JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 22)
The density of a solid metal sphere is determined by measuring its mass and its diameter. The
maximum error in the density of the sphere is $$\left( {{x \over {100}}} \right)$$ %. If the relative errors in measuring the mass
and the diameter are 6.0% and 1.5% respectively, the value of x is_______.
Answer
1050
Explanation
$$\rho $$ = $${M \over V}$$ = $${M \over {{4 \over 3}\pi {{\left( {{D \over 2}} \right)}^3}}}$$
$$ \Rightarrow $$ $$\rho $$ = $${6 \over \pi }M{D^{ - 3}}$$
For maximum error
$${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$$
= 6 + 3 $$ \times $$ 1.5
= 10.5 %
= $$\left( {{{1050} \over {100}}} \right)$$ %
$$ \therefore $$ x = 1050
$$ \Rightarrow $$ $$\rho $$ = $${6 \over \pi }M{D^{ - 3}}$$
For maximum error
$${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$$
= 6 + 3 $$ \times $$ 1.5
= 10.5 %
= $$\left( {{{1050} \over {100}}} \right)$$ %
$$ \therefore $$ x = 1050
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