JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 6)
A physical quantity z depends on four
observables
a, b, c and d, as z = $${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$$. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :
a, b, c and d, as z = $${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$$. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :
13.5 %
14.5%
16.5%
12.25%
Explanation
z = $${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$$
$$ \Rightarrow $$ $${{dz} \over z} \times 100 = \left( {2{{da} \over a} + {2 \over 3}{{db} \over b} + {1 \over 2}{{dc} \over c} + 3{{d\left( d \right)} \over d}} \right) \times 100$$
% error in z
= $$\left( {2 \times 2 + {2 \over 3} \times 1.5 + {1 \over 2} \times 4 + 3 \times 2.5} \right) $$ %
= ( 4 + 1 + 2 + 7.5 ) %
= 14.5 %
$$ \Rightarrow $$ $${{dz} \over z} \times 100 = \left( {2{{da} \over a} + {2 \over 3}{{db} \over b} + {1 \over 2}{{dc} \over c} + 3{{d\left( d \right)} \over d}} \right) \times 100$$
% error in z
= $$\left( {2 \times 2 + {2 \over 3} \times 1.5 + {1 \over 2} \times 4 + 3 \times 2.5} \right) $$ %
= ( 4 + 1 + 2 + 7.5 ) %
= 14.5 %
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