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CBSE CLASS XI

1. m = 1 gm = 1/1000 kg

So, the separation between the particles is 1 m.

2. 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
3. The force of attraction between the two charges
   
   The force of attraction is equal to the weight
   
   For example, Assuming m= 64 kg,
   
4. mass = 50 kg, r = 20 cm = 0.2 m

   

5. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of the force,




6. 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



7. Suppose the height is h.
   At earth station F = GMm/R²
   M = mass of earth
   m = mass of satellite
   R = Radius of earth



8. Two charged particle placed at a sehortion 2m. exert a force of 20m.

   

9. The force between the earth and the moon,

   

10. Charge on proton = 1.6 × 10^–19

    

    mass of proton = 1.732 × 10^–27 kg

    

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

    

12. 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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