JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

A student performs an experiment for determination of

\mathrm{g}\left( { = \frac{{4{\mathrm{\pi }^2}l}}{{{T^2}}}} \right),\mathrm{\;}l \approx 1\mathrm{m},

, and he commits an error of

\Delta l.

. For T he takes the time of n oscillations with the stop watch of least count ∆T and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate?

View answer and explanation

Correct option: D

\Delta L = 0.5,\;\Delta T = 0.05,\;n = 50

Option D (

\Delta L = 0.5, \Delta T = 0.05, n = 50

) is correct because it minimizes the percentage error in g by having a smaller error in time period measurement and a larger number of oscillations.

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Assertion :

\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

has the unit

\mathrm{ms}^{-1} \text {. }

Reason: Acceleration has the dimensions of

\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

The method of dimensional analysis can be used to derive which of the following relations? Statement-I: Dimensions of constant of proportionality of constants can be derived from dimensional method
Statement-II: Numerical value of constant of proportionality can be found from experiments only The dimensions of torque are The physical quantity having the dimensions

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right]

is A quantity X is given by

{\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;

, where

{\varepsilon _0}

is the permittivity of 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