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On hydrolysis of starch, we finally get
(A) Glucose and Fructose
(B) Glucose
(C) Fructose
(D) Sucrose

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Hint:Starch is known as main storage of polysaccharide of the plant. It is made up of monosaccharide glucose. It is one of the essential dietary sources for human beings. The starch is a polymer which is insoluble in alcohol and cold water. The starch is further divided into two components amylose and amylopectin.

Complete answer:As starch is divided into amylase and amylopectin, so amylase is a water soluble component. It contains 15% to 20% of starch in it. It is long linear and helical in shape with 200 to 1000 \[\alpha D-glucose\] units of monosaccharide linked to each other by \[{{C}_{1}}-{{C}_{4}}\]linkage.
The other component of starch is amylopectin which is an insoluble component in water. It contains 80% to 85% of starch in it. It is a chain which is branched in shape. It is made up of \[\alpha D-glucose\] units of monosaccharide linked by \[{{C}_{1}}-{{C}_{4}}\]linkage, though the branching occurs by \[{{C}_{1}}-{{C}_{6}}\]linkage.
On hydrolysis of starch, the product which we get is glucose. But the hydrolysis occurs at 393K temperature and under pressure of 2 to 3 bar. This can be seen by the following equation.
                                                  \[{{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}+n{{H}_{2}}O\xrightarrow[393K,2-3bar]{{{H}^{+}}}n{{C}_{6}}{{H}_{12}}{{O}_{6}}\]

Glucose is an anomer because the two cyclic forms of glucose differ only in its configuration of the hydroxyl group at \[{{C}_{1}}\] called anomeric carbon or anomers.

Hence the correct option is (B).

Note:Glucose have the aldehyde group but it does not give 2,4 DNP test which shows the presence of carbonyl group and Schiff's test also. Glucose has two crystalline forms and they are alpha and beta. The alpha form is obtained on crystallizing glucose with concentrated sulphuric acid at the temperature of 303K. while the beta form is obtained by crystallizing glucose from hot and saturated solutions at a temperature of 371K.