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The time period of a thin bar magnet is T. It is cut into 'n' equal parts by cutting it normal to its length. What will be the time period of each of them in the same field? |
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Answer» Solution :The expression for time period of a magnet `T= 2pi SQRT((I)/(MB))` `:. T prop sqrt((I)/(M))`........... (1) [`:.` B is same] But `I=m((l^(2) + b^(2))/(12))`, where m is mass For a THIN BAR magnet breadth `b lt lt l` , So neglect `b^(2)` in the above expression . ` :. ~~ (ml^(2))/(12) or I prop ml^(2)` .......... (2) From eq.s (1) and (2) , `T prop sqrt((ml^(2))/(M))` `:. (T_2)/(T_1) = (l_2)/(l_1) sqrt((m_2)/(m_1) xx (M_1)/(M_2))` But `l_2 = (l_1)/(n) , m_2 = (m_(1))/(n) and M_2 = (M_1)/(n)` `:. (T_2)/(T_1) = (1)/(n)=(1)/(n) sqrt((1)/(n) xx n) = (1)/(n) implies T_2 = (T)/(n)` |
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