convert 1; newton (SI unit of force) into dyn | vedclass

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

Question

The dimensional equation of force is $[F]=\left[M^{1} L^{1} T^{-2}\right]$

if $n_{1}, u_{1}$, and $n_{2}, u_{2}$ corresponds to $SI$ and $CGS$ unit

Vedclass · Accepted Answer

$10^5$

Vedclass · Answer

The dimensional equation of force is $[F]=\left[M^{1} L^{1} T^{-2}ight]$

if $n_{1}, u_{1}$, and $n_{2}, u_{2}$ corresponds to $SI$ and $CGS$ units respectively, then $n_{2}=n_{1}\left[\frac{M_{1}}{M_{2}}ight]^{1}\left[\frac{L_{1}}{L_{2}}ight]^{1}\left[\frac{T_{1}}{T_{2}}ight]^{2}$$=1\left[\frac{k g}{g}ight]\left[\frac{m}{c m}ight]\left[\frac{s}{s}ight]^{-2}=1 	imes 1000 	imes 100 	imes 1=10^{5}$

$	herefore 1$ newton $=10^{5}$ dyne.

convert $1\; newton$ ($SI$ unit of force) into $dyne$ ($CGS$ unit of force)

Similar Questions

A quantity $f$ is given by $f=\sqrt{\frac{{hc}^{5}}{{G}}}$ where $c$ is speed of light, $G$ universal gravitational constant and $h$ is the Planck's constant. Dimension of $f$ is that of

Write and explain principle of homogeneity. Check dimensional consistency of given equation.

Using dimensional analysis, the resistivity in terms of fundamental constants $h, m_{e}, c, e, \varepsilon_{0}$ can be expressed as

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity ' $\eta$ ' flowing per second, through a tube of radius $r$ and length / and having a pressure difference $P$ across its ends, is

The mass of a liquid flowing per second per unit area of cross section of a tube is proportional to $P^x$ and $v^y$ , where $P$ is the pressure difference and $v$ is the velocity. Then, the relation between $x$ and $y$ is