QJEEM19Ph3335: JEE Main 2019 Physics - Class 11 JEEM19Ph3335 Solution: Expression for time in terms of G(universal gravitional…

Let t $\propto$ G^x h^y c^z

$\therefore$    [t] = [G]^x [h]^y [c]^z    . . . . . (1)

We know,

F = ${{G{M^2}} \over {{R^2}}}$

$\Rightarrow$   G = ${{F{R^2}} \over {{M^2}}}$

$\therefore$   [G] = ${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$

[G] = $[{M^{ - 1}}{L^3}{T^{ - 2}}]$

Also,

E = hf

$\therefore$   [h] = ${{[E]} \over {[F]}}$

= ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$

= $\left[ {M{L^2}{T^{ - 1}}} \right]$

[C] = $\left[ {{M^o}L\,{T^{ - 1}}} \right]$

From equation (1) we get,

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$

By comparing the power of M, L, T

$-$ x + y = 0

$\Rightarrow$   x = y

3x + 2y + z = 0

$\Rightarrow$   5x + z = 0 . . . . . (2)

$-$ 2x $-$ y $-$ z = 1

$\Rightarrow$    $-$ 3x $-$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = ${1 \over 2}$ = y and z = $-$ ${5 \over 2}$

$\therefore$   t $\propto$ $\sqrt {{{Gh} \over {{C^5}}}}$