Step 1: Recall the rule for geometrical isomerism.
Geometrical isomerism, the cis and trans kind, happens in alkenes when the double bond cannot rotate freely and each carbon of the double bond carries two different groups.
Step 2: Write the test.
For a double bond drawn as $ab\,C=C\,cd$, we need $a$ not equal to $b$ on one carbon and $c$ not equal to $d$ on the other. If any carbon has two identical groups, there is no geometrical isomerism.
Step 3: Test 1-Butene.
1-Butene is $CH_2=CH-CH_2-CH_3$. The end carbon carries two hydrogens, which are identical, so it fails the test.
Step 4: Test Propene and 2-Methylpropene.
Propene $CH_2=CH-CH_3$ also has a $=CH_2$ end with two identical hydrogens. 2-Methylpropene $(CH_3)_2C=CH_2$ has one carbon with two identical methyl groups. Both fail.
Step 5: Test 2-Butene.
2-Butene is $CH_3-CH=CH-CH_3$. Each double-bond carbon carries one hydrogen and one methyl group, which are different. So the test passes.
Step 6: Choose the answer.
Only 2-Butene satisfies the rule, so it shows cis and trans forms. \[ \boxed{\text{2-Butene}} \]