Subject

Physics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is

  • 4

  • 3

  • 2

  • 2

485 Views

12.

Two conductors have the same resistance a 0°C but their temperature coefficients o resistance are α1 and α2. The respective temperature coefficients of their series parallel combinations are nearly

  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space fraction numerator straight alpha subscript 1 space plus space straight alpha subscript 2 over denominator 2 end fraction
  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space straight alpha subscript 1 space plus straight alpha subscript 2
  • space straight alpha subscript 1 space plus straight alpha subscript 2 comma space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction
  • space straight alpha subscript 1 space plus straight alpha subscript 2 comma space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction
662 Views

13.

The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y= 0.02(m) sin open square brackets 2 straight pi open parentheses fraction numerator straight t over denominator 0.04 space left parenthesis straight s right parenthesis end fraction minus fraction numerator straight x over denominator 0.50 space left parenthesis straight m right parenthesis end fraction close parentheses close square brackets The tension in the string is 

  • 4.0 N

  • 12.5

  • 0.5 N

  • 0.5 N

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14.

A rectangular loop has a sliding connector PQ of length

  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 6 straight R end fraction comma space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space minus straight I subscript 2 space equals space fraction numerator straight B space calligraphic l straight v over denominator straight R end fraction comma space straight I space equals fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction


C.

straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction

A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I1 + I2


Applying Kirchhoff's law
I1R + IR -vBl = 0 .... (i)
I2R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I1 = I2 = VBI/3R

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15.

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one–fourth its initial value. Then the ratio t1/t2 will be

  • 1

  • 1/2

  • 1/4

  • 1/4

234 Views

16.

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by

544 Views

17.

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

  • fraction numerator VR subscript 1 straight R subscript 2 over denominator square root of straight R subscript 1 superscript 2 plus straight R subscript 2 superscript 2 end root end fraction space at space straight t space equals space 0 space and space straight V over straight R subscript 2 space at space straight t space equals space infinity
  • straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator straight V left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis over denominator straight R subscript 1 straight R subscript 2 end fraction space at space equals space infinity
  • straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator VR subscript 1 straight R subscript 2 over denominator square root of straight R subscript 1 superscript 2 space plus space straight R subscript 2 superscript 2 end root end fraction space at space straight t space equals space infinity
  • straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator VR subscript 1 straight R subscript 2 over denominator square root of straight R subscript 1 superscript 2 space plus space straight R subscript 2 superscript 2 end root end fraction space at space straight t space equals space infinity
163 Views

18.

A thin semi–circular ring of radius r has a positive charge q distributed uniformly over it. The net field straight E with rightwards arrow on top at the center O is

  • fraction numerator straight q over denominator 4 straight pi squared straight epsilon subscript 0 straight r squared end fraction space bold j with hat on top
  • negative fraction numerator straight q over denominator 4 straight pi squared straight epsilon subscript 0 straight r squared end fraction
  • negative fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction straight j with hat on top
  • negative fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction straight j with hat on top
797 Views

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19.

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

  • zero

661 Views

20.

Let there be a spherically symmetric charge distribution with charge density varying as  space straight rho left parenthesis straight r right parenthesis space equals space space straight rho subscript 0 space open parentheses 5 over 4 minus straight r over straight R close parentheses upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r ( r < R) from the origin is given by

  • fraction numerator 4 space straight pi space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 3 minus straight r over straight R close parentheses
  • fraction numerator begin display style space straight rho subscript 0 straight r end style over denominator begin display style 4 straight epsilon subscript 0 end style end fraction space open parentheses fraction numerator begin display style 5 end style over denominator begin display style 3 end style end fraction minus fraction numerator begin display style straight r end style over denominator begin display style straight R end style end fraction close parentheses
  • fraction numerator 4 space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 3 minus straight r over straight R close parentheses
  • fraction numerator 4 space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 3 minus straight r over straight R close parentheses
568 Views

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