CBSE Important Questions

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CBSE CLASS XI

  1. m = 1 gm = 1/1000 kg

So, the separation between the particles is 1 m.

  1. A man is standing on the surface of earth
    The force acting on the man = mg ………(i)
    Assuming that, m = mass of the man = 50 kg
    And g = acceleration due to gravity on the surface of earth = 10 m/s2
    W = mg = 50× 10= 500 N = force acting on the man
    So, the man is also attracting the earth with a force of 500 N
  2. The force of attraction between the two charges

    The force of attraction is equal to the weight

    For example, Assuming m= 64 kg,
  3. mass = 50 kg, r = 20 cm = 0.2 m

  4. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of the force,
  5. The body builder exerts a force = 150 N.
    Compression x = 20 cm = 0.2 m
    Total force exerted by the man = f = kx
    kx = 150
  6. Suppose the height is h.
    At earth station F = GMm/R2
    M = mass of earth
    m = mass of satellite
    R = Radius of earth
  7. Two charged particle placed at a sehortion 2m. exert a force of 20m.

  8. The force between the earth and the moon,

  9. Charge on proton = 1.6 × 10–19

    mass of proton = 1.732 × 10–27 kg

  10. The average separation between proton and electron of Hydrogen atom is r= 5.3 10–11m.

  11. The geostationary orbit of earth is at a distance of about 36000km.
    We know that, g' = GM / (R+h)2
    At h = 36000 km. g' = GM / (36000+6400)2

    [ taking g = 9.8 m/s2 at the surface of the earth] A 120 kg equipment placed in a geostationary satellite will have weight Mg` = 0.233 � 120 = 26.79 = 27 N]

Submitted By Mr. Avinash kumar
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