The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $$z = x/y.$$ If the errors in $$x,y$$ and $$z$$ are $$\Delta x,\Delta y$$ and $$\Delta z,$$ respectively, then
$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$

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 $$r$$?

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $$z = x/y.$$ If the errors in $$x,y$$ and $$z$$ are $$\Delta x,\Delta y$$ and $$\Delta z,$$ respectively, then
$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$

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.

In an experiment the initial number of radioactive nuclei is $$3000.$$ It is found that $$1000 \pm 40$$ nuclei decayed in the first $$1.0s.$$ For $$\left| x \right| < < 1.$$ $$\ln \left( {1 + x} \right) = x$$ up to first power in $$x.$$ The error $$\Delta \lambda ,$$ in the determination of the decay constant $$\lambda ,$$ in $${s^{ - 1}},$$ is

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $$\delta T = 0.01$$ seconds and he measures the depth of the well to be $$L=20$$ meters. Take the acceleration due to gravity $$g = 10m{s^{ - 2}}$$ and the velocity of sound is $$300$$ $$m{s^{ - 1}}$$. Then the fractional error in the measurement, $$\delta L/L,$$ is closest to