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 Multiple Choice QuestionsMultiple Choice Questions

1.

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 x 103Am-1.The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is

  • 30 mA

  • 60 mA

  • 3 A

  • 6 A


C.

3 A

For solenoid, the magnetic field needed to be magnetised the magnet,
B =μonl
Where, n = 100
l = 10 cm = 10/100 m  = 0.1m
3 space straight x space 10 cubed space equals space fraction numerator 100 over denominator 0.1 end fraction space straight x space l
l space equals 3 straight A

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2.

An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be:
(straight e to the power of 5 space approximately equal to space 150)

  • 100 mA

  • 67 mA

  • 6.7 mA

  • 0.67 mA


D.

0.67 mA

After long time inductor behaves as short-circuit.
At t = 0, the inductor behaves as short circuit.The current,

straight I subscript 0 space equals space straight E subscript 0 over straight R space equals space fraction numerator 15 straight V over denominator 0.15 space kΩ end fraction space equals space 100 space mA

As K2 is closed, current through the indicator starts decay which is given at any time t as
straight I space equals space straight I subscript 0 straight e to the power of fraction numerator negative tR over denominator straight L end fraction end exponent space equals space left parenthesis 100 space mA right parenthesis space straight e to the power of fraction numerator negative tx 15000 over denominator 3 end fraction end exponent
straight t space equals space 1 space ms
straight I space equals space left parenthesis 100 mA right parenthesis straight e to the power of fraction numerator 1 space straight x space 10 to the power of negative 3 end exponent space straight x space 15 space straight x 10 cubed over denominator 3 end fraction end exponent
straight I space equals space left parenthesis 100 space mA right parenthesis straight e space equals space 06737 space mA space or space straight I space equals space 0.67 space mA

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3.

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement- I: Higher the range, greater is the resistance of ammeter.
Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.

  • Statement – I is true, Statement – II is true, Statement – II is the correct explanation of statement- I.

  • Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I.

  • Statement – I is true, statement – II is false.

  • Statement – I is false, Statement – II is true


D.

Statement – I is false, Statement – II is true

For Ammeter, S = fraction numerator straight I subscript straight g straight G over denominator straight I minus straight I subscript straight g end fraction
So for I to increase, S should decrease, so additional S can be connected across it.

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4.

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to:

  • 1.6 x 10-8Ωm

  • 1.6 x 10-7Ωm

  • 1.6 x 10-6Ωm

  • 1.6 x 10-5Ωm


D.

1.6 x 10-5Ωm


vd = 25 x 10-4 m/s ⇒ n = 8 x 1028 /m3
we know that
J= nevd or I = nevd A
where, symbols have their usual meaning
rightwards double arrow space straight V over straight R space equals nev subscript straight d straight A
or space fraction numerator straight V over denominator begin display style ρL over straight A end style end fraction equals space nev subscript straight d straight A
or space straight V over ρL equals nev subscript straight d
straight rho space equals space fraction numerator straight V over denominator nev subscript straight d straight L end fraction
space equals space fraction numerator 5 over denominator 8 space straight x space 10 to the power of 28 space straight x space 16 space straight x space 10 to the power of negative 19 end exponent space straight x space 25 space straight x space 10 to the power of negative 4 end exponent space straight x space 0.1 end fraction space

straight rho space equals space 16 space straight x space 10 to the power of negative 5 end exponent space Ωm
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5.

In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly point C is disconnected from point A and connected to point B at time t= 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to 

  • fraction numerator straight e over denominator 1 minus straight e end fraction
  • 1

  • -1

  • fraction numerator 1 minus straight e over denominator straight e end fraction

C.

-1

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6.

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

  • zero

  • 2.9V

  • 13.3 V

  • 10.04 V


D.

10.04 V

Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω

Voltage across bulb before heater is switched on,straight V subscript 1 space equals space 120 over 246 space straight x space 240
Voltage across bulb after heater is switched on,straight V subscript 2 equals 120 over 54 straight x 48
A decrease in the voltage is V1 − V2 = 10.04 (approximately)

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7.

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

  • 3 V/m

  • 6 V/m

  • 9 V/m

  • 12 V/m


B.

6 V/m

Ε = cB
Where E = electric field,
B = magnetic field
c = speed of EM waves
= 3 × 108 × 20 × 10–9 = 6 V/m

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8.

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ’ with the vertical. If wires have mass λ per unit length then the value of I is: (g = gravitational acceleration)

  • sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root
  • 2 space sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root
  • 2 square root of πgL over straight mu subscript 0 tanθ end root
  • square root of πλgL over straight mu subscript 0 tanθ end root

B.

2 space sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root

consider free body diagram of the wire
As the wires are in equilibrium. they must carry current in opposite direction.



Here, straight F subscript straight B space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator italic 2 pi d end fraction, where l is the length of each wire are d is a separation between wires.
 From the diagram, d = 2L sin θ
T = cos θ = mg = λlg 
(in vertical direction) ..... (i)
 straight T space sin space straight theta space equals space straight F subscript straight B space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator 4 straight pi space straight L space sin space straight theta end fraction
(In horizontal direction)  ..... (ii)
From eqs. (i) and (ii)
fraction numerator straight T space sin space straight theta over denominator straight T space cos space straight theta end fraction space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator 4 πL space sin space straight theta space xλ l straight g end fraction
straight I space equals space square root of fraction numerator 4 πλLg space sin squared straight theta over denominator straight mu subscript straight theta space cos space straight theta end fraction end root
equals space 2 space sin space straight theta space square root of fraction numerator πλlg over denominator straight mu subscript straight o cos space straight theta end fraction end root

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9.

In the circuit shown, the current in the 1Ω resistor is:

  • 1.3 A from P to Q

  • 0 A

  • 0.13 A, from Q to P

  • 0.13 A, from P to Q


C.

0.13 A, from Q to P

Connect Point Q to ground and by applying Kirchhoff's laws
consider the grounded circuit as shown below,


Applying Kirchhoff's law at point Q,
Incoming current at Q = outgoing current from Q
fraction numerator straight V plus 6 over denominator 3 end fraction space plus straight V over 1 space equals space fraction numerator 9 minus straight V over denominator 5 end fraction
or space space straight V open square brackets 1 third plus 1 fifth plus 1 close square brackets space equals space 9 over 5 minus 2
or space straight V space open square brackets fraction numerator begin display style 5 plus 3 plus 15 end style over denominator 15 end fraction close square brackets space equals space fraction numerator 9 minus 10 over denominator 5 end fraction
or space straight V space open square brackets 23 over 15 close square brackets space equals space minus space 1 fifth space or space straight V space equals space minus space fraction numerator negative 3 over denominator 23 end fraction space equals space minus space 0.13 space straight V
Thus, current in the 1Ω resistance is 0.13 A, from Q to P

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10.

The current voltage relation of the diode is given by I = e(1000V/T-1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?

  • 0.2 mA

  • 0.02 mA

  • 05 mA

  • 0.05 mA


A.

0.2 mA

Given, I =(e1000V/T-1)mA
dv =±0.01 V 
T = 300 K
I = 5 mA
I = e1000V/T-1
I +1 = e1000V/T
Taking log on both sides, we get
log(l+T) = 1000V/T
On differentiating,
fraction numerator dl over denominator 1 plus 1 end fraction space equals space 1000 over straight T space dV
dl space equals space fraction numerator 1000 over denominator straight l plus straight T end fraction space equals space 1000 over straight T space dV
dl space equals space 1000 over straight T space straight x space left parenthesis straight l plus 1 right parenthesis space dV
rightwards double arrow space dl space equals space 1000 over 300 space straight x space left parenthesis 5 plus 1 right parenthesis space straight x 0.01 space equals space 0.2 space mA
So, error in the value of current is 0.2 mA 

283 Views