JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 25)
A quantity f is given by $$f = \sqrt {{{h{c^5}} \over G}} $$ where c is
speed of light, G universal gravitational
constant and h is the Planck's constant.
Dimension of f is that of :
Energy
Momentum
Area
Volume
Explanation
[h] = M1L2T–1
[C] = L1T–1
[G] = M–1L3T–2
[f] = $$\sqrt {{{M{L^2}{T^{ - 1}} \times {L^5}{T^{ - 5}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}}} $$ = M1L2T–2
[C] = L1T–1
[G] = M–1L3T–2
[f] = $$\sqrt {{{M{L^2}{T^{ - 1}} \times {L^5}{T^{ - 5}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}}} $$ = M1L2T–2
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