Question

If g is the acceleration due to gravity and $\lambda$ is wavelength then which physical quantity does $\sqrt {\lambda g}$ represent?

(A) Velocity

(B) Length

(C) Time

(D) Charge

Answer 1

Hint Dimensions of wavelength and acceleration due to gravity are $\left[ L \right]$and $\left[ {L{T^{ - 2}}} \right]$respectively. Substitute the data in the physical quantity expression$\sqrt {\lambda g}$. On simplifying we get the dimensional formula of velocity$\left[ {L{T^{ - 1}}} \right]$ .

Complete step-by-step answer

This problem can be easily analyzed by using the dimensional formulas.

Since, wavelength $\lambda$ is the distance up to which a wave is travelling, its measure of length$m$ . So, the dimension of $\lambda$is$\left[ L \right]$.

Acceleration due to gravity$g$ is just a special case of acceleration $m{s^{ - 2}}$and has the dimensions as $\left[ {L{T^{ - 2}}} \right]$

The physical quantity is given by,

$\sqrt {\lambda g}$

Substitute the dimensions of g and$\lambda$in the above expression.

$\sqrt {\lambda g} = \sqrt {L \times L{T^{ - 2}}}$

$\sqrt {\lambda g} = \sqrt {{L^2}{T^{ - 2}}}$

$\sqrt {\lambda g} = \left[ {L{T^{ - 1}}} \right]$

The dimensions we got are$\left[ {L{T^{ - 1}}} \right]$that of velocity $m{s^{ - 1}}$ .

Hence, the physical quantity represented is velocity and the correct option is A.

Note Alternative way to solve this is by writing the given quantities in terms of the known physical quantity relations.
$\sqrt {\lambda g} = \sqrt {\dfrac{v}{f}g}$

Since,$v = f\lambda$; f is frequency or $\left[ {{T^{ - 1}}} \right]$ and v is speed/velocity i.e. $\left[ {L{T^{ - 1}}} \right]$
$\sqrt {\lambda g} = \sqrt {vtg}$

Since,$t = \dfrac{1}{f}$
$\sqrt {\lambda g} = \sqrt {vt \times \dfrac{v}{t}}$

Since,$g = \dfrac{v}{t}$

$\sqrt {\lambda g} = \sqrt {{v^2}}$

$\sqrt {\lambda g} = v$

Therefore, the physical quantity derived is velocity.