d
Let the order of reaction with respect to \(\mathrm{A}\) and \(\mathrm{B}\) is \(\mathrm{x}\) and \(\mathrm{y}\) respectively. 

So, the law can be given as

\(R=k[A]^{x}[B]^{y}\) ...... \((i)\)

When the concentration of only \(\mathrm{B}\) is doubled, the rate is doubled, so

\(R_{1}=k[A]^{x}[2 B]^{y}=2 R\) ..... \((ii)\)

If concentrations of both the reactants \(A\) and \(B\) are doubled, the rate increases by a factor of \(8,\) so \(R^{\prime \prime}=k[2 A]^{x}[2 B]^{y}=8 R\)  .... \((i i i)\)

\(\Rightarrow k 2^{x} 2^{y}[A]^{x}[B]^{y}=8\, R\) ..... \((iv)\)

From Eqs. \((i)\) and \((ii)\), 

we get \(\Rightarrow \quad \frac{2 R}{R}=\frac{|A|^{x}|2 B|^{y}}{|A|^{x}|B|^{y}}\)

\(2=2^{y}\)

\(\therefore y=1\)

From Eqs. (i) and (iv), we get \(\Rightarrow \frac{8 R}{R}=\frac{2^{x} 2^{y}[A]^{x}|B|^{y}}{|A|^{x}[B]^{y}}\)

or \(8=2^{x} 2^{y}\)

Substitution of the value of \(y\) gives,

\(8=2^{x} 2^{1}\)

\(4=2^{x}\)

\((2)^{2}=(2)^{x}\)

\(x=2\)

Substitution of the value of \(x\) and \(y\) in Eq. \((i)\) gives,

\(R=k[A]^{2}[B]\)