Page 4 - English Class X_cbse new (FINAL).cdr
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S.                      Solution                                       Marking   S.                                            Solution                     Marking
No.                                                                     Scheme   No.                                                                         Scheme

 2. 	  F  =    1   q1q2                                                     [1]  4. (i)	 Nuclear forces are the strongest forces in nature.                           1
             4πε0   r2                                                                                                                                                 2
                                                                                      (ii)	 Nuclear forces are independent of charge.
                                                                                                                                                                      1
       ( )( )= 9 × 109 ×                                                                                                                                               2
                    2 × 10−7 3 × 10−7
                                                                        [1]
                        (0.30)2

       = 6 × 10−3N

                                                                        [2]                                                                                  [1]
                                                                                                                                                             [1]
3. The following graph shows the variation of coulomb force (F) versus           5. Let q1 = q,q2 = -4q and q3 = +2q and r = 10 cm = 0.10 m
     r.                                                                                                                                                      [1]
                                                                                      Total work done =                                                      [1]
     	 r → r0.                                                                                                                                               [1]
                        Y                                                               1 q1q2 + 1 q2q3 + 1 q3q1                                             [1]
                                                                                      4π ∈0 r 4π ∈0 r 4π ∈0 r
                                                                                                                                                             [5]
                                                                                      ( )= 1

                                                                                        4π ∈0 r
                                                                                                   q1q2 + q2q3 + q3q1

                                  r q2(2μC, -3μC)                       [2]
                   F
                                                                                      =     1   r  (q)(−4q) +   (−4q)(+2q)    + (+2q)(q)
                                   q1(1μC, 2μC)                                          4π ∈0

                                                                                      ( )=  1                                      1
                                                                                         4π ∈0  r  −4q2  − 8q2  +  2q2   =  4π ∈0  r  −10q2

                      OX                                                              ( )= 9 × 109 × −10 1.6 × 10−10 2 = −9 × 109 × 10 × 2.56 × 10−20 × 100
                                              r                                                            0.10                                  10

	                                                                                     = −23.04 × 10−9 = −2.304 × 10−8 J

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