Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations:

$a.60mL\frac{M}{10}HCl+40mL\frac{M}{10}NaOH$
$b.55mL\frac{M}{10}HCl+45mL\frac{M}{10}NaOH$
$c.75mL\frac{M}{5}HCl+25mL\frac{M}{5}NaOH$
$d.100mL\frac{M}{10}HCl+100mL\frac{M}{10}NaOH$

pH of which one of the them will be equal to 1?

The correct option is C c
$pH\text{of}75mL\frac{M}{5}HCl+25mL\frac{M}{5}NaOH$ will be equal to 1.

$\text{Total volume}=75mL+25mL=100mL$

$\text{Number of mmol of HCl}=75mL\times \frac{M}{5}=15mmol$

$\text{Number of mmol of NaOH}=25mL\times \frac{M}{5}=5mmol$

$\text{5 mmol of NaOH will neutralise 5 mmol of HCl}.$

$\text{15−5=10 mmol of HCl will remain}$

$\left[H^{+}\right]=\left[HCl\right]=\frac{10mmol}{100mL}=0.1M$

$pH=-log\left[H^{+}\right]$

$pH=-log0.1pH=1$
Hence, option (c) is correct.