JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 19)
Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
$$\sqrt {{{h{c^5}} \over G}} $$
$$\sqrt {{{{c^3}} \over {Gh}}} $$
$$\sqrt {{{Gh} \over {{c^5}}}} $$
$$\sqrt {{{Gh} \over {{c^3}}}} $$
Explanation
Let t $$ \propto $$ Gx hy cz
$$ \therefore $$ [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 $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$
$$ \therefore $$ [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 $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$
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