the refractive index of water measured by the relation mu

The refractive index of water measured by the relation $$\mu = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}}$$ is found to have values of 1.34, 1.38, 1.32 and 1.36; the mean value of refractive index with percentage error is

A. $$1.35 \pm 1.48\% $$

B. $$1.35 \pm 0\% $$

C. $$1.36 \pm 6\% $$

D. $$1.36 \pm 0\% $$

Answer : $$1.35 \pm 1.48\% $$

Solution :
The mean value of refractive index,
$$\eqalign{ & \mu = \frac{{1.34 + 1.38 + 1.32 + 1.36}}{4} = 1.35\,{\text{and}} \cr & \Delta \mu = \frac{{\left| {\left( {1.35 - 1.34} \right)} \right| + \left| {\left( {1.35 - 1.38} \right)} \right| + \left| {\left( {1.35 - 1.32} \right)} \right| + \left| {\left( {1.35 - 1.36} \right)} \right|}}{4} = 0.02 \cr & {\text{Thus}}\,\,\frac{{\Delta \mu }}{\mu } \times 100 = \frac{{0.02}}{{1.35}} \times 100 = 1.48 \cr} $$

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Basic Physics >> Unit and Measurement

Releted Question 1

The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)

A. $$ML{T^{ - 1}}$$

B. $$M{L^2}{T^{ - 2}}$$

C. $$M{L^{ - 1}}{T^{ - 2}}$$

D. $$M{L^2}{T^{ - 1}}$$

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Releted Question 2

A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-

A. resistance

B. charge

C. voltage

D. current

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A cube has a side of length $$1.2 \times {10^{ - 2}}m$$ . Calculate its volume.

A. $$1.7 \times {10^{ - 6}}{m^3}$$

B. $$1.73 \times {10^{ - 6}}{m^3}$$

C. $$1.70 \times {10^{ - 6}}{m^3}$$

D. $$1.732 \times {10^{ - 6}}{m^3}$$

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Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-

A. $${M^0}{L^0}{T^0}$$

B. $${M^{ - 1}}{L^{ - 1}}{T^{ - 1}}$$

C. $${M^0}{L^2}{T^0}$$

D. $${M^{ - 1}}{L^1}{T^2}$$

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The refractive index of water measured by the relation $$\mu = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}}$$ is found to have values of 1.34, 1.38, 1.32 and 1.36; the mean value of refractive index with percentage error is

Releted MCQ Question on
Basic Physics >> Unit and Measurement

The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)

A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-

A cube has a side of length $$1.2 \times {10^{ - 2}}m$$ . Calculate its volume.

Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-

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The refractive index of water measured by the relation $$\mu = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}}$$ is found to have values of 1.34, 1.38, 1.32 and 1.36; the mean value of refractive index with percentage error is

Releted MCQ Question on Basic Physics >> Unit and Measurement

The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)

A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-

A cube has a side of length $$1.2 \times {10^{ - 2}}m$$ . Calculate its volume.

Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-

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Releted MCQ Question on
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