Q)
(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
...
