Subject

Physics

Class

JEE Class 12

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JEE Physics 2004 Exam Questions

Multiple Choice Questions

61.

The binding energy per nucleon of deuteron (21H) and helium nucleus (42He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

  • 13.9 MeV

  • 26.9 MeV

  • 23.6 MeV

  • 19.2 MeV


C.

23.6 MeV

As given,straight H presubscript 1 superscript 2 space plus space straight H presubscript 1 superscript 2 space rightwards arrow for space of space subscript 2 He to the power of 4 space plus space energy
The binding energy per nucleon of a deuteron (1H2 ) = 1.1 MeV
∴ Total binding energy = 2× 1.1 = 2.2 MeV

The binding energy per nucleon of helium (2He4 ) = 7 MeV
∴ Total binding energy = 4× 7 = 28 MeV
Hence, energy released in the above process = 28 - 2× 2.2 = 28 - 4.4 = 23.6 MeV

551 Views

62.

According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Vs the frequency, of the incident radiation gives straight line whose slope

  • depends on the nature of the metal used

  • depends on the intensity of the radiation

  • depends both on the intensity of the radiation and the metal used

  • is the same for all metals and independent of the intensity of the radiation.


D.

is the same for all metals and independent of the intensity of the radiation.


Einstein's photoelectric equation is
K.E.max = hv − φ ............... (i)

The equation of line is
y = mx + c .............. (ii)

Comparing above two equations m = h, c = - φ

Hence, the slope of the graph is equal to Planck's constant (non-variable) and does not depend on the intensity of radiation.
193 Views

63.

Two spherical conductor B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion, between B and C is

  • F/4

  • 3F/4

  • 3F/8

  • F/8


C.

3F/8

straight F apostrophe space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator left parenthesis straight q divided by 2 right parenthesis left parenthesis 3 straight q divided by 4 right parenthesis over denominator straight d squared end fraction space equals space fraction numerator 3 straight F over denominator 8 end fraction
199 Views

64.

The angle of incidence at which reflected light totally polarized for reflection from air to glass (refractive index n), is

  • sin−1 (n

  • sin−1 (1/n)

  • tan−1 (1/n)

  • tan−1 (n)


D.

tan−1 (n)

Brewster’s law: According to this law the ordinary light is completely polarised in the plane of incidence when it gets reflected from transparent medium at a particular angle known as the angle of polarisation. n = tan ip.

286 Views

65.

An electromagnetic wave of frequency ν = 3.0 MHz passes from vacuum into a dielectric medium with permittivity ε = 4.0. Then

  • wavelength is doubled and the frequency remains unchanged

  • wavelength is doubled and frequency becomes half

  • wavelength is halved and frequency remains unchanged

  • wavelength and frequency both remain unchanged.


C.

wavelength is halved and frequency remains unchanged

In vacuum, ε0 = 1
In medium, ε = 4
So, refractive index

straight mu space equals space square root of straight epsilon divided by straight epsilon subscript 0 end root
space equals space square root of 4 divided by 1 end root space equals space 2
wavelength space straight lambda to the power of apostrophe space equals space straight lambda over straight mu space equals space straight lambda over 2
and space wave space velocity space straight v equals space straight c over straight mu space equals straight c over 2

Hence, it is clear that wavelength and velocity will become half but frequency remains unchanged when the wave is passing through any medium.

138 Views

66.

An α-particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of

  • 1 Å

  • 10−10 cm

  • 10−12 cm

  • 10−15 cm


C.

10−12 cm

At closest approach, all the kinetic energy of the α-particle will converted into the potential energy of the system, K.E. = P.E.

straight i. straight e. space 1 half space mv squared space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r end fraction
5 space MeV space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space left parenthesis 2 straight e right parenthesis space straight x space 92 straight e over denominator straight r end fraction space open parentheses therefore space 1 half space mv squared space equals space 5 space MeV close parentheses

rightwards double arrow space straight r space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space 2 space straight x 92 space straight x space left parenthesis 1.6 space straight x space 10 to the power of negative 19 end exponent right parenthesis squared over denominator 5 space straight x space 10 to the power of 6 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction
straight r space equals space 5.3 space straight x space 10 to the power of negative 14 end exponent space straight m space equals space 10 to the power of negative 12 space cm end exponent

1123 Views

67.

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be

  • 21/3:1

  • 1:31/2

  • 31/2:1

  • 1:21/3


B.

1:31/2

Law of conservation of momentum gives
m1 v1 = m2 v2

straight m subscript 1 over straight m subscript 2 space equals straight v subscript 2 over straight v subscript 1
space But space space straight m space equals space 4 over 3 πr cubed straight rho
or space straight m proportional to space straight r cubed
therefore space straight m subscript 1 over straight m subscript 2 space equals space fraction numerator straight r subscript 1 superscript 3 over denominator straight r subscript 2 superscript 3 end fraction space equals space straight v subscript 2 over straight v subscript 1
straight r subscript 1 over straight r subscript 2 space equals space open parentheses 1 half close parentheses to the power of 1 divided by 3 end exponent
straight r subscript 1 space colon straight r subscript 2 space equals space 1 space colon 2 to the power of 1 divided by 3 end exponent

222 Views

68.

The work function of a substance is 4.0 eV. Then longest wavelength of light that can cause photoelectron emission from this substance approximately

  • 540 nm

  • 400 nm

  • 310 nm,

  • 220 nm


C.

310 nm,

hc over straight lambda space equals straight W
straight lambda subscript longest space equals hc over straight W space equals space fraction numerator 6.6 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 4.0 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction
rightwards double arrow space straight lambda subscript longest space almost equal to space 310 space nm
350 Views

69.

The thermistors are usually made of

  • metals with low temperature coefficient of resistivity

  • metals with high temperature coefficient of resistivity

  • metal oxides with high temperature coefficient of resistivity ‘

  • semiconducting materials having low temperature coefficient of resistivity.


C.

metal oxides with high temperature coefficient of resistivity ‘

These are devices whose resistance varies quite markedly with temperature mean having high-temperature coefficient of resistivity. [Their name are derived from thermal resistors]. Depending on their composition they can have either negative temperature coefficient or positive temperature coefficient or positive temperature coefficient or positive temperature coefficient characteristics. The negative temperature coefficient types consist of a mixture of oxides of iron, nickel and cobalt with small amounts of other substance. The positive temperature coefficient types are based on barium titanate.

206 Views

70.

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is

  • infinite

  • five

  • three

  • zero


B.

five

For interference maxima, d sin θ = nλ Here d = 2λ
∴ sin θ = n/2 and is satisfied by 5 integral values of n (−2, −1, 0, 1, 2), as the maximum value of sin θ can only be 1.

270 Views

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