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One ferromagnetic substance has magnetic susceptibility chi_(m) at 27^(@) C temperature. At which temperature, it would become half ? |
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Answer» `600^(@)C` `therefore (chi_(m_(1) ) )/( chi_(m_(2)) ) = (T_(2) )/( T_(1))` `(2 chi_(m_(1) ) )/(chi_(m_(1) ) ) = (T_2)/( 273+ 27) (because chi_(m_(2) ) = (chi_(m_(1 )))/( 2))` `therefore T_(2) = 600 K` `therefore t_(2) = T_(2) = 273 = 600 - 273 = 327 ^(@) C` |
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