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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2007 Exam Questions

Multiple Choice Questions

1.

Angular momentum of the particle rotating with a central force is constant due to

  • Constant Force

  • Constant linear momentum

  • Zero Torque

  • Constant Torque


C.

Zero Torque

256 Views

2.

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc.The value of α is

  • 1/3

  • 1/2

  • 1/6

  • 1/4


A.

1/3

In this question distance of centre of mass of new disc is αR not α/R.

negative fraction numerator 3 straight M over denominator 4 end fraction space straight alpha space straight R space plus space straight M over 4 space straight R space equals 0
rightwards double arrow space straight alpha space equals space 1 third
313 Views

3.

A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is 

  • K

  • 0

  • K/2

  • K/4


D.

K/4

132 Views

4.

A point mass oscillates along the x-axis according to the law x = x0 cos (ωt - π/4). If the acceleration of the particle is written as a = A cos (ωt + δ), then

  • A = x0 , δ = – π/4

  • A = x0 ω2 , δ = π/4 

  • A = x0 ω2, δ = –π/4 

  • A = x0 ω2, δ = 3π/4


D.

A = x0 ω2, δ = 3π/4

straight v equals space minus straight x subscript 0 space straight omega space sin space left parenthesis ωt space minus straight pi divided by 4 right parenthesis
straight a space equals space minus space straight x subscript 0 straight omega squared space cos space open parentheses ωt space plus space straight pi space minus straight pi over 4 close parentheses
straight a space equals space straight A space cos space left parenthesis ωt space plus straight delta right parenthesis
straight A space space equals space straight x subscript 0 straight omega squared space semicolon space straight delta space equals space fraction numerator 3 straight pi over denominator 4 end fraction
454 Views

5.

The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10-2 cos πt metres. The time at which the maximum speed first occurs

  • 0.5 s

  • 0.75 s

  • 0.125 s

  • 0.325 s


A.

0.5 s

x = 2 x 10-2 cos πt
v = 0.02 π sinπt
v is maximum at t = 1/2 = 0.5 sec

609 Views

6.

A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000. N/m. The spring compresses by

  • 5.5 cmYes

  • 2.5 cm

  • 11 cm

  • 3.5 cm


A.

5.5 cmYes

187 Views

7.

A round uniform body of radius R, mass M and moment of inertia ‘I’, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

  • fraction numerator straight g space sin space straight theta over denominator 1 space plus space straight I divided by MR squared end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 plus space MR squared divided by straight I end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 minus straight I divided by MR squared end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 minus MR squared divided by straight I end fraction

A.

fraction numerator straight g space sin space straight theta over denominator 1 space plus space straight I divided by MR squared end fraction


Mg space sin space straight theta space minus space straight f space equals space Ma
fR space equals space straight I straight a over straight R
rightwards double arrow space straight a space equals space fraction numerator straight g space sin space straight theta over denominator open parentheses 1 plus begin display style straight I over MR squared end style close parentheses end fraction
363 Views

8.

A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (mass less) of spring constant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is stretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’.

  • mF/M

  • (M+m)F/m

  • mF/(m+ M)

  • MF (m+M)


C.

mF/(m+ M)

195 Views

9.

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

  • v0 + 2g + 3f

  • v0 + g/2 + f/3

  • v0 + g + f

  • v0 + g/2 + f


B.

v0 + g/2 + f/3

integral subscript 0 superscript straight x space dx space equals space integral subscript 0 superscript 1 space left parenthesis straight V subscript 0 space plus space gt space plus ft squared right parenthesis space dt
straight x space equals straight v subscript 0 space plus space straight g space open parentheses 1 half close parentheses space plus straight f open parentheses 1 third close parentheses
554 Views

10.

If gE and gm are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the moon/ electronic charge on the earth) to be

  • 1

  • 0

  • gE/gM

  • gM/gE


A.

1

654 Views

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