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 Multiple Choice QuestionsMultiple Choice Questions

1.

The Kirchhoff’s first law (∑ i= 0)   and second law (∑iR = ∑E), where the symbols have their usual meanings, are respectively based on

  • conservation of charge, conservation of energy

  • conservation of charge, conservation of momentum

  • conservation of energy, conservation of charge

  • conservation of energy, conservation of charge


A.

conservation of charge, conservation of energy

Kirchhoff's  Ist law or KCl states that the algebraic sum of current meeting at any junction is equal to zero. Thus, no charge has been accumulated at any junction i.e., the charge is conserved, and hence KCl is based on conservation of charge.

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2.

An energy source will supply a constant current into the load of its internal resistance is

  • equal to the resistance of the load.

  • very large as compared to the load resistance.

  • zero

  • zero


B.

very large as compared to the load resistance.

straight I space equals space fraction numerator straight E subscript straight o over denominator straight R plus straight r end fraction straight E over straight r space if space straight R less than less than straight r
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3.

A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio

  • 2

  • 1

  • 1/4

  • 1/4


A.

2

straight R subscript 1 space equals space fraction numerator straight rho subscript straight A calligraphic l subscript straight A over denominator πR subscript straight A superscript 2 end fraction
straight R subscript 2 space equals space fraction numerator straight rho subscript straight B calligraphic l subscript straight B over denominator πR subscript straight B superscript 2 end fraction
calligraphic l subscript straight A over calligraphic l subscript straight B space equals space fraction numerator straight rho subscript straight B straight R subscript straight A superscript 2 over denominator straight rho subscript straight A straight R subscript straight B superscript 2 end fraction space
equals space fraction numerator 2 straight rho subscript straight A space straight R subscript straight A superscript 2 over denominator straight rho subscript straight A 4 straight R subscript straight A superscript 2 end fraction
rightwards double arrow calligraphic l subscript straight A over calligraphic l subscript straight B space equals space 2 space space
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4.

The resistance of the series combination of two resistances is S. When they are joined in parallel through total resistance is P. If S = nP, then the minimum possible value of n is

  • 4

  • 3

  • 2

  • 2


A.

4

Let the resistance R1 and R2

So, S = R1+ R2;

straight P space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator straight R subscript 1 plus straight R subscript 2 end fraction
straight S space equals nP
straight R subscript 1 space plus space straight R subscript 2 space equals space fraction numerator nR subscript 1 straight R subscript 2 over denominator straight R subscript 1 plus straight R subscript 2 end fraction
left parenthesis straight R subscript 1 space plus straight R subscript 2 right parenthesis squared space equals space nR subscript 1 straight R subscript 2
If space straight R subscript 1 space equals space straight R subscript 2 comma space so space minimum space value space of space straight n space equals 4

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5.

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the length and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be

  • 3

  • 1/3

  • 8/9

  • 8/9


B.

1/3

straight I subscript 1 over straight I subscript 2 space equals space straight R subscript 1 over straight R subscript 2
left square bracket Current space divider space rule space since space voltage space is space same space in space parallel right square bracket
straight I subscript 1 over straight I subscript 2 space equals space straight L subscript 2 over straight L subscript 1 space straight x space fraction numerator straight r subscript 1 superscript 2 over denominator straight r subscript 2 superscript 2 end fraction
therefore comma space straight I subscript 1 over straight I subscript 2 space equals space 3 over 4 space straight x space open parentheses 2 over 3 close parentheses squared space equals space 1 third
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6.

Two thin, long parallel wires separated by a distance  ‘d’ carry a current of ‘i’ A in the same direction. They will

  • attract each other with a force of µ0i2/(2πd) 

  • repel each other with a force of µ0i2/(2πd)

  • attract each other with a force of µ0i2(2πd2 )

  • attract each other with a force of µ0i2(2πd2 )


A.

attract each other with a force of µ0i2/(2πd) 

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7.

In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

  • 50 cm

  • 80 cm

  • 40 cm

  • 40 cm


A.

50 cm

We have from meter bridge experiment,

straight R subscript 1 over straight R subscript 2 space equals space fraction numerator begin display style calligraphic l subscript 1 end style over denominator calligraphic l subscript 2 end fraction comma space where space calligraphic l subscript 2 space equals space left parenthesis 100 space minus space calligraphic l subscript 1 right parenthesis space c m
I n space t h e space f i r s t space c a s e comma space X divided by Y space equals space 20 divided by 80
I n space t h e space s e c o n d space c a s e space fraction numerator 4 X over denominator Y end fraction space equals space fraction numerator calligraphic l over denominator 100 minus calligraphic l end fraction
calligraphic l space equals space 50 space c m

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8.

The total current supplied to the circuit by the battery is

  • 1A

  • 2 A

  • 4 A

  • 4 A


C.

4 A

The given circle is written as
I = V/R
I = 6 V / 1.5 Ω
  = 4 A 

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9.

In a Wheatstone’s bridge, there resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridge to be balanced will be

  • straight P over straight Q space equals space fraction numerator straight R space left parenthesis straight S subscript 1 plus straight S subscript 2 right parenthesis over denominator straight S subscript 1 straight S subscript 2 end fraction
  • straight P over straight Q space equals space fraction numerator straight R over denominator straight S subscript 1 plus straight S subscript 2 end fraction
  • straight P over straight Q space equals space fraction numerator 2 straight R over denominator straight S subscript 1 plus straight S subscript 2 end fraction
  • straight P over straight Q space equals space fraction numerator 2 straight R over denominator straight S subscript 1 plus straight S subscript 2 end fraction

A.

straight P over straight Q space equals space fraction numerator straight R space left parenthesis straight S subscript 1 plus straight S subscript 2 right parenthesis over denominator straight S subscript 1 straight S subscript 2 end fraction
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10.

The electrochemical equivalent of a metal is 3.3 × 10−7 kg per coulomb. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 seconds will be

  • 19.8 × 10−7 kg

  • 9.9 × 10−7 kg

  • 6.6 × 10−7 kg

  • 6.6 × 10−7 kg


A.

19.8 × 10−7 kg

Mass of substance liberated at cathode m = zit
where, z = electro-chemical equivalent = 3.3× 10-7 kg/C
i = current flowing = 3A.
t = 2 sec
∴ m = 3.3× 10-7× 3× 2
= 19.8× 10-7 kg

140 Views