I and II are not geometrical isomers of each other because:

I_{1} \neq I_{2}

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I_{1} > I_{2}

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I_{2} > I_{1}

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they are geometrical isomers

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Solution

The correct option is D $I_{1} \neq I_{2}$
For geometrical isomerism distance between groups attached incase of cis- and trans- should be different.

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Similar questions

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

I_{3} = π / 2 \int 0 cos x d x,

A B C D

O

1, 2, 3

4

I_{1}, I_{2}, I_{3} & I_{4}

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x

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