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MathsHardMCQ2019 ยท 08 Apr Shift 2

Q84.If โˆซ ๐‘‘๐‘ฅ 2 = ๐‘ฅ๐‘“๐‘ฅ1 + ๐‘ฅ6 3 + ๐ถ, where ๐ถ is a constant of integration, then the function ๐‘“๐‘ฅ is equal to ๐‘ฅ31 + ๐‘ฅ6 3 (1) 3 (2) - 1 ๐‘ฅ2 2๐‘ฅ3 1 1 (3) - (4) - 6๐‘ฅ3 2๐‘ฅ2 ๐‘ฅ ๐‘ฅ

What This Question Tests

This question requires recognizing a suitable substitution for a complex integrand and then carefully comparing the resulting antiderivative with the given format to determine the function f(x).

Concepts Tested

Indefinite integralSubstitution method for integrationComparing integral forms

Formulas Used

โˆซ u^n du = u^(n+1)/(n+1) + C

๐Ÿ“š NCERT Sections This Tests

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2.1 Two charges 5 ร— 10โ€“8 C and โ€“3 ร— 10โ€“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.

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5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position โŠณ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )โˆ†xAnswer The initial kinetic energy of the bullet โˆ† x โ†’ 0 โˆ‘ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1ร—1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = โˆซF ( i 1 2 x mv f = 100 J where โ€˜limโ€™ stands for the limit of the sum when 2 โˆ†x tends to zero. Thus, for a varying force 2 ร— 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m sโ€“1 The speed is reduced by approximately 68% (not 90%). โŠณ

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Chemistry Class 11 ยท Chapter 3

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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 Lโ€“1 0.20 0.20 0.40 B/ mol Lโ€“1 0.30 0.10 0.05 r0/mol Lโ€“1sโ€“1 5.07 ร— 10โ€“5 5.07 ร— 10โ€“5 1.43 ร— 10โ€“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 ยฎ C + D Experiment [A]/mol Lโ€“1 [B]/mol Lโ€“1 Initial rate of formation of D/mol Lโ€“1 minโ€“1 I 0.1 0.1 6.0 ร— 10โ€“3 II 0.3 0.2 7.2 ร— 10โ€“2 III 0.3 0.4 2.88 ร— 10โ€“1 IV 0.4 0.1 2.40 ร— 10โ€“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 Lโ€“1 [B]/ mol Lโ€“1 Initial rate/ mol Lโ€“1 minโ€“1 I 0.1 0.1 2.0 ร— 10โ€“2 II โ€“ 0.2 4.0 ร— 10โ€“2 III 0.4 0.4 โ€“ IV โ€“ 0.2 2.0 ร— 10โ€“2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 sโ€“1 (ii) 2 minโ€“1 (iii) 4 yearsโ€“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 ยฎ 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 ร— [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol Lโ€“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).