JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

A physical quantity

\mathrm{T}=\frac{\sqrt{\mathrm{abc}^2}}{\mathrm{~d}^3 \mathrm{e}^{1 / 6}}

is determined by measuring a, b, c, d, e separately with the percentage error of

1 \%, 2 \%, 3 \%, 4 \%, 3 \%

respectively. Minimum amount of error is contributed by the measurement of :

View answer and explanation

Correct option: C

a, e

Option C (a, e) is correct because the percentage errors of a and e, when multiplied by their respective fractional powers in the formula for T, yield the smallest contributions to the overall error in T.

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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