The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 ~mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 ~mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2 % , the relative percentage error in the density is

Step-by-step solution

Least count of screw gauge = 0.5 50 =0.01 ~mm = r Diameter, r=2.5 ~mm +20 0.5 50 =270 ~mm r r = 0.01 2.70 or r r 100= 1 2.7 Now, density d= m V = m 4 3 ( r 2 )^3 Here, r is the diameter. aligned d d & 100= m m +3 ( r r ) 100 = & m m 100+3 ( r r ) 100 aligned aligned & =2 %+3 1 27 & =3.11 % aligned Analysis of Question (i) Question is moderately difficult. (ii) In practical part, questions in JEE are asked from vernier callipers and screw gauge.