The velocity of water waves v may depend upon | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The dimension of velocity $v$ is $[M^0 L^1 T^{-1}]$.The dimension of wavelength $\lambda$ is $[M^0 L^1 T^0]$.The dimension of acceleration due to

Vedclass · Accepted Answer

$v^2 \propto g \lambda$

Vedclass · Answer

(C) The dimension of velocity $v$ is $[M^0 L^1 T^{-1}]$.The dimension of wavelength $\lambda$ is $[M^0 L^1 T^0]$.The dimension of acceleration due to gravity $g$ is $[M^0 L^1 T^{-2}]$.The dimension of density $\rho$ is $[M^1 L^{-3} T^0]$.Let the relation be $v = k \lambda^a g^b \rho^c$,where $k$ is a dimensionless constant.Equating dimensions on both sides:$[L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$$[M^0 L^1 T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]$Comparing the powers of $M, L,$ and $T$:For $M$: $c = 0$For $T$: $-2b = -1 \implies b = 0.5$For $L$: $a + b - 3c = 1 \implies a + 0.5 - 0 = 1 \implies a = 0.5$Substituting the values,$v \propto \lambda^{0.5} g^{0.5} \rho^0$.Therefore,$v \propto \sqrt{g \lambda}$,which implies $v^2 \propto g \lambda$.

The velocity of water waves $v$ may depend upon their wavelength $\lambda$,the density of water $\rho$,and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as:

Similar Questions

If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid,the density $\rho$ of the liquid,and the acceleration due to gravity $g$,then the correct expression for $F$ can be:

The time dependence of a physical quantity $P$ is given by $P = P_0 e^{-\alpha t^2}$,where $\alpha$ is a constant and $t$ is the time. Then the constant $\alpha$ has:

In a particular system,the units of length,mass,and time are chosen to be $10 \; cm$,$10 \; g$,and $0.1 \; s$ respectively. The unit of force in this system will be equal to (in $; N$)?

What is the dimensional formula of $ab^{-1}$ in the equation $(P+\frac{a}{V^2})(V-b)=RT$,where letters have their usual meaning?

Consider the equation $\frac{1}{2} m v^{2} = m g h$,where $m$ is the mass of the body,$v$ is the velocity,$g$ is the acceleration due to gravity,and $h$ is the height. Is this equation dimensionally correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module