A cross between pea plant with white flowers (vv) and pea plant with violet flowers (VV) resulted in $F_2$ progeny in which ratio of violet (VV) and white (vv) flowers will be :

When a pure violet-flowered pea plant ($VV$) is crossed with a pure white-flowered pea plant ($vv$), the $F_1$ generation consists entirely of heterozygous violet-flowered plants ($Vv$) because violet ($V$) is dominant over white ($v$).

When these $F_1$ plants are self-pollinated ($Vv \times Vv$):

• The genotypic ratio in the $F_2$ generation is $1\text{ }VV : 2\text{ }Vv : 1\text{ }vv$.
• Since $V$ is dominant, both $VV$ (homozygous dominant) and $Vv$ (heterozygous) plants will produce violet flowers. Thus, $1 + 2 = 3$ parts of the progeny have violet flowers.
• Only the $vv$ (homozygous recessive) plants will produce white flowers, which is $1$ part of the progeny.
• Therefore, the phenotypic ratio of violet to white flowers in the $F_2$ generation is $3 : 1$.

Why other options are incorrect:

• Option (a) 1 : 1: This is the ratio obtained in a test cross (crossing $Vv$ with $vv$), not in the $F_2$ generation of a self-pollinated monohybrid cross.
• Option (b) 2 : 1: This ratio does not represent any standard phenotypic or genotypic ratio of a Mendelian monohybrid $F_2$ generation.
• Option (d) 1 : 3: This is the inverse ratio (white to violet), whereas the question asks for the ratio of violet to white flowers.