Solution

In the circuit , the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance,the value of the resistor R will be

100 Ω
200 Ω
1000 Ω
500 Ω

Solution

In the circuit shown below, if the resistance of voltmeter is 4 kΩ, then the error in the reading of voltmeter will be

50%
68%
17%
33.3%

Seven resistances, each of value 20 Ω, are connected to a 2V battery as shown in the figure. The ammeter reading will be

1/10 A
3/10 A
4/10 A
7/10 A.

Resistances 1 Ω, 2Ω and 3Ωare connected to form a triangle.If a 1.5 V cell of negligible internal resistance is connected across the 3Ω resistor, the current flowing through this resistor will be

0.25 A
0.5 A
1.0 A
1.5 A

Solution

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