Subject

Physics

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

31.

Two metal wires of identical dimensions are connected in series if σ1a nd σ2 are the conductivity of the metal wires respectively, the conductivity of the combination is 

  • fraction numerator 2 straight sigma subscript 1 straight sigma subscript 2 over denominator straight sigma subscript 2 space plus straight sigma subscript 2 end fraction
  • fraction numerator straight sigma subscript 1 space plus straight sigma subscript 2 over denominator 2 space straight sigma subscript 1 straight sigma subscript 2 end fraction
  • fraction numerator straight sigma subscript 1 space plus straight sigma subscript 2 over denominator straight sigma subscript 1 straight sigma subscript 2 end fraction
  • fraction numerator straight sigma subscript 1 space plus straight sigma subscript 2 over denominator straight sigma subscript 1 straight sigma subscript 2 end fraction
3220 Views

32.

A series R -C circuit is connected to an alternating voltage source. Consider tow situations:

1. When the capacitor is air filled.
2. When the capacitor is mica filled.

Current through resistor is i and voltage across capacitor is V then

  • Va < Vb

  • Va > Vb

  • ia >ib

  • ia >ib

971 Views

33.

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 Ω all connected in series. If the ammeter has a coil of resistance 480 Ω and a shunt 20 Ω then reading in the ammeter will be

  • 0.5 A

  • 0.25 A

  • 2 A

  • 2 A

852 Views

34.

An electron moves on a straight line path XY as shown. The abcd is a coil adjacent in the path of the electron. What will be the direction of the current, if any induced in the coil?

  • abcd

  • adcb

  • the current will reverse its direction as the electron goes past the coil

  • the current will reverse its direction as the electron goes past the coil

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35.

A potentiometer wire of length L and a resistance r are connected in series with battery of e.m.f. Eo and a resistance r1. An unknown e.m.f is balanced at a length l of the potentiometer wire. The e.m.f E will be given by

  • fraction numerator LE subscript straight o straight r over denominator Ir subscript 1 end fraction
  • fraction numerator straight E subscript straight o straight r over denominator left parenthesis straight r space plus straight r subscript 1 right parenthesis end fraction. space I over L
  • fraction numerator straight E subscript straight o straight I over denominator straight L end fraction
  • fraction numerator straight E subscript straight o straight I over denominator straight L end fraction
2517 Views

36.

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is 

  • 4/9

  • 9/4

  • 27/5

  • 27/5

1116 Views

37.

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is 

(h = planck's constant, c= speed of light)

  • hc/ 2λ

  • hc/λ

  • 2 hc/λ

  • 2 hc/λ

1828 Views

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38.

Light of wavelength 500 nm is incident on metal with work function 2.28 eV. The de - Broglie wavelength of the emitted electrons is 

  • <2.8 x 10-10

  • <2.8 x 10-9 m

  • > equal  to 2.8 x 10-9 m

  • > equal  to 2.8 x 10-9 m


B.

<2.8 x 10-9 m

As, energy of photon, E = hv

straight E space space equals space hc over straight lambda
rightwards double arrow space straight E space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 over denominator 500 space straight x space 10 to the power of negative 9 end exponent end fraction
rightwards double arrow space straight E space equals space fraction numerator 0.0397 space straight x space 10 to the power of negative 34 end exponent space straight x space 10 to the power of 8 over denominator 10 to the power of negative 9 end exponent end fraction space equals space 0.0397 space straight x space 10 to the power of negative 21 end exponent space straight J
equals space fraction numerator 0.0397 space straight x space 10 to the power of negative 21 end exponent over denominator 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction space equals space 0.0248 space straight x space 10 squared space eV
space equals space 2.48 space eV
According space to space Einstein apostrophe straight s space photoelectric space emission comma space
we space have
KE subscript max space equals space straight E space minus space straight W space equals space 2.48 space minus space 2.28 space equals space 0.2 space eV
For space de space minus space broglie space wavelength space of space the space emitted space electron comma

straight lambda subscript straight e space min end subscript space equals space fraction numerator 12.27 over denominator square root of KE subscript max space eV end root end fraction space equals space fraction numerator 12.27 over denominator square root of 0.2 end root end fraction

equals space 27.436 space straight A with straight o on top space space equals 27.436 space straight x space 10 to the power of negative 10 end exponent straight m
Thus comma space minimum space wavelength space space of space the space emitted space electron space is
straight lambda subscript min space equals space 2.7436 space straight x 10 to the power of negative 9 end exponent space straight m space
straight lambda greater or equal than space straight lambda subscript min

2828 Views

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39.

The input signal given to a CE amplifier having a voltage gain of 150 is straight V subscript straight i space equals space open parentheses 15 space straight t space plus space fraction numerator 2 straight pi over denominator 3 end fraction close parentheses The corresponding output signal will be

  • 300 space cos space open parentheses 15 space straight t space plus straight pi over 3 close parentheses
  • 75 space cos space open parentheses 15 space straight t space plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses
  • 2 space cos space open parentheses 15 space straight t space plus fraction numerator 5 straight pi over denominator 3 end fraction close parentheses
  • 2 space cos space open parentheses 15 space straight t space plus fraction numerator 5 straight pi over denominator 3 end fraction close parentheses
1178 Views

40.

A nucleus of uranium decays at rest into nuclei of thorium and helium. Then, 

  • the helium nucleus has more kinetic energy than the thorium nucleus

  • the helium nucleus has less momentum than the thorium nucleus

  • the helium nucleus has more momentum than the thorium nucleus

  • the helium nucleus has more momentum than the thorium nucleus

701 Views

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