QJEEAdv11Phe2c3: JEE Advanced 2011 Physics - Class 11 JEEAdv11Phe2c3 Solution: The density of a solid ball is to be determined in an…

Least count of screw gauge

$= {{Pitch} \over {No.\,of\,divisions\,on\,the\,circular\,scale}}$

$= {{0.5\,mm} \over {50}}$ = 0.01 mm

Diameter of ball, $D = MSR + CSR \times LC$

= 2.5 mm + (20)(0.01 mm) = 2.7 mm

Density, $\rho = {{Mass} \over {Volume}} = {M \over {{{4\pi } \over 3}{{\left( {{D \over 2}} \right)}^3}}}$

The relative error in the density is

${{\Delta \rho } \over \rho } = {{\Delta M} \over M} + {{3\Delta D} \over D}$

The relative percentage in the density is

${{\Delta \rho } \over \rho } \times 100 = \left[ {{{\Delta M} \over M} + {{3\Delta D} \over D}} \right] \times 100$

$= 2\% + {{3 \times 0.01} \over {2.7}} \times 100 = 2\% + 1.11\% = 3.11\%$