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JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1

The velocity of water waves (v) may depend on their
wavelength l, the density of water r and the acceleration
due to gravity, g. The method of dimensions gives the
relation between these quantities is

View answer and explanation

Correct option: B

{V^2} \propto g\lambda

Option B (

{V^2} \propto g\lambda

) is correct because dimensional analysis shows that the velocity of water waves is proportional to the square root of the product of gravity and wavelength.

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

\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}},\;

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{\Delta V}

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{\Delta t}

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