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MathsMediumMCQ2025 · 28 Jan Shift 1

Q8. Two number k1 and k2 are randomly chosen from the set of natural numbers. Then, the probability that the value of ik1 + ik2, (i = √−1) is non-zero, equals ⎪ ⎪ 2025 (28 Jan Shift 1) JEE Main Previous Year Paper (1) 1 (2) 3 2 4 (3) 1 (4) 2 4 3

What This Question Tests

This question tests the understanding of powers of 'i' and basic probability to find the chance that a sum of powers of 'i' is non-zero.

Concepts Tested

ProbabilityProperties of imaginary unit 'i'Counting outcomes

Formulas Used

P(E) = n(E)/n(S)

📚 NCERT Sections This Tests

2.1Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At

Physics Class 11 · Chapter 2

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2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

3.10In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured

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3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

8.6Isomerism Isomers And This Phenomenon Is Termed

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8.6 ISOMERISM isomers and this phenomenon is termed as position isomerism. For example, the The phenomenon of existence of two or more molecular formula C3H8O represents twocompounds possessing the same molecular alcohols: formula but different properties is known as isomerism. Such compounds are called OH as isomers. The following flow chart shows different types of isomerism. CH3CH2CH2OH CH3−CH-CH3 Propan-1-ol Propan-2-ol 8.6.1 Structural Isomerism Compounds having the same molecular (iii) Functional group isomerism: Two or formula but different structures (manners more compounds having the same molecular in which atoms are linked) are classified as formula but different functional groups structural isomers. Some typical examples are called functional isomers and this of different types of structural isomerism are phenomenon is termed as functional group given below: isomerism. For example, the molecular (i) Chain isomerism: When two or more formula C3H6O represents an aldehyde and compounds have similar molecular formula but a ketone: Isomerism Structural isomerism Stereoisomerism Chain Position Functional Metamerism Geometrical Optical isomerism isomerism group isomerism isomerism isomerism Reprint 2025-26 organic chemistry – some basic principles and techniques 271 in understanding the reactivity of organic O H compounds and in planning strategy for their   synthesis. CH3−C-CH3 CH3−CH2—C= O In the following sections, we shall learn Propanone Propanal some of the principles that explain how these (iv) Metamerism: It arises due to different reactions take place. alkyl chains on either side of the functional 8.7.1 Fission of a Covalent Bond group in the molecule. For example, C4H10O represents methoxypropane (CH3OC3H7) and A covalent bond can get cleaved either by : (i) ethoxyethane (C2H5OC2H5). heterolytic cleavage, or by (ii) homolytic cleavage.8.6.2 Stereoisomerism In heterolytic cleavage, the bond breaks The compounds that have the same in such a fashion that the shared pair of constitution and sequence of covalent bonds electrons remains with one of the fragments. but differ in relative positions of their atoms After heterolysis, one atom has a sextet or groups in space are called stereoisomers. electronic structure and a positive charge and This special type of isomerism is called as the other, a valence octet with at least one lone stereoisomerism and can be classified as pair and a negative charge. Thus, heterolytic geometrical and optical isomerism. + cleavage of bromomethane will give C H3 and

📋 Question Details

Chapter
Probability
Topic
Basic probability
Year
2025
Shift
28 Jan Shift 1
Q Number
Q8
Type
MCQ
NCERT Ref
Class 12 Mathematics Ch 13: Probability

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