Which will be major and minor alkene formed if E1 reaction occurs on 2-chloro-2,4,4-trimethylpentane?
The major product in this case is favoured due to Steric hindrance in the product. Drawing the full structure of the molecule in the reaction makes this clearer

We can compare the degree of steric hindrance in the major product and the minor product by using A-values. The Wikipedia article gives the following A-values for common substituents:
$$\begin{array}{|c|c|}
\hline
\text{Substituent} &amp; \text{A-value} \\
\hline 
\ce{-CH3} &amp; 1.7 \\
\ce{-CH2C(CH3)3} &amp; 2 \\
\ce{-C(CH3)3} &amp; &gt;4 \\
\hline
\end{array}$$
Clearly, the major product has a lower set of A-values for it's substituents (2 &amp; 1.7) compared to the minor product (4 &amp; 1.7). Also, the minor product does not have any scope for forming stereoisomers, as both the groups on one end of the double bond are methyl. If one of them was a hydrogen group, then the E isomer would be the major product.