AIEEE Minor Test #1

 

Ques 1 -12 [1.5 marks] / Ques 13-25 [3 marks] / Ques 26-40 x 4.5 marks -ve marks are one thirdTest duration 2 hour

Q1. The triangle with vertices (1, 5); (–3, 1) and (3, –5) is

(a)isosceles                (b) equilateral (c) right angled           (d) None of these

Q2. If the points (4, – 4), (– 4, 4) and (x, y) form an equilateral triangle then(a) (b) (c) (d) None of these

Q3. If (–4, 6), (2, 3) and (2, –5) are vertices of a triangle, then its incentre is (a)(–1, 2)           (b) (2, –1) (c) (1, 2)                  (d) (2, 1)

Q4.Circumcentre of a triangle whose vertex are (0, 0), (4, 0) and (0, 6) is

(a)        

(b) (0, 0)       (c) (2, 3)                        (d) (4, 6)

Q5.Orthocentre of a triangle whose vertex are (8, –2), (2, –2) and (8, 6) is

(a) (8, –2)           (b) (8, 6)            (c)                    

(d) (0, 0)

Q6.The area of a triangle with vertices (3, 8); (–4, 2) and (5, –1) is (a) 40.5                 (b) 36.5                         (c) 3.75                    (d) 37.5

Q7.If D, E, F are mid points of the sides AB, BC and CA of a triangle formed by the points A(5, -1) B(-7, 6) and C(1, 3), then area of DDEF is (a) 2/5               (b) 5/2                         (c) 5                               (d) 10

Q8. The point (4, 1) undergoes two successive transformations(i) Reflection about the line y = x(ii) Translation through a distance 2 units along the positive direction of x – axisThe final position of the point is given by the coordinates

(a) (4, 3)                 (b) (3, 4)              (c) (7/2, 7/2)         (d) (1, 4)

Q9.If A(c, 0) and B(– c, 0) are two points, then the locus of a point P which moves such that PA2+ PB2 = AB2 is (a) x2 – y2 =c2           (b) y2 = 4cx           (c) x2 + y2 = c2             (d) None of these

Q10. Let A(2, 3) and B(–4, 5) are two fixed points. A point P moves in such a way that DPAB = 12 sq. units, then its locus is

(a) x2 + 6xy + 9y2 + 22 x + 66y – 23 = 0 (b) x2 + 6xy + 9y2 + 22 x + 66y + 23 = 0(c)x2 + 6xy + 9y2 – 22 x – 66y – 23 = 0 (d) None of these

Q11.If sum of square of distances of a point from axes is 4, then its locus is (a) x + y = 2                (b) x2 + y2 = 16        (c) x + y = 4         (d) x2 + y2 = 4

Q12.The extremities of diagonal of a right-angled triangle are (2, 0) and (0, 2), then locus of its third vertex is (a) x2 + y2 – 2x – 2y = 0              (b) x2 + y2 + 2x – 2y = 0(c)x2 + y2 – 2x + 2y = 0              (d) x2 + y2 + 2x + 2y = 0

Q13.Keeping coordinate axes parallel, the origin is shifted to a point (1, –2), then transformed equation of x2 + y2 = 2 is (a) x2 + y2 + 2x – 4y + 3 = 0              (b) x2 + y2 – 2x + 4y + 3 = 0(c)x2 + y2 – 2x – 4y + 3 = 0           (d) x2 + y2 – 2x + 4y + 3 = 0

Q14.To remove xy term from the second degree equation 5x2 + 8xy + 5y2 + 3x + 2y + 5 = 0, the coordinates axes are rotated through an angle q, then q equals

(a) p/2             (b) p/4    (c) 3p/8         (d) p/8

Q15.The ratio in which the line y – x + 2 = 0 divides the line joining (3, – 1) and (8, 9) is (a) 2 : 3        (b) 3 : 2 (c) – 2 : 3         (d) – 3 : 2

Q16.The area of the triangle, formed by the straight lines 7x – 2y + 10 = 0, 7x + 2y – 10 = 0 and 9x + y + 2 = 0, is

(a)       

(b) (c)        

(d) None of these

Q17.Two vertices of a triangle are (3, – 1) and (– 2, 3) and its orthocenter is origin, the coordinates of the third vertex are

(a)       

(b) (c)        

(d)

Q18.The equation of the internal bisector of ÐBAC of DABC with vertices A(5, 2), B(2, 3) and C(6, 5) is (a) 2x + y + 12 = 0 (b) x + 2y – 12 = 0 (c) 2x + y – 12 = 0        (d) None of these

Q19.The equation of the straight line upon which the length of perpendicular from origin is  units and this perpendicular makes an angle of 75° with the positive direction of x – axis, is (a)          (b) (c)        (d) None of these

Q20.The image of the point (– 8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is (a) (16, – 2) (b)  (– 16, 2) (c) (16, 2)      (d) (– 16, – 2)

Q21.The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7

= 0 and 12x – 5y – 13 = 0 and perpendicular to the line 2x – 3y + 5 = 0 is

(a) 33x + 22y + 13 = 0 (b) 33x + 22y – 13 = 0               (c) 33x – 22y + 13 = 0 (d) None of these

Q22.If the family of lines x(a + 2b) + y(a + 3b) = a + b passes through the point for all values of a and b, then the coordinates of the points are (a) (2, 1)       (b) (2, – 1)               (c) (– 2, 1) (d) None of these

Q23.The value of k so that the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent, is (a) 7   (b) – 7            (c)  5 (d) – 5

Q24.Let P be the image of the point (– 3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through an angle 60° in the clockwise sense. The coordinates of point P with respect to the new axes are

(a)         

(b)

(c)      

(d) None of these

Q25.If for a variable line , the condition a–2 + b–2 = c–2 (c is a constant) is satisfied, then the locus of foot of the perpendicular drawn from origin to this is

(a)x2 + y2 =          

(b) x2 + y2 = 2c2       (c)  x2 + y2 = c2             (d) x2 – y2 = c2

Q26.If A and B are two sets, then A (A È B)¢ equals (a) A (b) B           (c)  f (d) none of these

Q27.Which of the following is an empty set ?(a)The set of prime numbers which are even

(b)The solution set of the equation  = 0 ; x Î R(c)(A x B) Ç (B x A), where A and B are disjoint (d)The set of real which satisfy x2 + ix + i – 1 = 0

Q28.If sets A and B are defined as : A = {(x, y) : y = , x ¹ 0, x Î R} ,  B = {(x, y) : y =-x, x Î R}, then (a) A Ç B = A (b) A Ç B = B    (c)  A Ç B = f               (d) none of these

Q29.Let R be reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then        (a) m ³ n (b) m £ n    (c)  m = n          (d) none of these

Q30.f(x) = | sin x | has an inverse if its domain is :

(a) [ 0, p ] (b) (c)            

(d) none of these

Q31.Consider the following equations(1) A – B = A – (A Ç B)           

(2) A = (A Ç B) È (A – B)(2) A – (B È C) = (A – B) È (A - C)

Which of these is / are correct (a) 1 and 3 (b) 2 only  (c)  2 and 3        (d) 1 and 2

Q32.If a is the set of the divisors of the number 15, B is the set of prime numbers smaller than 10 and C is the set of even numbers smaller than 9, then (A È C) Ç B is the set  (a) {1, 3, 5} (b) {1, 2, 3}   (c) {2, 3, 5}        (d) {2, 5}

Q33.x, y and z are rational numbers. Consider the following statements in this regard (1) x + 3 = y + z Þ x = y             (2) xz = yz Þ x = y

Which of the above statement(s) is / are correct ? (a) 1 alone (b) 2 alone (c) Both 1 and 2             (d) Neither 1 nor 2

Q34.If second term of an AP is 2 and 7th term is 22, then sum of 9 terms is (a) 126          (b) – 126 (c) 90        (d) 252

Q35.If Sn denotes the sum of the first n terms of an AP and S2n = 3Sn, then (a) 7     (b)  6 (c) 8      (d) 10

Q36.The sum of r terms of an AP is denoted by Sr and , then the ratio of the 7th term and 5th term of the AP is

(a) (b)          (c)   

(d)

Q37.A square is drawn by joining the mid points of the given square a third square in the same way and this process continues indefinitely. If a side of the first square is 16 cm, then the sum of the areas of all the squares (a) 128 sq cm (b) 256 sq cm (c) 512 sq cm   (d) 1024 sq cm

Q38.If the pth term of an AP is  and qth term is  then the sum of the first pq terms is

(a) 0              (b) (pq – 1)

(c) (pq + 1)

(d)

Q39.If the AM between pth and qth terms of an AP be equal to AM between rth and sth terms of the AP, then

(a) p + s = q + r           (b) p + q = r + s           (c) p + r = q + s (d) p + q + r + s = 0

Q40.The value of 5 + 55 + 555 + …… to n terms is

(a) 10n + 1 – 10        (b)         

(c) [10n + 1 – 1 + 9n]             

(d) (10n + 1 – 10)

Answers Question 1 2 3 4 5 6 7 8 9 1

011

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Answer c c a c a d b b c c d a a b a a b c a d a b b a c

Question

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Answer c c c a b d a d a b d c c b b                    

 

AIEEE Minor Test #2

 

Circle, Pairs of Straight Lines, Permutation and Combination

Q1. The equation of the circle, whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 and 3x + 4x – 5 = 0 and passes through the origin, is

(a) 17(x2 + y2) + 2x – 44y = 0 (b) 17(x2 + y2) + 2x + 44y = 0

(c) 17(x2 + y2) + 2x – 44y = 0     (d) None of these

Q2. The equation of the circle which touches the axis of y at a distance + 4 from the origin and cuts off an intercept 6 from the axis of x is

(a) x2 + y2 – 10x – 8y + 16 = 0 (b) x2 + y2 + 10x – 8y + 16 = 0

(c) x2 + y2 – 10x + 8y + 16 = 0 (d) None of these

Q3. A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. The equation of the circle with centre at (6, 5) and touching the above circle externally is

(a) x2 + y2 + 12x – 10y + 52 = 0 (b) x2 + y2 – 12x + 10y + 52 = 0

(c) x2 + y2 – 12x – 10y + 52 = 0  (d) None of these

Q4. The equation of the circle which has two normals (x – 1) (y – 2) = 0 and a tangent 3x + 4y = 6 is

(a) x2 + y2 – 2x – 4y + 4 = 0        (b) x2 + y2 + 2x – 4y + 5 = 0

(c) x2 + y2 = 5 (d) (x + 3)2 + (y – 4)2 = 5

Q5. The coordinates of the middle point of the chord which the circle x2 + y2 + 4x – 2y – 3 = 0 cuts off on the line y = x + 2, are

(a) (b) (c) (d)

Q6. The equation of tangent to the circle x2 + y2 = 25, which is inclined at an angle of 30° to the axis of x, is(a) (b) (c) (d)

Q7. The limiting points of the coaxal system determined by the circle x2 + y2 – 2x – 6y + 9 = 0 and x2 + y2 + 6x – 2y + 1 = 0 are

(a) (b) (c) (d) None of these

Q8.The number of common tangents to the circle x2 + y2 = 4 and x2 + y2 – 8x + 12 = 0 is (a) 1 (b) 2 (c) 3 (d) 4

Q9.The equation of the circle passing through (1, 0) and (0, 1) and having smallest possible radius is (a) x2 + y2 – x – y = 0 (b) x2 + y2 + x + y = 0 (c) x2 + y2 – 2x – y = 0 (d) x2 + y2 – x – 2y = 0

Q10.The radical axis of the circles, belonging to the coaxal system of circles whose limiting points are (1, 3) and (2, 6), is (a) x – 3y – 15 =  0 (b) x + 3y – 15 =  0 (c) x – 3y + 15 =  0 (d) 2x + 3y – 15 =  0

Q11. The equation of the circle, which touches the circle x2 + y2 – 6x + 6y + 17 = 0 externally and to which the lines x2 – 3xy – 3x + 9y = 0 are normal, is

(a) x2 + y2 – 6x – 2y – 1 = 0 (b) x2 + y2 + 6x – 2y + 1 = 0

(c) x2 + y2 – 6x – 2y + 1 = 0 (d) None of these

Q12.A line meets the coordinate axes in A and B, and a circle is circumscribing triangle AOB where O is the origin. If m, n are the distances of the tangents to this circle at the origin from the points A and B respectively, then the diameter of the circle is

(a) m(m + n) (b) n(m + n) (c) m – n (d)m + n

Q13. Four distinct points (2K, 3K) , (1, 0), (0, 1) and (0, 0) lie on a circle when

(a) all values of K are integral  (b) 0 < K < 1

(c) K < 0       (d) For two values of K

Q14.The length of the tangent from any point on the circle 15x2 + 15y2 – 48x + 64y = 0 to the two circles 5x2 + 5y2 – 24x + 32y + 75 = 0 and 5x2 + 5y2 – 48x + 64y + 300 = 0 are in the ratio of (a) 1 :  2        (b) 2 : 3         (c) 3 : 4         (d) none of theseQ15.The circles x2 + y2 + x + y = 0 and x2 + y2 + x – y = 0 intersect at an angle of (a) p/6 (b) p/4     (c) p/3        (d) p/2

Q16. The difference of the tangents of the angles which the lines x2(sec 2q - sin 2q) – 2xy tan q + y2 sin 2q = 0 with the x – axis is  (a) 2 tan q (b) 2  (c) 2 cot q       (d) sin 2 q

Q17. If the line y = cuts the curve x3 + y3 + 3xy + 5x2 + 3y3 + 4x + 5y -1 = 0 at the points A, B, C then OA . OB. OC is

(a) (b) (c) (d) None of these

Q18.A variable line drawn through the point (1, 3) meets the x-axis at A and y-axis at B. If the rectangle OAPB is completed, where ‘O’ is the origin, then locus of ‘P’ is

(a) (b) x + 3y = 1 (c) (d) 3x + y = 1

Q19.ABC is a right angled isosceles triangle, right angled at A(2, 1). If the equation of side BC is 2x + y = 3, then the combined equation of lines AB and AC is

(a) 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0 (b) 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0

(c) 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0   (d) None of these

Q20. If the line y = x is one of the angle bisector of the pair of lines ax2 + 2hxy + by2 = 0, then (a) a + b = 0 (b) h = 0       (c) a – b = 0      (d) None of these

Q21. If the sum of the slopes of the lines given by x2 – 2cxy – 7y2 = 0 is four times their product, then c has the value: (a) 1 (b) – 1             (c) 2              (d) – 2

Q22. If one of the lines given by 6x2 – xy + 4cy2 = 0 is 3x + 4y = 0, then c equals: (a) 1 (b) – 1             (c) 3                      (d) – 3

Q23. If the pair of straight lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 be such that each pair bisects the angle between the other pair, then (a) p = q       (b) p = – q      (c) pq = 1                     (d) pq = – 1

Q24. The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is

(a) (b) (c) (0, 0)       (d)

Q25. Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by

the circle x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then which of the following equations can represents L1? (a) x + y = 0   (b) x – y = 0 (c) x + 7y = 0                  (d) x – 7y = 0

Q26.The number of numbers are there between 100 and 1000 in which all the digits are distinct is (a) 648 (b) 548          (c) 448                          (d) none of these

Q27.The number of arrangements of the letters of the word ‘CALCUTTA’ is (a) 5040        (b) 2550              (c) 40320                        (d) 10080

Q28. How many different words can be formed with the letters of the word “PATLIPUTRA” without changing the position of the vowels and consonants? (a) 2160                        (b) 180                    (c) 720                              (d) none of these

Q29. How many different words ending and beginning with a consonant can be formed with the letters of the word ‘EQUATION’?(a) 720                            (b) 4320                         (c) 1440                       (d) none of these

Q30.The number of 4 digit numbers divisible by 5 which can be formed by using the digits 0, 2, 3, 4, 5 is (a) 36                      (b) 42                                 (c) 48                     (d) none of these

Q31.The number of ways in which 5 biscuits can be distributed among two children is (a) 32                    (b) 31                                  (c) 30                               (d) none of these

Q32.How many five-letter words containing 3 vowels and 2 consonants can be formed using the letters of the word “EQUATION” so that the two consonants occur together? (a) 1380              (b) 1420                     (c) 1440                             (d) none

Q33.If the letters of the word ‘RACHIT’ are arranged in all possible ways and these words are written out as in a dictionary, then the rank of this word is (a) 365                  (b) 702                       (c) 481                             (d) none of these

Q34.On the occasion of Dipawali festival each student of a class sends greeting cards to the others. If there are 20 students in the class, then the total number of greeting cards exchanged by the students is (a) 20C2                     (b) 2 . 20C2                       (c) 2 . 20P2                       (d) none of these

Q35.The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is (a) 18                 (b) 108                   (c) 432                              (d) 144

Q36.How many six digits numbers can be formed in decimal system in which every succeeding digit is greater than its preceding digit (a)  9P6           (b) 10P6                           (c) 9P3                               (d) none of these

Q37.How many ways are there to arrange the letters in the work GARDEN with the vowels in alphabetical order? (a)  120 (b) 240                           (c) 360                              (d) 480

Q38.A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is    (a) 216          (b) 240                           (c) 600                               (d) 3125

Q39.How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? (a) 16 (b) 36                                   (c) 60                          (d) 180

Q40.The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is (a) 40 (b) 60                                   (c) 80                                (d) 100

Answers Questio

n 1 2 3 4 5 6 7 8 9 10

11

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Answer a a c a a cd b a a b c d d a d b a c d c c d d c bc

Question

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Answer a a a b b c c c b b c c a c a                    

 

AIEEE Minor Test #3

 

Conic Section and Permutation combination

Q1.The focal distance of a point on a parabola y2 = 8x is 4, its coordinates are(a) (2, 4)                       (b)  (– 2, 4)             (c)   (– 2, – 4)   (d) none of these

Q2.The equation of the parabola whose focus is (3, – 4) and directrix is the line x + y – 2 = 0, is (a) x2 + 2xy + y2 – 8x + 20y + 46 = 0 (b) x2 – 2xy + y2 – 8x + 20y + 46 = 0(c) x2 – 2xy + y2 + 8x – 20y + 46 = 0 (d) None of these

Q3.The equation of the tangent to the parabola y2 = 6x at the point whose ordinate is 6, is (a) x + 2y + 6 = 0               (b) 2x – y + 6 = 0        (c)   x – 2y + 6 = 0         (d) None of these

Q4.The normal to the parabola y2 = 8x at (2, 4) meets the parabola again at (a) (18, 12)             (b) (18, – 12)                  (c)   (– 18, 12)               (d) None of these

Q5.The equation of the tangent to the parabola y2 = 16x inclined at an angle of 60° to x – axis, is (a) (b) (c) (d) None of these

Q6.The point of intersection of two perpendicular tangents to a parabola lies on the (a) axis                 (b) tangent at the vertex            (c)   directrix             (d) None of these

Q7.The pole of the line 2x = y with respect to the parabola y2 = 2x is

(a) (b) (c) (d) None of these

Q8.The locus of the poles of tangents to the parabola y2 = 4ax w.r.t. the parabola y2 = 4bx is the parabola

(a) (b) (c) (d)

Q9.The length of latus rectum of the parabola 4y2 + 2x – 20y + 17 = 0 is (a) 3                 (b) 6         (c)   1/2                     (d) 9

Q10.If the focal distance of a point on the parabola y2 = 4x is 6, then the coordinates of the point are (a) (b) (c) (d) None of these

Q11.The equation of the ellipse referred to its axes as the axes of coordinates with length of major axis 8 and

eccentricity , is (a) 4x2 + 3y2 = 48     (b) 3x2 + 4y2 = 48         (c)   4x2 + 3y2 = 24                 (d) None of these

Q12.The equation of the ellipse with focus (– 1, 1), directrix x – y + 3 = 0 and eccentricity , is (a) 7x2 + 2xy + 7y2 + 10x + 10y + 7 = 0 (b) 7x2 + 2xy + 7y2 + 10x – 10y + 7 = 0(c) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0 (d) None of these

Q13.The equation of the normal to the ellipse  at the end of the latus rectum in the first quadrant, is(a) x + ey – ae3 = 0  (b) x – ey + ae3 = 0       (c)   x – ey – ae3 = 0               (d) None of these

Q14.The condition that the line x cos a + y sin a = p may be a tangent to the ellipse is (a) a2 cos2 a + b2 sin2 a = p2 (b) a2 cos2 a + b2 sin2 a = 2p2

(c) a2 sin2 a + b2 cos2 a = p2     (d) None of these

Q15.If the normal at the end of a latus rectum of an ellipse passes through one extremity of the minor axis, then (a) e4 + e2 – 1 = 0        (b) e4 – e2 + 1 = 0        (c)   e4 – e2 – 1 = 0                (d) None of these

Q16.In how many ways can 5 beads out 7 different beads be strung into a string?(a) 504                              (b) 2520                  (c)   252          (d) none of these

Q17.A person has 12 friends, out of them 8 are his relatives. In how many ways can he invite his 7 friends so as to include his 5 relatives? (a) 8C3 x 4C2                           (b) 12C7         (c)   12C5 x 4C3                (d) none of these

Q18.It is essential for a student to pass in 5 different subjects of an examination then the no. of method so that he may failure (a) 31                         (b) 32                      (c)   10     (d) 15

Q19.The number of ways of dividing 20 persons into 10 couples is

(a) (b) 20C10        (c) (d) none of these

Q20.The number of words by taking 4 letters out of the letters of the word ‘COURTESY’, when T and S are always included are (a) 120                            (b) 720                         (c)   360               (d) none of these

Q21.The number of ways to put five letters in five envelopes when one letter is kept in right envelope and four letters in wrong envelopes are– (a) 40                      (b)  45            (c)   30         (d) 70

Q22.  is equal to

(a) 51C4               (b) 52C4            (c)   53C4                (d) none of these

Q23.A candidate is required to answer 6 out of 10 questions which are divided into two groups each containing 5 questions and he is not permitted to attempt more than 4 from each group. The number of ways in which he can make up his choice is(a) 100                          (b) 200                (c)   300                   (d) 400

Q24.Out of 10 white, 9 black and 7 red balls, the number of ways in which selection of one or more balls can be made, is (a) 881                      (b) 891           (c)   879                        (d) 892

Q25.The number of diagonals in an octagon are (a) 28              (b) 48                  (c)   20                 (d) none of these

Q26.Out of 10 given points 6 are in a straight line. The number of the triangles formed by joining any three of them is (a) 100             (b) 150   (c)   120                        (d) none of these

Q27.In how many ways the letters AAAAA, BBB, CCC, D, EE, F can be arranged in a row when the letter C occur at different places?

(a) (b) (c) (d) none of these

Q28.A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen. The number of ways of chosen P and Q so that P Ç Q = f is (a) 22n – 2nCn       (b) 2n (c)   2n – 1                 (d) 3n

Q29.A parallelogram is cut by two sets of m lines parallel to the sides, the number of parallelograms thus formed is

(a) (b) (c) (d)

Q30.Along a railway line there are 20 stations. The number of different tickets required in order so that it may be possible to travel from every station to every station is (a) 380                (b) 225 (c)   196              (d) 105

Q31.The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100 is (a) 5081                 (b) 6005        (c)   4851                    (d) none of these

Q32.The number of numbers less than 1000 that can be formed out of the digits 0, 1, 2, 3, 4 and 5, no digit being repeated, is (a) 130                      (b) 131             (c)   156                  (d) none of these

Q33.A variable name in certain computer language must be either a alphabet or alphabet followed by a decimal digit. Total number of different variable names that can exist in that language is equal to (a) 280                               (b) 290             (c)   286                (d) 296

Q34.The total number of ways of selecting 10 balls out of an unlimited number of identical white, red and blue balls is equal to (a) 12C2                       (b) 12C3           (c)   10C2                     (d) 10C3

Q35.Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of them get atleast two blankets equal to (a) 10C3                             (b) 9C3                  (c)   11C3                      (d) none of these

Q36.The number of ways in which three distinct numbers in AP can be selected from the set {1, 2, 3, …, 24}, is equal to (a) 66                                     (b) 132               (c)   198                         (d) none of these

Q37.The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is: (a) 5                                    (b) 21            (c)   38                            (d) 8C3

Q38.The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by: (a) 6! x 5!                         (b)  30       (c)   5! x 4!                     (d) 7! x 5!

Q39.If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr + 1 + nCr – 1 + 2 x nCr equals: (a) n + 2Cr                           (b) n + 2Cr + 1 (c)   n + 1Cr                   (d) n + 1Cr + 1

Q40.If the letters of the word SACHIN are arranged in all possible ways and these are written out as in dictionary, then the word SACHIN appears at serial number(a) 600                   (b)  601             (c)   602                      (d) 603

Answers Questio

n 1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

Answer a b c b a c a b c a b b c a a c a a d c b b b c c

Question

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40                    

Answer a a d d a c b c a a b b a b b                    

 

Circles and System of CirclesIntroduction

Most of the things that we see around us are circular. Sun, moon on full moon day bangle mery-go around, which you loved so much when you where a child; all happened to circles. Ever wondered how a circle can be represented mathematically; well no!!, then we will

tell you in this chapter. Also we will take about tangent normulas, chords which we all have hear about. So let us prove deep in to circles.

Definition: locus of a set of points equidistant from a fixed point Equation ofcircle - (x - h)2+ (y - k)2 = r 2

x2h 2 - 2hx + y2 - 2ky + k2- r2 = 0

x2+ y2 + 2gx + 2fy + C = 0

Center Genreal second degree equation-ax2+ by2 + 2hxy + 2yx +2fy +C = 0this equation refresent circle when,

a = b,h = 0 , g2 + f2 C

Equation of circle in different forms -(1) Centre(h1K) radius a :-

(x - h)2 + (y - k)2 = a2

standard form (when center is origin):-x2 + y2 = a2

(2).center (h,k)and pass through origin-x2 + y2 - 2hx - 2ky = 0

Why :-

here (x - h)2 + (y - k)2 = r2 = h2 + k2

x2+ y2 - 2hx - 22ky = 0 Center(h,k) and touches the axis of x-y

x2+y2- 2hx - 2ky + h2= 0(x - h)2 + (y - k)2 = k2

or, x2+ y2 - 2hx - l2ky + h2 = 0 (4)Center(h,k) and touches the axis of y-

x2 + y2-2hx -2ky + k2 = 0Why :-

From fig it is clear that radius will be h .

(x - h)2 + (y - k)2 = h2

or, x2+ y2- 2hx - 2ky +k2 = 0Circle which touches both the axis:-

Center will be(h, h)and radius will be h. But since center would be in any of the four quadrants its coordinates can be taken as

radius h.

Illustration -1-. Find the equation of circle passing through (-2, 3) and touching both the axes.

Solution- As the circle toucher both theaxed and lies in 2ndquadrant, lits centre is Where r is the radius , Distance of center from (- 2, 3) = radius .

the circle are :- (x + r)2 + (y - r)2 = r2

X2 + y2+ 2agx + 2fy +c = 0 ----equation of circle.

x intercept

Why :- let it cut the axis x ie y = 0in points (x2,0) and (x2, 0)

x1, x2 are the roots of x2 + 2yx + c = 0x1, x2 = - 2y , x1 . x2 = c

Intercelt = x2 - x1

= [(x2 + x1)2- 4x 1x2] 1/2 =

similarily .y intercept = (6) Circle whose diameter is the line joining two point A (s1, y1 ) and B (x2, y2)-

Diametric form:- (xx- x1)(x - x2) + (y - y1(y - y2) = 0

Why :-

Angle in a semicircle is a right angle m1m2 = - 1

or (x - x1)(x - x2) + (y - y1)(y - y2) = 0

Circles and System of CirclesIntroduction

(7) Parametric form-

general point of a circle if centre is (0,0) isparameter (radius)

Illustration -2- Find the equation a circle which touches the .y axis at (0, 4) and cuts an intercept of length .6 units on x axis .Solution- The equation of circle toching x = 0 at (0,4) can be taken as (x - 0)2 + (y - 42 ) + kx = 0

x2 + y2 + kx - 8y + 16 = 0the circle cuts x -axis point (x1, 0) .8 (x2, 0 )given by, x2 + kx + 16 = 0Xintercept difference of root of this quadratic equation 6 = | x2 - x1|

36 = (x2 + x1)2 - 4x1 .x2

36 = k2- 4 (16)

k = 10

Hence the required circle is ,x2 + y2 10 x - 8y + 16 = 0

Some natations in a circle-1) s = x2 + y2 + 2gx + 2fy + c

2) s1= x x12 + y1

2 + 2yx1 + 2fy1 + c

3) T = xx1 + yy1 + g(x + x1) + f (y + y1) + c

Standrad form-1)s = x2 + y2 - a 2

2)s =1x12 + y1 2 - a2

3) T = xx1 + yy1 - a2

* If s1> 0 point lies out side the circle * If s1< 0 point lies in side the circle

* If s1 0 point lies upon the circle

Why :-Let equation of circle be X2 + y2 + 2yx + 2fy + c = 0

having centre C ( -y, -f) and radius Let P (x1, y1) be any point then :-

P lies outside the circle if :-PC > r

=> x1 2 + y12 +2 yx 1 + 2fy1 + c > 0

P lies on the circle if :-PC = r

=> x1 2 + y12 +2gx1 +2fy 1 + c = 0

P lies inside the circle if :- PC < r

=> x12 + y1

2 +2gx1 2fy1 + c < 0

Dumb question :-How does PC > r leads to - x1

2+ y12 + 2gx1 + 2fy1 + c > 0

Ans-

         

and r = Now PC > r => PC2 > r2

=> (x1 +y)2+(y1 +f)2 > y2 + f 2- c => x1

2 + y12 2gx, + 2fy1 + c > 0

(1) A line L and a circle intersed in two point A and B .

=> d < r => Perpendi cular distance of line L from the centre of circle is less than the radius, and the length of te chors AB is :-

(2) A line L and +a circle touch each other at a point P.=> d = r

=> Perpendicular distance of L from the centre of circle = radius.(3) A line L and a circle may not intersect at all

=> .d > r => Perpendicular distance of line from the centre of circle is greater than the radius .

(4) A line y = mx + c touches circle x2 + y2 = a2 If :- perpendicular distance of line from centre of the circle

= radius of the circle

Illustration- For what value of m, will line y = mx does not intersect the circle x2+y2 + 20 X +20y + 20 = 0Solution- IF the line y = mx does not intersect the circle ; the perpendicular distance of the line from the centre of the circle must be

greater than its radius .

Centre of circle (-10, -10) ; radius distance of line mx - y = 0 from (-10, -10)

=> |m(-10)-(-10)|

=>(2m + 1) (m + 2) < 0=> -2 < m < - 1/2

Intersection of line with circle-Let the line be y = mx + d and circle is x2+ y2 +2gx + 2fy + c

thes x. Coordinate of their point of intersection are given by, (1 + m2)x2+ (2g + 2fm +2dm) x + d2+ 2fd + = o Why :-

When the two curves intersect, both the curves will be simultaneously satisfied.So y = mx + d can be replaced in

x2y2 + 2gx + 2fy + c =0=> x2 + (mx + d )2 + 2gx + 2f (mx + d) + c =0

=> (1 + m)2 + (2g + 2fm + 2dm) x + d2 + 2fd + c = 0if. (i) B2 - 4AC = 0 then line touches the circle.

(ii) B2 - 4AC = > 0 then the line intersect circle at 2 different point.(iii) B2 - 4AC = < 0 then no real intersecti takes place.

Illustration 4- Find the point on the circle x2 + y2

= 4 whose distance from the line 4x + 3y = 12 is 4/5 .Soluction- Let A,B be the point on x2/u> + y2 = 4 luing ar a distance 4/5 from 4x + 3y = 12

=> AB will be parallel to 4x + 3y = 12

distance between the two line is => C = 16, 8

=> the equation of AB is :- 4x + 3y = 8 4x + 3y = 16 the point A,B can be formed by sliving for point of intersection of x2 y2 = 4 with AB.

AB (4x + 3y - 8 = 0)

=>=> 25 x2 - 64x + 28 = 0

=> x = 2, 14/25 y = 0, 48/25

AB (4x + 3y - 16 = 0)

=> => 25 x2 - 128 x + 220 = 0 => D < 0 => no real roots

Hence these are two pointr on circle at distance 4/5 from liine A (2,0) . & B (14/25, 48/25)

Circles and System of CirclesIntroduction

Alter native method -

let P be the point on the circle x2 + y2= 4 distance . 4/5 from given line. the distsnce from line = 4/5

Solve for to get the point . Equation of tangent in general form is :-

xx1 + yy1+ g(x + x1) + f(y + y1) + c = 0

equation of tangent on standard form :- xx1 + yy1 a2 = 0

Why :-

Slope of tangent = -equation of tangent :-

y - y1 - (x - x1) (y - y1) (y1 + f) = - (x1 + y) (x - x1)

on solving we get,xx1 + yy1 + g (x + x1) + f (y + y1) + = 0

Equation of tangen T = 0 >

Dumb question- Why slope of tangent ? Ans - The slope of line Joining the centre to point of contact is

Now tangent os perpendicular to this line -

slope of tangent is - Note:- Golden rule to write equation of tangent is to replace. x2 xx1, y2 yy

12x x + x1, 2y = y + y1 in equation of circle where (x1, y1) is of contact.

Equation of tangent.

Lngth of tangent:-

length

Why :-let equation of circle be

x2 + y2 + 2gx + 2fy + c = 0 then center is c (-g, -f) and radius = f

length of tangent = PA

on solving we get ,

length of tangent

          

length Condition of or line y = mx + c to be a ltangent to x 2 + y2 = a2 -

Condition:- c2 = a2 (1 + m2)

Equation of tangent:-

y = mx

Why -?putting y = mx + c in x2 + y2 = a2

x2 + (mx + c)2 = a2

(1 + m2) x2 + 2mxc + c2 - a2 = 0

4m 2 - c2 -4 (c2 - a2 ) (1 + m2) = 0

c2 = a2(1 + m2)

Similarily when circle equation is -(x - h)2 + (y - k)2 = a2

equation of tangent with slope m

=> (y - k) = m (x - h)

Dumb question :- Why D is taken zero ?

Ans- Line is touching circle. It means on ksolving line and circle -

We will get only are solution -It means quadratic of x will lhave repeated roots if means D = 0

Illustration- Find the equation of two tangents drawn to the circle x2 + y2 - 2x + 4y = 0 from point (0, 1)Solution- let m be the slope of the tangent .

For true lengths here will be two values of m which are requited.As the regent pases through (0, 1) its equation will be .

(y - 1) = m(x - 0)mx - y + 1 = 0

Now the centre of circle (x2 + y2 - 2x + 4y = 0) is (1,-2) and radius r =So using the condition of tangencu distance of centre (1, - 2) from line = radius (r)

=> m = 2, - 1/2

equation of tangents are :-2x - y = 1 = 0 and x = 2y - 2 = 0

Illustration:- Find the equation mcircle passing through (-4, 3)an touching the line x + y = 2 and x - y = 2.Solution .- let (h, k) be the centre of the circle the ditance of the centre from th given line and the given point must be equation to

radius :-

<>br /> Consider => h + k - 2 = (h - k - 2)

Case(I)- h + k - 2 = h - k -2 => k = 0

=> (h - 2)2 = 2(h + 4)2 + 18=> h2 + 20h + 46 = 0

=> k = 10

radius = Circle is :

Case(II) - h + k - 2 = - (h - k - 2)=> k2 = 72 + 2 (k - 3)2

=> k2 = 12 k + 90 = 0 The equation has no real roots. Hence no circle is possible for h = 2 .

Hence only two circle are possible (k = 0)

Circles and System of CirclesIntroduction

Pair of tangents - T 2 = ss1

Equation of normal-The normal to a curve at any P of a curve is the straight line which passes through P and is perpendicular to the tangent at P .

The equation of normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at any point l(x1,y1) is :- y(x1 + g) - x (y1 + f) +fx1 + gy1 = 0

Why :-normal will be OP

slope OP =

equation of normal -

on sloving we get, y(x1 + y) - x(y1 + f) + fx1 - gy1 = 0

With respect to circle S = 0 Equation of chord of contact:-

T = 0 Why :-

Equation of circle S = 0 Whose, S = x2 + y2 - a2

Equation of AP => xx2 + yy2 - a2 = 0 Equation of BP => xx3 + yy3 - a2 = 0

Both passes through P .

x1x2+ y1y2 - a2 = 0 ..............(1)

x1x3+ y1y3 - a2 = 0 ..............(2)

Now consider

xx1+ yy1 - a2 = 0 ..............(3)

from (1), (2) and (3) it is lvery that clear that A and B lies on (3)

equation of AB is xx1+ yy1 - a2 = 0=> T = 0

Equation of chord lhaving mid point (x1, y1)Only one such chord is possible

Equation of chord

T - s1 = 0 xx1+ yy1 +g(x + x1) + f (y + y1) + c

= x12 + y1

2 + 2gx1 + 2fy1 + c

Slop of CP = -Equation of CP =>

(y - y1) = - (x - x1)on solving we get,

T = S1

T - S1 = 0Illustration- Find the co-ordinates of the point from which tangen are drawn to the circle x2 + y2 - 6x - 4y + 3= 0 such that mid point of its

chord of contact is (1, 1).Ans- S = x2 + y2 - 6x - 4y + 3

TAB = xx1 + yy1 - 3(x + x1) - 2(y1 + y1) + 3

(x1 - 3) x + (y1 - 2) y - 3x1 - 2y1 + 3= 0........(1)

Equation of AB T - S1 = 0on solving

2x + y = 3 ............(2)Comparing (1) and (2)

on solving, .x1 = - 1, y1 = 0

In previous illastration why .

Why not x1 - 3 =2 y x1 - 3 =2 ?

Solution - On comparing two equation - a1 x + b1y + c1 = 0 a2 x + b2y + c2 = 0

If both the above equation are of some line, then we get :-

Two circle touching each other - (a) External tiuch-

c1 c2 = r1 r2 Point P divides the line Joining c1 & c2 internally in th ratio

r1; r2

( b) Internal touch:-c1 c2 = | r1 - r2|

point P divider the line Joining c1 & c2 externally in the ratio r1 : r2.

Conclition :- |(r1 - r2)| < c1 c2 < r1 - r2

(a)One is completely inside other-Conclition :-c1c2 < r1 - r2

Illustration- Examine if the two circle x2 + y2 - 8y - 4 = 0 & x2 + y2 - 2x -4y = 0 touch each other find the point contact if they touch. Solution- For x2 + y2 -2x - 4y = 0 centre c1 (1, 2)

& x2 + y2 -8x - 4 = 0 centre c2 (0, 4)

using

Now c1c2 =

=> r2 - r1=

ParabolaIntroduction

Parabola is a curve of infinite extent and is not often obsened in real life as circles or shaight line are . However, the parabplaic curres are very important and have some vert good propertvies like the light raus reflected from parabolic surface passes through one foxed point. This property of parabola is luidely used in making lenses and in optics.The brridges also take parabolic shape have been centuries lod. So, now let us start studying this interesting curre in more detail.Parabola is locus of point which moves such that its distance from fixed point (focus) is always equat to its distance from fixed line (directrix) i.e. eccentricity, e = 1 .

Note: The distance from fixed point to the distance from fixed line is calle

General second degree equation: eccentncity.The equation ax2 by + 2hxy + 2gx + 2fy + c = 0

(1) Second degrel thrms make a perfict square. or n 2 = ab.Why ?

Let the fixed point be . and fixed line ax + by + 1 = 0 Let variable point be (h,k).Now,according to the definition of parabola.

the distance of (h,K) from ( ) is same as distance from line ax + by + 1 = 0.

So locus of On simplification, we get

(a2 + b2) =(ax + by + 1)2

The expression reduces to b2 x 2 + a2 y2 - 2abxy + .....linear terms .... + conjtant=(bx - ay)2 + ......linear terms + constant = 0. So, second deqree terms make perfect squarc.

Dumb Question: Why is h2= ab same as second term making a perfect squarc ?Ans The second degree terms in equation are ax2 + by2+ 2hxy .Now if h2 = ab

then

So, the two thing are same .

(2)

Illustration 1Ide ntify the locus of point which move such that its distance from given point and line is equal ?(i) (-3,7) is the point and x + 2y - 8 = 0 is given line .

Ans Now according to the question, let the point whose locus is to be determined be (x,y).

or,=>. 5 (x2 + 9 + bx + y2 + 49 -14y)= x24y2 + 64 -16 x - 32y +4xy = 4x2 + y2 - 4xy + 46x - 38y + 226 = 0

Considening 20 term, 4x2 y2 - 4xy = (2x - y)2 which is a perfect square.Now consider

So, the locus is a parabola

(ii) (-6,7) is point and x + 2y - 8 = 0 is given line .

Ans. Note taht (- 6,7) satisfies the given line .So, the locus is stratight line itself .Standared Equation of parabola.

Important properties(i) Vertex 0 (0,0) (ii) Focus s(9,0)(iii) Foot of directrix (-9,0)(iv) Directrix x + a = 0(v) Equation of catus rectum x = aand length of Latus retum = 4a .(vi) Axis y= 0.(vii) Extremiofies of latus retum (91 2a) & (91-2a)

Note : Two parabolas are said to be equal if their length of latus rectii are equal .

ParabolaIntroductionTypes of parabola

x + a = 0 y2 = 4ax, a > 0           Y2 = - 4ax, a > 0s

Transformation of parabola.(a) (y - R)2 = 4a (x - h) , a > 0The vcetex will be (h , k), openitng of parabola will be on = ve side of axis, axis will be || to x axis and dire ctrox well be || to y axis.

(b) (y - R2) = 4a (x-h), a < 0Same as above except, the opening of parabolanwill be on -ve side of x - axis .

s

(c) (x -h 2) = 4a (y - yR), a > 0 The vertex will be at (h,K) opening of parabola will be on +ve side of yaxis , axis will be || t y - axid and directrix || to x - axis

(d) (x -h 2) = 4a (y-K), a < 0Same ad above except, lthe opening of parabola will be on -ve side of y-axis .

Illustration 2.Find the vertex, axis, focus, and latus rectum of parabola 4y2+ 12x -x - 20y + 67 = 0Ans The equation can be written as y2- 5y = - 3x - 67/4 . i.e. (y-5/22) = - 3x - 64/2 + 25/4 .= - 3(x + 7/2)So, this a transformed parabola whose vertex is (-7/2, 5/2), the axis is y = 5/2.The length of latus rectum = |4a| = 3 and a = -3/4 .So, focis is (-7/2 - 3/4 1 5/52)= (-17/4 1 5/2).

Natations :For standard parabola (y2 = 4ax)1) S y2- 4ax 2) S1 y 1

2- 4ax1

3) T yy1-2a(x + x1) 4) F (at ,2 2at)

Position of a point (x1,y1) w. r. t y 2 = 4ax If S1 > 0 => Outside parabola. S1 < 0 => Inside parabola. S1 0 => On parabola.

Why ?Suppose a point is outside parabola .

s So, PL > PM.=> PL 2 PM2

or y12 > y2

2

or y12 > 4ax1( M lics oh parabola).

So, S1 > 0 => Point is outside parabola .Similarily when point is inside the parabola .S1 < 0 .

Dumb Question:- What dose inside and outside the parabola mean in a curre like parabola which is not closed ?

Ans :- the wird "outside" refers to the region from where tangent can be drawn . On other hand, the region from where tangent cannot be drawn is fefference as "in side" the parabola.

Par ame tric form.x = at2, y = 2at where t is a parameter represents the parametric form.(at2, 2at) is general point on parabola y2 = 4ax .

Illustration 2.Find the equations of the parabola if the extremeties of its latus ractum are (3,5) and (3,7).Ans. Now the length of latus ractuin is

So, 4a = 2 or a = 1/2 .Now middle point of catus ractum is (3+3/2, 5+7/2) = (3,6) which is focus of parabola.So the two cases as shown in figure below are possible.

Since the vertex is at a distance a away from the parabola the vertices are (7/2, 6) and (5/2,6).Now lBy transformation of parabolas the two parabolas possible are (y-6)2 = 4(1/2) (x-5/2)& (y-6)2 = 4(1/2) (x-7/2). So, equation of parabolas are (y-6)2 = 2 (x-5/2)& (y-6)2 =2 (x-7/2)

HYPERBOLA

IntroductionHyperbola is a very important lonic section. It has a wide use in further engineering courses. Hyperbola is a very special curve which is very rarely seen in day to day life. The new concepts such as touching the curve at infinity fills us with a great excitement to read this

chapter, so enjoy the curve named hyperbola by getting into the chapter and feel the touch of the line to the line to the curve at infinity.

DEFINATION:

The locus of point which moves in a plane such that its distance from a fixed point called focus is e times (e > 1) its distance from a fixed line called directrix.

EQUATION OF HYPERBOLA

The standard equation of hyperbola is:

    where   b2 = a2(e2-1)

Why ?

Let s be the focus & ZM the directix of the hyperbola. Draw SZ ZM & Divided SZ internally & externally in the ratio e:1 (e>1) & let A & A' be their internal & external points of division, then

    SA = eAZ     &  ...................(1)    SA'= eA'Z    ........................(2)

Points A & A' will lie on the hyperbola. Let AA' = ZA & take C, the mid point of AA' of AA' as origin. i.e.   CA = CA' = a

Let P be any point on the hyperbola.Then adding (1)& (2) we get

    SA + SA' = e( AZ + A'Z)(CS - CA) + (CS + CA') = eAA'

2cs + CA' - CA = e(za)But CA' = CA, Hence

    2CS = 2aeCS = ae

Thus focus S is ( ae, 0)Now subtrating (1) from (2) use get

    SA'- SA = C(A'Z - AZ)    AA' = e[( CA' + CZ) - (CA - CZ)]

    AA' = e[2CZ + CA' - CA] But   CA'= CA, then

    2a = e(2cz)CZ = a/e

Thus directrix is x = a/e

Now draw PM MZWe know by the definaton of hyperbola, that

=   SP = ePM=   (sp)2 = e2(pm)2

=   =   (x - ae)2 + y2 = (ex - a)2

=   x2 + a2e2 - 2aex + y2 = e2x2 + a2 - 2aex=   x2(e2 - 1) - y 2 = a2(e2 - 1)

=  

=   where b2 = a2(e2 - 1)

ILLUSTRATION : 1 .

Find the equation of hyperbola whose foci are (2, 4) & (10, 4) & eccentricity is 2.

Ans: We know that the center of hyperbola is the mid point of two foci i.e. coordinates of centers are (6, 4). We also know that the distance between two foci is 2ae.

i.e   (10 - 2)2 + (4 - 4)2 = (zae)2

2ac = 8 4a = 8a = 2

     b2 = a 2(e2 -1),     Hencea = 2

     b2 = 4(4 - 1) = 12 The required equation of hyperbola is

    

SOME TERM RELATED TO HYPERBOLA

(i) CENTER : This is the mid point of line joining the two foci. In standard form C is (0,0).

(ii) ECCENTRICITY : This is the ratio of the distances of any point on hyperbola from focus to the directrix. Here we have   &  b2 = a2(e2 - 1)

e =

(iii) FOCI: Foci consists of two points on the transvers axis of hyperbola whose mid point is the center of hyperbola. Coordinates: (Iae, 0)

(iv) DIRECTRICES : There are the two lines perpendicula to the transverse axis on oppsite sieds of centre. In standard form directrices are

    

(v) AXES : In standard form points A(a,0) & A'(-a, 0) are called the vertices of hyperbola line AA' is called transverse axis & perpendicular to this is called conjugate axis.

(vi) DOUBLE ORDINATES : This is the length of the chord of hyperbola with end points as a, a', which is perpendicular to the transverse axis.

If the abscissa of Q is h then

    

    

Hence coordinates of Q & Q' are & .

(vii) LATUS RECTUM: The double coordinates which pass through foci are called Latus rectums. The abscissa of Latus rectums are

Iae. Hence the coordinates are , & the length of Lotus

rectum is .

(viii) FOCAL CHORD : A chord of hyperbola that passes through focus is called the focal chord of hyperbola.

(ix) PARAMETRIC EQUATION OF HYPERBOLA: The parametric equations of given hyperbola

     are

    x = a sec

    y = b tan

why ?

Let be the hyperbola with centre C & transverse axis AA', Then the circle drawm with centre C & segment AA' as diameter is called auxiliary circle. Equation is:

    x2 + y 2 = a2

Let P be any point on hyperbola & Q on circle Draw PN perpendicular to axis & NQ be the tangent ot circle, then join CQ.

Let eccentric angle of P

since Q = (a cos , a sin )

        x = CN = a sec But P(x, y) lies on hyperbola, then

    

y = b tan

Hence the parametric equations of hyperbola are x= sec , y = tan

ILLUSTRATION : 2. Find the centre, foci, vertices, eccentricity, length, of axes, Latus Rectum & directrices of the hyperbola 3x2 - 2y2 + 12x + 4y + 4 =0

Ans: 3x2 - 2y2 + 12x + 4y + 4     = 3(x2 + 4x + 4 - 4) - 2 (y 2 - 2y + 1 - 1) + 4

     3(x+2)2 - 12 - 2(y - 1)2 + 2 + 4 = 03(x + 2)2 - 2 (y - 1)2 = 6

Centre = (- 2, 1)

Here a2 = 2     

       b2 = 3     

Hence  

foci = (± ae, 0) =

vertics = (± ae,0) =

eceentricity = length of axex = 2a   &   2b

                     =

Latus Rectum =

Directrices =

HYPERBOLA 

FOCAL DISTANCES OF A POINT:

Another Defination of hyperbola is that the difference of focal distances of any point on hyperbola is constant & equal to the length of transverse axis of the hyperbola.

Why ?

Let the hyperbola be  

We know that Distance from focus = (Distance from Directrix)

Hence Distance of point P(x1, y2) from S,(ae,0) is

     SP = ePM = e = ex1 - a

Similarly S'P = e(PM') = e = ex1 + aHence S'P - SP = 2a

                     = Transverse axis

Hence Hyperbola is the Locus of a point which moves in a plane such that the difference of its distances from two fixed points i.e. foci is constant.

POINT AND HYPERBOLA

The point (x1, y2) lies outside, on ,or inside the hyperbola accordingly as < , = or > 0

Why ?Let P = (x1, y2) & Q = (x1 y1)

Draw QL perpendicular to x axis then     QL > PL

y1 > y2

Adding on both sides

    

But

Hence

When point lies outside. Similarly we can prove that when point lies on of inside the hyperbola Then

LINE AND A HYPERBOLA

The line y = mx + c will cut the hyperbola in two points, one point or will not cut accordingly as c2 >, = or < a2m2 - b2

Why ?Let the line be y = mx + c ............................................. (1)

& the hyperbola   .......................................... (2)Eliminating y form (1) & (2) we get

     x2(a2m2 - b2) + 2mca2x + a2(b2 + c2) = 0

This is a quadratic is x, hence Discriminant = b2 - 4ac

                  = 4m2c2a4 - 4(a2m2 - b2)(a2)(b2 + c2)                  = c2 + b2 - a2m2

Line will cut in two points if D > 0c2 + b2 - a2m2 > 0 c2 >a2m2 - b2

Line will touch the parabola if D = 0i.e. c2 + b2 - a2m2 = 0 c2 = a2m2 - b2

Line will not be touch or be touch a chord of parabola if D < 0 i.e. c2 + b2 - a2m2 < 0

c2 < a2m2 - b2

ILLUSTRATION : 3.

For what values of K will the line y = 3x + K be a chord to the hyperbola

Ans: For this hyperbola we have     a2 = 9   &   b2 = 45

From the equation of given line      m = 3   &   c = k

Hence we know that for a line to be a chord to hyperbola c2 > a2m2 - b2

i.e. K2 > 9 x 9 - 45  K2 > 36

  K2 - 36 > 0  (K - 6)(K + 6) > 0

Hence

EQUATIONS OF TANGENT

(a) POINT FORM:

The equation of tangent to the hyperbola at (x 1y1) is

i.e. T = 0 where T =

Why ? The equation of hyperbola is:

     Differentiating w.r.t.x. we get

    

    

Equation of tangent when it passes through (x1y1) is

(y - y1) =

a2y1y - a2y12 = b2x1x - b2x1

2

Dividing whole equation by a2b2 we get

    

But as (x1, y1) lies on hyperbola

is the requurired equation

(b) PARAMETRIC FORM:

The equation of tangent to hyperbola at

(a sec , b tan ) is

    

Why ?We have to paranetric equations of hyperbola as

     x = a sec & y = b tan

Differentiating both equations we get dx = a sec tan d & dy = b sec2 dHence dividing these equations we get,

  

Hence the equations of tangent is (y - b tan ) =

ay sin cos - ab sin2 = cos bx - ab

say sin cos - bx cos = - ab cos2

  is required equation.

SLOPE FORM:

The equations of tangent to the hyperbola of slope m is y = mx ± & coordinates of points of contact are

    

Why ?

Let the line with slope m be      y = mx + c, tangent to

     Eliminating y from these two equations we get

    (a2m2 - b2)x2 + 2 mca2x + a2(c2 + b2) = 0This is a quadratic equations in x hence for only one solution D should be zero.

  D = b2 - 4ac = 4m2c2a4 - 4(a2m2 - b2(a2b2 + c2) = 0c2 = a2m2 - b2

c = Hence the equation of tangent is:

     y = mx .................................. (1)Now, we also know that the equation of tangent at (x1, y1) is

     ......................................... (2)Comparing (1) & (2) as they are the same equation we get

    

&

    

Hence the coordinates of point of contact are .

3-D GEOMETRYIntroduction

Three dimensional geometry developed accordance to Einsteins field equations. It is useful in several branches of science like it is useful in Electromagnetism. It is used in computer alogorothms to construct 3D models that can be interactively experinced in virtual reality fashion. These models are used for single view metrology. 3-D Geometry as carrier of information about time by Einstein. 3-D

Geometry is extensively used in quantum & black hole theory.

Section Formula:

(1) Integral division: If R(x, y, z) is point dividing join of P(x1, y1, z1) & Q(x2, y2, z3) in ratio of m : n.

Then,   x = , y = , z =

(2) External division: Coordinates of point R which divides join of P(x1, y1, z1) & Q(x2, y2, z2) externally in ratio m : n are  

Illustration: Show that plane ax + by + cz + d = 0 divides line joining (x1, y1, z1) & (x2, y2, z2) in ratio of Ans: Let plane ax + by + cz + d = 0 divides line joining (x1, y1, z1) & (x2, y2, z2) in ratio K : 1

Coordinates of P

    

must satisfy eq. of plane.     ax + by + cz + d = 0

[Dumb Question: Why coordinates of P satisfy eq. of plane ?  Ans: Point P lies in the plane so, it satisfy eq. of plane.]

    a(Kx2 + x1) + b(Ky2 + y1) + c(Kz2 + z1) + d(K + 1) = 0

K(ax2 + by2 + cz2 + d) + (ax1 + by1 + cz1 + d) = 0

K = -

Direction Cosines:

Let is a vector , , inclination with x, y & z-axis respectively. Then cos , cos & cos are direction cosines of . They denoted by

       , , direction angles.

& lies   0 , ,

Note: (i) Direction cosines of x-axis are (1, 0, 0)

Direction cosines of y-axis are (0, 1, 0) Direction cosines of z-axis are (0, 0, 1)

(ii) Suppose OP be any line through origin O which has direction l, m, n

(r cos , r cos , r cos ) where OP = r

coordinates of P are (r cos , r cos , r cos ) or   x = lr, y = mr, z = nr

(iii) l2 + m2 + n2 = 1

Proof:   | | = | | =

           | |2 = x2 + y2 + z2 = l2| |2 + m2| |2 + n2| |2

        l2 + m2 + n2 = 1

(iv)

Direction ratios: Suppose l, m, & n are direction cosines of vector & a, b, c are no.s such that a, b, c are proportional to l, m, n. These a, b, c are c/d direction ratios.

       = k

Suppose a, b, c are direction ratios of vector having direction cosines l, m, n. Then,

       =   l = a, m = b, n = c      l2 + m2 + n2 = 1

       a2 2 + b2 2 + c2 2 = 1   = ±

       l = ± , m = ± , n = ±

Note: (i) If having direction cosines l, m, n. Then, l = , m = , n =

(ii) Direction ratios of line joining two given points (x1, y1, z1) & (x2, y2, z2) is (x2 - x1, y2 - y1, z2 - z1)

(iii) If direction ratio's of are a, b, c

       =

(iv) Projection of segment joining points P(x1, y1, z1) and Q(x2, y2, z2) on a line with direction cosines l, m, n, is:        (x2 - x1)l + (y2 - y1)m + (z2 - z1)n

Illustration: If a line makes angles , & with coordinate axes, prove that sin2 + sin2 + sin2 = 2

Ans: Line is making , & with coordinate axes.

Then, direction cosines are l = cos , m = cos & n = cos .But   l2 + m2 + n2 = 1

       cos2 + cos2 + cos2 = 1

       1 - sin2 + 1 - sin2 + 1 - cos2 = 1

    sin2 + sin2 + sin2 = 2

Angle b/w two vectors in terms of direction cosines & direction ratios:

(i) Suppose & are two vectors having d.c's l1, m1, n1 & l2, m2, n2 respectively. Then,

         &  

Dumb Question: How ?

Ans:   But   l12 + m1

2 + n12 = 1

       

So,  

   cos =

   cos = l1l2 + m1m2 + n1n2

(ii) IF a1, b1, c1 and a2, b2, c2 are d.r.s of & . Then

         &  

       cos =

       cos =

Note: (i) If two lines are then,              cos = 0   or   l1l2 + m1m2 + c1c2 = 0

             or a1a2 + b1b2 + c1c2 = 0

(ii) If two lines are || then

     cos = 1   or  

     or  

Single Option Question

1.A source of emf E = 10 V and having negligible internal resistance isconnected to a variable resistance. The resistance varies as shown in Figure. The total charge

that has passed through the resistor R during the time interval t1 to t2 is

(A) 40 loge4 (B) 30 loge3 (C) 20 loge2 (D) 10 loge2

2. A cylindrical tube contains two ideal gases with equal masses. The molar masses of the gas are M1M= 32 and M2 = 28 respectively. The gases are seperated by one movable

partition A and another fixed partition B as shown in Figure. The movable partition can move freely without friction inside the cylinder. The ratio of the heights of the gases inside the

cylinder in equilibrium is

(A) 0.875 (B) 1.143 (C) 0.9 (D) 0.6

3. Two masses m1 and m2 initially infinite distance apart are at rest. Then velocities of approach when the seperation between them becomes r, is given by v1 = km2 and v2 = km1 respectively. If k is a constant for a given distance r, then the value of k is

(A) (B)

(C) (D)

4. One end of a string of length L is tied to the ceiling of a lift accelerating upwards with an acceleration g/2. The linear mass density of the string is (x) = 0 x1/2 where, x is measured from the bottom. The time taken by a pulse to reach from bottom to top is

(A) (B) (C) (D)

5. The number of emission lines of atomic hydrogen spectrum between 400 nm to 700 nm according to Bohr' s theory is(A) 1 (B) 4 (C) 3 (D) 6

6. A lake is lit by an under water isotropic lamp. The refractive index of water is 4/3. If the surface of the lake is covered by a layer of

refractive index , the fraction of light totally internally reflected at water-oil surface is (Assume that the lake is absolutely calm and that both the oil and water are 100% transparent).(A) 0.25 (B) 0.5 (C) 0.75 (D) 1

Multiple Option Question

7. The Figure shows the displacement of a particle going along the x-axis as a function of time. The force acting on the particle is zero in the region.

(A) AB (B) BC(C) CD (D) DE

8. Seven identical rods of material of thermal conductivity k are connected as shown in Figure. All the rods are of identical length l and cross section area A. If the one end B is kept at 100oC and the other end is kept at 0oC, what would be the temperature of the junctions C,

D and E ( C, D and E) in the steady state?

(A) C > E > D

(B) E = 50oC and D = C (C) E = 50oC, C = 62.5oC and D = 37.5oC (D) E = 50oC, C = 60oC and D = 40oC

9. In the circuit shown the potential difference across 3 F capacitor is V and the equivalent capacitance between A and B is CAB?

(A) CAB = 4 F (B)CAB = F

(C) V = 20 V (D) V = 40 V

10. A thin biconvex lens with their radius of curvature 20 cm and 30 cm is placed with its principal axis first along a beam of parallel red light and then along a beam of parallel blue light. If the local length for the mean colour is 23.1 cm, B = 1.528 and B = 1.504, the seperation of foci for red and blue light is f. The dispersive power of the material of the lens is . Then

(A) f = 1.08 cm (B) f = 2.16 cm (C) = 0.47 (D) = 0.047

Take

PASSAGE

The techniques of solving the motion of a body with respect to a non-inertial frame of reference is by introducing the concept of PSeudo forces.

Suppose a frame of reference S' moves with constant acceleration 0 with respect to an inertial frame S. The acceleration of the particle P measured with respect to S' is PS' and that with respect to S is PS. The acceleration of S' with respect to S is S'S = 0

Since S' is translating with respect to S, we have

PS = PS' - S'S = PS' - SS'

or = PS - 0

m = m PS - m 0 where m is the mass of the particle.

Since S is an inertial frame m PS is equal to the sum of all the forces acting on P. Writing this sum as , we get m = - m 0

This equation relates the acceleration of the particle and forces acting on it. Comparing it with equation = m where acceleration is

measured in an inertial frame the acceleration of frame with respect to an inertial frame comes into the equation. Newton' s law is not valid for such non-inertial frame. An extra term - m has to be added to the sum of the forces acting on the particle before using the equation. It can be noted that in this extra term m is the mass of particle under cosideration and is the acceleration of working frame reference with respect to the inertial frame. We are so familiar with Newton' s laws that we would like to use the same equation for this frame also. This can be done if we call (- m ) a force acting on the particle. Then while preparing the list of forces acting on the

particle. We include all real forces acting on P and include an imaginary force - m this is called pseudoforce. This force is to be included because we are discussing the motion from a non-inertial frame and still want to use Newton' s law.

11. A body of mass m is placed over a smooth inclined plane of inclination , which is placed over a lift which is moving up with an acceleration a0. Base length of the inclined plane is L. Calculate the velocity of the block with respect to lift at the bottom. If it is allowed to slide down from the top of the plane from rest.

(A) (B)

(C) (D)

12. A particle of mass m is placed on the smooth face of an inclined plane of mass M and slope which is free to slide on a smooth horizontal plane in a direction perpendicular to

the edge. If the particles slides with an acceleration 'a', the acceleration of the inclined plane towards right A will be

(A) (B) (C) (D)

13. A ball is suspended on a thread from the ceiling of a car. The brakes are applied and the speed of car changes from 5 m/sec to 5/3 m/sec during the time interval of 3 seconds. Find the angle that the thread with the ball will deviate from vertical ? Take g = 10 m/sec2.

(A) (B) (C) (D)

PASSAGE

The rate at which sun radiates is 39 x 1026 W. All the radiated energy is generated by the proton-proton cycle.

14. In each proton-proton cycle, the number of protons used is(A) 2 (B) 4 (C) 8 (D) 16

15. If m(1H1) = 1.007825 u, m (4He2) = 4.002603 u the energy released in the proton-proton cycle is nearly (1 u = 981.5 MeV) (A) 26.7 eV (B) 26.7 MeV (C) 2.67 eV (D) 2.67 MeV

16.The rate at which hydrogen is being consumed in the core of sun is (A) 3.05 x 1011 kg/s (B) 6.1 x 1011 kg/s (C) 3 x 105 kg/s (D) 6.1 x 105 kg/s

Subjective Question

17. A circuit draws a power of 550 W from a source of 220 V/50 Hz. The power factor of the circuit is 0.8 and the current lags in phase behind the voltage. To make power factor 1, what capacitance is to be connected in the circuit? Express answer in microfarads.

18. A hollow steel sphere, weighing 200 kg is floating on water. A weight of 10 kg is to be placed on it in order to submerge when the temperature is 20oC. How much less weight is to be placed when temperature increases to 25oC? Express the weight in grams.

Match the following (with only one choice correct)

19. Column 1   Column 2 (A) Capacitance (P) C2J-1

(B) Inductance (Q) NA-1m-1

(C) Magnetic Induction (R) VsA-1

(D) Special Resistance (S) .m

20. Column 1   Column 2 (A) Interference (P) Colours in peacock feathers (B) Diffraction (Q) Spherical wavefront (C) Point source (R) Huygen's principle (D) Secondary wavelets (S) Study of crystal structure

Completed the test? Click here for Answers

Solution

Single Option Question

1. (D) From the graph, resistance varies uniformly at the rate of 1 ohm/sec.

  R2 = R1 + t   ......(1)

Now    

where t is the time  

Integrating, from (1)

q

  = E [loge (R1 + R2 - R1) - log R1]

  = E loge = 10 loge = 10 loge 2

2. (A) For an ideal gas PV =

  MV =

Here P1 = P2 in equilibrium   T1 = T2 and   m1 = m2 (given)

MV = constant  

Hence M1 V1 = M2 V2

i.e., 32 x r2 h1 = 28 x r2 h2

= = = 0.875

3. (D) v1 = km2; v2 = km1 (given)By the law of conversation of energy,

Substituting for v1 and v2,

4. (B) Consider a small element dx from the bottom of string.Weight of this element = [ (x) dx] (g + g/2)

Tension at this element

 

 

Velocity of transverse wave at this element

If l be the length of the string, then speed of transverse wave at lower end = 0

( tension is zero) speed of transverse wave at upper end =

 

=

5. (B) For = 700 nm

  = 1.77 eV

  For = 400 nm

  = 3.1 eV

1.77 eV < Ei - Et < 3.1 eV

E6 - E2 = 3.023 eV

E5 - E2 = 2.84 eV

E4 - E2 = 2.55 eV E3 - E2 = 1.89 eV

Only four lines are possible.

6. The critical angle for water-oil boundary is

Fraction of light escaping from water surface

 

 

fraction of light totally reflected at water-oil surface = 0.75.

Multiple Option Question

7. (A) and (C)The displacement is proportional to time in the region AB and CD in the Figure. Hence it is moving with constant velocity. So the force acting on the particle in these two regions is zero.

8. (A), (C)This problem can be solved similar to Electric Circuit Problem.Let R1, R2, R3, R4, R5, R6 and R7 be the rate of heat flow through AE, EB, AC, CD, CE, ED and DB respectively.

SinceR1 = R2

R5 = R6

E = 50oC

R3 = R4 + R5 = R7

R4 + R6 = R7

.....(1)

.....(2)

C + D = 100    

2 C - 2 D = 50 C = 62.5oC     D = 37.5oC

C > E > D    

9. (A) and (D)Equivalent capacitance of 3 F and 6 F = 2 F Equivalent capacitance CAB = 2 + 2 = 4 F Charge flowing in the 3 F capacitor = 60 x 2 = 120 coulombs

potential difference across

10. (A), (D) For red colour

r = 1.504, R1 = 0.20 m, R2 = - 0.3 m

= (1.504 - 1) = 4.2

b = 1.528

For blue colour

Dispersive power

PASSAGE

11. (C) Acceleration along the plane with respect to lift is, a = (a0 + g) sin Initial velocity = 0

v2 = u2 + 2as

12. (A) As the particle moves down the inclined plane, let a be its acceleration. Now the inclined plane moves horizontally towards right with acceleration A.

Considering the vertical component of forces acting on the particle.

  mg - R cos = ma sin .....(1)

For horizontal components the motion of the particle is considered at the accelerated frame. So a horizontal fictious forces= acts opposite to A.

This pseudo force mA acts to the left.  R sin + mA = ma cos    R sin = m(a cos - A) .....(2)

Now consider the force on inclined plane. The forces are shown in Figure.  R sin = MA .....(3)  R cos + Mg = S .....(4)

From equations (2) and (3), we get

13. (A) The situation is shown in Figure. When the car moves towards right with acceleration a the rope carrying the mass makes an angle .

Forces acting on the body are:(1) Weight mg downwards.(2) Tension T in the rope.(3) Pseudoforce ma towards left.

Since the body is in equilibrium under the action of three forces, we can apply condition of equilibrium.T cos - mg = 0; T sin = ma

tan = (or) a = g tan

Acceleration,

= 10 tan (or) tan =

14. (B) In proton-proton cycles, it amounts to the combinations of four protons and two electrons to form an -particle, two neutrino' s and six gamma rays.

15. (B) In proton-proton cycle, 4 protons 4 (4He + 2e-) + 2v + 6 The energy released in the reaction

Q = (mi - mf)c2 = [4m(1H) - m(4He)]c2   = [4 x 1.0078 - 4.0026]931.5  = 26.7 MeV

16. (B) Energy available within the sun for every four protons consumed = 26.7 MeV

= 6.4 x 1014 J/kg

Hydrogen consumption rate

= 3.9 x 1026 W, = 6.4 x 1014 J/kg

= 6.1 x 1011 kg/s

Subjective Question

17.

where This is because current lags behind this voltage.

or

But

 

Impedance of the circuit after inserting the capacitance is given by

power factor

 

But for power factor to be equal to 1,

 

or

 

 

  = 75 F

18. At the instant of submergence.Total mass of sphere and weight placed on it = mass of water displaced.

mass of water displaced at 20oC = (200 + 10)kg = 210 kg

and volume of the sphere = volume of water displaced by it.

volume of sphere at 20oC is

where 0 is density of water at 20oC.volume of sphere at 25oC becomes equal to V = V0(1 + 3 2 )

 

Density of water at 25oC becomes equal to   = 0 = (1 - ) = 0.99925 0

Mass of water to be displaced at 25oC in order to submerge the sphere = V. = 209.874 kg Required difference of weight to be placed on it.

  = 210 - 209.874 = 0.126 kg = 1269

Match the following (with only one choice correct)

19. Column I   Column II  (A) (P)  (B) (R)  (C) (Q)  (D) (S)

20. Column I   Column II   (A) (P)  (B) (S)  (C) (Q)  (D) (R)

 

- Test duration is 2 hours.

 

Question Type A : Single Answer is Correct

1. Which of the following statement is incorrect?

(a) C — D bond vibrates more slowly than a C — H bond.(b) C — D bond has lower zero point energy than a C — H bond.(c) In aromatic electrophilic sulphonation primary isotopic effect occurs.(d) C — D bond has higher vibrational energy than C — H bond.

(a) (b)

(c) (d)

Here product A is

(a) (b)

(c) (d)

4. Among the given gem diols, which is/are stable with respect to their corresponding carbonyls?

(i) CCl3CH(OH)2 (ii) (iii) (iv)

Correct options are

(a) i, ii, iii (b) i, ii (c) i, iii, iv (d) iii, iv

Rearranged alkene product after rearrangement at low temperature will be mainly,

(a) (b) (c) (d)

6. Which of the following alcohols would you expect to form a carbocation most readily in sulphuric acid

(a) (b) (c) (d)

7. Which of the following will accept H+ ion with fastest rate ?

(a) (b) (c) (d)

8. In the reaction

(a) (b) (c) (d) None of these

9. Which of the following compounds does not give Diel’s Alder reaction in presence of ethylene -

(a) (b) (c) (d)

Correct option are

(a) i, iii (b) i, iv (c) ii, iii (d) iii, iv

10. The correct order of increasing basic nature for the following compounds is:

(a) IV < I < III < II (b) I < II < III < IV (c) IV < III < II < I (d) II < IV < I < III

11. A solution of 0.4 mole of KI (100% dissociated) in 1000 g water freezes at T1°C. Now to this solution 0.2 mole HgI2 is added and the resulting solution freezes at T2°C. Which of the following is correct?

(a) T1 = T2 (b) T1 < T2 (c) T1 > T2 (d) can’t be predicted

12. There is mixture of Cu(II) chloride and Fe(II) sulphate. The best way to separate the metal ions from this mixture in qualitative

analysis is by treating it with :(a) hydrogen sulphide in mild acidic medium, where only Cu(II) sulphide will be precipitated(b) ammonium hydroxide buffer, where only Fe(II) hydroxide will be precipitated(c) hydrogen sulphide in mild acidic medium, where only Fe(II) sulphide will be precipitated(d) ammonium hydroxide buffer, where only Cu(II) hydroxide will be precipitated

13. Borax structure contains :

(a) two BO4 groups and two BO3 groups (b) four BO4 groups only

(c) four BO3 groups only (d) three BO4 and one BO3 groups

14. Gaseous ethylene, C2H4 reacts with hydrogen gas in the presence of platinum catalyst to form ethane

A mixture of C2H4 and H2 known only to contain more H2 than C2H4 has a pressure of 52 mm in an unknown volume. After the gas had been passed over a platinum catalyst, its pressure was 34 mm in the same volume and at the same temperature. Fraction of the ethylene molecules in the original mixture :

(a) 0.222 (b) 0.654 (c) 0.346 (d) 0.421

15. In a solid, oxide ions are arranged in ccp. One sixth of the tetrahedral voids are occupied by the cations A while one third of the octahedral voids are occupied by the cations B. The formula for the compound is :

(a) ABO3 (b) A2B3O4 (c) ABO2 (d) A3B2O4

16. In which of the following case, increase in concentration of ion cause increase in Ecell ?

(a) Pt (H2) | H+ (aq) (b)

(c) Pt | quinhydrone | H+ (aq) (d) Ag | Ag+ (aq)

17. Which of the following is not optically active?

(a) [Co(en)3]3+ (b) [Cr(Ox)3]3– (c) cis-[CoCl2(en)2]+ (d) trans-[CoCl2(en)2]+

18. Water and chlorobenzene are immiscible liquids. Their mixture boils at 90°C under a reduced pressure of 9.58 × 104 Pa. The vapour pressure of pure water at 90°C is 7.03 × 104 Pa and molecular weight of chlorobenzene is 112.5. The percent of chlorobenzene in the distillate is equal to :

(a) 55 (b) 67 (c) 78 (d) 81

9. Given that If Fe2+, Fe3+ and Fe solid are kept together then:

(a) Fe3+ increases (b) Fe3+ decreases

(c) Fe2+ / Fe3+ remains uncharged (d) Fe2+ decreases

20. A colourless water soluble solid ‘X’ on heating gives equimolar quantities of Y and Z. Y gives dense white fumes HCl and Z does so with NH3. Y gives brown precipitate with Nessler’s reagent and Z gives white precipitate with nitrates of Ag+, Pb2+ and Hg+. ‘X’ is:

(a) NH4Cl (b) NH4NO3 (c) NH4NO2 (d) FeSO4

Question Type B : More than ONE may be correct

21. Which of the following statements is/are correct?(a) Compounds of S, Se and Te with oxygen are generally tetravalent, but fluorine brings out maximum oxidation state of +6.(b) The higher oxidation states become less stable as we go down the group.(c) The +4 state shows both oxidizing and reducing properties but in +6 state the compounds are only oxidizing.(d) These compounds are typically volatile.

22. Potassium manganate can be converted into potassium permanganate by oxidation

(a) with chlorine (b) with ozone (c) with CO (d) electrolytically.

23. A radioactive element X decays by the sequence and with half lives given below :

Which of the following statements about this system are correct?(a) Maximum amount of Y present at any time is less than 50% of the initial amount of X(b) Atomic number of X and Z are same(c) The mass number of Y is greater than X.(d) The mass number of Z is less than X.

24. Which of the following statements are correct?(a) The probability of finding a 4d electron right at the nucleus is zero.(b) For all values of n, the p-orbitals have the same shape, but the overall size increases as n increases, for a given atom.(c) A 2px atomic orbital consists of two lobes of electron density.(d) There is no probability of finding a p-electron right at the nucleus.

25. Which alkyl halide undergo E2 elimination

(a) (b) (c) (d)

26.A = Standard free energy of 1 mol of N2O4

C = Standard free energy of 2 mol of NO2

B = Equilibrium point

Following the graph, which of the statement/s is/are correct?(a) The absolute conversion of N2O4 into NO2 (N2O4 2NO2) is spontaneous(b) The dynamic conversion (N2O4 2NO2) at equilibrium point is spontaneous(c) Formation of N2O4 is more spontaneous than dissociation(d) Both process dissociation as well as formation are equally spontaneous

27. What could be the correct statements about idealized Carnot’s heat engine working between temperatures T2 (source) and T1 (sink)?

(a) Efficiency , which is maximum(b) Complete conversion of heat into work is possible i.e., can be unity(c) Power of the engine is maximum(d) Power of the engine is zero.

28. Addition of which of the following will separate the cations from an aqueous solution of Mg(ClO4)2 and AgNO3 ?

(a) Na2CO3 (b) NaF (c) KSCN (d) Na2C2O4.

29. An isothermal vessel contains SO2(g) in equilibrium with Cl2(g) as

. At equilibrium, total pressure is 4 atm when volume is 1 L. If the volume is increased to 2 L, the new equilibrium pressure in the vessel is P. Then

(a) P < 4 atm (b) P > 2 atm (c) P = 2 atm (d) P < 2 atm.

30. Pick up the false statement.(a) In the fluorite structure (CaF2), the Ca2+ ions are located at the lattice points and the fluoride ions fill all the tetrahedral holes in the ccp crystal.(b) In the antifluorite structure (Li2O, Rb2S) the cations are located at the lattice points and anions fill the tetrahedral holes in the ccp structure.(c) The radius of a metal atom is taken as half the nearest metal-metal distance in a metallic crystal.(d) One tetrahedral void per atom is present in hcp structure.

Question Type C : Comprehention Based Questions

PASSAGE– I

Conjugation occurs when there is alternative double and single bond like 1, 3-butadiene other type of delocalisation is - bond, -lone pair and -vaccant orbital. As the number of resonanting structure increases, stability of system also increases. In any organic reactions as the stability of intermediate species increases reactivity of initial compound increases by process of neutralisation.

31. Which of the following anion is most stable due to delocalisation?

(a) (b) (c) (d)

32. What are the stability order of the following intermediates?

(a) II > I > IV > III (b) I > III > II > IV (c) I > IV > III > II (d) I > IV > II > III

33. Which of the following ion is most stable ?

(a) (b) (c) (d)

Passage – II

When halogenation of alkenes occur, intermediate formation of cyclic halonium takes place by ionic mechanism. When cyclopentene reacts with bromine in presence of CCl4 anti addition occurs and gives trans isomer so addition of halogen to an alkene is a stereospecific reaction.

34.

Intermediate forms are

(a) Hydroxonium ion (b) Chloronium ion (c) Hydronium ion (d) none

35. Trans-2-butene on reacting with Br2 in CCl4 forms mainly

(a) ± 2,3-dibromobutane (b) + 2,3-dibromobutane (c) – 2,3-dibromobutane (d) meso 2,3-dibromobutane

36. The product obtain in the reaction is

(a) (b) (c) (d)

37.Final product is

(a) meso 2,3-butanediol (b) 2-butanol (c) ± 2,3-butanediol (d) 2-butanone

Passage – III

The reaction of an alcohol with an alkyl halide proceed by an SN1 mechanism, provided the halide is tertiary or provides a resonance - stabilized carbocation neutral alcohol molecules are generally not potent enough as nucleophilies to react 1° or 2° alkyl halides by an SN2 mechanism. However if the alcohol is first converted to its conjugate base, its nucleophilicity and basicity are vastly increased so alkoxide ions do engage in SN2 ether forming reactions, provided the alkyl halide is 1°.

38. Which ether cannot prepared by willamson’s syntheris

(a) CH2 == CH — O — CH == CH2 (b) (CH3)3C — O — C(CH3)3

(c) CH3 — O — C2H5 (d) (CH3)3C — O — CH3

39. When ROH reacts with SOCl2 and no added base, which is correct statement for reaction(a) Retention of configuration occurs(b) This is an example of SNi mechanism(c) The Cl of the anion of the ion pair attacks the front side of the C+, with retention of configuration(d) In this reaction if weakbase pyridine is present inversion of configuration occurs.

Passage – IV

Athletes sometimes upset the buffering of their blood through a process called hyperventilation. In the excitement and anxiety of a contenst, they breathe more rapidly and deeply they need to. When hyperventilation occurs, a person expels more CO2 than necessary, upsetting the carbonic acid equilibrium. According to Le-Chateliers principle, as the CO2 is exhaled, more H2CO3 decomposes to replace the CO2 in the reaction :

As H2CO3 is used, the equilibrium between carbonic acid and hydrogen carbonate is upset and HCO3– is consumed in replacing the

H2CO3.

Eventually, the HCO3– concentration drops to the point at which it is insufficient to maintain the blood pH at a safe level. Because using

HCO3–, also consumes H3O+, the blood pH is raised (becmes more basic). One response of the body to this condition is a construction

of the cerebral blood vessels. As the blood flow to the brain is reduced, the individual becomes dizzy and can lapse into unconciousness. At that point, the body’s reflex mechanisms usually restore normal breathing and the blood pH returns to normal.

40. To maintain a pH of 7.4 for blood at normal condition which is 2 M in H2CO3 (at equilibrium). What volume of 5 M NaHCO3 solution is required to mix with 10 mL of blood?

(a) 78.36 mL (b) 102.00 mL (c) 52.71 mL (d) 89.01 mL

41. The administration of medicine containing H2PO4– in significant amount will :

(a) increase the exhaling of CO2 (b) decrease the exhaling of CO2

(c) doesn’t affect the liberation of CO2 (d) none of these

42. Dizziness of athlete is caused basically due to increased :

(a) ionization of carbonic acid (b) decrease the exhaling of CO2

(c) ionization of hydrogen carbonate ion (d) hydration of hydrogen carbonate ion

Passage – V

The amount of sulphur present in steel samples can be determined by conversion of the sulphur to hydrogen sulphide which is subsequently estimated by titration with iodine. In this latter process the H2S is converted to elemental and the iodine reduced to iodide ion. Iodine solutions used in volumetric analysis is standardised prior to use as iodine is volatile and the solutions decrease in concentration with storage. One way of standardising iodine solutions is by titrating against a known mass of arsenic (III) oxide. One representation of this reaction is as follows:

43. In the standardisation method essentially sodium hydrogen carbonate is added in excess to the solution of arsenic (III) oxide is :(a) the impurity of iodine gets precipitated out with Na(b) NaHCO3 combines with H+ drives equilibrium completely in forward direction(c) NaHCO3 drives the equilibrium backward releasing H+, thus maintains pure iodine(d) NaHCO3 produces NaI and thus prevents the volatility.

44. During volumertric analysis chemist finds that the H2S liberated from a 5.00 g sample of steel requires 1.90 mL of 0.049 M iodine solution to reach an end point in the titration. The percentage of sulphur in the steel sample is :

(a) 0.90% (b) 1.20% (c) 0.06% (d) 6.0%

Passage – VI

(i) A white amorphous powder (A) on heating strongly gives a colourless non-combustible gas (B) and solid (C).(ii) The gas (B) turns lime water milky and turbidity disappears with the passage of excess of gas.(iii) The solution of (C) in dilute HCl, (D), gives a white precipitate with aqueous solution of K4Fe(CN)6.(iv) (D) gives a white precipitate (E) on passing H2S in presence of excess of NH4OH.

45. Compound (A) (white amorphous powder) is

(a) AlCl3 (b) ZnCO3 (c) BaSO4 (d) SrSO4

46. Gas (B) which turns lime water milky is

(a) SO2 (b) SO3 (c) CO2 (d) Cl2.

47. Compound (D) is

(a) Al(OH)3 (b) ZnCl2 (c) BaCl2 (d) Sr(OH)2.

Question type D : Column Matching Type Questions

48. Column - I Column - II  (a) On increasing pressure (P) Viscosity of liquid increases, L increase  (b) On increasing temperature (Q) VL decreases, L decreases  (c) On dilution (R) Viscosity of water decreases  (d) On increasing mass (S) None of these

The matching grid :

(a) (b) (c) (d)

49. Following formation reactions are given at 25°C. Match the column I with the appropriate defect in

formation equation

Column - I Column - II

(a) H2(g) + Br2(l) 2HBr(g) (P) elements are not in standard state(b) 1/2 I2(s) + 1/2Br2(I) IBr(s) (Q) defect in stoichiometric coefficients

(c) O2(g) + 2H2(g) 2H2O(I) (R) Pressure is not mentioned(d) S(monoclinic) + O2(g) SO2(g) (S) no defect

The matching grid :

(a) (b) (c) (d)

50. Match the column (I) & (II) such that the options from table II can satisfies table (I)

Column - I Column - II

(a)

(b)

(c)

(d)

The matching grid :

(a) (b) (c) (d)

 

Answer Sheet

1. (d) 11. (b) 21. (a, b, c, d) 31. (b) 41. (b)2. (b) 12. (a) 22. (a, b, d) 32. (c) 42. (b)3. (d) 13. (a) 23. (b, d) 33. (c) 43. (b)4. (a) 14. (c) 24. (a, b, c, d) 34. (b) 44. (c)5. (d) 15. (a) 25. (a, b) 35. (d) 45. (b)6. (b) 16. (b) 26. (b, c) 36. (c) 46. (c)7. (d) 17. (d) 27. (a, d) 37. (a) 47. (b)8. (c) 18. (b) 28. (b, c) 38. (a, b, c, d) 48. (a) P,R (b) Q,R (c) Q (d) S9. (d) 19. (b) 29. (a, b) 39. (a, b, c, d) 49. (a) Q,R (b) S (c) P,Q,R (d) P,R

10. (a) 20. (a) 30. (b, d) 40. (a) 50. (a) P,Q (b) Q (c) P,R (d) R,S

Solution

2.

3.

5.

6. Answer is (b), because middle -OH undergoes protonation to form stable intermediate tropolium cation.

7. (d) accept easily because lower 'N' atom lone pair is not delocalised in the ring and -NH2 group exerts +M effect.

9. Diel' s Alder reaction never given by traniod form so (c) and (d) cannot give.

10. Basic strength of nitrogen containing compound given by donation of electron so +I and +R effect increases basic nature while -I and -R effect decreases basic strength. So order is (a).

11. 2KI + HgI2 K2HgI4   .4 .2 .2

Total Initial moles = .8after addition of HgI2 total number of ions

  = .6[2K+[HgI4]-2] after complex formation number of ions decreases So depression in freezing point will be less.

  T2 > T1

12.(Cu+2) It is IInd group radical so its get precipitated first due to lower solubility product (Ksp of CuS = 10-44).

13.Formula of Borax is Na2B4O7 . 10H2O

14. Mixture contains C2H4 + H2

  x y(Excess)  x + y = 52 mm - (1)

  y - x + x = 32 mm - (2) By equation (1) and (2)

  x = 18 mm y = 34 mm

So fraction of C2H4 = = .35

15. Given in the Question that number of oxide ions = x number of octahedral voids = xnumber of tetrahedral voids = 2x

no. of cation A = = x/3

no. of cation B   = x/3   oxide : A ! B    x x/3 x/3    3 1 1 So formula ABO3

17. Trans complex is optically inactive due to molecular symatry.

18. In steam distillation where both are immisible liquids we apply formula

  = 2.27 So In a total weight of 3.37 gm, weight of chlorobenzene = 2.27

 % of =

= 67.4 %

19.

Go must be negative so Fe+3 decreases

20. NH3 + K2HgI4 +KOH NH2 - H2O - HgI

  (y)brown ppt. of

millan' s base ofiodide.

22. 2K2MnO4 + Cl2 2KMnO4 + 2KCl2K2MnO4 + H2O + O3 2KMnO4 + 2KOH + O2

23.

25. E2 Elimination is straeospecific it required trans leaving group for anti elimination.

26. As Go for N2O4 2NO2 is KJ/mole so process is nonspontaneous. But at equilibrium point BN2O4 is dissociated upto equilibrium point spontaneous as

 Go = - 0.86 KJ

Again, formation of N2O4 upto equilibrium point is fairly spontaneous. Hence, dynamic conversion N2O4 into equilibrium mixture and 2NO2 into equilibrium mixture both are spontaneous.

31. (b) is maximum stable due to highest -I and -R effect so maximum neutralisation.

32. Stability increases by neutralisation so answer is (c)

33. Due to back bonding of lone pair of nitrogen (c) is maximum stable.

34. Formation of cyclic halonium ion takes place.

35. Addition of halogen to alkene is always anti in nature.

36. In the presence of strong base potassium ter. butoxide formation of electrophile : CBr2 takes place which undergo addition.

38. (a) and (b) is cannot possible. In case of (a) intermediate cation is not possible while in case of (b) elimination reaction occurs.

40. PH

= PKa + log   K = 1.28 x 106 given   Ka = 7.8 x 10-7

  7.4= - log 7.8 x 10-7 + log

  v = 78.36 ml

42. H2CO3 H2O + CO2 This is dissociation of carbonic acid.

43. As2O3 + 2I2 As2O5 + 4H+ + 4I- There is an equilibrium.

NaHCO3 reacts with , produced as a product and so causes the equilibrium to move completely to the forward and thus ensure complete reaction of AS2O3.

44. 1 mole H2S = 1 mole I2 In titration requires 1.9 ml of .049 m I2 solution number of moles of I2 used = .049 x .0014 = 9.46 x 10-5 mol I2

  = 9.46 x 10-5 mol H2S Mass of sulphur in 5 gm sample of 5HCl = 9.46 x 10-5 x 32 = .003gm

% of sulphur in 5 gm sample

Superposition of waves Introduction

1. Wave Motion:

a) The sound waves can also be called as elastic waves and medium is needed for their propagation. Their velocity is different for different media.

b) The velocity of sound waves depends on the property of inertia and coefficient of elasticity of that medium.

c) In this type of motion the energy is transferred from one point to other by the propagation of disturbance in the medium. The motion of disturbance is called the wave motion.

d) In simple harmonic wave the particles of the medium executes simple harmonic motion. Each particle of the medium executes simple harmonic motion at all the times, though the phase of its vibration depends on its position and time.

On the basis of vibration of the particle waves are classified in two types:

(a)   Transverse waves.

(b)   Longitudinal waves.

(a) Transverse waves:

The particles of medium vibrate in a direction perpendicular to the direction of propagation of wave. Their polarization is possible. Example: Vibration of string, the surface waves produced on the surface of solid and liquid.

Fig (1)

(b) Longitudinal waves:

Vibrations of the particles of the medium are in the direction of wave propagation wave proceeds in

form of compression and rarefaction. At places of compression the pressure and density are maximum, while at places of rarefaction those are minimum.

Example: Sound waves, waves is gases, waves is solids.

Fig (2)

Introduction

2) Some important points related to Wave Motion:

1) Amplitude: The maximum displacement of a vibrating particle of the medium from the mean position.

2) Time period (T): The time taken by the particle to complete one vibration.

3) Frequency: The number of vibration in one second by the particle n = 1/T.

Angular Frequency

4) Wavelength : The distance between two consecutive particles vibrating in the same phase.

Or

The distance between two consecutive crests or trough.

Or

The distance between two consecutive compressions or rarefaction.

Here v is the velocity of wave.

is the wave propagation vector which is written as:

, being the unit vector along the direction of wave propagation then,

Fig (3)

3) Equations for the progressive Simple Harmonic Waves:

1) The equation of a wave moving along the direction of positive X-axis is

Where ‘a’ is the amplitude, is the phase difference and ‘y’ is the displacement of a particle.

2) The equation of a progressive wave moving along the direction of negative X-axis is

3) The general equation of wave moving along X-axis is:

4) Other equation for a progressive wave moving along the direction of positive X-axis is:

Is propagation constant or wave vector

Points to remember: Wave equation can be shown by any of the following relations:

All these waves are also called as ‘sinusoidal’ waves.

4) Necessary Condition for two particles to be in same phase or in opposite phase:

5) Intensity of waves, Energy Density:

1) Intensity (I): Amount of energy flowing per second from unit area of the medium in a direction perpendicular to the propagation of wave.

2) For a given medium ( and v are constants)

3) Unit of Intensity: Joule/m2 second or watt/m2.

4) Energy density (E): Energy per unit volume of the medium

5) For second waves the intensity obeys the inverse square law. If the distance between the source and the point

of observation is ‘x’ then

6) Superposition of Waves:

1) If two or more waves propagate in a medium simultaneously, then the resultant displacement of the particle of the medium is given by the vector sum of displacements produced by the individual wave.

2) If the displacement produced by different waves are respectively

Then the resultant displacement of particle is given by,

3) Resultant displacement depends on

a) Amplitude

b) Phase difference

c) Frequencies

d) Direction of propagation of waves

4) Different phenomena occurring as a result of superposition of waves are

a) Interference

b) Beats

c) Stationary waves

d) Lissajous figures.

7) Interference:

1) As a result of superposition of two waves of same frequency traveling in the same direction simultaneously, the phenomenon of intensity of resultant wave becoming maximum and minimum is called as interference.

2) Equation of resultant wave formed by superposition of two waves:

Amplitude of the resultant wave

3) Intensity of the resultant wave

I1 and I2 are separate intensities of the superposing waves.

4) Law of conservation of energy is obeyed in interference also, only the energy of the medium is redistributed.

8) Constructive and Destructive Interference:

1) When superposing waves are in the same phase then

The amplitude of the resultant wave

In this condition the interference is called ‘constructive interference’

2) For if is zero or even multiple of ; , where n = 0, 1, 2, ---------------- or path difference of waves.

3) Then in above condition, the intensity of resultant wave is maximum.

4) If the interfering waves are in opposite phase then , and path difference

In this condition the amplitude of the resultant wave

Interference of this kind is called the ‘destructive interference’. In this condition the resultant intensity is also minimum.

5) Ratio of maximum intensity and minimum intensity of a wave is

6) To observe clear interference:

a) Phase difference of the waves must be fixed.

b) Amplitude of the waves must be equal.

9) Beats:

1) When two progressive sound waves of nearly equal frequencies superpose while traveling in the same direction, then the intensity of the resultant sound increases and decreases with time.

2) The increase and decrease (waxing and waning) of intensity of sound occurs with a regular interval. This regular waxing and waning of sound is called Beats. One decrease and one increase together make one BEAT.

3) Number of beats produced in one second is called to be beat frequency.

4) If the frequencies of the waves are m and n (m>n), then beat frequency = m-n. This difference must not be greater than 10Hz, otherwise beats are not heard.

5) Beat period = 1/ N = 1/ (m-n).

6) If the waves producing beats are

Then the equation of the resultant wave is

And the amplitude of the wave is;

10) APPLICATION OF BEATS:

1) In the tuning of radio receiver.

2) In manufacturing the electronic oscillators of less and stable frequencies.

3) In detecting poisonous gases produced in mines.

4) In the tuning of musical instruments.

5) In determining unknown frequency of tuning fork: Let the known frequency of the tuning fork be ‘m’, unknown frequency be ‘n’ and the number of beats produced be ‘N’ while playing these two together.

(a) If by putting some wax on the arm of unknown frequency tuning fork:

If the beat frequency decreases then n = m+N.

If the beat frequency increases then n = m-N.

(b) If by putting wax on the arm of known frequency tuning fork:

If the beat frequency decreases, then n = m-N.

If the beat frequency increases, then n = m+N.

(c) If the arms of unknown frequency tuning fork are filled then:

If the beat frequency increase n = m+N.

If the beat frequency decrease n = m-N.

11) Reflection of sound waves when wave posses from one medium to the other:

(a) Reflection from a rigid wall or a denser medium:

1) Phase change of .

2) Change of in the path difference

3) No change in the nature of wave.

4) Compression is reflected as compression and rarefaction as a rarefaction.

5) Node is always formed at rigid surface

6) Direction of wave is changed.

(b) Reflection from rare medium:

1) No change in the phase.

2) No change in the path difference

3) No change in the nature of wave.

4) Compression is reflected as rarefaction while rarefaction as compression.

5) Antinode is formed at the surface of a rare medium.

6) Direction of the wave is changed.

12) stationary waves:

1) This is the wave produced by the superposition of two identical waves traveling along the same straight line but in opposite direction.

2) Energy is not transferred by these waves in the medium. It is only redistributed.

3) If be the propagation wave along +ve x-axis and be the progressive wave along –ve x-axis, then for the resultant stationary wave.

Introduction

4) Its amplitude is this depends on position x.

5) Antinodes: At these points the amplitude of the vibrating particles is maximum

and the change in pressure and the density is minimum.

6) For the positions of antinodes:

7) Nodes: At these points the amplitude of particles is minimum and the change in pressure and density is maximum.

13) for stationary waves:

1) Waves produce nodes and antinodes at regular points in limited medium.

2) The distance between to consecutive nodes/ antinodes is .

3) Distance between a node and an adjacent antinode is .

4) Particles situated between two nodes execute simple harmonic motion whose amplitude is different but frequencies are same.

5) Amplitude depends on the position of the particle; maximum amplitude is obtained at antinodes and zero amplitude at the nodes.

6) The particles situated between two consecutive nodes vibrate in the phase with different amplitudes while the particles situated on either side of a node vibrate in the opposite phase.

14) vibration in a stretched string:

1) Transverse progressive wave is produced in a stretched string.

2) Node is always formed at fixed end of string.

3) Velocity of the waves produced in a stretched string is

Where T = tension and m = mass per unit length of string.

Given density of material of string is d; radius is r, then;

4) For a string fixed at both the ends, nodes are formed at the ends.

5) If the length of a string is L and p loops are formed in it, then the frequency of string

(a)   When p=1, then . In this condition string vibrates in one loop. is

called the fundamental frequency. This is also called first harmonic.

(b)   If p =2, then . In this condition is called the second harmonic or the first

overtone.

(c)   If p=3, then is called third harmonic or the second overtone.

(d)   In strings all harmonics are produced and their ratio is ;

(e)   Some harmonics are shown in the figure

Fig (4)

15) laws of vibration of a stretched string:

1) Law of length: where T and m are constants.

2) Law of tension: where L and m are constants.

3) Law of mass: where L and T are constants.

4) Law of radius: where d, L and T are constants.

5) Law of density: where r, T and L are constants.

On the basis of above Laws the formulae of frequency of vibration of string are:

M being the mass hanged on string.

16) melde’s experiment:

1) In a vibrating string of fixed length, the product of number of loops of loops in a vibrating string and square root of tension is a constant or

2)

3) In longitudinal vibration system the frequency of tuning fork is given by;

= 2 ´ (vibration frequency of string)

4) In this experiment vibrations of string are always transverse, but in longitudinal vibration system; the vibrations of arms of the tuning fork are along the direction of string. This experiment is also based on the stationary (transverse) waves.

17) vibration of air columns in pipes:

The pipe which contains air and in which sound vibrations are produced is called organ pipe.

1) CLOSED PIPE:

(a) One end of the pipe of this kind is closed and the other end is open.

(b) Node is formed at closed end and antinodes at the open end. In this pipe, number of antinodes and nodes are the same.

(c) Closed end of pipe reflects the compression as compression and rarefaction as rarefaction open end of the pipe reflects compression as rarefaction while rarefaction as compression.

(d) For a pipe of length L, the frequency of fundamental node of stationary waves produced. Fundamental frequency , is same as first harmonic while other are multiples of this frequency

wavelength . Frequency of third harmonic or the first overtone

wavelength

Frequency of the second overtone or the fifth harmonic

, wavelength

Introduction

These are shown in the following figure:

Fig (5)

(a)   Fundamental node or first harmonic .

(b)   First overtone or third harmonic .

(c)   Second overtone or fifth harmonic .

Therefore the ratio of overtone is 3:5:7.

(e) It is clear that the only odd harmonics are produced in this pipe:

2) Vibration of an open pipe:

(a) These pipes are open at both ends where antinodes are formed. At these ends compression is reflected as rarefaction while rarefaction as compression. Here the number of antinodes is more than that of nodes.

(b) For a pipe of length L, the frequency of fundamental node and wavelength

First overtone or second harmonic frequency , wavelength

Second overtone or third harmonic frequency wavelength

These are shown in the following figure:

Fig (6)

(a)   Fundamental node or first harmonic

(b)   First overtone or second harmonic

(c)   Second overtone or third harmonic

Clearly in open pipe all the harmonics are produced. The ratio of the frequencies is:

In this condition the ratio of overtone is 2:3:4:5: …………

End correction: e = 0.6r, where ‘r’ is the radius of pipe. Therefore for a closed pipe the effective length of the air column = L+e = L + 0.6r.

For open pipe the effective length of the air column is L+2e = L+1.2r.

For a closed pipe with end correction

By comparison:

(For fundamental only)

2) In open pipe all the harmonic are obtained while in closed pipe only odd harmonics are obtained.

3) The sound produced in open organ tube is pleasing and that of the closed organ tube is less pleasing.

18) resonance tube:

Resonance: If the frequency of a tuning fork used is same as the frequency of vibration of air column in the tube, then the intensity of sound becomes maximum. This is said to be the condition of resonance.

It is an example of a closed pipe in which the length of the air column can be changed by adjusting the level of water.

Application:

To determine the velocity of sound and the frequency of tuning fork

Formula:

If the first and second resonance lengths are L1 and L2 then

19)

(a) Velocity of sound (Longitudinal wave) in elastic medium:

Where E = Coefficient of elasticity of medium, d = density

For a solid medium:

, where E=Y = Young’s coefficient of elasticity

In a liquid medium:

, where E=K= Coefficient of volume elasticity.

In gaseous medium:

M = molecular, P = pressure, T = temperature.

(b) In stretched vibrating wire or a string, the velocity of a transverse wave:

20) factors affecting the velocity of sound in the gas:

5) At 00C velocity of sound in dry air = 332 m/sec, while the velocity sound in water is 1440 m/sec. At t0 C the velocity of sound will be (for low values of t, t/273<<1).

Vt = 330+0.61 m/sec.

6) In atomic weights of two gases are M1 and M2 respectively then

21) features of musical sound:

1) Loudness:

The quality of sound on the basis of which sound is said to be high or low. It depends on the.

a)      shape of the source

b)      intensity of sound

unit of L is “phone” when I is measured in decibel.

The amplitude of a roaring lion is more than sound produced by a mosquito.

2) Pitch:

On the basis of this characteristic sound is said to be sharp or dull. Pitch increases with the frequency of sound. For example pitch of roaring lion is less than that of sound of mosquito.

3) Quality:

On the basis of this property the sound of same loudness and pitch can be differentiated. The basis of this differentiation is number of harmonics present in the sound, relative intensity etc. On increasing the number of overtones, sweetness of sound increases. Example: Whistle’s sound (closed pipe) is less sweet than that of flute (open pipe)

22) some important information:

1) Shock waves: A body moving with a supersonic velocity leaves behind itself a conical disturbance region. Disturbance of this kind is called a short wave. These have a large amount of energy which can damage the buildings.

2) Frequency of infrasonic waves is less than 20 Hz.

3) Frequency of ultrasonic waves is more than 20000 Hz.

4) Mach number = speed of any vehicle or body/ speed of sound.

5) Vibration of tuning fork: when tuning fork is sounded by striking its one end on rubber pad, then

(a)   The ends of prongs vibrate in and out while the stem vibrates up and down or vibrations of the

prongs are transverse and that of the stem is longitudinal. Generally tuning fork produces fundamental

node.

(b)   At the free end of a fork antinodes are formed. At the place where stem is fixed antinodes is formed. In

between these antinodes, nodes are formed.

Fig (7)

(c)   Frequency of tuning fork :

t = thickness of tuning fork.

l = length of arm of fork.

E = Coefficient of elasticity for the material of fork.

d = density of the material of a fork.

(d) Frequency of tuning fork decrease with increase in temperature.

(e) Increasing the weight, the frequency of a tuning fork decreases while on filing the prongs near stem, the

frequency decreases.