1

JEE Advanced 2018 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-1

In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $$[E]$$ and $$[B]$$ stand for dimensions of electric and magnetic fields respectively, while $$\left[ {{\varepsilon _0}} \right]$$ and $$\left[ {{\mu _0}} \right]$$ stand for dimensions of the permittivity and permeability of free space respectively. $$\left[ L \right]$$ and $$\left[ T \right]$$ are dimensions of length and time respectively. All the quantities are given in $$SI$$ units.

The relation between $$[E]$$ and $$[B]$$ is

The relation between $$[E]$$ and $$[B]$$ is

2

JEE Advanced 2018 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-1

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 $$r$$?

$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$

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 $$r$$?

3

JEE Advanced 2018 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-1

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

$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$

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

4

JEE Advanced 2017 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-0.75

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

Questions Asked from Units & Measurements (MCQ (Single Correct Answer))

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