Subject

Physics

Class

JEE Class 12

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JEE Physics 2012 Exam Questions

Multiple Choice Questions

21.

Truth table for system of four NAND gates as shown in figure is


A.

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22.

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slits. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference φ is given by

  • straight I subscript straight m over 9 left parenthesis 4 space plus 5 space cos space straight ϕ right parenthesis
  • straight I subscript straight m over 3 open parentheses 1 space plus space 2 space cos squared space straight ϕ over 2 close parentheses
  • straight I subscript straight m over 5 open parentheses 1 space plus 4 space cos squared straight ϕ over 2 close parentheses
  • straight I subscript straight m over 9 open parentheses 1 plus 8 space cos squared space straight ϕ close parentheses

D.

straight I subscript straight m over 9 open parentheses 1 plus 8 space cos squared space straight ϕ close parentheses

Let A1 = A0, A2 = 2A0
If amplitude of resultant wave is A then

straight A squared space equals space straight A subscript 1 superscript 2 space plus space straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2 space cos space straight ϕ
 For maximum intensity

straight A subscript max superscript 2 space equals space straight A subscript 1 superscript 2 space plus space straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2
therefore fraction numerator straight A squared over denominator straight A subscript max superscript 2 end fraction space equals space fraction numerator straight A subscript 1 superscript 2 space plus straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2 space cos space straight ϕ over denominator straight A subscript 1 superscript 2 space plus straight A subscript 2 superscript 2 space plus space 2 straight A subscript 1 straight A subscript 2 end fraction
space equals space fraction numerator straight A subscript 0 superscript 2 space plus space 4 straight A subscript 0 superscript 2 space plus space 2 left parenthesis straight A subscript 0 right parenthesis left parenthesis 2 straight A subscript 0 right parenthesis cos space straight ϕ over denominator straight A subscript 0 superscript 2 space plus space 4 straight A subscript 0 superscript 2 space plus space 2 left parenthesis straight A subscript 0 right parenthesis left parenthesis 2 straight A subscript straight o right parenthesis end fraction
straight I over straight I subscript straight m space equals space fraction numerator 5 space plus 4 space cos space straight ϕ over denominator 9 end fraction space equals space fraction numerator 1 space plus space 8 space cos squared space left parenthesis straight ϕ divided by 2 right parenthesis over denominator 9 end fraction

673 Views

23.

Assume that a neutron breaks into a proton and an electron. The energy released during this process is(Mass of neutron = 1.6725 x 10–27kg; mass of proton = 1.6725 x 10–27kg; mass of electron = 9 x 10–31kg)

  • 0.73 MeV

  • 7.10 MeV

  • 6.30 MeV

  • 5.4 MeV


A.

0.73 MeV

increment straight m space equals space left parenthesis straight m subscript straight p plus straight m subscript straight e right parenthesis minus straight m subscript straight n
space equals space 9 space straight x 10 to the power of negative 31 end exponent space kg.
Energy space released space equals space left parenthesis 9 space straight x space 10 to the power of negative 31 end exponent space kg right parenthesis space straight c squared space joules
space equals space fraction numerator 9 space straight x 10 to the power of negative 31 end exponent space straight x space left parenthesis 3 space straight x space 10 to the power of 8 right parenthesis squared over denominator 1.6 space straight x space 110 to the power of negative 13 end exponent end fraction space space MeV
space equals space 0.73 space MeV
1351 Views

24.

An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film? 

  • 7.2 m

  • 2.4 m

  • 3.2 m

  • 5.6 m


D.

5.6 m

Case I: u = –240cm, v = 12, by Lens formula

1 over straight f space equals space 7 over 80
Case space II colon space 12 minus 1 third space equals space 35 over 3 space open parentheses normal space shift space equals space 1 minus 2 over 3 space equals space 1 third close parentheses
straight f equals space 7 over 80
straight u space equals space 5.6

522 Views

25.

Proton, Deuteron and alpha particle of the same kinetic energy is moving in circular trajectories in a constant magnetic field. The radii of the proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relations is correct?

  •  rα = rp= rd

  •  rα = rp< rd

  •  rα > rd> rp

  •  rα = rd> rp


B.

 rα = rp< rd

For charged particle moving with a speed v, in magnetic field B, on a circular track of radius

straight r space equals space mv over qB space equals space fraction numerator square root of 2 km end root over denominator qB end fraction
open parentheses therefore space mv space equals space straight p space and space straight K space equals fraction numerator straight p squared over denominator 2 straight m end fraction space rightwards double arrow space straight p space equals space square root of 2 km end root close parentheses
rightwards double arrow space straight r space proportional to space fraction numerator square root of straight m over denominator straight q end fraction
or space straight m subscript straight d space equals 2 straight m subscript straight p space and space straight q subscript straight d space equals space straight q subscript straight p semicolon
straight m subscript straight alpha space equals space 4 straight m subscript straight p space and space straight q subscript straight alpha space equals space 2 straight q subscript straight p
rightwards double arrow straight r subscript straight p colon straight r subscript straight d colon straight r subscript straight alpha space equals space fraction numerator square root of straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 2 straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 4 straight m subscript straight p end root over denominator 2 straight q subscript straight p end fraction
space equals space 1 colon space square root of 2 colon 1
straight r subscript straight alpha space equals space straight r subscript straight p space less than thin space straight r subscript straight d

808 Views

26.

A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation, a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc than the variation of the magnetic induction at the centre of the disc will be represented by the figure


A.


Consider ring like the element of the disc of radius r and thickness dr.
If σ is charge per unit area, then charge on the element
dq = σ(2πr dr)
current ‘i’ associated with rotating charge dq is
straight i space equals space fraction numerator left parenthesis dp right parenthesis straight w over denominator 2 straight pi end fraction space equals space straight sigma space straight w space straight r space dr
Magnetic field dB at center due to element

dB space equals space fraction numerator straight mu subscript 0 straight i over denominator 2 straight r end fraction space equals space fraction numerator straight mu subscript 0 σωdr over denominator 2 end fraction
straight B subscript net space equals space integral dB space equals space fraction numerator straight mu subscript straight o σω squared over denominator 2 end fraction space integral subscript 0 superscript straight R dr space equals space fraction numerator straight mu subscript straight o σωR over denominator 2 end fraction
rightwards double arrow space straight B subscript net space equals fraction numerator straight mu subscript straight o Qω over denominator 2 πR end fraction
So if Q and w are unchanged then
straight B subscript net space proportional to 1 over straight R

817 Views

27.

A radar has a power of 1 Kw and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 x 106 m) is

  • 80 km

  • 16 km

  • 40km

  • 64 km


A.

80 km

Maximum distance on earth where object can be detected is d, then

(h + R)2 = d2 + R2

⇒ d2 = h2 + 2Rh
Since, h<<R,
⇒d2 =2hR
straight d equals square root of 2 left parenthesis 500 right parenthesis left parenthesis 6.4 space straight x 10 to the power of 6 right parenthesis end root equals 80 space km

321 Views

28.

The hydrogen atom is excited from ground state to another state with the principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be

  • 2

  • 3

  • 3

  • 4


D.

4

Number of spectral lines from a state n to ground state is

fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction space equals space fraction numerator 4 left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction space equals space 6

218 Views

29.

This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements
Statement 1: Davisson – Germer experiment established the wave nature of electrons.
Statement 2: If electrons have wave nature, they can interfere and show diffraction.

  • Statement 1 is false, Statement 2 is true

  • Statement 1 is true, Statement 2 is false

  • Statement 1 is true, Statement 2 is the correct explanation for statement 1

  • Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1


C.

Statement 1 is true, Statement 2 is the correct explanation for statement 1

Davisson – Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction

155 Views

30.

A cylindrical tube, open at both ends, has a fundamental frequency, f, in the air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

  • f

  • f/2

  • 3f/4

  • 2f


A.

f

Initially for open organ pipe, fundamental frequency,
straight v subscript straight o space equals space fraction numerator straight v over denominator 2 l end fraction space equals straight f
But when it is half dipped in water, then it becomes closed organ pipe of length l/2. In this case fundamental frequency.
straight v subscript straight c space equals space fraction numerator straight v over denominator 4 l end fraction space equals space fraction numerator straight v over denominator 4 begin display style l over 2 end style end fraction space equals space fraction numerator straight V over denominator 2 l end fraction space equals straight f

313 Views

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