Step 1: Understand the oxidation reaction.
ConcHNO$_3$ is a strong oxidizing agentIt oxidizes both terminal groups of an aldoseThat means the aldehyde group $-CHO$ at C-1 and the primary alcohol group $-CH_2OH$ at C-6 are both converted into carboxylic acid groups
\[
\text{Aldohexose} \xrightarrow{\text{conc. HNO}_3} \text{Aldaric acid}
\]
Step 2: Identify the structure of the product.
After oxidation, the product has identical terminal groups on both ends
\[
\text{HOOC}-(\text{CHOH})_4-\text{COOH}
\]
Since both ends are now identical, the molecule can become meso if it has internal symmetry
Step 3: Condition for meso-dicarboxylic acid.
A meso compound must contain stereocentres but should also have an internal plane of symmetryTherefore, the arrangement of $-OH$ groups in the Fischer projection must be internally symmetrical after oxidation
Step 4: Count total stereoisomers of glucose.
Glucose is an aldohexose and has four chiral centres at C-2, C-3, C-4 and C-5
\[
\text{Number of stereoisomers} = 2^4 = 16
\]
Step 5: Check which oxidized acids become meso.
On oxidation, only those aldohexoses give meso aldaric acids in which the configuration becomes symmetrical about the middle of the carbon chainThe relevant stereochemical patterns are those which have internal compensation after conversion of both ends into $-COOH$ groups
Step 6: Identify the number of such stereoisomers.
Among the 16 stereoisomers of glucose, 4 stereoisomers give aldaric acids having internal plane of symmetryThese acids are meso because optical activity is cancelled due to internal compensation
Step 7: Conclusion.
Hence, the number of stereoisomers of glucose that give meso-dicarboxylic acid on oxidation with concHNO$_3$ is
\[
\boxed{4}
\]