Question

The zero-error in the spring balance shown and the correct weight of the solid, suspended from it, are equal, respectively, to :



(1) + 2g wt; 19g wt
(2) + 2g wt; 15g wt
(3) − 2g wt; 19g wt
(4) − 2g wt; 15g wt

Answer 1

Total number of the divisions on the scale = 30

Least count$=\frac{30}{30}=1\text{g}$

As the zero mark of the scale is leading from its main position by two divisions, so the zero error of the scale = $2\times 1=2\text{g}$

Weight according to the reading = 17 g wt

Actual weight = Main reading $-$ zero error = 17 g wt $-$ 2 g wt = 15 g wt

Hence, the correct answer is option 2.