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 Multiple Choice QuestionsMultiple Choice Questions

1.

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then 

  • the magnetic field is zero only on the axis of the pipe

  • the magnetic field is different at different points inside the pipe

  • the magnetic field at any point inside the pipe is zero

  • the magnetic field at all points inside the pipe is the same, but not zero


C.

the magnetic field at any point inside the pipe is zero

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2.

The phase difference between the alternating current and emf is π/2. Which of the following cannot be the constituent of the circuit?

  • C alone

  • R.L

  • L.C

  • L alone


B.

R.L

0<phase difference for R-L circuit < π/2 

644 Views

3.

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2µF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is

  • 4 × 10−3 V

  • 2.5 × 10−2 V

  • 40 V

  • 250 V


D.

250 V

straight i space equals space 100 over 1000 space equals space 0.1 space straight A
straight V subscript straight L space equals space straight V subscript straight C space equals space fraction numerator 0.1 over denominator 200 space straight x space 2 space straight x space 10 to the power of negative 6 end exponent end fraction space equals space 250 space straight V
560 Views

4.

In a series LCR circuit, R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is

  • 305 W

  • 210 W

  • zero

  • 242 W


D.

242 W

The given circuit is under resonance as XL = XC. Hence, power dissipated in the circuit is 

P = V2/R = 242 W

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5.

Alternating current can not be measured by D.C. ammeter because

  • A.C. cannot pass through D.C.

  • A.C. changes direction

  • average value of current for complete cycle is zero

  • D.C. ammeter will get damaged.


C.

average value of current for complete cycle is zero

The full cycle of alternating current consists of two half cycles. For one half, currently is positive and for second half, current is negative. Therefore, for an a.c. cycle, the net value of current average out to zero. While for the half cycle, the value of current is different at different points. Hence, the alternating current cannot be measured by D.C. ammeter.

352 Views

6.

A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the circuit will be

  • 0.8

  • 0.4

  • 1.25

  • 0.125


A.

0.8

cosϕ space equals space straight R over straight Z space equals space 12 over 15 space equals 4 over 5 space equals space 0.8
174 Views

7.

In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be

  • 50 V

  • 50√2 V

  • 100 V

  • 0 V(zero)


D.

0 V(zero)

In an LCR series a.c. circuit, the voltage across inductor L leads the current by 900 and the voltage across capacitor C lags behind the current by 900. Hence, the voltage across LC combination will be zero.

579 Views

8.

Two voltameters one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are z1 and z2 respectively the charge which flows through the silver voltameter is

  • fraction numerator straight q over denominator 1 plus begin display style straight z subscript 1 over straight z subscript 2 end style end fraction
  • fraction numerator straight q over denominator 1 plus begin display style straight z subscript 2 over straight z subscript 1 end style end fraction
  • straight q straight z subscript 1 over straight z subscript 2
  • straight q straight z subscript 2 over straight z subscript 1


B.

fraction numerator straight q over denominator 1 plus begin display style straight z subscript 2 over straight z subscript 1 end style end fraction
straight q subscript 1 straight Z subscript 1 space equals space straight q subscript 2 straight Z subscript 2
straight q space equals space straight q subscript 1 space plus straight q subscript 2
therefore space straight q subscript 2 space equals space fraction numerator straight q over denominator 1 plus begin display style straight Z subscript 2 over straight Z subscript 1 end style end fraction
171 Views

9.

In an a.c. circuit the voltage applied is E = E0 sinπt. The resulting current in the circuit is I = I0 sin open parentheses ωt space minus space straight pi over 2 close parentheses.The power consumption in the circuit is given by

  • straight P space equals space fraction numerator straight E subscript 0 straight I subscript 0 over denominator square root of 2 end fraction
  • P = zero

  • straight P space equals fraction numerator straight E subscript 0 straight I subscript 0 over denominator 2 end fraction
  • straight P space equals space square root of 2 straight E subscript 0 straight I subscript 0

B.

P = zero

204 Views

10.

In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

  • 4L

  • 2L

  • L/2

  • L/4


C.

L/2

In the condition of resonance XL = XC
or

ωL =1/ ωC  = .................. (i)
Since, resonant frequency remains unchanged

So, square root of LC = constant

or LC = constant
∴ L1 C1 = L2 C2 ⇒ L×C = L2× 2C
L2 = L/2

304 Views