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Q2/7: A particle covers half of a straight track with speed v0. The remaining half is covered in two equal time intervals with speeds v1 and v2. The average speed of the particle for the entire journey is:
Q1/7: A particle moves along a straight line such that its displacement x at time t is given by x^2 = t^2 + 1. The acceleration of the particle at any instant is:
📚 Today's Topic: Kinematics
🔬 Physics
⏳ 7 MCQs | ~60-90s gaps
📅 Get ready!
🧠 Memory Trick — Three Dimensional Geometry
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🔤 The Mnemonic:
Passengers Dislike Crowded Metros
📖 What It Stands For:
This mnemonic helps you perfectly recall the complex formula for the Shortest Distance (SD) between two skew lines, r = a1 + λb1 and r = a2 + μb2:
• P = Points given on the lines (a1 and a2)
• D = Difference between those points: (a2 - a1)
• C = Cross product of the direction vectors: (b1 × b2)
• M = Magnitude of that cross product: |b1 × b2| (which goes in the denominator)
💡 How to Use It:
When a JEE question asks for the shortest distance between two skew lines, write down the components step-by-step using the mnemonic:
1. Calculate the Point Difference: Vector d = a2 - a1
2. Calculate the Cross product: Vector c = b1 × b2
3. Take the dot product of the top two: (a2 - a1) • (b1 × b2)
4. Divide by the Magnitude of the cross product: |b1 × b2|
Formula: SD = | (a2 - a1) • (b1 × b2) | ÷ |b1 × b2|
#MemoryTrick #JEETricks
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Q5/5: In paper chromatography, if the solvent front travels 10 cm and the compound spot travels 6 cm, what is the retardation factor (Rf) of the compound?
Q4/5: Kjeldahl's method cannot be used for the estimation of nitrogen in which of the following compounds?
Q3/5: Which of the following carbocations is expected to be the most stable?
Q2/5: Arrange the following carboxylic acids in decreasing order of their acidic strength: (I) CF3COOH, (II) CCl3COOH, (III) CHCl2COOH, (IV) CH3COOH.
Q1/5: What is the correct IUPAC name of the compound CH3-CH(OH)-CH2-CO-CH3?
📚 Today's Topic: Some Basic Principles of Organic Chemistry
🔬 Chemistry
⏳ 5 MCQs | ~60-90s gaps
📅 Get ready!
🧠 Memory Trick — Organic Compounds Containing Halogens (Haloalkanes and Haloarenes)
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🔤 The Mnemonic:
To master the confusing differences between SN1 and SN2 nucleophilic substitution mechanisms, remember these two simple characters:
• For SN1: "3 Polar Rabbits"
• For SN2: "1 Aprotic Invader"
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📖 What It Stands For:
1. SN1 Mechanism ➔ "3 Polar Rabbits"
• 3 = 3° (Tertiary) alkyl halides react fastest (due to the stability of the 3° carbocation intermediate: 3° > 2° > 1°).
• Polar = Favored by Polar Protic solvents (like H2O, EtOH, NH3) which stabilize the carbocation and the leaving group via hydrogen bonding.
• Rabbits (R) = Racemisation occurs (gives a mixture of retention and inversion products because the nucleophile can attack the planar carbocation from either side).
2. SN2 Mechanism ➔ "1 Aprotic Invader"
• 1 = 1° (Primary) alkyl halides react fastest (due to minimum steric hindrance: 1° > 2° > 3°).
• Aprotic = Favored by Polar Aprotic solvents (like DMSO, Acetone, DMF) which do not cage the nucleophile, leaving it highly reactive.
• Invader (I) = Inversion of configuration (Walden Inversion) occurs because the strong nucleophile acts like an "invader" attacking strictly from the backside.
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💡 How to Use It:
When a JEE/NEET question asks you to predict the mechanism or product for a reaction like:
2-bromobutane reacting in the presence of Acetone and sodium iodide (NaI)
• Identify the solvent: Acetone is a polar aprotic solvent.
• Recall "1 Aprotic Invader" ➔ This points directly to the SN2 mechanism.
• Therefore, the reaction will proceed in a single step with complete Inversion of configuration at the chiral carbon, without forming any carbocation intermediates!
#MemoryTrick #JEETricks
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🧠 Memory Trick — Chemical Bonding and Molecular Structure
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The Mnemonic:
"Small Cats Love All Caviar"
What It Stands For:
• Small Cation = High polarizing power (high charge density pulls electron cloud easily).
• Large Anion = High polarizability (outer electrons are far from the nucleus and easily distorted).
• Caviar = Covalent Character (Fajan's Rules).
How to Use It:
Use this trick to instantly solve JEE/NEET questions on covalent character, melting points, or solubility in organic solvents:
• Example 1 (Cation variation): Compare the covalent character of BeCl2, MgCl2, and CaCl2. Since Be²⁺ is the Smallest Cation, BeCl2 has the highest Covalent character.
• Example 2 (Anion variation): Compare LiF, LiCl, LiBr, and LiI. Since I⁻ is the Largest Anion, LiI has the maximum Covalent character and therefore the lowest melting point.
#MemoryTrick #JEETricks
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🌟 MCQ Session Complete!
━━━━━━━━━━━━━━━━━━
📊 Score yourself out of 7 in the comments below! 🎯
✨ Keep learning, keep growing.
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