JEE Advanced PYQ: Units & Measurements – Units of Physical Quantities

(1)

$A \to p, q$

$Using F = \frac{G M_{e} M_{s}}{r^{2}} \Rightarrow G M_{e} M_{s} = F r^{2} = N m^{2} = k g \frac{m}{s^{2}} \times m^{2} = k g m^{3} s^{- 2}$

$Also (volt) (coulomb) (metre) = (joule) (metre) (∵ volt = joule/coulomb)$

$= (N - m) (m) = N m^{2} = k g m^{3} s^{- 2}$

$B \to r, s$

$Using v_{rms} = \sqrt{\frac{3 R T}{M}} \Rightarrow v_{rms}^{2} = \frac{3 R T}{M}$

$Unit of \frac{3 R T}{M} is m^{2} s^{- 2}$

$Also (farad) {(volt)}^{2} {(k g)}^{- 1} = (joule) k g^{- 1} (∵ u = \frac{1}{2} C V^{2})$

$= N \cdot m k g^{- 1} = k g m s^{- 2} m k g^{- 1} = m^{2} s^{- 2}$

$C \to r, s$

$Using F = q v B \Rightarrow v^{2} = \frac{F^{2}}{q^{2} B^{2}}$

$∴$ $Unit of v^{2} is m^{2} s^{- 2}, which is further equal to F V^{2} k g^{- 1}$

$D \to r, s$

$Reason: Escape velocity v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow v_{e}^{2} = \frac{2 G M}{R}$

$∴$ $Unit of \frac{G M}{R} is m^{2} s^{- 2}$

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	Column I		Column II
(A)	$G M_{e} M_{s},$ $G – universal gravitational constant,$ $M_{e} – mass of the earth,$ $M_{s} – mass of the Sun$	(p)	$(volt) (coulomb) (metre)$
(B)	$\frac{3 R T}{M},$ $R – universal gas constant,$ $T – absolute temperature,$ $M – molar mass$	(q)	$(kilogram) {(metre)}^{3} {(second)}^{- 2}$
(C)	$\frac{F^{2}}{q^{2} B^{2}},$ $F – Force,$ $q – charge,$ $B – magnetic field$	(r)	${(metre)}^{2} {(second)}^{- 2}$
(D)	$\frac{G M_{e}}{R_{e}},$ $G – universal constant,$ $M_{e} – mass of the earth,$ $R_{e} – radius of the earth$	(s)	${(farad)(volt)}^{2} {(kg)}^{- 1}$