sanskrit-varnamala-and-sandhi-sa-l2 MCQ — 9 अभ्यास प्रश्न उत्तर सहित

Explanation

The key is the guṇa join devendra. In deva + indra the boundary brings a against i, and guṇa-saṃdhi states a/ā + i/ī → e, so the join is devendra and the rule is guṇa. The devindra-as-dīrgha choice keeps a wrong form and mislabels it, since dīrgha needs two like simple vowels, not a + i. The devaindra-as-vṛddhi choice fails because vṛddhi requires a following e/ai/o/au, not i. The devyindra-as-yaṇ choice fails because yaṇ needs the first vowel to be i/u/ṛ, but here it is a. Only the devendra–guṇa pairing matches the correct form with the correct rule name, which is exactly what Aśok must verify before ticking.

Explanation

The key is that both statements are correct. Statement (i): in su + āgata the vowel u meets a dissimilar vowel ā; yaṇ-saṃdhi turns i/ī→y, u/ū→v, ṛ→r before a dissimilar vowel, so u→v gives svāgata — correct. Statement (ii): in ne + ana the diphthong e is followed by a vowel; ayādi-saṃdhi turns e→ay, ai→āy, o→av, au→āv before a vowel, so e→ay gives nayana — correct. The only-(i)-true choice and the only-(ii)-true choice each reject one genuinely valid statement by confusing the yaṇ direction (semivowel from i/u/ṛ) with the ayādi direction (semivowel-glide from e/ai/o/au); the neither-true choice wrongly rejects both. Karan must check the first vowel's class to keep yaṇ and ayādi apart.

Explanation

The key is the mapping 1-dīrgha, 2-vṛddhi, 3-yaṇ. (1) bhānu + udaya: u meets u, two like simple vowels merge to ū → bhānūdaya, dīrgha-saṃdhi. (2) eka + eka: a is followed by e, and a/ā + e/ai → ai → ekaika, vṛddhi-saṃdhi. (3) iti + ādi: i is followed by the dissimilar vowel ā, so i → y → ityādi, yaṇ-saṃdhi. The 1-guṇa/2-yaṇ/3-vṛddhi mapping mislabels the u+u join as guṇa and the i+ā join as vṛddhi; the 1-vṛddhi/2-guṇa/3-dīrgha mapping swaps every pair; the 1-yaṇ/2-dīrgha/3-guṇa mapping wrongly calls bhānu+udaya yaṇ though no i/u/ṛ becomes a semivowel there. The deciding habit, taught to Aśok, is to test the second vowel's identity and class for each pair separately.

Explanation

The key is that both statements are correct. Statement (i): in namaḥ + te the visarga is preceded by a and followed by the voiceless dental t; the visarga surfaces as s here, giving namaste — the stated reason is correct. Statement (ii): in vāk + īśa the word-final voiceless velar k stands before a vowel (a voiced environment) and undergoes voicing to g, giving vāgīśa — also correct. Both joins and both reasons are textbook-standard. The only-(i)-true choice and the only-(ii)-true choice each reject one valid statement by isolating the visarga rule from the consonant-voicing rule; the neither-true choice wrongly rejects both even though each reason is independently sound. The discipline drilled with Karan is to check the voicing of the following sound for each statement independently.

Explanation

The key is the dīrgha join vidyālaya. In vidyā + ālaya the boundary brings ā against ā — two identical simple long vowels — and dīrgha-saṃdhi fuses identical or homogeneous simple vowels into their single long form, giving vidyālaya. The guṇa label vidyaalaya is wrong because guṇa requires a/ā plus a different vowel (i/u/ṛ), not ā meeting ā; the vṛddhi form vidyailaya overshoots since vṛddhi needs a following e/ai/o/au; the yaṇ form vidyyālaya applies only when i/u/ṛ precedes a dissimilar vowel, and no i/u/ṛ exists here. None of those triggering environments is present, so only dīrgha applies. The deciding clue, taught to Devikā, is to check whether the two vowels are the same simple class first.