Q76.The number of real values of λ for which the system of linear equations, 2x + 4y −λz = 0 , 4x + λy + 2z = 0 and λx + 2y + 2z = 0 , has infinitely many solutions, is: (1) 3 (2) 1 (3) 2 (4) 0 Q77. ⎧ 0 cos x −sin x ⎫ π If S = x ∈[0, 2π] : sin x 0 cos x = 0 , then ∑x ∈S tan( 3 + x) is equal to: ⎨ ⎬ ⎩ cos x sin x 0 ⎭ (1) 4 + 2√3 (2) −4 -2 √3 (3) −2 + √3 (4) -2 −√3 |x| < 12 , x ≠0, is equal to:
What This Question Tests
This problem tests the condition for a homogeneous system of linear equations to have infinitely many solutions, which requires the determinant of the coefficient matrix to be zero.
Concepts Tested
Formulas Used
For a homogeneous system Ax=0 to have infinitely many solutions, det(A) = 0
📚 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).
13.4 — 1.23 351
Physics Class 12 · Chapter 13
13.4 1.23 351 Reprint 2025-26 Physics 13.5 (i) Q = –4.03 MeV; endothermic (ii) Q = 4.62 MeV; exothermic 56 – 2m 28 Al = 26.90 MeV; not possible. 13.6 Q = m ( 26 Fe ) ( 13 ) 13.7 4.536 × 1026 MeV 13.8 About 4.9 × 104 y 13.9 360 KeV CHAPTER 14 14.1 (c) 14.2 (d) 14.3 (c) 14.4 (c) 14.5 (c) 14.6 50 Hz for half-wave, 100 Hz for full-wave Reprint 2025-26 Bibligraphy BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987). 3 Advanced Physics, Tom Duncan, John Murray (2000). 4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition, Addison—Wesley (2004). 6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965). 8 Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 – Electricity and Magnetism (E.M. Purcell) c. Vol. 3 – Waves and Oscillations (Frank S. Crawford) d. Vol. 4 – Quantum Physics (Wichmann) e. Vol. 5 – Statistical Physics (F. Reif ) 9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967). 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960). 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15 Physics, Patrick Fullick, Heinemann (2000). 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998). 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18 University Physics, Harris Benson, John Wiley (1996). 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 21 Physics, Hans C. Ohanian, W.W. Norton (1989). Reprint 2025-26 Physics 22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996). 23 Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997). 25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987). 26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988). 27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005). 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005). GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965). 7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12 Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005). 16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). 354 Reprint 2025-26
2.5 — Potential Due To A System Of Charges
Physics Class 11 · Chapter 2
2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…, rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge q1 is 1 q1 V1 = 4 πε0 r1P where r1P is the distance between q1 and P. Similarly, the potential V2 at P due to q2 and V3 due to q3 are given by 1 q 2 1 q 3 V 2 = , V 3 = 4 πε0 r2P 4 πε0 r3P where r2P and r3P are the distances of P from charges q2 and q3, respectively; and so on for the potential due to other charges. By the FIGURE 2.6 Potential at a point due to a superposition principle, the potential V at P due system of charges is the sum of potentials to the total charge configuration is the algebraic due to individual charges. sum of the potentials due to the individual charges V = V1 + V2 + ... + Vn (2.17) 51 Reprint 2025-26 Physics 1 q1 q 2 q n = + + ...... + (2.18) 4 πε0 r1P r2 P rnP If we have a continuous charge distribution characterised by a charge density r (r), we divide it, as before, into small volume elements each of size Dv and carrying a charge rDv. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution. We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by 1 q V = (r ≥ R ) [2.19(a)] 4 πε0 r where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is 1 q V = [2.19(b)] 4 πε0 R Example 2.2 Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Solution Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7). FIGURE 2.7 Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have 1 3 × 10 – 8 2 × 10 –8 − x × 10 –2 4 πε0 (15 − x ) × 10 –2 = 0 where x is in cm. That is, 3 2 − = 0 2.2 x 15 − x which gives x = 9 cm. If x lies on the extended line OA, the required condition is 3 2 − = 0 EXAMPLE x x − 15 Reprint 2025-26 Electrostatic Potential and Capacitance which gives x = 45 cm Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the EXAMPLE formula for potential used in the calculation required choosing potential to be zero at infinity. 2.2 Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively. Electric potential, equipotential-sufaces-12584/ FIGURE 2.8 equipotential (a) Give the signs of the potential difference VP – VQ; VB – VA. (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B. surfaces: (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Solution 1 (a) As V ∝ , VP > VQ. Thus, (VP – VQ) is positive. Also VB is less negative r than VA . Thus, VB > VA or (VB – VA) is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive. http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field- (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. EXAMPLE (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A. 2.3 53 Reprint 2025-26 Physics 2.6 EQUIPOTENTIAL SURFACES An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8): 1 q V = 4 πεo r This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential FIGURE 2.9 For a surface: there is no potential difference between any two points on the single charge q surface and no work is required to move a test charge on the surface. (a) equipotential The electric field must, therefore, be normal to the equipotential surface surfaces are at every point. Equipotential surfaces offer an alternative visual picture spherical surfaces in addition to the picture of electric field lines around a charge centred at the configuration. charge, and (b) electric field lines are radial, starting from the charge if q > 0. FIGURE 2.10 Equipotential surfaces for a uniform electric field. For a uniform electric field E, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11. FIGURE 2.11 Some equipotential surfaces for (a) a dipole, 54 (b) two identical positive charges. Reprint 2025-26 Electrostatic Potential and Capacitance 2.6.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + dV, where dV is the change in V in the direction of the electric field E. Let P be a point on the surface B. d l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|dl. This work equals the potential difference VA–VB. Thus, |E|d l = V – (V + dV)= – dV V i.e., |E|= −δ (2.20) δl Since dV is negative, dV = – |dV|. we can rewrite FIGURE 2.12 From the Eq (2.20) as potential to the field. δV δV E = − = + (2.21) δl δl We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.
📋 Question Details
- Chapter
- Matrices & Determinants
- Topic
- System of linear equations
- Year
- 2017
- Shift
- 08 Apr Online
- Q Number
- Q76
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 4: Determinants
More from this Chapter
Q75.The number of values of k for which the linear equations 4x + ky + 2z = 0; kx + 4y + z = 0; 2x + 2y + z = 0 possess a non-zero solution is (1) 2 (2) 1 (3) zero (4) 3
Q77.Statement 1: If the system of equations x + ky+ 3z = 0, 3x + ky −2z = 0, 2x + 3y −4z = 0 has a nontrivial solution, then the value of k is 31 . Statement 2: A system of three homogeneous equations in three variables 2 has a non trivial solution if the determinant of the coefficient matrix is zero. JEE Main 2012 (26 May Online) JEE Main Previous Year Paper (1) Statement 1 is false, Statement 2 is true. (2) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. (3) Statement 1 is true, Statement 2 is true,, (4) Statement 1 is true, Statement 2 is false. Statement 2 is not a correct explanation for Statement 1.
Q80.The number of values of k, for which the system of equations : (k + 1)x + 8y = 4k JEE Main 2013 (07 Apr) JEE Main Previous Year Paper kx + (k + 3)y = 3k −1 has no solution, is : (1) 2 (2) 3 (3) Infinite (4) 1
Q81.If y = sec(tan−1 x), then dxdy at x = 1 is equal to (1) 1 (2) √2 (3) 1 (4) 1 √2 2 x