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ฟิสิกส์โอลิมปิก วิทยาศาสตร์โอลิมปิก ข้อสอบแข่งขัน ข้อสอบชิงทุน => GRE - Physics => Topic started by: conantee on September 20, 2010, 09:15:34 PM



Title: GR9277.048
Post by: conantee on September 20, 2010, 09:15:34 PM
48. The magnitude of the force F on an object can be determined by measuring both the mass m of an object and the magnitude of its aceleration a, where F = ma. Assume that these measurements are uncorrelated and normally distributed. If the standard deviations of the measurements of the mass and the accelerations are \sigma_m and  \sigma_a, respectively, then \sigma_F/F is
(A)  (\frac{\sigma_m}{m})^2 + (\frac{\sigma_a}{a})^2
(B)  (\frac{\sigma_m}{m} + \frac{\sigma_a}{a})^\frac{1}{2}
(C)  [(\frac{\sigma_m}{m})^2 + (\frac{\sigma_a}{a})^2]^\frac{1}{2}
(D)  \frac{\sigma_m \sigma_a}{ma}
(E)  \frac{\sigma_m}{m} +\frac{\sigma_a}{a}


Title: Re: GR9277.048
Post by: gons on September 22, 2010, 08:23:33 PM
ตอบ (C)

จาก \frac{\text{d}F}{F} = \frac{\text{d}a}{a} + \frac{\text{d}m}{m}

จะได้ \left( \frac{\Delta F}{F} \right)^2 = \left( \frac{\Delta a}{a} \right)^2 + \left( \frac{\Delta m}{m} \right)^2 + \frac{\Delta m \Delta a}{ma} \approx \left( \frac{\Delta a}{a} \right)^2 + \left( \frac{\Delta m}{m} \right)^2

\sigma _m  = \left\langle {\left( {\Delta m} \right)^2 } \right\rangle ,\sigma _a  = \left\langle {\left( {\Delta a} \right)^2 } \right\rangle ,\sigma _F  = \left\langle {\left( {\Delta F} \right)^2 } \right\rangle

\therefore \frac{{\sigma _F }}{F} = \sqrt {\left( {\frac{{\Delta a}}{a}} \right)^2  + \left( {\frac{{\Delta m}}{m}} \right)^2 }








Title: Re: GR9277.048
Post by: conantee on September 26, 2010, 02:24:34 AM
The correct answer is (C) krub. This question is straightforward. No trick as far as I know.

The keywords are "uncorrelated" and "normally distributed". The uncertainty is thus related by square relation.
That is, (\frac{\sigma_F}{F})^2 = (\frac{\sigma_m}{m})^2 + (\frac{\sigma_a}{a})^2, not just \frac{\sigma_F}{F} =\frac{\sigma_m}{m} + \frac{\sigma_a}{a}

66 of 100 people answer this question correctly.