From the equation tan theta = frac{rg}{v^2},o | vedclass

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

Question

(C) The dimension of $	an 	heta$ is $[M^0 L^0 T^0]$ as it is a dimensionless quantity.For the right-hand side,the dimensions are: $[r] = [L]$,$[g] =

Vedclass · Accepted Answer

Dimensionally correct only

Vedclass · Answer

(C) The dimension of $	an 	heta$ is $[M^0 L^0 T^0]$ as it is a dimensionless quantity.For the right-hand side,the dimensions are: $[r] = [L]$,$[g] = [L T^{-2}]$,and $[v^2] = [L^2 T^{-2}]$.Thus,the dimension of $\frac{rg}{v^2}$ is $\frac{[L] [L T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$.Since both sides have the same dimensions,the equation is dimensionally correct.However,the correct physical formula for the angle of banking is $	an 	heta = \frac{v^2}{rg}$.Therefore,the given equation is numerically incorrect.

From the equation $\tan \theta = \frac{rg}{v^2}$,one can obtain the angle of banking $\theta$ for a cyclist taking a curve (the symbols have their usual meanings). Then say,it is

Similar Questions

The position of a body with acceleration '$a$' is given by $x = K{a^m}{t^n}$,where $t$ is time. Find the dimensions of $m$ and $n$.

The speed of ripples $(v)$ on a water surface depends on surface tension $(\sigma)$,density $(\rho)$,and wavelength $(\lambda)$. Then the square of speed $(v^2)$ is proportional to

The frequency $(v)$ of an oscillating liquid drop may depend upon the radius $(r)$ of the drop,the density $(\rho)$ of the liquid,and the surface tension $(s)$ of the liquid as: $v = r^{a} \rho^{b} s^{c}$. The values of $a, b,$ and $c$ respectively are:

Frequency $(n)$ is a function of density $(\rho)$,length $(a)$,and surface tension $(T)$. Then its value is:

If the speed of light $(c)$,acceleration due to gravity $(g)$,and pressure $(p)$ are taken as the fundamental quantities,then the dimension of the gravitational constant $(G)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module