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(chatquestion)...(u). 1iele The errol ll AR= (AR; + the equivalent resistance along with error is mce is given by AR2) = (3 + 4) Q=72 Length of the cube, L = 1.2 x10-2 m Volume of the cube, V = (1.2 x10-2, As the result can have only two therefore, on rounding off, we get, (800 7) 2. 63. (a) : Here, R1 = (100 ± 3)2; R2= (200 ± 4) Q The equivalent resistance in parallel combination is %3D %3D 01 cm 70. (c): Radius of the sphere, = ct precise 3 200 =66.7 Q 3 Volume of the sphere, R R Rg Rp The error in equivalent resistance is given by %3D 100 200 200 4 Tr: 3 x 3.14 x (1 %3D Rounded off upto 3 significant AR AR AR, R R R Rp + AR2 R2 71. (a): Here, mass of the box Mass of one gold piece, m1 = Mass of other gold piece, m2 ;AR, = AR %3D %3D R 66.7 =D3 100 2. 66.7 +4 200 = 1.8 2 . Total mass = m + my +r %3D = 2.3 kg + 0.02015 kg + As the result is correct onl %3D O1mm Hence, the equivalent resistance along with error in Jum parallel combination is (66.7 ± 1.8) Q. therefore, on rounding off, Total mass = 2.3 kg

asked Feb 15, 2020 by balaji.thirumalai

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