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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2006 Exam Questions

Multiple Choice Questions

1.

A force of negative straight F straight k with hat on top acts on O, the origin of the coordinate system. The torque about the point (1, −1) is

  • negative straight F left parenthesis straight i with hat on top space minus straight j with hat on top right parenthesis
  • straight F left parenthesis straight i with hat on top space minus straight j with hat on top right parenthesis
  • negative straight F left parenthesis straight i with hat on top space plus straight j with hat on top right parenthesis
  • straight F left parenthesis straight i with hat on top space plus straight j with hat on top right parenthesis

C.

negative straight F left parenthesis straight i with hat on top space plus straight j with hat on top right parenthesis
straight tau with rightwards arrow on top space equals space left parenthesis negative straight i with hat on top plus straight j with hat on top right parenthesis space straight x space left parenthesis negative straight F straight k with hat on top right parenthesis
equals negative straight F space left parenthesis straight i with hat on top space plus straight j with hat on top right parenthesis
229 Views

2.

A wire elongates by 

  • 2

  • zero


B.

210 Views

3.

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

  • Mg left parenthesis square root of 2 minus 1 right parenthesis
  • Mg left parenthesis square root of 2 plus 1 right parenthesis
  • Mg square root of 2
  • fraction numerator Mg over denominator square root of 2 end fraction

A.

Mg left parenthesis square root of 2 minus 1 right parenthesis
straight F calligraphic l space sin space 45 space equals space M g left parenthesis calligraphic l minus calligraphic l space cos space 45 right parenthesis
F space equals space M g left parenthesis square root of 2 minus 1 right parenthesis
285 Views

4.

The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is

  • 3.3 × 10−3 J/m3

  • 4.58 × 10−6 J/m3

  • 6.37 × 10−9 J/m3

  • 81.35 × 10−12 J/m3


B.

4.58 × 10−6 J/m3

straight U subscript av space equals space straight epsilon subscript 0 straight E subscript rms superscript 2 space equals space 4.58 space straight x space 10 to the power of negative 6 end exponent space straight J divided by straight m cubed
straight E subscript rms superscript 2 space equals space 4.58 space straight x space 10 to the power of negative 6 end exponent space straight J divided by straight m cubed
244 Views

5.

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω′

  • fraction numerator ωm over denominator left parenthesis straight m plus 2 straight M right parenthesis end fraction
  • fraction numerator straight omega space left parenthesis straight m space plus space 2 straight M right parenthesis over denominator straight m end fraction
  • fraction numerator straight omega space left parenthesis straight m minus 2 straight M right parenthesis over denominator left parenthesis straight m plus 2 straight M right parenthesis end fraction
  • fraction numerator ωm over denominator left parenthesis straight m plus straight M right parenthesis end fraction

A.

fraction numerator ωm over denominator left parenthesis straight m plus 2 straight M right parenthesis end fraction
straight L subscript straight i space equals space straight L subscript straight f
mR squared straight omega space equals space left parenthesis mR squared space plus space 2 MR squared right parenthesis straight omega apostrophe
straight omega apostrophe space equals space open parentheses fraction numerator mω over denominator straight m plus 2 straight M end fraction close parentheses
251 Views

6.

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the force of gravity during the time the particle goes up is

  • 0.5 J

  • -0.5 J

  • −1.25 J

  • 1.25 J


C.

−1.25 J

negative mgh space equals negative space mg open parentheses fraction numerator straight v squared over denominator 2 straight g end fraction close parentheses space equals space minus space 1.25 space straight J
270 Views

7.

The potential energy of a 1 kg particle free move along the x-axis is given by
straight V left parenthesis straight x right parenthesis space equals space open parentheses straight x to the power of 4 over 4 minus straight x squared over 2 close parentheses space straight J

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

  • 2

  • 3 divided by square root of 2
  • square root of 2
  • 1 divided by square root of 2

B.

3 divided by square root of 2
kE subscript max space equals space straight E subscript straight T minus straight U subscript min
straight U subscript min space left parenthesis plus-or-minus 1 right parenthesis space equals negative 1 fourth straight J
KE subscript max space equals space 9 divided by 4 space straight J thin space
rightwards double arrow space straight U space equals space fraction numerator 3 over denominator square root of 2 end fraction space straight J
349 Views

8.

A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms−1 . The kinetic energy of the other mass is

  • 144 J

  • 96 J

  • 288 J

  • 192 J


C.

288 J

m1v1 = m2v2

KE space equals space 1 half straight m subscript 2 straight v subscript 2 superscript 2 space
space equals space 1 half space straight x space 4 space straight x space 144 space equals space 288 space straight J

506 Views

9.

Four point masses, each of value m, are placed at the corners of a square ABCD of side A. The moment of inertia through A and parallel to BD is

  • m

  • 2m

  • √m

  • 3m


D.

3m

I = 2m (

616 Views

10.

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity ‘v’ that varies as v= α√x . The displacement of the particle varies with time as

  • t3

  • t2

  • t

  • t1/2


B.

t2

straight v equals space straight alpha square root of straight x
dx over dt space equals space straight alpha square root of straight x space space space space open parentheses therefore space straight v space equals space dv over dt close parentheses
fraction numerator dx over denominator square root of straight x end fraction space equals space straight alpha space dt
Perform space integration
integral subscript 0 superscript straight x fraction numerator dx over denominator square root of straight x end fraction space equals space integral subscript 0 superscript straight t straight alpha space dt
because space at space straight t space equals space 0 comma space straight x space equals space 0 space and space let space at space any space time space straight t comma space particle space is space at space straight x right square bracket
rightwards double arrow right enclose space fraction numerator straight x to the power of 1 divided by 2 end exponent over denominator 1 divided by 2 end fraction end enclose subscript 0 superscript straight x space equals space αt
rightwards double arrow space straight x to the power of 1 divided by 2 end exponent space equals space straight alpha over 2 straight t
rightwards double arrow space straight x space equals space straight alpha squared over 4 straight x space straight t squared space rightwards double arrow space straight x proportional to space straight t squared
898 Views

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