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Chapter 10 MCQs For AMU/JMI XI-Dip.Engg. Entrance Test

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Chapter 10  : Gravitation    


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  1. Mohd Qasim .Gravitation sheet
    Ques 21 .The Gravitation for of attraction between two masses ,.,.,..................gravitational force of attraction between them

    ReplyDelete
    Replies
    1. F= α = G.m1.m2/r^2 -------------------1
      r’ = (100-60)% of r = 0.4r
       F’ = G.m1.m2/(0.4r)^2 = (1/0.4)^2.G.m1.m2/r^2
       F’ = 6.25. G.m1.m2/r^2 = 6.25α (from 1)

      Delete
  2. Mohd Qasim Gravitation Sheet
    Ques 36 .An object of mass........
    ....Neglect the air resistance.

    ReplyDelete
    Replies
    1. 36. Ans: (b)
      u = 0, m=m, t=t
      h = ut-1/2 at^2
      h = 0 - 1/( 2) 10 x t^2
      H = -5t^2
      Height of the tower = 5t^2

      Delete
  3. Mohd Qasim Gravitation Sheet.
    Ques 50. Calculate the .........mass is 20 kg

    ReplyDelete
  4. Mohd Qasim Gravitation Sheet.
    Ques 51 . Calculate the,........mass is 20 kg

    ReplyDelete
    Replies
    1. Sol. 51, Ans (b)
      Gravity on moon = one-sixth of gravity n earth = 10/6 m/s2
      Weight on moon = mass x gravity at moon
      = 20 x 10/6 = 33.33N

      Delete
  5. Mohd Qasim Gravitation Sheet.
    Ques 54 .When a person........on the ground.

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    Replies
    1. Ans (a)
      thrust is the weight of the person and it will not change if he standing or lay down.
      so, the thrust will be the same in both cases

      Delete
  6. Mohd Qasim Gravitation Sheet.
    Ques 62 .In Q61 what ..............table leg.

    ReplyDelete
    Replies
    1. 62. Ans: (d)
      Load on the table = 600N
      Load on one leg = 600/4 = 150N
      Area of each leg = 25cm2
      Pressure increase on each leg = 150/25 = 6N/cm2 = 6000N/m2

      Delete
  7. Mohd Qasim Gravitation Sheet.
    Ques 64 . Three identical bricks .......
    ..........bricks will be

    ReplyDelete
    Replies
    1. 64. Ans (a)
      Thrust = Weight =mg = 30 x 9.8 =29.4

      Delete
  8. Mohd Qasim Gravitation Sheet.
    Ques 68 .The foundation of the buildings.

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    Replies
    1. Pressure on the ground = weight of the building/ Area of footing of foundation.
      The load of the building comes on the columns and then to the foundation and finally to the ground.
      Increasing the area of footing decreases the pressure on the ground.
      Therefore, foundations decrease the pressure on the ground due to the weight of the building
       Ans is (C )

      Delete
  9. Sir Q-31 ka ans d ho ga shayad lekin answersheet me c diya he

    ReplyDelete

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