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(D) Given: Number of cells in each row $(n)$ = $10$,Number of rows $(m)$ = $5$,$emf$ of each cell $(E)$ = $1.5 \,V$,Internal resistance of each cell $

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$0.68$

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(D) Given: Number of cells in each row $(n)$ = $10$,Number of rows $(m)$ = $5$,$emf$ of each cell $(E)$ = $1.5 \,V$,Internal resistance of each cell $(r)$ = $1 \,\Omega$,External resistance $(R)$ = $20 \,\Omega$.In a mixed grouping of cells,the total $emf$ of the battery is equal to the $emf$ of one row,which is $nE = 10 	imes 1.5 = 15 \,V$.The total internal resistance of the battery is given by $r_{eq} = \frac{nr}{m} = \frac{10 	imes 1}{5} = 2 \,\Omega$.The current $i$ flowing through the external resistor $R$ is given by the formula:$i = \frac{nE}{R + \frac{nr}{m}}$Substituting the values:$i = \frac{15}{20 + 2} = \frac{15}{22} \approx 0.68 \,A$.

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