Explanation

(c)
Let the order with respect to A and B is x and y respectively.
Hence,
$Rate\ r = \lbrack A\rbrack^{x}\lbrack B\rbrack^{y}$ …(i)
On doubling the concentration of A, rate increases 4 times,
$4r = \lbrack A\rbrack^{x}\lbrack B\rbrack^{y}$ … (ii)
From Eqs. (i) and (ii)
$\frac{1}{4} = \left( \frac{1}{2} \right)^{x}$
$\therefore\ \ \$X=2
$\therefore\$ order with respect to A is 2
If concentration of A and B both are doubled,
$8r = \lbrack 2A\rbrack^{x}\lbrack 2B\rbrack^{y}$ … (iii)
From Eqs. (i) and (iii), we get
$\frac{1}{8} = \frac{1}{(2)^{x}}\ .\ \frac{1}{(2)^{y}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because x = 2\rbrack$
$\frac{1}{8} = \frac{1}{(2)^{2}}\ .\ \frac{1}{(2)^{y}}$
$\frac{1}{8} = \frac{1}{{4 \times 2}^{y}}$
$2^{y} = 2$
$\therefore\$Y=1
Hence , differential rate equation is
$r{\propto \lbrack A\rbrack}^{2}\lbrack B\rbrack^{1}$ or $\frac{dC}{dt} = kC_{A}^{2} \times C_{B}$
[Where, $C_{A}\$and $C_{B}$=concentrations of A and B]