JEE Advance - Physics (2018 - Paper 1 Offline - No. 16)
$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$
The series expansion for $${\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}},$$ to first power in $$\Delta y/y.$$ is $$1 \pm \left( {\Delta y/y} \right).$$ The relative errors in independent variables are always added. So the error in $$z$$ will be
$$$\Delta z = z\left( {{{\Delta x} \over x} + {{\Delta y} \over y}} \right).$$$
The above derivation makes the assumption that $$\Delta x/x < < 1,$$ $$\Delta y/y < < 1.$$ Therefore, the higher powers of these quantities are neglected.
Consider the ratio $$r = {{\left( {1 - a} \right)} \over {1 + a}}$$ to be determined by measuring a dimensionless quantity $$a.$$ If the error in the measurement of $$a$$ is $$\Delta a\left( {\Delta a/a < < 1.} \right.$$ then what is the error $$\Delta r$$ in determining $$r$$?
Explanation
Given, ratio $$r = {{1 - a} \over {1 + a}}$$ ..... (1)
Given, error in measurement of a is $$\Delta$$a ($$\Delta$$a/a << 1)
Taking natural log of Eq. (1), we get
$$\ln r = \ln \left( {{{1 - a} \over {1 + a}}} \right)$$
$$ \Rightarrow \ln r = \ln (1 - a) - \ln (1 + a)$$
$${{\Delta r} \over r} = {{\Delta a} \over {1 - a}} + {{\Delta a} \over {1 + a}} = {{(1 + a)\Delta a + (1 - a)\Delta a} \over {(1 - a)(1 + a)}}$$
$$ \Rightarrow {{\Delta r} \over r} = {{2\Delta a} \over {(1 - a)(1 + a)}}$$
$$ \Rightarrow \Delta r = r\,.\,{{2\Delta a} \over {(1 - a)(1 + a)}}$$
Substituting value of r, we get
$$\Delta r = {{1 - a} \over {1 + a}}{{2\Delta a} \over {(1 - a)(1 + a)}} = {{2\Delta a} \over {{{(1 + a)}^2}}}$$
Therefore, error $$\Delta$$r in determining r is $${{2\Delta a} \over {{{(1 + a)}^2}}}$$
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