Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II:
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 $)(\text { metre })^{3}(\text { second })^{-2}$
C $\frac{F^{2}}{q^{2} B^{2}}$
$F -$ force,
$q -$ charge,
$B -$ magnetic field.
R (metre) $^{2}(\text { second })^{-2}$
D $\frac{G M_{e}}{R_{e}}$
$G$ - universal gravitational constant,
$M_{e}$ - mass of the Earth,
$R_{e}-$ radius of the Earth.
S $($ farad $)(\text { volt })^{2}( kg )^{-1}$

JEE AdvancedJEE Advanced 2007

Solution:

$( A ) \rightarrow( P ),( Q )$
Quantity given in option $(A)$ has a dimension of force $\times(\text { length })^{2}$;
quantities given in options $(P)$ and $(Q)$ has the same dimensions:
$G M_{e} M_{s}=F r^{2}$ or its unit is $kgm ^{3} s ^{-2}$.
(B) $\rightarrow( R ),( S )$
We have $\frac{3 R T}{M}=C_{\max }^{2} \Rightarrow $ the unit $m^{2} s^{-2}$,
which is nothing but energy per unit mass; the dimensions of this quantity is $L / T^{2}$;
quantities given in options $(R)$ and $(S)$ have the same dimension.
$( C ) \rightarrow( R ),( S )$
We have $\frac{F^{2}}{q^{2} B^{2}}=v^{2} \Rightarrow $ the unit $m^{2} s^{-2}$,
that is, the condition is same as that of option (B).
$(D) \rightarrow(R),(S)$
It is square of the escape velocity of the Earth:
$\frac{G M e}{R e}=V_{g}=\frac{W}{m} \Rightarrow $ the unit $; m^{2} s^{-2}$
The dimensions are same as that of option (B).