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 Multiple Choice QuestionsMultiple Choice Questions

91.

A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 15 V battery and an external resistance 40 Ω. A secondary cell of emf E in the secondary circuit is balanced by 240 cm long potentiometer wire. The emf E of the cell is

  • 2.4 V

  • 1.2 V

  • 2.0 V

  • 3 V


92.

In the following circuit, the current flowing through 1 kΩ resistor is

            

  • 0 mA

  • 5 mA

  • 10 mA

  • 15 mA


93.

The current I through 10 Ω resistor in the circuit given below is

           

  • 50 mA

  • 20 mA

  • 40 mA

  • 80 mA


94.

In the figure shown below, the terminal voltage across E2 is

        

  • 12 V

  • 12.66 V

  • 11.34 V

  • 11.66 V


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95.

The drift velocity of the electrons in a copper wire of length 2 m under the application of a potential difference of 220 V is 0.5 ms-1. Their mobility (in m2V-1s-1)

  • 2.5 × 10-3 

  • 2.5 × 10-2

  • 5 × 102

  • 5 × 10-3


96.

When two resistances R1 and R2 are connected in series, they consume 12 W power. When they are connected in parallel, they consume 50 W power. What the ratio of the powers of R1 and R2 ?

  • 1/4

  • 4

  • 3/2

  • 3


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97.

In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, heat developed in 2 Ω must be about (in Js-1)

     

  • 25

  • 20

  • 30

  • 35


C.

30

The simplified circuit is shown below :

             

From figure,

 15 I1 = 5 I2  or  I2 = 15 I15 = 3I1  I = I1 + I2 = I23 + I2 = 4I23              ........ (i)But,  I22 × 5 = 42or             I22 = 425 = 8.4

Putting value of l2 in Eq. (i), we get

         I = 43 × 8.4

Therefore, heat dissipated across 2 Ω

= I2 × 2= 169 × 8.4 × 2 30 Js-1


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98.

When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be 540 cm. If the balancing length becomes 500 cm when the cell is short circuited with 1 Ω, the internal of the cell is

  • 0.08 Ω

  • 0.04 Ω

  • 1.0 Ω

  • 1.08 Ω


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99.

A resistor 30 Ω, inductor of reactance 10 Ω and capacitor of reactance 10 Ω are connected in series to an AC voltage source e = 3002 sin ωt. The current in the circuit is

  • 102 A

  • 10 A

  • 3011 A

  • 30/ 11 A


100.

The resistance of a wire at room temperature 30˜C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [The  temperature coefficient of resistance of the material of the wire is 0.002/°C]

  • 36°C

  • 83°C

  • 63°C

  • 33°C


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