61. The thermo e.m.f E in volts of a certain thermocouple is found to  vary with temperature difference θ in °C between the two junctions according to the relation
    

    E - 30θ -	θ²
    	15

    
    The neutral temperature for the thermocouple will be ​
    

1. 1. 30° C ​

   
   
   

   2. 450° C ​

   
   
   

   3. 400 ° C ​

   
   
   

   4. 225° C

   
   
   

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1. View Hint View Answer Discuss in Forum

   E = 30Ω -	θ²
   	15

   
   For neutral Temperature, de/dθ = 0
   0 = 30 - (2/15)θ
   ∴ 0 = 15 x 15 = 225° C
   Hence, neutral temperature is 225°C.

   Correct Option: D

   E = 30Ω -	θ²
   	15

   
   For neutral Temperature, de/dθ = 0
   0 = 30 - (2/15)θ
   ∴ 0 = 15 x 15 = 225° C
   Hence, neutral temperature is 225°C.

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62. Consider the following two statements: ​
    (A)​Kirchhoff's junction law follows from the conservation of charge.
    (B)​Kirchhoff's loop law follows from the conservation of energy. ​
    Which of the following is correct?​
    

1. 1. ​Both (A) and (B) are wrong ​

   
   
   

   2. (a) is correct and (B) is wrong

   
   
   

   3. (a) is wrong and (B) is correct

   
   
   

   4. Both (A) and (B) are correct

   
   
   

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1. View Hint View Answer Discuss in Forum

   Junction law follows from conservation of charge and loop law is the conservation of energy
   

   Correct Option: D

   Junction law follows from conservation of charge and loop law is the conservation of energy
   

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63. In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367×10^–3 kg / C)​

1. 1. ​1.76 × 10^–3 kg ​

   
   
   

   2. 9.67 × 10^–3 kg ​

   
   
   

   3. 17.61 × 10^–3 kg ​

   
   
   

   4. 3.67 × 10^–3 kg

   
   
   

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1. View Hint View Answer Discuss in Forum

   I =	P	=	100 × 103	A =	105	A
   	V		125		60

   
   E.C.E. = 0.367 × 10^-6 kg C ^-1
   Charge per minute = (I × 60) C
   

   =	105 × 60	C =	6 × 106	C
   	125		125

   
   ∴Mass liberated
   

   =	6 × 106	× 0.367 × 10-6
   	125

   
   

   =	6 × 1000 × 10-3	= 17.616 × 10-3kg
   	125

   
   

   Correct Option: C

   I =	P	=	100 × 103	A =	105	A
   	V		125		60

   
   E.C.E. = 0.367 × 10^-6 kg C ^-1
   Charge per minute = (I × 60) C
   

   =	105 × 60	C =	6 × 106	C
   	125		125

   
   ∴Mass liberated
   

   =	6 × 106	× 0.367 × 10-6
   	125

   
   

   =	6 × 1000 × 10-3	= 17.616 × 10-3kg
   	125

   
   

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64. In the circuit shown in the figure, if potential at point A is taken to be zero, the potential at point B is​​
    

1. 1. –1V ​

   
   
   

   2. + 2V ​

   
   
   

   3. –2V ​

   
   
   

   4. + 1V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Current from D to C = 1A ​
   ∴ V_D – V_C = 2 × 1 = 2V ​ 
   V_A = 0
   ∴ V_C = 1V, 
   ∴ V_D – VC = 2 ​
   ⇒ V_D – 1 = 2
   ∴ V_D = 3V ​
   ∴ V_D – V_B = 2
   ∴ 3 – V_B = 2  ∴ V_B = 1V

   Correct Option: D

   Current from D to C = 1A ​
   ∴ V_D – V_C = 2 × 1 = 2V ​ 
   V_A = 0
   ∴ V_C = 1V, 
   ∴ V_D – VC = 2 ​
   ⇒ V_D – 1 = 2
   ∴ V_D = 3V ​
   ∴ V_D – V_B = 2
   ∴ 3 – V_B = 2  ∴ V_B = 1V

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65. A thermocouple of negligible resistance produces an e.m.f. of 40 µV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1µA/div, is employed with the termocouple. The smallest value of temperature difference that can be detected by the system will be​​

1. 1. 0.5°C ​

   
   
   

   2. 1°C ​

   
   
   

   3. 0.1°C ​

   
   
   

   4. 0.25°C

   
   
   

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1. View Hint View Answer Discuss in Forum

   1 division = 1µA ​
   Current for 1°C = 40μV/10 = 4μA ​
   1μA =1/4 °C = 0.25°C.

   Correct Option: D

   1 division = 1µA ​
   Current for 1°C = 40μV/10 = 4μA ​
   1μA =1/4 °C = 0.25°C.

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