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Physics Semester 2  STPM  Chapter 14 Electric Current
                Answer: B
                   Volume is constant,
                        A 1  l 1  = A 2  l 2
                        A 1   =   l 2
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                        A 2   l 1
                              ρl
                          R  =    ⇒ R ∝   l
                              A        A
                                ⇒ R 1  ∝   l 1
                                       A 1
                                     R 2  ∝   l 2
                                        A 2
                        \   R 2   =   l 2  ×   A 1
                           R 1  A 2  l 1
                             =  l 2   ×   A 1
                                l 1  A 2
                             = 1 2 2
                                l 2
                                l 1
                              R 2   =  1  2 + 0.001 2 2
                          20        2
                              R 2  = 20.02 Ω

                Question 3                                                            Semester

                  A hollow copper rod has an inner radius of 0.65 cm and an outer radius of
                  1.0 cm as shown in the diagram.
                                    P                                                 2
                                                                      Q
                                     O
                           0.65 cm
                                                      20 cm
                                     1.0 cm
                  A current flows from P to Q. If the length of the rod is 20 cm and its resistivity
                  is 1.7 × 10  Ω m, what is the resistance of the copper rod?
                          –8
                  A  3.4 × 10 Ω                  C  1.8 × 10 Ω
                                                          –4
                           –9
                  B  1.9 × 10 Ω                  D  2.9 × 10 Ω
                           –5
                                                          –3

                Answer: B
                       ρ l
                   R =
                        A
                                          2    2
                   Cross-sectional area, A = πr 2  – πr 1
                                      = π(1 × 10 )  – π(0.65 × 10 )
                                                             –2 2
                                               –2 2
                                      = 1.8143 × 10  m 2
                                                 –4
                                                233


         014 STPM Q&A Physics T2.indd   233                                  17/02/2022   10:12 AM
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