Young's modulus of elasticity Y is expres | vedclass

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

Question

$Y=c^\alpha h^\beta G^\gamma$

$\mathrm{ML}^{-1} \mathrm{~T}^{-2}=\left(\mathrm{LT}^{-1}\right)^\alpha\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^\b

Vedclass · Accepted Answer

$\alpha=7, \beta=-1, \gamma=-2$

Vedclass · Answer

$Y=c^\alpha h^\beta G^\gamma$

$\mathrm{ML}^{-1} \mathrm{~T}^{-2}=\left(\mathrm{LT}^{-1}ight)^\alpha\left(\mathrm{ML}^2 \mathrm{~T}^{-1}ight)^\beta\left(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight)^\gamma$

$1=\beta-\gamma$      $. . . . . (1)$

$-1=\alpha+2 \beta+3 \gamma$     $. . . . .(2)$ 

$\frac{-2=-\alpha-\beta-2 \gamma}{-3=\beta+\gamma}$   $. . . .(3)$

$\frac{1=\beta-\gamma}{-2=2 \beta} \Rightarrow \beta=-1, \gamma=-2$

$-1=\alpha-2-6 \quad 	herefore \alpha=7 $

Young's modulus of elasticity $Y$ is expressed in terms of three derived quantities, namely, the gravitational constant $G$, Planck's constant $h$ and the speed of light $c$, as $Y=c^\alpha h^\beta G^\gamma$. Which of the following is the correct option?

Similar Questions

Let $[ {\varepsilon _0} ]$ denote the dimensional formula of the permittivity of vacuum. If $M =$ mass, $L=$ length, $T =$ time and $A=$ electric current, then:

If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

The dimensions of $K$ in the equation $W = \frac{1}{2}\,\,K{x^2}$ is

If force $({F})$, length $({L})$ and time $({T})$ are taken as the fundamental quantities. Then what will be the dimension of density

The position of a particle at time $t$ is given by the relation $x(t) = \left( {\frac{{{v_0}}}{\alpha }} \right)\,\,(1 - {e^{ - \alpha t}})$, where ${v_0}$ is a constant and $\alpha > 0$. The dimensions of ${v_0}$ and $\alpha $ are respectively