Answers to 8 June 2008 Practice paper

Mathematics

1. if |z-1| /|z-4| = 2, the locus of z = x+iy in Argand plane will be:

a. parabola
b. hyperbola
c. circle
d. straight line

Answer (c)

z = x+iy
| (x+iy-1) |/| (x+iy-4) | = 2

(x-1) ² + y² = 4[(x-4) ² +y²]

x² -2x+1+y² = 4{x-8x+16+y²]
x² -2x+1+y² = 4x²-32x+64+4y²
3x²-30x+63+3y² = 0
x²+y²-10x+21 = 0
Equation represents a circle with (5,0) as centre and SQRT(5²-21) as radius.
Radius is 2






2.

Let a,b,c be three non-coplanar vectors and p,q,r are vectors defined by the relations

p = b×c/[abc]
q = c×a/[abc]
r = a×b/[abc]

then the value of the expression
[(a+b).p +(b+c).q +(c+a).r] is equal to

a. 3
b. 1
c. 0
d. 2



Answer: (a)

[(a+b).p +(b+c).q +(c+a).r] = [1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)]

The property of distributivity of scalar product over vector addition is
a.(b+c) = a.b+a.c and
(b+c).a = b.a+b.c

[1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)] =

[1/(abc)][( a .(b×c)+ b.(b×c)+b .( c×a ) + b.(b×c)+c.( a×b) +a.( a×b)]

If a┴b, a.b = 0

Hence b.(b×c), and b.(b×c) and a.( a×b) are zero as .(b×c) will be a vector perpendicular to vectors b and c.

Hence the expression reduces to [1/(abc)][( a .(b×c)+ b .( c×a ) + c.( a×b)]

As a .(b×c) = b .( c×a ) = c.( a×b) = (abc)
The expression reduces to 3/(abc) /(abc) = 3



Physics
3. A particle moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

a. zero
b. 1/SQRT(2) m/s² towards north-west
c. ½ m/s² toward north-west
d. ½ m/s² towards north



Answer (b)

The average acceleration must have an westward component and northward component to make the velocity towards east as zero and to make velocity towards north as 5 m/s

Hence it will act in the direction of northwest.

Westward component of average acceleration = (0-5)/10 = -½ m/s²
Northward component of average of acceleration = (5-0)/10 = ½ m/s²

Hence magnitude of average acceleration = Sqrt( (-1/2) ²+(1/2) ²) = 1/SQRT(2)



Chemistry

4. Which of the following has maximum number of atoms?

a. 24 g of C (M=12 g mol^-1
b. 23 g of Na (M = 23 g mol^-1)
c. 48 g of S (M = 32 b mol^-1)
d. 108 g of Ag (M = 108 mol ^-1)



Answer: (a)

Number of moles (N) = m/M
Where m = given mass
M = molar mass


Hence Number of moles of carbon = 2
Number of moles of Na = 1
Number of moles of S = 1.5
Number of moles of Ag = 1

So 24 g of C having 2 moles has the maximum number of atoms.



5. The wave number of the shortest wavelength transition in the Balmer series of atomic hydrogen is.

a. 27420 cm^-1
b. 28420 cm^-1
c. 29420 cm^-1
d. 12186 cm^-1


Answer: (a)

Wave number = 1/ λ = R(1/nf² - 1/ni²)
Wave length is shortest, when energy emitted is highest. This will happen when the electron jumps from n = ∞ to n=2.

So wave number = 109677 cm^-1(1/2² - 1/∞²)
= 109677(1/4)
= 27419.25 cm^-1