Q88.If y = y(x), y ∈[0, π2 ) is the solution of the differential equation sec y dxdy −sin(x + y) −sin(x −y) = 0, with y(0) = 0, then 5y′( π2 ) is equal to _____. JEE Main 2021 (27 Jul Shift 1) JEE Main Previous Year Paper → → →
What This Question Tests
The question involves solving a first-order differential equation by separating variables and using trigonometric identities, along with an initial condition to find the particular solution.
Concepts Tested
Formulas Used
sin(A+B) + sin(A-B) = 2sin A cos B
∫ sec y dy = ln|sec y + tan y|
∫ cos x dx = sin x
📚 NCERT Sections This Tests
3.10 — In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 · Chapter 3
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).
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
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.
3.21 — The Following Data Were Obtained During The First Order Thermal
Chemistry Class 11 · Chapter 3
3.21 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2 Cl 2 g SO 2 g Cl 2 g Experiment Time/s–1 Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.
📋 Question Details
- Chapter
- Differential Equations
- Topic
- Solving first-order differential equations
- Year
- 2021
- Shift
- 27 Jul Shift 1
- Q Number
- Q88
- Type
- Numerical
- NCERT Ref
- Class 12 Mathematics Ch 9: Differential Equations
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