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Generally IEโ increases left to right, but Al (3pยน) has lower IEโ than Mg (3sยฒ) because:
- Mg has a full 3sยฒ subshell (extra stability)
- Al's 3p electron is shielded by 3s electrons and easier to remove
Matching logic:
- Lysine is an amino acid โ detected by ninhydrin (Q), which gives a purple colour with primary amines/amino acids.
- Furfural is an aldehyde from furan โ detected by 1-naphthol (P) via the Tollens-type or colour reaction specific to furfural.
- Benzyl alcohol (PhCH$_2$OH) is a primary alcohol โ detected by ceric ammonium nitrate (S), which gives a red colour with alcohols.
- Styrene (PhโCH=CH$_2$) has a C=C bond โ oxidized by KMnO$_4$ (R), which gets decolourized.
Correct match: (A)โ(Q), (B)โ(P), (C)โ(S), (D)โ(R)
Denver Developmental Screening Test (DDST-II) milestones at 6 months:
| Domain | Expected at 6 Months |
|---|---|
| Gross Motor | Sits with support, rolls over (front to back), bears weight on legs |
| Fine Motor | Transfers objects hand to hand, reaches for objects, raking grasp |
| Language | Babbles, monosyllables (ba, da, ma), laughs, squeals |
| Personal-Social | Recognizes faces, smiles spontaneously, feeds self cracker |
Key milestones timeline:
- 3 months: Head control, social smile
- 6 months: Sits with support, transfers objects โ
- 9 months: Stands holding furniture, pincer grasp developing
- 12 months: Walks with support, 1โ2 words
- 18 months: Walks alone, 10+ words
- 24 months: 2โ3 word sentences, runs
Red flags at 6 months: No head control, no social smile, not reaching for objects โ warrant immediate developmental evaluation.
A bag has 4 red and 6 black balls. One ball is drawn, observed, and returned along with 2 more balls of the same colour. A second ball is then drawn. Find the probability that the second ball drawn is red.
Initially: 4 red, 6 black (10 total).
Case 1 โ First ball is red (prob $= \frac{4}{10}$):
Bag becomes: 6 red, 6 black (12 total). $P(\text{red 2nd}) = \frac{6}{12} = \frac{1}{2}$.
Case 2 โ First ball is black (prob $= \frac{6}{10}$):
Bag becomes: 4 red, 8 black (12 total). $P(\text{red 2nd}) = \frac{4}{12} = \frac{1}{3}$.
Total: $\dfrac{4}{10}\cdot\dfrac{1}{2}+\dfrac{6}{10}\cdot\dfrac{1}{3} = \dfrac{2}{10}+\dfrac{2}{10} = \dfrac{4}{10} = \dfrac{2}{5}$
The velocity-time graph for a body projected vertically upward is best described as a:
For a body projected vertically upward, only gravity acts on it โ a constant acceleration $g$ downward. From $v = u - gt$, velocity changes linearly with time, producing a straight line on a $v$-$t$ graph with slope $-g$.
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