(5) The equation of family of circles, which touch y − y1 = m(x − x1) at (x1,y1) for any finite m is
(x − x1)2 + (y − y1)2 + λ {(y − y1) − m(x − x1)} = 0
And if m is infinite, the family of circles is (where λ is a parameter) (x1,y1)
(x − x1)2 + (y − y1)2 + λ (x − x1) = 0
y–y1=m(x–x1)
(6) Equation of the circles given in diagram is (x − x1)(x − x 2 ) + (y − y1)(y − y2 ) ± cotθ {(x − x1)(y − y2 )
− (x − x 2 )(y − y1)} = 0
θ
(x1,y1) (x2,y2)
θ
Example : 44 The equation of the circle through the points of intersection of x 2 + y 2 − 1 = 0, x 2 + y 2 − 2x − 4y + 1 = 0 and touching
Solution : (c)
the line x + 2y = 0 is (c) x 2 + y 2 − x − 2y = 0 [Roorkee 1989]
(a) x 2 + y 2 + x + 2y = 0 (b) x 2 + y 2 − x + 20 = 0
(d) 2(x 2 + y 2 ) − x − 2y = 0
Family of circles is x 2 + y 2 − 2x − 4y + 1 + λ (x 2 + y 2 − 1) = 0
x2 + y2 2 x − 1 4 λ y + 1 − λ =0
− 1+ λ + 1 + λ
Centre is 1 , 2 and radius = 1 1 λ 2 + 1 2 λ 2 − 1 − λ = 4 + λ2
1 + λ 1 + λ + + 1 + λ (1 + λ)2
Since it touches the line x + 2y = 0, Hence Radius = perpendicular distance from centre to the line = 1 λ + 1 4 λ
1+ +
12 + 22
= 4 + λ2 = 4 + λ2 ⇒ 5 = 4 + λ2 ⇒ λ = ± 1
1+λ
(1 + λ)2
λ = −1 cannot be possible in case of circle, so λ = 1. ∴ Equation of circle is x 2 + y 2 − x − 2y = 0
Example : 45 The equation of the circle through the points of intersection of the circles x 2 + y 2 − 6x + 2y + 4 = 0,
Solution : (b)
x 2 + y 2 + 2x − 4y − 6 = 0 and with its centre on the line y = x
(a) 7x 2 + 7y 2 + 10x − 10y − 12 = 0 (b) 7x 2 + 7y 2 − 10x − 10y − 12 = 0
(c) 7x 2 + 7y 2 − 10x + 10y − 12 = 0 (d) 7x 2 + 7y 2 + 10x + 10y + 12 = 0
Equation of any circle through the points of intersection of given circles is
(x 2 + y 2 − 6x + 2y + 4) + λ (x 2 + y 2 + 2x − 4y − 6) = 0 ⇒ x 2 (1 + λ) + y 2 (1 + λ) − 2x (3 − λ) + 2y (1 − 2λ) + (4 − 6λ) = 0
or, x2 + y2 − 2x (3 − λ) + 2y (1 − 2λ) + (4 − 6λ) = 0 …..(i)
(1 + λ) (1 + λ) (1 + λ)
Its centre = 3 − λ , 2λ − 1 lies on the line y = x . Then 2λ − 1 = 3−λ ⇒ 2λ − 1 = 3 − λ { λ ≠ −1 }
1+λ 1+λ
1 + λ 1+λ
⇒ 3λ = 4 ⇒ λ = 4
3
Substituting the value of λ = 4 in (i), we get the required equation as 7x 2 + 7y 2 − 10x − 10y − 12 = 0 .
3
Radical Axis.
The radical axis of two circles is the locus of a point which moves such that the lengths of the tangents drawn
from it to the two circles are equal.
Consider, S ≡ x 2 + y2 + 2gx + 2 fy + c = 0 …..(i) and S' ≡ x 2 + y 2 + 2g1 x + 2 f1y + c1 = 0 …..(ii)
Let P (x1,y1) be a point such that |PA | = | PB |
⇒ (x 2 + y12 + 2gx1 + 2 fy1 + c) = (x12 + y12 + 2g1 x1 + 2 f1y1 + c1 )
1
On squaring, x12 + y12 + 2gx1 + 2 fy1 + c = x12 + y12 + 2g1 x1 + 2 f1y1 + c1
⇒ 2(g − g1) x1 + 2( f − f1)y1 + c − c1 = 0
∴ Locus of P (x1,y1) is 2(g − g1) x + 2(f − f1)y + c − c1 = 0
P(x1,y1) P(x1,y1)
A B AB
C1 C2 C1 C2
S=0 S′=0
which is the required equation of radical axis of the given circles. Clearly this is a straight line.
(1) Some properties of the radical axis
(i) The radical axis and common chord are identical : Since the radical axis and common chord of two circles
S = 0 and S' = 0 are the same straight line S – S' = 0, they are identical. The only difference is that the common
chord exists only if the circles intersect in two real points, while the radical axis exists for all pair of circles
irrespective of their position (Except when one circle is inside the other).
Radical axis Common tangent Common chord
C1 C2 C1 C2 C1 C2
Non intersecting circles Touching circles Intersecting circles
(ii) The radical axis is perpendicular to the straight line which joins the centres of the circles :
Consider, S ≡ x 2 + y2 + 2gx + 2 fy + c = 0 …..(i)
and S1 ≡ x 2 + y 2 + 2g1x + 2 f1y + c1 = 0 …..(ii)
Since C1 ≡ (−g,− f) and C2 ≡ (−g1, − f1) are the centres of the circles P(x1,y1)
(i) and (ii), then slope of the straight line C1C2 = − f1 + f = f − f1 = m1 (say)
− g1 + g g − g1
B
Equation of the radical axis is, 2(g − g1) x + 2(f − f1)y + c − c1 = 0 A R S′=0
C1 Q C2
Slope of radical axis is − (g − g1) = m2 (say). ∴ m1m2 = −1
(f − f1 ) S=0
Hence C1C2 and radical axis are perpendicular to each other.
(iii) The radical axis bisects common tangents of two circles : Let AB be the common tangent. If it meets the
radical axis LM in M then MA and MB are two tangents to the circles. Hence MA = MB, since length of tangents
are equal from any point on radical axis. Hence radical axis bisects the common tangent AB.
A MB T T
C1 L C2 C1 A B C2 C1 C2 A
If the two circles touch each other externally or internally then A and B coincide. In this case the common
tangent itself becomes the radical axis.
(iv) The radical axis of three circles taken in pairs are concurrent : Let the equations of three circles be
S1 ≡ x 2 + y2 + 2g1x + 2 f1y + c1 = 0 …..(i)
S2 ≡ x 2 + y 2 + 2g2 x + 2 f2y + c2 = 0 …..(ii)
S3 ≡ x 2 + y 2 + 2g3 x + 2 f3y + c3 = 0 …..(iii)
The radical axis of the above three circles taken in pairs are given by
S1 − S2 ≡ 2x (g1 − g2 ) + 2y( f1 − f2 ) + c1 − c2 = 0 …..(iv)
S2 − S3 ≡ 2x (g2 − g3 ) + 2y( f2 − f3 ) + c2 − c3 = 0 …..(v)
S3 − S1 ≡ 2x (g3 − g1) + 2y( f3 − f1) + c3 − c1 = 0 .....(vi)
Adding (iv), (v) and (vi), we find L.H.S. vanished identically. Thus the three lines are concurrent.
(v) If two circles cut a third circle orthogonally, the radical axis of the two circles will pass through the centre
of the third circle or
The locus of the centre of a circle cutting two given circles orthogonally is the radical axis of the two circles.
Let S1 ≡ x 2 + y2 + 2g1x + 2 f1y + c1 = 0 …..(i)
S2 ≡ x 2 + y 2 + 2g2 x + 2 f2y + c2 = 0 …..(ii)
S3 ≡ x 2 + y 2 + 2g3 x + 2 f3y + c3 = 0 …..(iii)
Since (i) and (ii) both cut (iii) orthogonally, ∴ 2g1g3 + 2 f1 f3 = c1 + c3 and 2g2g3 + 2 f2 f3 = c2 + c3
Subtracting, we get 2g3(g1 − g2) + 2 f3( f1 − f2 ) = c1 − c2 …..(iv)
Now radical axis of (i) and (ii) is S1 − S2 = 0 or 2x (g1 − g 2 ) + 2y ( f1 − f2 ) + c1 − c 2 = 0
Since it will pass through the centre of circle (iii)
∴ − 2g3(g1 − g2 ) − 2 f3( f1 − f2 ) + c1 − c2 = 0 or 2g3(g1 − g2 ) + 2 f3( f1 − f2 ) = c1 − c2 …..(v)
which is true by (iv).
Note : Radical axis need not always pass through the mid point of the line joining the centres of the two
circles.
4.22 Radical Centre.
The radical axes of three circles, taken in pairs, meet in a point, which is called their radical centre. Let the
three circles be S1 = 0 .....(i) , S2 = 0 .....(ii) , S3 = 0 .….(iii) S1=0 L
Let OL, OM and ON be radical axes of the pair sets of circles S2=0
{S1 = 0, S2 = 0}, {S3 = 0, S1 = 0} and {S2 = 0, S3 = 0} respectively. O
Equation of OL, OM and ON are respectively
S1 − S2 = 0 .....(iv) , S3 − S1 = 0 .....(v), S2 − S3 = 0 .....(vi) M S3=0 N
Let the straight lines (iv) and (v) i.e., OL and OM meet in O. The equation of any straight line passing
through O is (S1 − S2) + λ (S3 − S1) = 0 where λ is any constant
For λ = 1 , this equation become S2 − S3 = 0 , which is by (vi), equation of ON.
Thus the third radical axis also passes through the point where the straight lines (iv) and (v) meet.
In the above figure O is the radical centre.
(1) Properties of radical centre A
(i) Co-ordinates of radical centre can be found by solving the equations FE
S1 = S2 = S3 = 0 I
(ii) The radical centre of three circles described on the sides of a triangle BDC
as diameters is the orthocentre of the triangle :
Draw perpendicular from A on BC. ∴ ∠ADB = ∠ADC = π / 2
Therefore, the circles whose diameters are AB and AC passes through D and A. Hence AD is their radical
axis. Similarly the radical axis of the circles on AB and BC as diameter is the perpendicular line from B on CA and
radical axis of the circles on BC and CA as diameter is the perpendicular line from C on AB. Hence the radical axis
of three circles meet in a point. This point I is radical centre but here radical centre is the point of intersection of
altitudes i.e., AD, BE and CF. Hence radical centre = orthocentre.
(iii) The radical centre of three given circles will be the centre of a fourth circle which cuts all the three circles
orthogonally and the radius of the fourth circle is the length of tangent drawn from radical centre of the three given
circles to any of these circles.
Let the fourth circle be (x − h)2 + (y − k)2 = r 2, where (h, k) is centre of this circle and r be the radius. The
centre of circle is the radical centre of the given circles and r is the length of tangent from (h, k) to any of the given
three circles.
Example : 46 The gradient of the radical axis of the circles x 2 + y 2 − 3x − 4y + 5 = 0 and 3x 2 + 3y 2 − 7x + 8y + 11 = 0 is
Solution : (b)
[MP PET 2000]
Example : 47
(a) 1 (b) − 1 (c) − 1 (d) − 2
3 10 2 3
Equation of radical axis is S1 − S2 = 0
S1 ≡ x2 + y2 − 3x − 4y + 5 = 0 , S2 ≡ x2 + y2 − 7 x + 8y + 11 =0
3 3 3
∴ Radical axis is −2x − 20y + 4 = 0 .
Hence, gradient of radical axis = − 1
10
The equations of three circles are x 2 + y 2 − 12x − 16y + 64 = 0, 3x 2 + 3y 2 − 36x + 81 = 0 and x 2 + y 2 − 16x + 81 = 0.
The coordinates of the point from which the length of tangents drawn to each of the three circles is equal [Rajasthan PET 2002]
(a) 33 , 2 (b) (2, 2) (c) 2, 33 (d) None of these
4 4
Solution : (d) The required point is the radical centre of the three given circles
Now, S1 − S2 = 0 ⇒ − 16y + 37 = 0 , S2 − S3 = 0 ⇒ 4 x − 54 = 0 and S3 − S1 = 0 ⇒ − 4 x + 16y + 17 = 0
Solving these equations, we get x = 54 , y = 37 ⇒ x = 27 , y = 37 . Hence the required point is 27 , 37 .
4 16 2 16 2 16
Example : 48 The equation of the circle, which passes through the point (2a, 0) and whose radical axis is x = a with respect to the
2
circle x 2 + y 2 = a 2 , will be [Rajasthan PET 1999]
(a) x 2 + y 2 − 2ax = 0 (b) x 2 + y 2 + 2ax = 0 (c) x 2 + y 2 + 2ay = 0 (d) x 2 + y 2 − 2ay = 0
Solution : (a) Equation of radical axis is x = a ⇒ 2x − a = 0
2
Equation of required circle is x 2 + y 2 − a 2 + λ (2x − a) = 0
It is passes through the point (2a, 0) , ∴ 4a 2 − a 2 + λ (4a − a) = 0 ⇒ λ = − a
∴ Equation of circle is x 2 + y 2 − a 2 − 2ax + a 2 = 0 ⇒ x2 + y2 − 2ax = 0
Co-Axial System of Circles.
A system (or a family) of circles, every pair of which have the same radical
axis, are called co-axial circles. S+λP=0 S+λP=0
(1) The equation of a system of co-axial circles, when the equation of the
radical axis and of one circle of the system are P ≡ lx + my + n = 0 and S+λP=0
S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0 respectively, is S + λP = 0 (λ is an arbitrary S=0 P=0
constant).
(2) The equation of a co-axial system of circles, where the equation of any two circles of the system are
S1+λS2=0 S1+λ( S1+S2)=0 S1+λ( S1–S2)=0
S1+λ( S1–S2)=0 S2=0
S1=0 S2=0 S1=0 S1–S2=0
S1 ≡ x 2 + y 2 + 2g1 x + 2 f1y + c1 = 0 and S2 ≡ x 2 + y 2 + 2g 2 x + 2 f2y + c 2 = 0
Respectively, is S1 + λ (S1 − S2) = 0 , (λ ≠ −1) or S2 + λ1 (S1 − S2) = 0 , (λ1 ≠ −1)
Other form S1 + λS2 = 0, (λ ≠ −1)
(3) The equation of a system of co-axial circles in the simplest form is x 2 + y2 + 2gx + c = 0 , where g is
variable and c is a constant.
Limiting Points.
Limiting points of a system of co-axial circles are the centres of the point circles belonging to the family
(Circles whose radii are zero are called point circles).
(1) Limiting points of the co-axial system : Let the circle is x 2 + y2 + 2gx + c = 0 …..(i)
where g is variable and c is constant.
∴ Centre and the radius of (i) are (−g, 0) and (g 2 − c) respectively. Let g 2 − c = 0 ⇒ g = ± c
Thus we get the two limiting points of the given co-axial system as ( c, 0) and (− c, 0)
Clearly the above limiting points are real and distinct, real and coincident or imaginary according as c>, =, <0
(2) System of co-axial circles whose two limiting points are given : Let (α, β) and (γ , δ ) be the two
given limiting points. Then the corresponding point circles with zero radii are
(x − α)2 + (y − β )2 = 0 and (x − γ )2 + (y − δ )2 = 0
or x 2 + y 2 − 2αx − 2βy + α 2 + β 2 = 0 and x 2 + y 2 − 2γ x − 2δ y + γ 2 + δ 2 = 0
The equation of co-axial system is (x 2 + y 2 − 2αx − 2βy + α 2 + β 2 ) + λ (x 2 + y 2 − 2γ x − 2δy + γ 2 + δ 2 ) = 0
where λ ≠ −1 is a variable parameter.
⇒ x 2(1 + λ) + y2(1 + λ) − 2x (α + γλ) − 2y(β + δλ) + (α 2 + β 2) + λ (γ 2 + δ 2) = 0
or x2 y2 2(α + γλ) x 2 (β + δλ) y (α 2 + β 2) + λ (γ 2 + δ 2) 0
(1 + λ) (1 + λ) (1 + λ)
+ − − + =
Centre of this circle is (α + γλ) , (β + δλ) …..(i)
(1 + λ) (1 + λ)
For limiting point, radius = (α + γλ)2 (β + δλ )2 (α 2 + β 2 ) + λ (γ 2 +δ 2) 0
(1 + λ)2 (1 + λ) (1 + λ)
+ 2 − =
After solving, find λ . Substituting value of λ in (i), we get the limiting point of co-axial system.
(3) Properties of limiting points
(i) The limiting point of a system of co-axial circles are conjugate points with respect to any member of the
system : Let the equation of any circle be x 2 + y2 + 2gx + c = 0 ….(i)
Limiting points of (i) are ( c, 0) and (− c, 0) . The polar of the point ( c, 0) with respect to (i) is
x c + y.0 + g (x + c ) + c = 0 or x c + g (x + c ) + c = 0 or (x + c ) (g + c ) = 0 or x + c = 0 and it
clearly passes through the other limiting point (− c, 0) . Similarly polar of the point (− c, 0) with respect to (i) also
passes through ( c, 0). Hence the limiting points of a system of co-axial circles are conjugate points.
(ii) Every circle through the limiting points of a co-axial system is orthogonal to all circles of the system :
Let the equation of any circle be x 2 + y2 + 2gx + c = 0 …..(i)
where g is a parameter and c is constant. Limiting points of (i) are ( c, 0) and (− c, 0)
Now let x 2 + y2 + 2g′x + 2 f ′y + c′ = 0 …..(ii)
be the equation of any circle. If it passes through the limiting points of (i), then c + 2g′ c + c′ = 0 and
c − 2g′ c + c′ = 0 . Solving, we get c′ = −c and g′ = 0
From (ii), x 2 + y2 + 2 f ′y − c = 0 .....(iii)
where c is constant and f ′ is variable. Applying the condition of orthogonality on (i) and (iii) i.e.,
2g1g2 + 2 f1 f2 = c1 + c2 we find that 2 × g × 0 + 2 × 0 × f ′ = c − c i.e., 0 = 0
Hence condition is satisfied for all values of g′ and f ′ .
Example : 49 The point (2, 3) is a limiting point of a co-axial system of circles of which x 2 + y 2 = 9 is a member. The coordinates of the
other limiting point is given by [MP PET 1993]
(a) 18 , 27 (b) 9 , 6 (c) 18 , − 27 (d) − 18 , − 9
13 13 13 13 13 13 13 13
Solution : (a) Equation of circle with (2, 3) as limiting point is (x − 2)2 + (y − 3)2 = 0
or (x 2 + y 2 − 9) − 4 x − 6y + 22 = 0 or (x 2 + y 2 − 9) − λ (2x + 3y − 11) = 0 represents the family of co-axial circles.
c = λ, 3λ , r= λ2 + 9λ2 − 11λ + 9 . For limiting points r = 0 ⇒ 13λ2 − 44λ + 36 = 0 ⇒ λ = 18 , 2
2 4 13
∴ The limiting points are (2, 3) and 18 , 3 18 or 18 , 27 .
13 2 13 13 13
Example : 50 In the co-axial system of circle x 2 + y 2 + 2gx + c = 0 where g is a parameter, if c > 0. Then the circles are
Solution : (d)
(a) Orthogonal (b) Touching type (c) Intersecting type [Karnataka CET 1999]
(d) Non intersecting type
The equation of a system of circle with its centre on the axis of x is x 2 + y 2 + 2gx + c = 0 . Any point on the radical axis is (0, y1)
Putting, x = 0, y = ± − c
If c is positive (c >0), we have no real point on radical axis, then circles are said to be non-intersecting type.
Image of the Circle by the Line Mirror.
Let the circle be x 2 + y2 + 2gx + 2 fy + c = 0 and line mirror lx + my + n = 0 . In this condition, radius of circle
remains unchanged but centre changes. Let the centre of image circle be (x1, y1).
Slope of C1C2 × slope of lx + my + n = −1 …..(i) (–g,–f) r r
C1 C2
and mid point of C1(−g, − f) and C2(x1, y1) lie on lx + my + n = 0
i.e., l x1 − g + m y1 − f + n = 0 …..(ii) Given circle Image circle
2 2
lx+my+n=0
Solving (i) and (ii), we get (x1, y1)
∴ Required image circle is (x − x1)2 + (y − y1)2 = r 2 , where r = (g 2 + f 2 − c)
Example : 51 The equation of the image of the circle x 2 + y 2 + 16x − 24y + 183 = 0 by the line mirror 4 x + 7y + 13 = 0 is
Solution : (d)
(a) x 2 + y 2 + 32x − 4y + 235 = 0 (b) x 2 + y 2 + 32x + 4y − 235 = 0
(c) x 2 + y 2 + 32x − 4y − 235 = 0 (d) x 2 + y 2 + 32x + 4y + 235 = 0
The given circle and line are x 2 + y 2 + 16x − 24y + 183 = 0 …..(i) and 4 x + 7y + 13 = 0 …..(ii)
Centre and radius of circle (i) are (– 8, 12) and 5 respectively. Let the centre of the image circle be (x1, y1).
Then slope of C1C2 × slope of 4 x + 7y + 13 = −1
⇒ y1 − 12 × − 4 = −1 or 4y1 − 48 = 7x1 + 56 (–8,12) (x1,y1)
x1 + 8 7 C1 5 5 C2
or 7x1 − 4y1 + 104 = 0 …..(iii)
and mid point of C1C2 i.e., x1 − 8 , y1 + 12 lie on 4 x + 7y + 13 = 0 , 4x+7y+13=0
2 2
then 4 x1 − 8 + 7 y1 + 12 + 13 = 0 or 4 x1 + 7y1 + 78 = 0 …..(iv)
2 2
Solving (iii) and (iv), we get (x1, y1 ) = (−16, − 2)
∴ Equation of the image circle is (x + 16)2 + (y + 2)2 = 5 2 or x 2 + y 2 + 32x + 4y + 235 = 0
Some Important Results.
(1) Concyclic points : If A, B, C, D are concyclic then OA.OD = OC.OB , where O′ be the centre of the circle.
(2) Equation of the straight line joining two points α and β on the circle A
x2 + y2 = a2 B
Required equation is x cos α +β + y sin α +β = a cos α −β O O′
2 2 2
D
C
(3) The point of intersection of the tangents at the point P (α) and Q (β) on the circle x 2 + y2 = a2 is
a cos α + β a sin α + β Q(β)
2 2 M
, a α–β
cos α − β cos α − β α a P(α)
2 2 O β
(4) Maximum and Minimum distance of a point from the circle :
Let any point P (x1, y1) and circle x 2 + y2 + 2gx + 2 fy + c = 0 …..(i) A P(x1,y1)
The centre and radius of the circle are P
C (−g, − f) and (g 2 + f 2 − c) respectively.
C(–g,–f)
The maximum and minimum distance from P (x1, y1) to the circle (i) are
PB = CB + PC = r + PC and PA = | CP − CA| = | PC − r | (P inside or outside) B
where r = (g 2 + f 2 − c)
(5) Length of chord of contact is AB = 2LR A S=0
(R2 + L2 ) R
LO
and area of the triangle formed by the pair of tangents and its chord of
R
RL3
contact is = R 2 + L2 LB
Where R is the radius of the circle and L is the length of tangent from P (x1, y1) on S=0. Here L = S1 .
(6) Length of an external common tangent and internal common tangent to two circles is given by
A Lex B
Length of external common tangent Lex = d 2 − (r1 − r2)2 (r1–r2)
and length of internal common tangent
r1 d B′ r2 r2
r1 Lin C2
Lin
r1+r2 A′ r2
Lin = d 2 − (r1 + r2 )2 [Applicable only when d > (r1 + r2) ] A″
where d is the distane between the centres of two circles i.e., |C1C2 | = d and r1 and r2 are the radii of two
circles.
(7) Family of circles circumscribing a triangle whose sides are given by L1 = 0; L2 = 0 and L3 = 0 is given
by L1L2 + λL2L3 + µL3 L1 = 0 provided coefficient of xy = 0 and coefficient of x 2 = coefficient of y2.
Equation of the circle circumscribing the triangle formed by the lines ar x + br y + cr = 0, where r = 1, 2, 3, is
a12 + b12 a1 b1 A
a1x + b1y + c1
a 2 + b22 L1=0 L3=0
2
a2 b2 = 0
a2x + b2y + c2
a 2 + b32 B L2=0 C
3
a3 b3
a3 x + b3y + c3
(8) Equation of circle circumscribing a quadrilateral whose sides in order are represented by the lines
L1 = 0, L2 = 0, L3 = 0 and L4 = 0 is given by A
L1L3 + λL2 L4 = 0 L3=0 D
provided coefficient of x 2 = coefficient of y2 and coefficient of xy = 0
L4=0 L2=0
B
L1=0
C
(9) Equation of the circle circumscribing the triangle PAB is A
(x − x1) (x + g)+ (y − y1) (y + f) = 0 O(0,0)
where O(−g,− f) is the centre of the circle x 2 + y2 + 2gx + 2 fy + c = 0 P(–8,2) B
(10) Locus of mid point of a chord of a circle x 2 + y 2 = a 2 , which subtends an angle α at the centre is
x 2 + y 2 = (a cosα / 2)2
(11) The locus of mid point of chords of circle x 2 + y2 = a2 , which are making right angle at centre is
x2 + y2 = a2
2
(12) The locus of mid point of chords of circle x 2 + y2 + 2gx + 2 fy + c = 0 , which are making right angle at
origin is x 2 + y2 + gx + fy + c / 2 = 0.
(13) The area of triangle , which is formed by co-ordinate axes and the tangent at a point (x1, y1) of circle
x 2 + y 2 = a 2 is a4 / 2x1y1
(14) If a point is outside, on or inside the circle then number of tangents from the points is 2, 1 or none.
(15) A variable point moves in such a way that sum of square of distances from the vertices of a triangle
remains constant then its locus is a circle whose centre is the centroid of the triangle.
(16) If the points where the line a1x + b1y + c1 = 0 and a2 x + b2y + c2 = 0 meets the coordinate axes are
concyclic then a1a2 = b1b2.
Example : 52 If mi , 1 , i = 1, 2, 3, 4 are concylic points, then the value of m1. m2. m3. m4 is [Karnataka CET 2002]
Solution : (a) mi
(d) None of these
Example : 53 (a) 1 (b) – 1 (c) 0
Solution : (a)
Let the equation of circle be x 2 + y 2 + 2gx + 2 fy + c = 0
Since the point mi , 1 lies on this circle
mi
∴ mi 2 + 1 + 2gmi + 2f +c =0 ⇒ mi 4 + 2gmi 3 + cmi 2 + 2 fmi + 1 = 0
mi 2 mi
Clearly its roots are m1, m2 , m3 and m4 , ∴ m1. m2. m3. m4 = product of roots = 1 =1
1
Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X
on the circumference of the circle, then 2r equals [IIT Screening 2001]
(a) PQ . RS (b) PQ + RS (c) 2 PQ . RS (d) PQ 2 + RS 2
2 PQ + RS 2
tanθ = PQ = PQ
PR 2r
S Q
Also tan π −θ = RS X P
2 2r
π/2
i.e. cot θ = RS R θr πC/2-rθ
2r
∴ tanθ . cot θ = PQ . RS
4r 2
⇒ 4r 2 = PQ . RS ⇒ 2r = (PQ) (RS)
Computing and binary operations
Introduction.
The modern digital computer or simply a computer is a general purpose electronic machine which can process
a large amount of information at a very high speed. A computer can perform millions of computations in a few
minutes. It can also perform arithmetical and logical operations.
PRINTER MAGNETIC
OUTPUT TAPE
UNIT MEMORY
DISPLAY ALU
DEVICE
MEMORY
INPUT CONTROL MAGNETIC
UNIT UNIT DISK
MEMORY
PUNCHED PUNCHED CONSOLES MAGNETIC
CARDS PAPER OR DRUM
TAPE TERMINALS MEMORY
BLOCK DIAGRAM OF A GENERAL PURPOSE COMPUTER
A computer has five major components :
(1) Input unit (2) Memory unit (3) Control unit
(4) Arithmetic logical unit (5) Output unit
(1) Input unit : The input unit is the means where the user communicates data or information to the
computer.
(2) Memory unit : The memory unit stores instructions, data and intermediate results. It supplies, when
required, the stored information to the other units of the computer.
(3) Control unit : The control unit controls all the activities in the computer by sending electronic command
signals to other components of the computer.
(4) Arithmetic Logical Units (ALU) : ALU is the unit where the arithmetic and logical (e.g., less than,
greater than) computations are carried out. Control unit and ALU taken together is called Central Processing
Unit (CPU).
(5) Output unit : The output unit receives the stored result from the memory unit converts it into a form. The
user can understand and produces it in the desired format.
A computer may have more than one input and output units. For example, printer and display screen are two
different output units attached to the same computer.
Memory.
Our aim is to see how we can use the computer to solve some problems. For that purpose, it is useful to know
a little more about main memory. From the users point of view, main memory can be thought of as a collection of
compartments (or locations) as shown in fig. (i) Each compartment is assigned a number called its address (starting
with zero as shown in the fig. (ii). The total number of compartments gives us the size of the memory.
01 2
34
Main memor as a collection of compartments (locations)
0 12
Bits in a memory location
Each compartment of memory (as well as a register in ALU) consists of sub-compartments fig. (ii). Each sub-
compartment can store either a zero or a 1. Any information to be stored inside a computer is put using zeros and
1’s. The digits 0 and 1 are called binary digits (bits in short). The acronym bit is formed by taking the letter b from
the word ‘binary’ and the letters i, t from the word ‘digit’. Similarly, we have the acronym dit for decimal digit, hit
for hexadecimal digit etc. The number system that uses only two digits is called binary number system. Computers
use binary number system for computation.
Algorithms.
An algorithm is defined as a finite set of rules, which gives a sequence of operations for solving a specific type
of problem.
In other words, algorithm is a step-by-step procedure for solving problems.
An algorithm has following five important features:
(1) Finiteness (2) Definiteness (3) Completeness (4) Input and (5) Output
(1) Finiteness: An algorithm should always terminate after a finite number of steps.
(2) Definiteness: Each step of algorithm should be precisely defined. This means that the rules should be
consistent and unambiguous.
(3) Completeness : The rules must be complete so that the algorithm can solve all problems of a particular
type for which the algorithm is designed.
(4) Input : An algorithm has certain inputs.
(5) Output : An algorithm has certain outputs which are in specific relation to the inputs.
An important consideration for an algorithm concerns its efficiency. Some algorithms are far more efficient
than others in that, when programmed, one may require fewer steps or perhaps less memory than another and will
therefore, be more satisfactory or economical in actually solving problems on a computer. We shall often deal with
considerations of this type in the subsequent work.
In the development of an algorithm, sequence, selection and repetition (or interaction) play an important role.
(1) Sequence : Suppose that we want to find the value of the expression a3 + 4ab + b2 for given values of a
and b. Algorithm (i.e., step by step procedure) for achieving this will consist of steps given in fig. to be carried out
one after the other. 1. Get the value of a
2. Get the value of b
3. Calculate a3 , call it S
4. Calculate 4ab , call it T
5. Calculate b2 , call it V
6. Find the sum S + T + V , call it M
7. Write the value of M as answer.
Steps of an algorithm to evaluate a 3 + 4ab + b 2 , given the values of a, b
This algorithm, you will agree, is very straightforward, consisting of simple steps which are to be carried out
one after the other. We say that such an algorithm is a sequence of steps, meaning that
(i) At a time only one step of the algorithm is to be carried out.
(ii) Every step of the algorithm is to be carried out once and only once; none is repeated and none is omitted.
(iii) The order of carrying out the steps of the algorithm is the same as that in which they are written.
(iv) Termination of the last step of the algorithm indicates the end of the algorithm.
Here afterwards we shall follow the convention that (i) the successive steps in a sequence will be written on
successive lines and hence (ii) steps will not be necessarily numbered as they are in fig.
(2) Selection: An algorithm which consists of only a sequence, is not sufficient for solving any type of
problem. Let us consider the problem of solving an equation of the type m + nx = r (where m, n, r are given
integers) for integral values of x. We immediately use laws of algebra to find x = (r − m) ÷ n,n ≠ 0 . Let us call an
algorithm that works for only some (not necessarily all) possible sets of input values, a semi-algorithm.
Semi-algorithm (for the above problem) :
Step 1: Get the values of m, r and n.
Step 2: Subtract m from r, call this difference b.
Step 3: Divide b by n; print this result as the value of x.
The above steps are certainly efficient, As an example, let m = 9,n = 5 and r = 24 , in which case we have
9 + 5x = 24 . Then in step 2, we have 24 −9 =b i.e., b = 15 and in step 3, we have b = 15 = 3 , and so we print x = 3.
n 5
The above steps have two fatal flaws, however. First, if n equals 0, then either m = r and x can have any
integral value, or m ≠ r and no solution is possible i.e., there is no integer x which may satisfy the given equation.
Second, if there is a non-zero remainder when b is divided by n then again there is no integer x which may satisfy
the given equation. So we must modify our algorithm to deal with all such situations as may arise. Given below is
the modified algorithm which suits all the possible situations that may arise.
Step 1 : Get the value of m, n and r
Step 2 : If n = 0 and m = r
then go to step 7
else go to step 3
Step 3 : If n = 0 and m ≠ r
then go to step 6
else go to step 4
Step 4 : Subtract m from r, call this difference b (i.e.,
b =r−m)
Step 5: Divide b by n;
If there is a remainder
then go to step 6
else print the value of b , which is the
n
required value of x.
Step 6: Print ‘No integer satisfies this equation’. Stop
Step 7 : Print ‘Any integer satisfies this equation’. Stop
Algorithm to solve an equation of the form
m + nx = r, where m, n, r, x are integers
The above algorithm provides the person or computer that will execute the algorithm with an ability to choose
the step to be carried out depending upon the values of m, n and r (and subsequently, the value of b). This ability is
called selection. The power of selection is that it permits that different paths could be followed, depending upon the
requirement of the problem, by the one who executes the algorithm.
In the above algorithm, selection is expressed by using the special words ‘if, ‘then’, ‘else’. Further, it may be
noted that all that is written using these special words (once) constitutes one step. Note the way it is written. Nothing
appears below the word ‘if’ till that step is over. This is known as indentation. The words ‘then’ and ‘else’ come with
exactly same indentation with respect to the word ‘if’.
(3) Iteration or Repetition : In forming an algorithm certain steps are required to be repeated before
algorithm terminates after giving an answer. This is known as iteration or repetition.
Let us consider the problem of finding the just prime number greater than a given positive integer. The
following list of steps shows the step by step procedure to be followed for solving the problem.
Consider the given integer
add 1 to it
test new number for primeness
S if it is prime,
then write it down and stop
else add 1 to it
S test new number for primeness
if it is prime,
then write it down and stop
else add 1 to it
We see in the above procedure that the steps
“add 1 to it S test new number for primeness
test new number for primeness
if it is prime if.......
Algorithm for finding a prime number
greater then a given positive integer
then write it down and stop ”
are repeated again and again till (after a finite number of repetitions) we get a prime number and print it. If
this sequence (which involves a decision also) is denoted by S, then S is repeated again and again till, we get the
result and print the result. This is technically known as iteration or repetition. The way of writing adopted in fig.
poses a problem as we do not know the number of times S is repeated. This number depends upon the given
positive integer. The difficulty presented above is overcome by introducing a new way of writing iterations in
algorithms. The algorithm shown in fig. is (in new ways) then written as shown below
Consider the given number
repeat
add 1 to it
until the new number is prime
write the new number
Or
Consider the given number
add 1 to it
while the new number is not prime
do add 1 to it
write the new number
Two different ways of writing
iteration occuring in fig.
Pseudo language.
The languages used by human beings for talking and writing among themselves are called natural languages.
Expression in a natural language can be ambiguous.
Computer, being a machine, requires that there should be no ambiguity at all when we give instructions to it.
Languages used to communicate with a computer are known as programming languages.
We shall use meaningful mnemonic variable names, assignment, symbol ← constructions employing If-then-
else, Repeat-until, While-Do and other constructions employing word For for writing an algorithm. We shall
also require instructions to input data in an algorithm as well as instructions to output computed results from an
algorithm. All these will constitute our language to present any algorithm. This language will not resemble in total
with any actual existing programming language but will have desirable characteristics of a good programming
language. We shall call it a pseudo-language.
Pseudo language Constructs.
(1) If-then-else construct : The general form of this construct which is used to provide selection of actions is
If condition
then step 1
else step 2
when an instruction using this type of construct is executed, condition determines which of the step 1 and step
2 is to be executed. If condition is true, step 1 is executed, otherwise (i.e., if condition is not true) step 2 is executed.
A particular case of this construct does not have the word else. The general form of this construct is
If condition
then step
Clearly when this form is used, no action is taken when condition is false, and step is executed when condition
is true. In other words the following constructs are equivalent as they do the same thing.
If condition If condition
then step and then step
else do nothing
(2) Repeat until construct : The general form of this construct is
Repeat
Part of the algorithm
until condition
This construct is used when repetition of certain action is required. Note that “Part of algorithm” is always
executed at least once as the condition is tested at the end, unlike the 'while-do' construct where the condition is
tested in the beginning.
(3) While-do-construct : This construct is an alternative to the 'repeat-until' construct. The general form of
this construct, which is also used to provide repetition of instruction is
while condition
do T
Where T is a sequence of instructions. When this construct is executed, condition is evaluated first. If the
condition is true, the sequence T of instructions is executed and the condition is evaluated again and so on. If the
condition is false, execution of T is skipped and the execution of algorithm proceeds with the portion that appears
after T. Thus condition is tested again and again till it is false. Every execution of T modifies some variables in the
algorithm and eventually after some repetitions, the condition becomes false. This completes the execution of the
'while-do-construct', the execution proceeds to the portion appearing after T.
(4) For construct : When we know in advance how many times a part of the algorithm is to be executed we
use 'for construct', whose general form is
For identifier = initial value to test value by increment Do S.
The word, For, To, By and Do are reserved words for this construct. Initial value gives the starting value that
the identifier should take, when the S is executed. The value of identifier is increased by the increment after each
execution. The execution of S continues until the value of identifier exceeds the test value.
Example: 1 An algorithm must terminate in [DCE 1997]
Solution: (c) (a) One iteration [DCE 1995]
Example: 2
Solution: (d) (b) One step (d) None of these
Example: 3
Solution: (c) (c) Finite number of steps
Example: 4
Solution: (c) (d) Finite number of steps but sometimes in infinite number of steps
Example: 5
Solution: It is obvious.
Example: 6 The WHILE-DO control structure executes the loop at least
Solution
(a) Thrice (b) Twice (c) Once
Example: 7
Solution: It is obvious.
Comment:
The control structure IF-THEN is a [DCE 1994]
(a) Multiple selection (b) Double selection (c) Single selection (d) None of these
1 time
It is obvious. [DCE 1996]
REPEAT-UNTIL control operation executes the loop at least (d) None of these
(a) 3 times (b) 2 times (c)
It is obvious.
Write an algorithm to find the first prime number greater than given number using the fact that even integers (except 2) are
not prime
Step I get n
Step II If n is even
then m ← n − 1
else m ← n
Step III repeat m ← m + 2
test m for primeness
until m is prime
Step IV Output m
Write an algorithm to find n ! for given n
Step I get n
Step II If n < 0
then output “factorial is not defined”
Step III If n = 0
then Fact ← 1
Step IV
Step V Fact ← 1
For I = 1 to n
do Fact ← Fact * I
I ← I +1
Step VI Output Fact
Write an algorithm to multiply two matrices
Step I get A, B
A (i, j) and B(i, j) are m × n and n × p matrices respectively,
Step II For i = 1 to m
do
For j = 1 to p
do
C (i, j) ← 0
For k = 1 to n
do
C(i, j) ← C(i, j) + A(i, k)* B(k, j)
Step III Output C
Comment : C = C (i, j) is an m × p matrix.
Example: 8
Write an algorithm to find the solution of a system of simultaneous linear equations.
Solution:
a1 x + a2y + a3 z = d1
b1x + b2y + b3z = d2
c1x + c2y + c3z = d3
Step I get a (1), a(2), a(3), d(1), b(1),
b(2), b(3), d(2), c (1), c(2), c (3), d(3), ε
Step II I ← 1, X(1) ← 0, Y(1) ← 0, Z(1) ← 0
Step III Repeat
X(I + 1) ← 1 [d(1) − a(2)Y(I) − a (3) Z (1)]
a(1)
Y(I + 1) ← 1 [d(2) − b(1)X(I ± 1) − b(3)Z(1)]
b(2)
Z(I + 1) ← c 1 [d(3) − c(1)
(3)
Until I ← I +1
| X(I) − X(I − 1)|< ε
| Y(I) − Y(I − 1)|< ε
| Z(I) − Z(I − 1)|< ε
Step IV Output X(I), Y(I), Z(I) .
Flow Charts (Presentation of Algorithm)
A graphic representation of an algorithm is called a ‘flow-chart’. A flow chart constitutes a schematic and
pictorial representation of the sequence of steps, which are to be executed in solving a problem.
A flow-chart consists of some boxes linked by arrows. In each box, some instruction to be carried out is
mentioned. Arrows on the lines connecting the boxes indicate the direction, in which we should proceed.
The boxes are of different shapes. Each particular shape is associated with a specific type of instruction as
shown in fig.
Flow-chart conventions:
While drawing a flow-chart, the following conventions are observed.
(i) The general direction of flow is from left to right and from top to bottom.
(ii) Only one flow-line should leave a process symbol.
(iii) Only one flow line should enter a decision box and atleast two lines must leave it.
(iv) A flow line that goes in upward direction, completes and iteration (or repetition) or a loop.
Basic operations and flow-charts:
The three basic operations are: (1) Sequence (2) Selection (3) Iteration
The selection of a flow-chart corresponding to an iteration, or the Repeat-Until construct or the While-Do
construct gives rise to a cycle, usually called a loop. There are two types of loops :
(i) When an operation is repeated, a fixed number of times, whatever the value of the variables involved may
be, then the corresponding section of the flow diagrams gives rise to a fixed loop.
(ii) When the number of times an iteration is to be carried out depends upon the values of the variables, then
the corresponding section of the flow diagrams gives rise to variable loop. This loop is also known as backward
jump. Examples of Use
BASIC SYMBOLS
Start Stop
TERMINAL
Read New Write the
Value for X value of
INPUT OUT PUT
SYMBOLS
This flow line This block
is entered
can be omitted
if X = Y 15 times
each pass
COMMENT
ANNOTATION
1 If these appear on a flow 1A
chart they represent the same
CONNECTOR
X←Z+4 X=
A←B×C
B2 − C2
PROCESS
No Is Yes Is X
X=Y <=>
X< Y X>
DECISION
X=
FLOW CHART SYMBOLS
Example: 9 Write an algorithm and flowchart to obtain H C F of two given positive integers using Euclid's algorithm
Solution: We know Euclid's algorithm. Therefore we can express it in pseudo language :
START get M, N
OBTAIN M, N. comment M, N are two given positive integers, M > N
NUM ← M
DEN ← N
QUOT ← integral part of NUM/DEN
REM ← NUM – QUOT * DEN
while REM ≠ 0
do NUM ← DEN
DEN ← REM
QUOT ← integral part of NUM/DEN
REM ← NUM – QUOT * DEN
output DEN
comment HCF is the value of denominator when remainder is zero.
NUM ← M
DEN ← N DEN ← REM
QUOT ← INTEGRAL PART
OF RATIO NUM / DEN
REM ← NUM – QUOT *
DEN
IS NO
REM = 0? NUM ← DEN
Yes STOP
OUTPUT DEN Flow Chart
Number system.
(1) Decimal system : Number system which we use in our daily life is the decimal system. In decimal system
we use the digits namely 0, 1, 2,.......8, 9 and with the help of these 10 digits we are able to write any rational
number. The decimal system is a place-value system, meaning thereby that the value represented by a digit
depends upon the place of the digit within the numeral. The value assigned to consecutive places in the decimal
system are 104 ,103,......100 ,10−1,10−2..... (from left to right)
Example : Number 3864. 342 can be written as
3864.342 = 3 × 103 + 8 × 102 + 6 × 101 + 4 × 100 + 3 × 10–1 + 4 × 10–2 + 2 × 10–3
As ten basic symbols are used for representing the numbers, ten is called the base of the system and the
system is called base-ten system or decimal system.
(2) Binary number system : The number system for which the base is two is called the binary system. In
this system numbers are represented with the help of two basic symbols namely 0 and 1. The values assigned to
consecutive places in the system are (when expressed in the decimal system)..... 24 , 23 , 22 , 21, 20 , 2−1, 2−2.... where
20 place is the unit place. The binary numeral can be converted into the decimal numeral and vice-versa.
(3) Octal number system : As the name implies this is base eight (23 ) system. The numerals are written
with the help of eight basic symbols namely 0, 1, 2,......,7. The value (expressed in the decimal system) assigned to
consecutive places are ....... 8 3 , 8 2, 81, 8 0 , 8 −1, 8 −2..... , where 8 0 place is the unit's place. The procedures for
converting a decimal numeral into an octal numeral and the other way round are similar to the procedures
discussed in connection with binary system.
(4) Hexadecimal system : As the name implies, this is the base sixteen system. The numerals are written
with the help of sixteen symbols, namely 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Note that the symbol A
represents ten (in decimal system). Similarly B, C, D, E, F represent respectively the numbers 11, 12, 13, 14 and 15.
The value assigned to consecutive places are ........, 162 ,161,160 ,16−1..... (as expressed in decimal system), where
160 place is unit's place.
Example: 10 74.1875 in binary is (b) 1001010.0011 (c) 1001010.0101 [DCE 2001]
Solution: (b)
(a) 10010010.0011 (d) 1001100.0101
For integral part
2 74 −
2 37 0
2 18 1
290
241
220
10
1
= (1001010)2
For fractional part
0.1875 = (0.0011)2
×2
0.3750
×2
0×.75020
1.5000
×2
1.0000
Therefore 74.1875 in binary is (1001010.0011)2
Example: 11 (1101101)2 is (b) 121 (c) 109 (d) 92 [DCE 2001]
Solution: (c) (a) 81 (d) 82 [DCE 1999]
Example: 12
Solution: (c) (1101101)2 = 1× 26 + 1× 25 + 0 × 24 + 1× 23 + 1× 22 + 0 × 21 + 1× 20
= 64 + 32 + 0 + 8 + 4 + 0 + 1 = 109.
What is the octal equivalent of binary number 111001
(a) 69 (b) 70 (c) 71
(111001)2 =
111 001
1 × 22 + 1 × 21 + 20 0 × 22 + 0 × 21 + 1 × 20
71
= (71)8
Example: 13 The decimal equivalent of (264)8 is [DCE 1995]
Solution: (a)
Example: 14 (a) 180 (b) 170 (c) 166 (d) None of these
Solution: (b)
(264)8 = (.......)10 [DCE 1994]
= 2 × 82 + 6 × 81 + 4 × 80 = 128 + 48 + 4 = 180. (d) None of these
What is the hexadecimal equivalent of decimal number 785.
(a) 1A2 (b) 311 (c) AB5
(785)10 = (........)16
16 785
16 49 1
31
3
Thus (785)10 = (311)16 .
Computing and binary operations
Definition.
A binary operation on a non-empty set A is a mapping which associates with each ordered pair (a, b) of
elements of A, a uniquely defined element c ∈ A. This is a mapping from the product set A × A to A. Symbolically,
a map : A × A → A, is called a binary operation on the set A.
The image of the element (a, b) ∈ A × A is denoted by a * b. If a set A is closed with respect to the
composition, then we say that * is a binary operation on the set A.
Let, a ∈ N, b ∈ N ⇒ a + b ∈ N for all a, b ∈ N.
Multiplication on N is also a binary operation, since a ∈ N, b ∈ N ⇒ a × b ∈ N for all a, b ∈ N
But subtraction on N is not a binary operation, since 3 ∈ N, 5 ∈ N but 3 – 5 = – 2 ∉ N.
Note : It is obvious that addition as well as multiplication are binary operations on each one of the sets Z
(of integer), Q (of rational number), R (of real number) and C (of all complex number).
Subtraction is a binary operation on each of the sets Z, Q, R and C. But it is not binary operation on N.
Division is not a binary operation on any of sets N, Z, Q, R and C.
Types of Binary Operation
(1) Commutative binary operation : A binary operation * on a set S is said to be commutative if
a * b = b * a for all a, b ∈ S
Addition and multiplication are commutative binary operations on Z but the subtraction is not a commutative
binary operation, since 2 – 3 ≠ 3 – 2.
(2) Associative binary operation : A binary operation * on a set S is said to be associative if
(a * b) * c = a * (b * c) for all a, b, c ∈ S
Addition and multiplication are associative binary operations on N, Z, Q, R and C. But subtraction is not an
associative binary operation on Z, Q, R and C.
(3) Distributive binary operation : Let * and o be two binary operations on a set S. Then * is said to be
(i) Left distributive over o if a * (b o c) = (a * b) o (a * c) for all a, b, c ∈ S;
(ii) Right distributive over o if (b o c) * a = (b * a) o (c * a) for all a, b, c ∈ S.
If * is both left and right distributive over o, then * is said to be distributive over o.
Example : The multiplication (⋅) on Z is distributive over addition (+) on Z, since
a ⋅ (b + c) = a ⋅ b + a ⋅ c and (b + c) ⋅ a = b ⋅ a + c ⋅ a for all a, b, c ∈ Z.
But addition is not distributive over multiplication.
Identity and Inverse elements
(1) Identity element : Let * be a binary operation on a set S. An element e ∈ S is said be an identity
element for the binary operation * if a * e = a = e * a for all a ∈ S.
For addition on Z, 0 is the identity element, since a + 0 = a = 0 + a for all a ∈ Z.
For multiplication on R, 1 is the identity element, since 1 × a = a = a × 1 for all a ∈ R.
(2) Inversible element for a binary operation with identity : An element a of a set A is said to be
inversible for a binary operation * with identity e if ∃ b ∈ A such that a * b = e = b * a.
Also, then b is said to be an inverse of a and is denoted by a–1. The inversible elements in A are also called the
units in A. The identity element is always inversible and is its own inverse, since e * e = e * e = e. Thus e–1 = e.
Composition Table
A binary operation on a finite set can be completely described by means of a table known as a composition
table. Let S = {a1, a2,....., an} be a finite set and * be a binary operation on S. Then the composition table for * is
constructed in the manner indicated below.
We write the elements a1, a2, ….. ,an of the set S in the top horizontal row and the left vertical column in the
same order. Then we put down the element ai * aj at the intersection of the row headed by ai (1 ≤ i ≤ n) and the
column headed by aj(1 ≤ j ≤ n) to get the following table.
* a1 a2 ….. ai ….. aj ….. an
a1 a1 * a1 a1 * a2 ….. a1 * ai ….. a1 * aj ….. a1 * an
a2 a2 * a1 a2 * a2 ….. a2 * ai ….. a2 * aj ….. a2 * an
ai ai * a1 ai * a2 ….. ai * ai ….. ai * aj ….. ai * an
aj aj * a1 aj * a2 ….. aj * ai ….. aj * aj ….. aj * an
an an * a1 an * a2 ….. an * ai ….. an * aj ….. an * an
From the composition table we infer the following results :
(1) If all the entries of the table are elements of set S and each element of S appears once and only once in
each row and in each column, then the operation is a binary operation. Sometimes we also say that the binary
operation is well defined which means that the operation * associates each pair of elements of S to a unique
element of S, i.e. S is closed under the operation *.
(2) If the entries in the table are symmetric with respect to the diagonal which starts at the upper left corner of
the table and terminates at the lower right corner, we say that the binary operation is commutative on S, otherwise
it is said to be not commutative on S.
(3) If the row headed by an element say aj, coincides with the row at the top and the column headed by aj
coincides with the column on extreme left, then aj is the identity element for the binary operation * on S.
(4) If each row except the topmost row or each column except the left most column contains the identity
element then every element of S is invertible with respect to *. To find the inverse of an element say aj, we consider
row (or column) headed by ai. Then we determine the position of identity element e in this row (or column). If e
appears in the column (or row) headed by aj, then ai and aj are inverse of each other.
It should be noted that the composition table is helpless to determine associativity of the binary operation.
This has to be verified for each possible trial.
Example: 1 Let S be a finite set containing n elements. Then the total number of binary operations on S is [EAMCET 1992]
Solution: (c)
Example: 2 (a) nn (b) 2n2 (c) nn2 (d) n2
Solution: (c)
Example: 3 Since a binary operation on S is a function from S × S to S, therefore the total number of binary operations on S is the
Solution: (a)
total number of functions from S × S to S, which is nn2 .
Example: 4
Solution: (a) The identity element for the binary operation * defined by a * b = ab , a, b ∈ Q0 (the set of all non-zero rational numbers) is
2
(a) 1 (b) 0 (c) 2 (d) None of these
Let e be the identity element for the binary operation * on Q0 defined by a*b = ab
2
Then, a * e = a = e * a for all a ∈ Q0
⇒ ae =a for all a ∈ Q0 ⇒ e = 2.
2
Let z be the set of integers and o be a binary operation on z defined as a o b = a + b − ab for all a, b ∈ z . The inverse of an
element a (≠ 1) ∈ z is
(a) a (b) a (c) a −1 (d) None of these
a −1 1−a a
Let e be the identity element for the binary operation o defined on z given by a o b = a + b − ab
Then a o e = a = e o a for all a ∈ z
⇒ a + e − ae = a for all a ∈ z ⇒ e(1 − a) = 0 for all a ∈ z ⇒ e = 0 .
So, 0 is the identity element for the binary operation o and z.
Let x be the inverse of a ∈ z . Then, a o x = x o a = 0
⇒ a + x − ax = 0 ⇒ x(1 − a) = −a ⇒ x = a (a ≠ 1)
a −1
Thus, a is the inverse of a (≠ 1) ∈ z .
a −1
* is defined on the set of real numbers by a * b = 1 + ab . Then the operation * is [CET 1991]
(a) Commutative but not associative (b) Associative but not commutative
(c) Neither commutative nor associative (d) Both commutative and associative
We have a * b = 1 + ab = 1 + ba = b * a
So, * is commutative on R.
For any, a, b, c ∈ R, we have (a * b) * c = (1 + ab) * c = 1 + (1 + ba)c = 1 + c + abc
and a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc
∴ (a * b)* c ≠ a * (b * c)
So, * is not associative on R.
Correlation and regression
Introduction .
“If it is proved true that in a large number of instances two variables tend always to fluctuate in the same or in
opposite directions, we consider that the fact is established and that a relationship exists. This relationship is called
correlation.”
(1) Univariate distribution : These are the distributions in which there is only one variable such as the
heights of the students of a class.
(2) Bivariate distribution : Distribution involving two discrete variable is called a bivariate distribution. For
example, the heights and the weights of the students of a class in a school.
(3) Bivariate frequency distribution : Let x and y be two variables. Suppose x takes the values
x1, x 2 ,....., xn and y takes the values y1, y2,....., yn, then we record our observations in the form of ordered pairs
(x1, y1), where 1 ≤ i ≤ n,1 ≤ j ≤ n . If a certain pair occurs fij times, we say that its frequency is fij .
The function which assigns the frequencies fij ’s to the pairs (xi , y j ) is known as a bivariate frequency distribution.
Example: 1 The following table shows the frequency distribution of age (x) and weight (y) of a group of 60 individuals
Solution:
x (yrs) 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65
y (yrs.)
45 – 50 258 3 0
50 – 55 1 3 6 10 2
55 – 60 0 2 5 12 1
Then find the marginal frequency distribution for x and y.
Marginal frequency distribution for x
x 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65
19 25 3
f3 10
Marginal frequency distribution for y
y 45 – 50 50 – 55 55 – 60
f 18 22 20
Covariance .
Let (x1, xi );i = 1,2,....., n be a bivariate distribution, where x1, x 2 ,....., xn are the values of variable x and
y1, y2 ,....., yn those of y. Then the covariance Cov (x, y) between x and y is given by
1 n x)(yi or 1 n where, 1 n and 1 n are
n n n i=1 n i=1
(xi (x i yi
i=1 i=1
∑ ∑ ∑ ∑Cov(x, y) = y) y yi
− − Cov (x, y) = − x y) x = xi =
means of variables x and y respectively.
Covariance is not affected by the change of origin, but it is affected by the change of scale.
Example: 2 ∑ ∑ ∑Covariance (x, y) between x and y, if x = 15 , y = 40 , x.y = 110, n = 5 is [DCE 2000]
(a) 22 (b) 2 (c) – 2 (d) None of these
∑ ∑Solution: (c) Given, x = 15, y = 40
∑ x.y = 110, n = 15
∑ ∑ ∑ ∑ ∑ ∑We know that, 1 n 1 n 1 n 1 x.y 1 x 1 y
Cov(x, y) = n i =1 xi .yi − n i =1 xi n i =1 yi = n − n n
= 1 (110) − 15 40 = 22 − 3×8 = −2 .
5 5 5
Correlation .
The relationship between two variables such that a change in one variable results in a positive or negative
change in the other variable is known as correlation.
(1) Types of correlation
(i) Perfect correlation : If the two variables vary in such a manner that their ratio is always constant, then
the correlation is said to be perfect.
(ii) Positive or direct correlation : If an increase or decrease in one variable corresponds to an increase or
decrease in the other, the correlation is said to be positive.
(iii) Negative or indirect correlation : If an increase or decrease in one variable corresponds to a decrease
or increase in the other, the correlation is said to be negative.
(2) Karl Pearson's coefficient of correlation : The correlation coefficient r(x, y) , between two variable x
Cov(x, y) or Cov(x, y) , r(x, y) = ∑ ∑ ∑nnxiyi − n xi n yi
Var(x) Var(y) σ xσ y i=1 i=1 i=1
and y is given by, r(x, y) =
2 2
∑ ∑ ∑ ∑nnxi2 − n xi n n yi 2 − n yi
i=1 i=1 i=1 i=1
r = ∑(x − x)(y − y) = ∑ dxdy .
∑(x − x)2 ∑(y − y)2 ∑ dx 2 ∑ dy2
∑ dxdy − ∑ dx.∑ dy
(3) Modified formula : r = n , where dx = x − x;dy = y − y
dx 2 dy 2
(∑ ) (∑ )
∑ ∑
dx 2 − n dy2 − n
Also rxy = Cov(x, y) = Cov(x, y) .
σ xσ y var(x). var(y)
Example: 3 For the data
Solution: (a)
x: 4 7 8 3 4
y: 5 8 6 3 5
The Karl Pearson’s coefficient is [Kerala (Engg.) 2002]
(a) 63 (b) 63 (c) 63 (d) 63
94 × 66 94 66
Take A = 5, B = 5
xi yi ui = xi − 5 vi = yi − 5 ui2 vi2 uivi
4 5 –1 0 1 0 0
782399 6
863111 3
3 3 –2 –2 4 4 4
4 5 –1 0 0 0 0
Total ∑ ∑ ∑ ∑ ∑ui = 1
vi = 2 ui2 = 19 vi2 = 14 uivi = 13
∑ ∑(∑ ) ∑∑ ∑ (∑ ) r(x,y) = uivi − 1 ui vi 13 − 1 × 2 = 63 .
n 5
=
1 ui 2 1 vi 2 12 22 94 66
ui2 − n vi − n 19 − 5 14 − 5
Example: 4 Coefficient of correlation between observations (1, 6),(2, 5),(3, 4), (4, 3), (5, 2), (6, 1) is
Solution: (b)
Example: 5 [Pb. CET 1997; Him. CET 2001; DCE 2002]
Solution : (d) (a) 1 (b) – 1 (c) 0 (d) None of these
Example: 6 Since there is a linear relationship between x and y, i.e. x + y = 7
Solution: (b)
∴ Coefficient of correlation = – 1.
The value of co-variance of two variables x and y is − 148 and the variance of x is 272 and the variance of y is 131 .
3 3 3
The coefficient of correlation is
(a) 0.48 (b) 0.78 (c) 0.87 (d) None of these
We know that coefficient of correlation = Cov(x, y)
σ x .σ y
Since the covariance is – ive.
∴ Correlation coefficient must be – ive. Hence (d) is the correct answer.
The coefficient of correlation between two variables x and y is 0.5, their covariance is 16. If the S.D of x is 4, then the S.D.
of y is equal to [AMU 1988, 89, 90]
(a) 4 (b) 8 (c) 16 (d) 64
We have, rxy = 0.5, Cov(x, y) = 16 . S.D of x i.e., σ x = 4 , σ y = ?
We know that, r(x, y) = Cov(x, y)
σ x .σ y
0.5 = 16 ; ∴ σy =8.
4 .σ y
Example: 7 ∑ ∑ ∑For a bivariate distribution (x, y) if x = 50 , y = 60 , xy = 350, x = 5, y = 6 variance of x is 4, variance of y is 9,
then r(x, y) is [AMU 1991; Pb. CET 1998; DCE 1998]
(a) 5/6 (b) 5/36 (c) 11/3 (d) 11/18
Solution: (a) ∑x = x ⇒ 5 = 50 ⇒n = 10 .
n
n
∑∴ Cov (x,y) = xy − x.y = 350 − (5)(6) = 5.
n 10
∴ r(x, y) = Cov(x, y) = 5 9 = 5 .
σ x .σ y 4. 6
Example: 8 A, B, C, D are non-zero constants, such that
(i) both A and C are negative. (ii) A and C are of opposite sign.
If coefficient of correlation between x and y is r, then that between AX + B and CY + D is
(a) r (b) – r (c) A r (d) − A r
C C
Solution : (a,b) (i) Both A and C are negative.
Now Cov(AX + B,CY + D) = AC Cov.(X,Y)
σ AX +B =| A|σ x and σ CY +D =|C|σ y
Hence ρ (AX + B,CY + D) = AC.Cov ( X, Y ) = | AC | ρ (X, Y) = ρ (X,Y) = r, ( AC > 0)
(| A |σ x )(| C |σ y ) AC
(ii) ρ(AX + B, CY + D) = | AC | ρ(X, Y) , ( AC < 0)
AC
= AC ρ(X, Y ) = −ρ(X,Y) = −r .
− AC
Rank Correlation .
Let us suppose that a group of n individuals is arranged in order of merit or proficiency in possession of two
characteristics A and B.
These rank in two characteristics will, in general, be different.
For example, if we consider the relation between intelligence and beauty, it is not necessary that a beautiful
individual is intelligent also.
Rank Correlation ∑6 d 2
: ρ =1− , which is the Spearman's formulae for rank correlation coefficient.
n(n 2 − 1)
∑Where d 2 = sum of the squares of the difference of two ranks and n is the number of pairs of
observations.
∑ ∑ ∑ ∑Note : We always have, di = (xi − yi) = xi − yi = n(x) − n(y) = 0 , (x = y)
If all d's are zero, then r = 1 , which shows that there is perfect rank correlation between the variable
and which is maximum value of r.
If however some values of xi are equal, then the coefficient of rank correlation is given by
6 ∑ d 2 + 1 (m 3 − m)
12
r = 1−
n(n2 − 1)
where m is the number of times a particular xi is repeated.
Positive and Negative rank correlation coefficients
Let r be the rank correlation coefficient then, if
• r > 0 , it means that if the rank of one characteristic is high, then that of the other is also high or if
the rank of one characteristic is low, then that of the other is also low. e.g., if the two
characteristics be height and weight of persons, then r > 0 means that the tall persons are also
heavy in weight.
• r = 1 , it means that there is perfect correlation in the two characteristics i.e., every individual is getting
the same ranks in the two characteristics. Here the ranks are of the type (1, 1), (2, 2),....., (n, n).
• r < 1 , it means that if the rank of one characteristics is high, then that of the other is low or if the
rank of one characteristics is low, then that of the other is high. e.g., if the two characteristics be
richness and slimness in person, then r < 0 means that the rich persons are not slim.
• r = −1 , it means that there is perfect negative correlation in the two characteristics i.e, an
individual getting highest rank in one characteristic is getting the lowest rank in the second
characteristic. Here the rank, in the two characteristics in a group of n individuals are of the type
(1, n), (2,n − 1),....., (n,1) .
• r = 0 , it means that no relation can be established between the two characteristics.
Important Tips
• If r = 0 , the variable x and y are said to be uncorrelated or independent.
• If r = −1 , the correlation is said to be negative and perfect.
• If r = +1, the correlation is said to be positive and perfect.
• Correlation is a pure number and hence unitless.
• Correlation coefficient is not affected by change of origin and scale.
• If two variate are connected by the linear relation x + y = K , then x, y are in perfect indirect correlation. Here r = −1 .
• If x, y are two independent variables, then ρ (x + y, x − y) = σ 2 − σ 2 .
x y
σ 2 + σ 2
x y
∑ ∑(∑ ) ∑∑ ∑ (∑ )• r(x,y) = uivi − 1 ui. vi vi 2 , where ui = xi − A, vi = yi − B .
n
ui2 − 1 ui 2 vi2 − 1
n n
Example: 9 Two numbers within the bracket denote the ranks of 10 students of a class in two subjects
Solution: (b)
(1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1). The rank of correlation coefficient is [MP PET 1996]
Example : 10
Solution: (c) (a) 0 (b) – 1 (c) 1 (d) 0.5
∑d2
Rank correlation coefficient is r = 1 − 6. n(n2 − 1) , Where d = y − x for pair (x, y)
∴ ∑ d2 = 92 + 72 + 52 + 32 + 12 + (−1)2 + (−3)2 + (−5)2 + (−7)2 + (−9)2 = 330
Also n = 10 ; ∴ r = 1 − 6 × 330 = −1 .
10(100 − 1)
Let x1, x2, x3,....., xn be the rank of n individuals according to character A and y1, y2,......, yn the ranks of same
individuals according to other character B such that xi + yi = n + 1 for i = 1, 2,3,....., n . Then the coefficient of rank
correlation between the characters A and B is
(a) 1 (b) 0 (c) – 1 (d) None of these
xi + yi = n + 1 for all i = 1, 2,3,....., n
Let xi − yi = di . Then, 2xi = n + 1 + di ⇒ di = 2xi − (n + 1)
nn n
∑ ∑ ∑∴ di2 = [2xi − (n + 1)]2 = [4 xi2 + (n + 1)2 − 4 xi(n + 1)]
i=1 i=1 i =1
∑ ∑ ∑n
di 2
i =1
n n xi = 4 n(n + 1)(2n + 1) + (n)(n + 1)2 − 4(n + 1) n(n + 1)
6 2
=4 xi2 + (n)(n + 1)2 − 4(n + 1)
i =1 i =1
∑n di 2 = n(n2 − 1) .
3
i =1
∑∴ r = 1 − 6 di2 = 1 − 6(n)(n2 − 1) i.e., r = −1 .
n(n2 − 1) 3(n)(n2 − 1)
Regression
Linear Regression .
If a relation between two variates x and y exists, then the dots of the scatter diagram will more or less be
concentrated around a curve which is called the curve of regression. If this curve be a straight line, then it is
known as line of regression and the regression is called linear regression.
Line of regression: The line of regression is the straight line which in the least square sense gives the best fit
to the given frequency.
Equations of lines of Regression .
(1) Regression line of y on x : If value of x is known, then value of y can be found as
y − y = Cov(x, y) (x − x) or y − y = r σ y (x − x)
σx
σ 2
x
(2) Regression line of x on y : It estimates x for the given value of y as
x−x = Cov(x, y) (y − y) or x−x =r σx (y − y)
σy
σ 2
y
(3) Regression coefficient : (i) Regression coefficient of y on x is byx = rσ y = Cov(x, y)
σx
σ 2
x
(ii) Regression coefficient of x on y is b xy = rσ x = Cov(x, y) .
σy
σ 2
y
Angle between Two lines of Regression .
Equation of the two lines of regression are y − y = byx (x − x) and x − x = bxy(y − y)
We have, m1 = slope of the line of regression of y on x = byx = r. σ y
σx
m2 = Slope of line of regression of x on y = 1 = σy
bxy r.σ x
σy − rσ y (σ y − r 2σ y )σ (1 − r 2)σ xσ y
rσ x
∴ tanθ = ± m2 − m1 = ± σx = ± x = ± .
1 + m1m2
1 + rσ y . σy rσ 2 + rσ 2 r(σ 2 +σ 2 )
rσ x x y x y
σx
Here the positive sign gives the acute angle θ , because r 2 ≤ 1 and σ x ,σ y are positive.
∴ tanθ = 1 − r2 . σ xσ y .....(i)
r +σ
σ 2 2
x y
Note : If r = 0 , from (i) we conclude tanθ = ∞ or θ = π / 2 i.e., two regression lines are at right angels.
If r = ±1 , tanθ = 0 i.e., θ = 0 , since θ is acute i.e., two regression lines coincide.
Important points about Regression coefficients bxy and byx .
(1) r = byx.bxy i.e. the coefficient of correlation is the geometric mean of the coefficient of regression.
(2) If byx > 1 , then bxy < 1 i.e. if one of the regression coefficient is greater than unity, the other will be less than unity.
(3) If the correlation between the variable is not perfect, then the regression lines intersect at (x, y) .
(4) byx is called the slope of regression line y on x and 1 is called the slope of regression line x on y.
b xy
(5) byx + bxy > 2 byxbxy or byx + bxy > 2r , i.e. the arithmetic mean of the regression coefficient is greater than
the correlation coefficient.
(6) Regression coefficients are independent of change of origin but not of scale.
(7) The product of lines of regression’s gradients is given by σ 2 .
σ y
2
x
(8) If both the lines of regression coincide, then correlation will be perfect linear.
(9) If both byx and bxy are positive, the r will be positive and if both byx and bxy are negative, the r will be negative.
Important Tips
• If r = 0 , then tanθ is not defined i.e. θ = π . Thus the regression lines are perpendicular.
2
• If r = +1 or −1 , then tan θ = 0 i.e. θ = 0. Thus the regression lines are coincident.
• If regression lines are y = ax + b and x = cy + d , then x = bc + d and y= ad + b .
1 − ac 1 − ac
• If byx, bxy and r ≥ 0 then 1 (bxy + byx ) ≥ r and if bxy, byx and r≤0 then 1 (bxy + byx ) ≤ r .
2 2
• Correlation measures the relationship between variables while regression measures only the cause and effect of relationship between the variables.
• If line of regression of y on x makes an angle α , with the +ive direction of X-axis, then tanα = byx .
• If line of regression of x on y makes an angle β , with the +ive direction of X-axis, then cot β = bxy .
Example : 11 The two lines of regression are 2x − 7y + 6 = 0 and 7x − 2y + 1 = 0 . The correlation coefficient between x and y is
Solution: (b)
[DCE 1999]
Example: 12
Solution: (c) (a) – 2/7 (b) 2/7 (c) 4/49 (d) None of these
Example: 13
Solution: (c) The two lines of regression are 2x − 7y + 6 = 0 .....(i) and 7x − 2y + 1 = 0 ......(ii)
If (i) is regression equation of y on x, then (ii) is regression equation of x on y.
We write these as y= 2 x + 6 and x = 2 y − 1
7 7 7 7
∴ byx = 2 , bxy = 2 ; ∴ byx .bxy = 4 < 1 , So our choice is valid.
7 7 49
∴ r2 = 4 ⇒ r = 2 . [ byx > 0, bxy > 0 ]
49 7
Given that the regression coefficients are – 1.5 and 0.5, the value of the square of correlation coefficient is
[Kurukshetra CEE 2002]
(a) 0.75 (b) 0.7
(c) – 0.75 (d) – 0.5
Correlation coefficient is given by r 2 = byx .bxy = (−1.5)(0.5) = −0.75 .
∑ ∑ ∑ ∑In a bivariate data x = 30, y = 400 , x2 = 196, xy = 850 and n = 10 .The regression coefficient of y on x is
[Kerala (Engg.) 2002]
(a) – 3.1 (b) – 3.2 (c) – 3.3 (d) – 3.4
Cov(x, y) = 1 ∑ xy − 1 ∑ x. ∑ y = 1 (850) − 1 (30)(400) = −35
n n2 10 100
2 1 x2 ∑x 2 196 30 2
x n n 10 10
Var(x) = σ = ∑ − = − = 10.6
byx = Cov(x, y) = −35 = – 3.3.
Var (x) 10.6
Example: 14 If two lines of regression are 8x − 10y + 66 = 0 and 40x − 18y = 214 , then (x,y) is [AMU 1994; DCE 1994]
Solution: (b)
Example: 15 (a) (17, 13) (b) (13, 17) (c) (– 17, 13) (d) (– 13, – 17)
Solution: (a) Since lines of regression pass through (x, y) , hence the equation will be 8x − 10y + 66 = 0 and 40x − 18y = 214
Example: 16
On solving the above equations, we get the required answer x = 13, y = 17 .
Solution: (c)
The regression coefficient of y on x is 2 and of x on y is 4 . If the acute angle between the regression line is θ , then
3 3
tanθ = [DCE 1995]
(a) 1 (b) 1 (c) 2 (d) None of these
18 9 9
2 4 bxy − 1 4 − 3 1
3 3 byx 3 2 18
byx = , bxy = . Therefore, tanθ = = = .
bxy 4/3
1+ byx 1+ 2/3
If the lines of regression of y on x and x on y make angles 30o and 60o respectively with the positive direction of X-axis,
then the correlation coefficient between x and y is [MP PET 2002]
(a) 1 (b) 1
2 2
(c) 1 (d) 1
3 3
Slope of regression line of y on x = byx = tan 30o = 1
3
Slope of regression line of x on y = 1 = tan 60o = 3
bxy
⇒ bxy = 1 . Hence, r = bxy.byx = 1 1 = 1.
3 3 3 3
Example: 17 If two random variables x and y, are connected by relationship 2x + y = 3 , then rxy = [AMU 1991]
(a) 1 (b) – 1 (c) – 2 (d) 3
Since 2x + y = 3
Solution: (b)
∴ 2x + y = 3 ; ∴ y − y = −2(x − x) . So, byx = −2
Also x − x = − 1 (y − y) , ∴ bxy = − 1
2 2
∴ rx2y = byx .bxy = (−2) − 1 = 1 ⇒ rxy = −1 . ( both byx ,bxy are –ive)
2
Standard error and Probable error.
(1) Standard error of prediction : The deviation of the predicted value from the observed value is known
∑as the standard error prediction and is defined as Sy = (y − yp)2
n
where y is actual value and yp is predicted value.
In relation to coefficient of correlation, it is given by
(i) Standard error of estimate of x is Sx = σ x 1 − r 2 (ii) Standard error of estimate of y is Sy = σ y 1 − r 2 .
(2) Relation between probable error and standard error : If r is the correlation coefficient in a sample of
n pairs of observations, then its standard error S.E. (r) = 1 − r 2 and probable error P.E. (r) = 0.6745 (S.E.)=
n
0.6745 1 − r 2 . The probable error or the standard error are used for interpreting the coefficient of correlation.
n
(i) If r < P.E.(r), there is no evidence of correlation.
(ii) If r > 6P.E.(r) , the existence of correlation is certain.
The square of the coefficient of correlation for a bivariate distribution is known as the “Coefficient of
determination”.
Example: 18 If Var(x) = 21 and Var(y) = 21 and r = 1 , then standard error of y is
Solution: (a) 4
(a) 0 (b) 1 (c) 1 (d) 1
2 4
Sy = σ y 1 − r 2 = σ y 1 − 1 = 0 .
Dynamics
(1) Displacement : The displacement of a moving point is its change of position. To know the displacement
of a moving point, we must know both the length and the direction of the line joining the two positions of the
moving point. Hence the displacement of a point involves both magnitude and direction i.e., it is a vector quantity.
(2) Speed : The speed of a moving point is the rate at which it describes its path. The speed of a moving
particle or a body does not give us any idea of its direction of motion; so it is a scalar quantity. In M.K.S. or S.I.
system, the unit of speed is metre per second (m/s).
Average speed : The average speed of moving particle in a time-interval is defined as the distance travelled
by the particle divided by the time interval.
If a particle travels a distance s in time-interval t, then Average speed = s
t
Note : The average speed of a particle gives the overall “rapidity” with which the particle moves in the given
interval of time.
Instantaneous speed : The instantaneous speed or simply speed of a moving particle is defined as the rate
of change of distance along its path, straight or curved.
If s is the distance travelled by a particle along its path, straight or curved, in time t, then
Instantaneous speed = ds or speed at time t= ds
dt dt
Note : The average speed is defined for a time interval and the instantaneous speed is defined at a
particular instant.
Velocity and Acceleration.
(1) Velocity : The velocity of a moving point is the rate of its displacement. It is a vector quantity.
Let a particle starting from the fixed point O and moving along the straight line OX describes the distance
OP = x in time t. If the particle describes a further distance PQ = δx in time δx
x PQ
δt , then δx is called the average velocity of the particle in time δt . O X
δt
And lim δx dx =v is called the instantaneous velocity (or simply
δt = dt
δt →0
velocity) of the particle at time t.
Note : Average velocity in time t is the mean of the initial and final velocity.
(2) Acceleration : The rate of change of velocity of a moving particle is called its acceleration. It is a vector quantity.
The acceleration of a moving point at time t at distance x is given by, f = v2 − v1 ⇒ f = dv = d2x = v. dv
t dt dt 2 dx
In M.K.S. or S.I. system, the unit of acceleration is m/sec2.
It should be noted that a > 0 if v increases with time and a < 0 if v decreases with time. A negative
acceleration is called retardation.
Clearly, retardation means decrease in the velocity.
Example: 1 The average speed of a bicycle over a journey of 20 km; if it travels the first 10 km. at 15 km/hr and the second 10 km. at
Solution: (a)
10 km/hr, is
(a) 12 km/hr (b) 10 km/hr (c) 15 km/hr (d) None of these
Time taken by the bicycle to travel the first 10 km = 10 hour = 2 hour
15 3
Time taken by the bicycle to travel the second 10 km = 10 hour = 1 hour
10
Total distance travelled = 10 + 10 = 20 km
Total time taken = 2 +1 = 5 hour.
3 3
∴ Average speed of bicycle = Distance = 20 = 60 = 12 km/hr.
Time 5/3 5
Example: 2 If a particle moves in a straight line according to the formula, x = t3 − 6t 2 − 15t , then the time interval during which the
velocity is negative and acceleration is positive, is
(a) [0, 5] (b) (2,∞) (c) (2, 5) (d) None of these
Solution: (c) We have x = t 3 − 6t 2 − 15t
∴ v = dx = 3t 2 − 12t − 15 and a = d2x = 6t − 12
dt dt 2
Now v < 0 and a > 0 ⇒ 3(t 2 − 4t − 5) < 0 and 6(t − 2) > 0 ⇒ −1 < t < 5 and t > 2 ⇒ t ∈ (2,5) .
Example: 3 A particle moves in a fixed straight path so that S = 1 + t . If v is the velocity at any time t, then its acceleration is
Solution: (a)
Example: 4 (a) − 2v3 (b) − v3 (c) − v2 (d) 2v3
Solution: (c) We have s = 1 + t
⇒ ds = 2 1 ⇒ d2s = −1 = −1 1 t 3 ⇒ d2s = −1 2 ds 3 ⇒ d2s = −2v3 .
dt 1+ t dt 2 4(1 + t)3 / 2 4 1+ dt 2 4 dt dt 2
Two particles are moving with uniform velocities u and v respectively along X and Y axes, each directed towards the
origin. If the particles are at distances a and b from the origin, the time at which they will be nearest to each other will be
equal to
(a) au (b) bu (c) au + bv (d) au
u2 + v2 u2 + v2 u2 + v2 bv
Let the two particles be at A and B respectively at t = 0 and at time t their positions be P and Q respectively.
Then, AP = ut and BQ = vt
∴ OP = a − ut and OQ = b − vt Y
Now, PQ2 = OP 2 + OQ2 ⇒ PQ2 = (a − ut)2 + (b − vt)2 B vt
Q
⇒ d (PQ 2 ) = −2u(a − ut) − 2v(b − vt) and d2 (PQ 2 ) = 2(u2 + v2)
dt dt 2
b – vt
For maximum or minimum value of PQ2 , we must have
d (PQ 2 ) = 0 ⇒ −2u(a − ut) − 2v(b − vt) = 0 ⇒ t = au + bv O P ut A X
dt u2 + v2
Clearly, d2 (PQ 2 ) > 0 for all t. Hence, PQ is least at t = au + bv .
dt 2 u2 + v2
Resultant and Components of Velocities.
(1) Resultant of velocities : The resultant of two velocities possessed B C
by a particle is given by the law of parallelogram of velocities. v
R
Parallelogram law of velocities : If a moving particle has two
simultaneous velocities represented in magnitude and direction by the two
sides of a parallelogram drawn from an angular point, then their resultant is α u
represented in magnitude and direction by the diagonal of the parallelogram θ
passing through that point. A
O
Let a particle at O has simultaneously two velocities u and v represented in magnitude and direction by OA
and OB respectively of the parallelogram OACB. The diagonal OC represents the resultant velocity.
Let R be the resultant velocity.
If ∠AOB = α and ∠AOC = θ , then the magnitude of their resultant velocity is R = u2 + v2 + 2uv cosα
and, the direction of this resultant velocity makes an angle θ with the direction of u such that tan θ = v sin α
u + v cos α
Case I : When α = 0 , then Rmax = u + v
Case II : When α = π , then Rmin =|u − v|
Case III : When α = π / 2 , i.e., u and v are at right angle to each other, then
R= u2 + v2 and θ = tan −1 v .
u
Case IV : When u = v , then R = 2u cos α and θ = α .
2 2
Note : The angle made by the direction of the resultant velocity with the direction of v is given by
tan−1 v u sin α α .
+ u cos
If the direction of resultant velocity R makes an angle θ with the direction of u , then
sin θ = v sin α and cosθ = u + v cosα
R R
(2) Components of a velocity in two given directions : If components of a velocity V in two given
directions making angles α and β with it, then by the law of parallelogram of velocities the required components
of V are represented by AB and AC = BD along AX and AY respectively. Y
AB BD AD i.e., AB = BD = AD CD
sin ADB = sin BAD = sin ABD sin β sinα sin(α + β ) Vβ
∴ AB = AD. sin β β ) and BD = AD. sinα β ) β π–(α+β)
sin(α + sin(α + α B
X
A
Hence, the components of velocity V in the directions making angle α and β are V sin β and V sinα respectively.
sin(α + β ) sin(α + β )
Important Tips
• If components of velocity V are mutually perpendicular, then β = 90o − α . Thus components of V are V cosα along AX and V sin α
along AY.
Example: 5 Drops of water falling from the roof of the tunnel seems from the window of the train to be falling from an angle making
tan−1 1 from the horizontal. It is known that their velocities are 24 decimeter/sec. Assuming the resistance of air negligible,
2
what will be the velocity of the train [UPSEAT 1999]
(a) 42 decimeter/sec. (b) 48 decimeter/sec. (c) 45 decimeter/sec. (d) 44 decimeter/sec.
Solution: (b) Let the velocity of train is v decimeter/sec.
Example: 6
Let the real velocity of drops in magnitude and direction may be represented by OB and the velocity of train in opposite
direction is represented by OD, then OB = 24 , OD = BC = v
Completing parallelogram OBCD, OC will represent the relative velocity of v Oθ
drops. D Direction Train
If the drops are seen in a direction making an angle θ with the horizontal, i.e., of drop 90–θ
∠COD = θ , then θ = tan −1 1 or tanθ = 1 or cotθ =2 24
2 2
CB
OB BC 24 v
In ∆OBC , from sine formula, sinθ = sin(90 − θ ) or sinθ = cosθ or
v = 24 cotθ = 24 × 2 = 48 decimeter/second.
If u and v be the components of the resultant velocity w of a particle such that u = v = w , then the angle between the
velocities is
(a) 60o (b) 150o (c) 120o (d) 30o
Solution: (a) Let α be the angle between the components of the resultant velocity w.
Example: 7
Then w2 = u2 + v2 + 2uv cosα ⇒ u2 = 2u2(1 + cosα) ⇒ 4 cos 2 α =1 ⇒ cos α = 1 ⇒ α = 60o .
2 2 2
A man swims at a speed of 5 km/hr. He wants to cross a canal 120 metres wide, in a direction perpendicular to the
direction of flow. If the canal flows at 4 km/hr, the direction and the time taken by the man to cross the canal are
(a) tan−1 3 , 2.4 min . (b) π − tan−1 3 ,144 sec . (c) tan−1 1 , 100 sec . (d) None of these
4 4 2
Solution: (b) Suppose the man swims in a direction making an angle α with the direction of current.
Since the man wants to cross the canal in a direction perpendicular to the direction of flow.
Therefore, tan π = 4 5 sinα ⇒ 4 + 5 cosα = 0 ⇒ cosα = −4 D
2 + 5 cosα 5
⇒ α = cos−1 −4 ⇒ α =π − cos−1 4 ⇒ α =π − tan−1 3 BC
5 5 4
120 m
Let V be the resultant velocity, then
5 km/hr V
V 2 = 42 + 52 + 2 × 4 × 5 cosα α
V2 = 16 + 25 + 40 × −4 =9 ⇒ V = 3km / hr. O 4 km/hrA
5
Time taken by the man to cross the canal = 120 hr = 1 hr = 60 min . = 2.4 min. = 144 sec .
3 ×1000 25 25
Relative Velocity.
Let A and B are two bodies in motion. The velocity of A relative to B is the velocity with which A appears to
move as viewed by B i.e., velocity of A relative to B = Resultant of velocity of A and reversed velocity of B.
Apparent velocity of A as seen from B means velocity of A relative to B.
The relative velocity of A with respect to B is also called the velocity of A
relative to B and is denoted by V AB and V AB = V A − V B . V→A V→AB →
If two bodies A and B are moving with velocities of magnitudes VA and → VB
VB
VB respectively inclined at an angle α with each other, then
V AB = V 2 + VB2 − 2VA VB cos α
A
If the direction of the relative velocity of A with respect to B makes an angle θ with the direction of velocity of
B, then tanθ = VA sinα
VB − VA cosα
Case I : When the particles A and B move parallel to each other in the same direction with velocities u and v
respectively. Here α = 0 , then VAB = (u − v) in the direction of A A VA = u
VBA = (v − u) in the direction of B B VB = v
Case II : When the particles A and B move parallel to each other in opposite directions with velocities u and
v respectively. Here α = π , then VAB = (u + v) in the direction of A. A VA = u
VBA = (v + u) in the direction of B. B VB = v
Note : True velocity of A = Resultant of the relative velocity of A with respect to B and the true velocity of
B
i.e., V A = V AB + V B [Since V AB = V A − V B ]
Example: 8 A train of length 200 m travelling at 30 m/sec. overtakes another of length 300 m travelling at 20 m/sec. The time taken by
Solution: (b)
the first train to pass the second is [MNR 1988; UPSEAT 2000]
Example: 9
Solution: (c) (a) 30 sec. (b) 50 sec. (c) 10 sec. (d) 40 sec.
Distance covered by the first train to pass the second = (200 + 300)m = 500 metres.
Since both the trains are travelling in the same direction. ∴ Velocity of first train relative to second = (30 – 20) = 10 m/sec.
∴ Time taken by the first train to pass the second = 500 = 50 second.
10
To a man running at a speed of 20 km/hr, the rain drops appear to be falling at an angle of 30o from the vertical. If the
rain drops are actually falling vertically downwards, their velocity in km/hr. is [UPSEAT 2001; MNR 1994]
(a) 10 3 (b) 10 (c) 20 3 (d) 40
Velocity of rain relative to man = Actual velocity of rain – Velocity of man 20(Rev O 20(Act.
Resolving horizontally and vertically,
V cos 30o = u and V sin 30o = 20 i.e., V = 20 , ∴ V = 40 30°
2
∴40 3 =u ⇒ u = 20 3 km / hr . Actua
2
V(Rel.)
Example: 10 A man is walking at the rate of 3 km/hr and the rain appears to him falling vertically with a velocity of 3 3 km / hr . If the
Solution: (d)
actual direction of the rain makes an angle θ with vertical, then θ = [MNR 1995]
(a) 15o (b) 30o (c) 45o (d) 60o
Let V M denotes the velocity of man. Then, VM =|V M |= 3km / hr. It appears to the man that rain is falling vertically at the
rate of 3 3km / hr . This means that the relative velocity of rain with respect to O Vm= 3km/hr. A
man is 3 3km / hr , in a direction perpendicular to the velocity of the man. Let θ
V RM denote the relative velocity of rain with respect to man and V R be the
velocity of rain. VRm=3√3Km/hr VR
Let OA = V R , OB = V RM . Complete the parallelogram OACB.
Then, the diagonal OC represents true velocity VR of the rain. BC
∴ VR2 = VM2 + VR2M + 2VM .VRM cos 90o [Using w2 = v2 + u2 + 2uv cos θ ]
⇒ VR = 32 + (3 3)2 + 2 × 3 × 3 3 × 0 = 6km / hr.
Let θ be the angle between the direction of velocity of man and velocity of rain. Then,
tanθ = VRM sin 90o ⇒ tanθ = 33 = 3
VM + VRM cos 90o 3
Thus, the actual velocity of the rain is 6 km/hr in a direction making an angle of 60o with the direction of motion of the man.
Example: 11 The particles start simultaneously from the same point and move along two straight lines, one with uniform velocity u and
the other from rest with uniform acceleration f . Let α be the angle between their directions of motion. The relative
velocity of the second particle w.r.t. the first is least after a time [AIEEE 2003]
(a) u sinα (b) f cosα (c) u sinα (d) u cosα
f u f
Solution: (d) After t, velocity = f × t A
u
VBA = ft + (−u)
B
VBA = f 2t 2 + u2 − 2 fut cos α Oα
f
For max. and min., d (VB2A ) = 2f 2t − 2 fu cosα = 0
dt
t = u cos α
f
Rectilinear motion with uniform Acceleration.
A point or a particle moves in a straight line, starting with initial velocity u, and moving with constant
acceleration f in its direction of motion. If v be its final velocity at the end of time t and s be its distance at the
instant, from its starting point, then the equations describing the motion of the particle are,
s
t=0 t X
P
Velocity = u
(i) v = u + ft (ii) s = ut + 1 ft 2 (iii) v2 = u2 + 2 fs
2
(iv) If sn is the distance travelled by the particle in nth second, then sn =u+ 1 f(2n − 1)
2
Important Tips
• If a particle moves in a straight line with initial velocity u m/sec. and constant acceleration f m/sec2, then distance travelled in t seconds
is given by, s = ut + 1 ft 2 = 1 [2ut + ft 2] = 1 [u + (u + ft)] t = u + v . t, where v = u + ft
2 2 2 2
= (Average velocity) × t {Average velocity = u+v }
2
Example: 12 A body is moving in a straight line with uniform acceleration. It covers distances of 10 m and 12 m in third and fourth
Solution: (d) seconds respectively. Then the initial velocity in m/sec. is
[UPSEAT 1994]
(a) 2 (b) 3 (c) 4 (d) 5
Let initial velocity is u m/sec. and acceleration is f m / sec2 .
So, u+ 1 f(2 × 3 − 1) = 10 or u+ 5 f = 10 .....(i) and u+ 1 f(2 × 4 − 1) = 12 or u+ 7 f = 12 ......(ii)
2 2 2 2
Subtracting (i) from (ii), we get 0 + 2 f = 2 or f = 2m / sec2
2
Substituting value of f in equation (i) , u+ 5 × 2 = 10 or u + 5 = 10 or u = 5m / sec .
2
Example: 13 Two trains A and B, 100 kms apart, are travelling to each other with starting speed of 50 km/hr for both. The train A is
Solution: (d) accelerating at 18 km/hr2 and B is decelerating at 18 m/h2. The distance where the engines cross each other from the initial
Example: 14 position of A is [UPSEAT 2000]
Solution: (c)
Example: 15 (a) 50 kms (b) 68 kms (c) 32 kms (d) 59 kms
Solution: (d)
Let engine of train A travel x km and cross engine of train B in t hours.
u = 50km / hr ,
Acceleration f = 18km / hr 2 , so s = 50t + 1 ×18t 2 , s = 50t + 9t2 .....(i)
2
So distance travelled by engine of train B will be (100 − s) in t hour.
u = 50km / hr., Acceleration = –18 km/hr
∴ 100 − s = 50t − 1 ×18t 2 ⇒ 100 − s = 50t − 9t2 …..(ii)
2
Adding (i) and (ii), we get 100 = 100t,∴ t = 1 hour. From (i), s = 50 × 1 + 9 × 12 = 50 + 9 = 59kms .
A particle is moving with a uniform acceleration. If during its motion, it moves x, y and z distance in pth, qth and rth seconds
respectively, then [MNR 1987]
(a) (q − r)x + (r − p)y + (p − q)z = 1 (b) (q − r)x + (r − p)y + (p − q)z = −1
(c) (q − r)x + (r − p)y + (p − q)z = 0 (d) (q + r)x + (r + p)y + (p + q)z = 0
Let u be the initial velocity and f be the acceleration of the particle.
Then s pth = u + 1 f (2 p − 1) = x , sqth = u + 1 f (2q − 1) = y , srth = u + 1 f (2r − 1) = z
2 2 2
Now, multiplying (i) by (q − r), (ii) by (r − p) , (iii) by (p − q) and adding, we get x(q − r) + y(r − p) + z(p − q) = 0 .
A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with
constant acceleration f and in the second part with constant retardation r. The value of t is given by
[AIEEE 2003]
(a) 2s 1 + 1 (b) 2s (c) 2s(f + r) (d) 2s 1 + 1
f r f r
1 + 1
f r
Portion OA, OB corresponds to motion with acceleration ‘f’ and retardation ‘r’ respectively.
Area of ∆OAB = s and OB = t . Let OL = t1, LB = t2 and AL = v vA
s = 1 OB.AL = 1 t.v and v= 2s
2 2 t
Also, f = v , t1 = v = 2s and r = v , t2 = v = 2s v
t1 f tf t2 r tr
t = t1 + t2 = 2s + 2s ; t = 1 + 1 2s ⇒ t= 2s 1 + 1 . O t1 L t2 B
tf tr f r t f r t
Motion Under Gravity.
When a body is let fall in vacuum towards the earth, it will move vertically downward with an acceleration
which is always the same at the same place on the earth but which varies slightly from place to place. This
acceleration is called acceleration due to gravity. Its value in M.K.S. system is 9.8 m / sec 2 , in C.G.S. system
981cm / sec 2 and in F.P.S. system 32 ft / sec 2 . It is always denoted by g.
(1) Downward motion: If a body is projected vertically downward from a point at a height h above the
earth’s surface with velocity u, and after t second its velocity becomes v, the equation of its motion are v = u + g t ,
h = ut + 1 gt 2 , v2 = u2 + 2gh , st =u + 1 g (2t − 1). In particular, if the body starts from rest or is simply let fall
2 2
or dropped, then, v = gt , h = 1 gt 2 , v2 = 2gh . ( u = 0)
2
(2) Upward motion : When a body be projected vertically upward from a point on the earth’s surface with
an initial velocity u and if the direction of the upward motion is regarded as +ve, the direction of acceleration is –ve
and it is, therefore, denoted by – g. The body thus moves with a retardation and its velocity gradually becomes
lesser and lesser till it is zero. Thus, for upward motion, the equations of motion are,
v = u − gt , h = ut − 1 gt 2 , v2 = u2 − 2gh , st =u − 1 g (2t − 1).
2 2
(3) Important deductions
(i) Greatest height attained : Let H be the greatest height. From the result, v 2 = u 2 − 2gh .
We have, 0 = u2 − 2gH ∴ H = u2 . Hence greatest height (H) = u2 .
2g 2g
(ii) Time to reach the greatest height : Let T be the time taken by the particle to reach the greatest height.
From the result, v = u − gt
We have, 0 = u − gT i.e., T = u . Therefore time to reach the greatest height (T) = u
g g
(iii) Time for a given height : Let t be the time taken by the body to reach at a given height h. Then,
h = ut − 1 gt 2 ⇒ gt 2 − 2ut + 2h = 0 . Hence, t1 = u − u2 − 2gh and
2 g
t2 = u + u2 − 2gh A
g
Clearly t1 and t2 are real; if u2 ≥ 2gh . If u2 = 2gh , P
t1 h u
then t1 = t2 = u , which is the time taken to reach the highest point.
g O
So, let u2 > 2gh .
The lesser value of t1 gives the time when the body is going up i.e., it is time from O to P and the larger value
t2 gives the total time taken by the body to reach the highest point and then coming back to the given point, i.e., it is the
time from O to A and then A to P.
(iv) Time of flight : It is the total time taken by the particle to reach the greatest height and then return to the
starting point again. When the particle returns to the starting point, h = 0 . Therefore, from the result, h = ut − 1 gt 2 .
2
We have, 0 = ut − 1 gt 2 or t = 0 or 2u
2 g
t = 0 corresponds to the instant with the body starts, and t = 2 u/g corresponds to the time when the particle
after attaining the greatest height reaches the starting point. ∴ Time of flight = 2u .
g
Example: 16 A stone is dropped from a certain height which can reach the ground in 5 sec. If the stone is stopped after 3 seconds of its
fall and then time taken by the stone to reach the ground for the remaining distance is [MNR 1985; UPSEAT 2000]
(a) 2 seconds (b) 3 seconds (c) 4 seconds (d) None of these
Solution: (c) Let distance travelled in 5 sec. is h1 metre. ∴ h1 = 0 + 1 g × 5 × 5 or h1 = 25 g
2 2
Let distance travelled in 3 seconds is h2 metre. So, h2 = 0 + 1 g × 3 × 3 or h2 = 9 g
2 2
So remaining distance h = h1 − h2 or h= 25 g − 9 g or h = 16 g = 8 g .
2 2 2
Let time taken by the stone to reach the ground for the remaining distance 8g is t second.
∴ 8g =0+ 1 gt 2 or t2 = 16, ∴ t = 4 sec.
2
Example: 17 A man in a balloon, rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 seconds after the balloon is let go
from the ground. The greatest height above the ground reached by the ball is [MNR 1986]
(a) 14.7 m (b) 19.6 m (c) 9.8 m (d) 24.5 m
Solution: (a) Velocity after 2 seconds = (4.9)2 = 9.8 m/sec.
Example: 18 Distance covered in 2 seconds = 1 ft 2 = 1 (4.9)(2)2 = 9.8m
Solution: (b) 2 2
Again after the release of ball, velocity of the ball = 9.8 m/sec.
v = 0 ; Using v2 = u2 − 2gh , we get 0 = (9.8)2 − 2(9.8)h ⇒ h= 9.8 = 4.9 metre.
2
Hence greatest height attained = 9.8 + 4.9 = 14.7 m.
A particle is dropped under gravity from rest from a height h(g = 9.8m / sec 2) and then it travels a distance 9h in the last
25
second. The height h is [MNR 1987; UPSEAT 2000]
(a) 100 metre (b) 122.5 metre (c) 145 metre (d) 167.5 metre
Sn = u+ 1 f(2n − 1) = 1 g(2n − 1) (u = 0, f = g)
2 2
9h = 1 g(2n − 1) and h= 1 gn2
25 2 2
∴ 9 . 1 gn2 = 1 g(2n − 1) ⇒ 9 n2 = (2n − 1) ⇒ 9n2 = 50n − 25
25 2 2 25
⇒ 9n2 − 50n + 25 = 0 ⇒ n = 50 ± 2500 − 900 = 50 ± 40 ; n = 5 or 5
18 18 9
Since 5 < 1 , ∴ rejecting this value. We have n=5 ; ∴ h= 1 (9.8) × 25 = 122.5 metre .
9 2
Laws of Motion.
(1) Newton’s laws of motion
(i) First law : Every body continues in its state of rest or of uniform motion in a straight line except when it is
compelled by external impressed forces to change that state.
(ii) Second law : The rate of change of momentum is proportional to the impressed force, and takes place in
the direction of the straight line in which the force acts.
d (mv) ∝ F ⇒ m dv ∝ F ⇒ m dv = KF ⇒ F = mf (for K = 1)
dt dt dt
Force = mass × acceleration. The S.I. unit of force is Newton and C.G.S. unit of force is dyne.
1 Newton = 1 kg-m/sec2, 1 Dyne = 1 gm-cm/sec2
(iii) Third law : To every action there is an equal and opposite reaction, or the actions and reaction are
always equal and opposite.
Note : 1 Newton = 105 Dynes.
The action and reaction do not act together on the same body or the same part of the body.
(2) Weight : The weight of a body is the force with which it is attracted by the earth towards its centre.
For a body of mass m, the weight W is given by W = mg (By Newton's second law of motion)
Note : Let W1 and W2 be the weights of two bodies of masses m1 and m2 respectively at a place on the
earth then, W1 = m1g and W2 = m2 g ⇒ W1 = m1
W2 m2
1 gm.wt. = g dynes = 981 dynes
1 kg. wt. = g Newtons = 9.81 N.
(3) Momentum of a body : It is the quantity of motion in a body and is equal to the product of its mass (m) and
velocity (v) with which it moves. Thus, momentum of the body is mv. The units of momentum are gm-cm/sec or kg-m/sec.
Momentum = Mass × Velocity = m.v.
(4) Impulse of a body : The impulse of a force in a given time is equal to the product of the force and the
time during which it acts. The impulse of a force F acting for a time t is therefore F.t.
Example: 19 A train weighing W tons is moving with an acceleration f ft/sec2. When a carriage of weight w tons is suddenly detached
Solution: (c)
from it, then the change in the acceleration of the train is
(a) Wf ft / sec2 (b) W ft / sec 2 (c) wf ft / sec 2 (d) w w ft / sec 2
W −w W −w W −w W−
Mass of train = W × 2240 lbs.
∴ Pull of the engine = W × 2240 × f Poundals
When a carriage of mass W tons is detached, mass of the train = (W – w) tons = (W – w) . 2240 lbs
Pull is the same as before i.e., W × 2240 × f Poundals.
∴ New acceleration = f1 = W × 2240 × f
(W − w)2240
f1 = Wf ft / sec 2
W −w
Change in acceleration = f1 − f = Wf − f = W −W + w f = wf .
W −w W −w W−
w
Example: 20 A hockey stick pushes a ball at rest for 0.01 second with an average force of 50 N. If the ball weighs 0.2 kg, then the
Solution: (b) velocity of the ball just after being pushed is [UPSEAT 1998]
Example: 21
(a) 3.5 m/sec. (b) 2.5 m/sec. (c) 1.5 m/sec. (d) 4.5 m/sec.
If v m/sec. is the velocity of the ball just after being pushed, then 50 × 0.01 = 0.2 × v [ Impulse = change of momentum]
v = 2.5 m/sec.
A mass m is acted upon by a constant force P lb.wt., under which in t seconds it moves a distance of x feet and acquires a
velocity v ft/sec. Then x is equal to [Roorkee 1994]
(a) gP (b) mg (c) gt 2 (d) mv2
2mt 2 2v2P 2Pm 2gP
Solution: (d) Force = P lb.wt.= Pg poundals and mass = m lbs
∴ f = F = Pg ft / sec 2
m m
Now initial velocity = 0, final velocity = v ft/sec.
Distance = x and time = t
Hence x = v2 − u2 = v2 − 0 = mv2 .
2f 2Pg
2 Pg
m
Example: 22 An engine and train weight 420 tons and the engine exerts a force of 7 tons. If the resistance to motion be 14 lbs. wt. per
ton, then the time, the train will take to acquire a velocity of 30 m/hr. from rest is [BIT Ranchi 1994]
(a) 2.2 m (b) 2.6 m (c) 2.8 m (d) 3 m
Solution: (a) Effective force = Pull of the engine – Resistance
= 7 tons – 420 × 14 lbs = 7 × 2240 lbs – 5880 lbs = 9800 lbs = 9800 × 32 Poundals
Also acceleration = P = 9800 × 32 = 1 ft / sec2
m 420 × 2240 3
Again initial velocity = 0 and final velocity = 30 m.p.h. = 44 ft/sec.
Use v=u+ ft, we get 44 =0+ 1 t
3
∴ t = 132 second = 2 12 minute; t = 2 1 minute = 2.2 minute.
60 5
Motion of a Body released from a Balloon or a Lift.
(1) When a lift is ascending with uniform acceleration of f m/sec2 and after t second a body is dropped from it,
then at the time when the body is dropped:
(i) Initial velocity of the body is same as that of the lift and is in the same direction. So, the velocity of the
body is ft m/sec.
(ii) Initial velocity of the body relative to the lift = Velocity of the body – Velocity of the lift = ft – ft = 0
(iii) Acceleration of the body = g m/sec2 in downward direction.
(iv) Acceleration of the lift = f m/sec2 in upward direction.
(v) Acceleration of the body relative to the lift
= Acceleration of the body – Acceleration of the lift = g – (– f) = f + g in downward direction.
(2) When a lift is ascending with uniform acceleration of f m / sec 2 and after t second a body is thrown
vertically upward with velocity v m / sec , then at that time, we have the following :
(i) Initial velocity of the body = v + velocity of lift = v + ft, in upward direction
(ii) Initial velocity of the body relative to the lift = Velocity of the body – Velocity of lift = (v + ft) – ft = v m/sec.
(iii) Acceleration of the body relative to the lift in vertically downward direction is (f + g) m/sec2.
(3) When a lift is descending with uniform acceleration f m / sec2 and after time t a body is dropped from it.
Then at that time, we have the following
(i) Velocity of the body = Velocity of the lift = ft m/sec in downward direction
(ii) Acceleration of the body relative to the lift in downward direction
= Acceleration of the body – Acceleration of lift = g − f m / sec 2
Apparent weight of a Body resting on a moving Horizontal plane or a Lift.
Let a body of mass m be placed in a lift moving with an acceleration f and R is the normal reaction, then
(1) When the lift is rising vertically upwards : Effective force in upward direction = Sum of the external
forces in the same direction.
⇒ mf = R − Mg ⇒ R = m(f + g)
Clearly, the pressure R exerted by the body on the plane is greater than the actual weight mg of the body.
This pressure is also known as apparent weight.
If a man of mass m is standing on a lift which is moving vertically upwards with an acceleration, then
R = m(g + f) ⇒ R = mg1 + f R
g f
⇒ R = (weight of the man) 1 + f
g
⇒ R = Weight of the man + f (weight of the man) mg
g
Thus, the apparent weight of the man is f times more than the actual weight.
g
(2) When the lift is descending vertically downwards : Effective force in downward direction = Sum of
the forces in the same direction.
R
⇒ mf = mg − R ⇒ R = m(g − f) ......(ii)
f
mg
Clearly, the pressure exerted by the body on the plane is less than its actual weight when the plane moves
vertically downwards.
If a man of mass m in is standing on a lift, which is moving vertically downward with an acceleration f, then
the pressure R = m(g − f) = mg1 − f ⇒ R = weight of the man – f (weight of the man)
g g
Thus, the apparent weight of the man is f times less than his actual weight.
g
Note : Effective force stopping a falling body = F − mg R
Sand
If R be the resistance of sand on a body of mass m falling in sand,
then effective force = R − mg . mg
Important Tips
If the plane moves vertically upward with retardation equal to g i.e., f = −g , then from R = m(f + g) , we get R = 0 . Thus there is no
pressure of the body on the plane when the plane rises vertically with retardation equal to g.
If the plane moves down freely under gravity i.e., with acceleration equal to g, then from R = m(g − f) , we get R = 0 . Thus there is no
pressure of the body on the plane, when it moves vertically downwards with an acceleration equal to g.
Example: 23 A paracute weighing 112 lbs. wt. falling with a uniform acceleration from rest, describes 16 ft. in the first 4 sec. Then the
Solution: (c) resultant pressure of air on the parachute is [BIT Ranchi 1995]
(a) 85 lbs. wt. (b) 95 lbs. wt. (c) 105 lbs. wt. (d) 115 lbs. wt.
Let f be the uniform acceleration. Here u = 0 , s = 16 , t = 4 sec .
Since s = ut + 1 ft2 , ∴ 16 = 0 + 1 f(4)2 . ∴ f = 2 ft / sec2
2 2
Equation of motion is mf = mg − R (Downward)
∴ R = m(g − f) = 112 (32 – 2) = 112 × 30 Poundals = 112 × 30 lbs. wt. = 105 lbs. wt.
32
Dynamics
Example: 24 A man falls vertically under gravity with a box of mass ‘m’ on his head. Then the reaction force between his head and the
Solution: (c)
Example: 25 box is [MNR 1990; UPSEAT 2000]
Solution: (d)
Example: 26 (a) mg (b) 2 mg (c) 0 (d) 1.5 mg
Solution: (c) Let reaction be R. Since the man falls vertically under gravity ∴ f = g
From mg − R = mf , we have mg − R = mg . ∴ R = 0 .
A man of mass 80 kg. is travelling in a lift. The reaction between the floor of the lift and the man when the lift is ascending
upwards at 4 m/sec2 is
(a) 1464.8 N (b) 1784.8 N (c) 1959.8 N (d) 1104.8 N
When the lift is ascending, we have R = mg + ma = m(g + a) = 80(9.81 + 4) = 80(13.81) = 1104.8 N.
A particle of mass m falls from rest at a height of 16 m and is brought to rest by penetrating 1 m into the ground. If the
6
average vertical thrust exerted by the ground be 388 kg. wt, then the mass of the particle is
(a) 2 kg (b) 3kg (c) 4kg (d) 8kg
Let f m / sec2 be the retardation produced by the thrust exerted by the ground and let v m / sec . be the velocity with
which the mass m penetrates into the ground. Then, v2 = 02 + 2g × 16 [ v2 = u2 + 2gh]
⇒ v = 2 × 9.8 ×16 = 4 19.6 = 17.70 m / sec . u=0
16 m
Since the mass m penetrates through 1 m in the ground.
6 f
v=0
Therefore, its final velocity is zero.
So, 02 = v2 − 2 fs ⇒ 0 = (4 19.6 )2 − 2f × 1 1/6 m
6
⇒ f = 16 × 19.6 × 6 = 940.8 m / sec 2 ⇒ (388 − m) = 940.8 m [ g = 9.8 m / sec 2 ]
2 9.8
⇒ 388 − m = 96m ⇒ m= 388 = 4. Hence, the mass of the particle is 4 kg.
97
Motion of two particles connected by a String.
(1) Two particles hanging vertically : If two particles of masses m1 and m2 (m1 > m2 ) are suspended
freely by a light inextensible string which passes over a smooth fixed light pulley, then the particle of mass m1(> m2 )
will move downwards, with Acceleration f = (m1 − m2 )g
m1 + m2
Tension in the string T = 2m1m2 g f f
m1 + m2
T
Pressure on the pulley = 2T = 4m1m2 g T
m1 + m2
m2g m1g
(2) One particle on smooth horizontal table : A particle of mass m1 attached at one end of light
inextensible string is hanging vertically, the string passes over a pulley at the end of a smooth horizontal table, and a
particle of mass m2 placed on the table is attached at the other end of the string. Then for the system,
Acceleration f = m1 g f
m1 + m2 TT
Tension in the string T = m1m2 g fT
m1 + m2 T
Pressure on the pulley = T 2= 2m1m2 g . m1g m2g
m1 + m2
(3) One particle on an inclined plane : A particle of mass m1 attached at one end of a light inextensible
string is hanging vertically, the string passes over a pulley at the end of a smooth plane inclined at an angle α to the
horizontal, and a particle of mass m2 placed on this inclined plane is attached at the other end of the string. If the
particle of mass m1 moves vertically downwards, then for the system,
Acceleration f = (m1 − m2 sin α) g ff
m1 + m2
Tension in the string T = m1m2 (1 + sin α ) g m2g sinα
m1 + m2
α m2g m1g
Pressure on the pulley = {2(1 + sinα)}T = 2m1m2 g(1 + sinα)3 / 2
m1 + m2
Example: 27 The sum of two weights of an Atwood's machine is 16 lbs. The heavier weight descends 64 ft in 4 seconds. The value of
Solution: (a)
weights in lbs are [Roorkee 1992]
Example: 28
Solution: (d) (a) 10, 6 (b) 9, 7 (c) 8, 8 (d) 12, 4
Let the two weights be m1 and m2 (lbs.)
(m1 > m2), ∴ m1 + m2 = 16 ......(i)
Now, the distance moved by the heavier weight in 4 seconds from rest is 64 ft.
If acceleration = f, then 64 = 0+ 1 f (4)2 ⇒ f = 8 ft / sec 2
2
But f = m1 − m2 g ⇒ m1 − m2 × 32 = 8 ⇒ m1 − m2 = 4 ......(ii)
m1 + m2 16
From (i) and (ii), m1 = 10 lbs, m2 = 6 lbs .
Two weights W and W′ are connected by a light string passing over a light pulley. If the pulley moves with an upward
acceleration equal to that of gravity, the tension of the string is [UPSEAT 1995]
(a) WW ′ (b) 2WW ′ (c) 3WW ′ (d) 4WW ′
W + W′ W + W′ W + W′ W + W′
Suppose W descends and W′ ascends. If f is the acceleration of the weights relative to the pulley, then since the pulley
ascends with an acceleration f, the actual acceleration of W and W′ are (f − g) (downwards) and (f + g) (upwards)
respectively. Let T be tension in the string.
∴ Wg − T = W(f − g) and T − W′g = W′(f + g) or g − T = f −g and T − g = f +g
W W′
Subtracting, We get T + T − 2g = 2g or T = 4WW ′ kg. wt.
W W′ W + W′
Example: 29 To one end of a light string passing over a smooth fixed pulley is attached a particle of mass M, and the other end carries
Solution: (c)
a light pulley over which passes a light string to whose ends are attached particles of masses m1 and m2 . The mass M will
remain at rest or will move with a uniform velocity, if [MNR 1993]
(a) 111 (b) 21 1 (c) 411 (d) 811
M = m1 + m2 M = m1 + m2 M = m1 + m2 M = m1 + m2
If the mass M has no acceleration, then Mg = T1
But T1 = 2T2 = 2 2m1m2 g = 4m1m2 g
m1 + m2 m1 + m2
∴ Mg = 4m1m2 g or 4 = m1 + m2 = 1 + 1 T2 T2 T1
m1 + m2 M m1m2 m1 m2 mg
m2g
m1g
Impact of Elastic Bodies.
(1) Elasticity : It is that property of bodies by virtue of which they can be compressed and after compression
they recover or tend to recover their original shape. Bodies possessing this property are called elastic bodies.
(2) Law of conservation of momentum : It states “the total momentum of two bodies remains constant
after their collision of any other mutual action.” Mathematically m1.u1 + m2 .u2 = m1.v1 + m2 .v2
where m1 = mass of the first body, u1 = initial velocity of the first body, v1 = final velocity of the first body.
m2,u2,v2 = Corresponding values for the second body.
(3) Coefficient of restitution : This constant ratio is denoted by e and is called the coefficient of restitution
or coefficient of elasticity. The values of e varies between the limits 0 and 1. The value of e depends upon the
substances of the bodies and is independent of the masses of the bodies. If the bodies are perfectly elastic, then
e = 1 and for inelastic bodies e = 0 .
(4) Impact : When two bodies strike against each other, they are said to have an impact. It is of two kinds :
Direct and Oblique.
(5) Newton's experimental law of impact : It states that when two elastic bodies collide, their relative
velocity along the common normal after impact bears a constant ratio of their relative velocity before impact and is
in opposite direction.
If u1 and u2 be the velocities of the two bodies before impact along the common normal at their point of
contact and v1 and v2 be their velocities after impact in the same direction, v1 − v2 = −e(u1 − u2 )
Case I : If the two bodies move in direction shown in diagram given below, then
(a) v1 − v2 = −e(u1 − u2 ) and (b) m1v1 + m2v2 = m1u1 + m2u2
u1 m1 v1 u2 m2 v2
A B
Case II : If the direction of motion of two bodies before and after the impact are as shown below, then by the
laws of direct impact, we have
(a) v1 − v2 = −e[u1 − (−u2 )] or v1 − v2 = −e(u1 + u2 )
and (b) m1v1 + m2v2 = m1u1 + m2 (−u2 ) or m1v1 + m2v2 = m1u1 − m2u2
u1 m1 v1 u2 m2 v2
Case III : If the two bodies move in directions as shown below, then by the laws of direct impact, we have
(a) v1 − (−v2 ) = −e[−u1 − (−u2 )] or v1 + v2 = −e(u2 − u1 )
and (b) m1v1 + m2 (−v2 ) = m1(−u1 ) + m2 (−u2 ) or m1v1 − m2v2 = −(m1u1 + m2u2 )
u1 m1 v1 u2 m2 v2
A B
(6) Direct impact of two smooth spheres : Two smooth spheres of masses m1 and m2 moving along
their line of centres with velocities u1 and u2 (measured in the same sense) impinge directly. To find their velocities
immediately after impact, e being the coefficient of restitution between them.
Let v1 and v2 be the velocities of the two spheres immediately after impact, measured along their line of
centres in the same direction in which u1 and u2 are measured. As the spheres u1
are smooth, the impulsive action and reaction between them will be along the
common normal at the point of contact. From the principle of conservation of u2
momentum, m1 m2
m1v1 + m2v2 = m1u1 + m2u2 …..(i) v2
v1
Also from Newton’s experimental law of impact of two bodies,
v2 − v1 = e[u1 − u2 ] …..(ii)
Multiplying (ii) by m2 and subtracting from (i), we get
⇒ (m1 + m2 )v1 = (m1 − em2 )u1 + m2u2 (1 + e)
∴ v1 = (m1 − em2 )u1 + m2u2 (1 + e) …..(iii)
m1 + m2
(7) Oblique impact of two spheres : A smooth spheres of mass m1 , impinges with a velocity u1 obliquely
on a smooth sphere of mass m2 moving with a velocity u2 . If the direction of motion before impact make angles α
and β respectively with the line joining the centres of the spheres, and if the coefficient of restitution be e, to find
the velocity and directions of motion after impact,
Let the velocities of the sphere after impact be v1 and v2 in directions inclined at angles θ and φ
respectively to the line of centres. Since the spheres are smooth,
there is no force perpendicular to the line joining the centres of the
two balls and therefore, velocities in that direction are unaltered. u1 u2
v1 sinθ = u1 sinα …..(i) α m1 β m2
Oθ O′ φ
v2 sinφ = u2 sin β …..(ii)
v1 v2
And by Newton’s law, along the line of centres,
v2 cos φ − v1 cosθ = −e(u2 cos β − u1 cos α) …..(iii)
Again, the only force acting on the spheres during the impact is along the line of centres. Hence the total
momentum in that direction is unaltered.
∴m1v1 cosθ + m2v2 cos φ = m1u1 cos α + m2u2 cos β …..(iv)
The equations (i), (ii), (iii) and (iv) determine the unknown quantities.
(8) Impact of a smooth sphere on a fixed smooth plane : Let a smooth sphere of mass m moving with
velocity u in a direction making an angle α with the vertical strike a fixed smooth horizontal plane and let v be the
velocity of the sphere at an angle θ to the vertical after impact.
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