(D) For $0.1 \, M \, H_2SO_4$,the concentration of $H^+$ ions is $[H^+] = 2 \times 0.1 = 0.2 \, M$. $pH_1 = -\log(0.2) = -(\log 2 + \log 10^{-1}) = -0

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(D) For $0.1 \, M \, H_2SO_4$,the concentration of $H^+$ ions is $[H^+] = 2 \times 0.1 = 0.2 \, M$. $pH_1 = -\log(0.2) = -(\log 2 + \log 10^{-1}) = -0.3 + 1 = 0.7$. For $0.1 \, N \, H_2SO_4$,the normality is given as $0.1 \, N$,which means $[H^+] = 0.1 \, M$. $pH_2 = -\log(0.1) = 1$. The ratio of $pH$ is $pH_1 : pH_2 = 0.7 : 1 = 7 : 10$.