STATEMENT-1 : D-glucose is dextrorotatory where as L-glucose is laevorotatory.
STATEMENT-2 : D-compounds are always dextrorotatory and L-compounds are always laevorotatory.

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is False

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STATEMENT-1 is False, STATEMENT-2 is True

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Solution

The correct option is C STATEMENT-1 is True, STATEMENT-2 is False
D-glucose is dextrorotatory where as L-glucose is laevorotatory.
However the general conclusion that D-compounds are always dextrorotatory and L-compounds are always laevorotatory is incorrect.
Because there are some dextrorotatory D-compounds and laevorotatory L compounds.
Thus STATEMENT-1 is True, STATEMENT-2 is False.

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

I

\to L = I \to ω

II

\to τ = \frac{d \to L}{d t}

(p \lor q) \land \sim p

\sim p \land q

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