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
Showing posts with label weekly practice sets. Show all posts
Showing posts with label weekly practice sets. Show all posts
Wednesday, June 18, 2008
Saturday, June 7, 2008
IIT JEE June 8 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
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
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
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)
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
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
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
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
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)
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
Solutions to June 1 2008 practice set
1. State whether the statement is true or false
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
Answer: true
At the top of the path, the vertical component of velocity is zero and the particle has only horizontal component. Hence the speed (the magnitude of the velocity) is minimum at the top of the path.
2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is
a. three
b. two
c. one
d. infinite
e. none of these
Answer (b)
If vector (x,y,z) is a unit vector perpendicular to (1,1,0) their scalar product is zero
=> x +y = 0
=> y = -x
If vector (x,y,z) is a unit vector perpendicular to (0,1,1) their scalar product is zero
=> y+z = 0
=> y = -z
As it is a unit vector x²+y²+z² = 1
=> 3y² = 1
=> y = ±1/SQRT(3)
This gives two value of y and hence there are two vectors
3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be
a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these
Answer: (b) and (d)
Assume z1 = a+ib, a is positive means a>0,
z2 = c+id, d<0
|z1| = |z2|
=> a²+b² = c²+d²
(z1+z2)/(z1-z2) = [(a+c) + i(b+d)]/ [(a-c) + i(b-d)]
Taking the multiplicative inverse of denominator and multiplying the numerator
= [(a+c) + i(b+d)]* [(a-c) - i(b-d)]/[(a-c) ² + (b-d) ²]
= [(a²-c²)+(b²-d²)]+i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]
=[{(a²+b²)-(c²+d²)}+i{(b+d)(a-c)-(a+c)(b-d)}]/ [(a-c) ² + (b-d) ²]
As a²+b² = c²+d² the first term of the numerator is zero.
Hence
(z1+z2)/(z1-z2)
= i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]
So this is a pure imaginary number.
If b+d as well as a+c are equal zero, the expression can be zero also.
4. If two compounds have the same empirical formula but different molecular formulae they must have ------
a. different percentage composition
b. different molecular weights
c. same viscosity
d. same vapour density
answer (b)
Different molecular formulae but same empirical formula imply same empirical mass but different molecular masses
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
Answer: true
At the top of the path, the vertical component of velocity is zero and the particle has only horizontal component. Hence the speed (the magnitude of the velocity) is minimum at the top of the path.
2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is
a. three
b. two
c. one
d. infinite
e. none of these
Answer (b)
If vector (x,y,z) is a unit vector perpendicular to (1,1,0) their scalar product is zero
=> x +y = 0
=> y = -x
If vector (x,y,z) is a unit vector perpendicular to (0,1,1) their scalar product is zero
=> y+z = 0
=> y = -z
As it is a unit vector x²+y²+z² = 1
=> 3y² = 1
=> y = ±1/SQRT(3)
This gives two value of y and hence there are two vectors
3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be
a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these
Answer: (b) and (d)
Assume z1 = a+ib, a is positive means a>0,
z2 = c+id, d<0
|z1| = |z2|
=> a²+b² = c²+d²
(z1+z2)/(z1-z2) = [(a+c) + i(b+d)]/ [(a-c) + i(b-d)]
Taking the multiplicative inverse of denominator and multiplying the numerator
= [(a+c) + i(b+d)]* [(a-c) - i(b-d)]/[(a-c) ² + (b-d) ²]
= [(a²-c²)+(b²-d²)]+i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]
=[{(a²+b²)-(c²+d²)}+i{(b+d)(a-c)-(a+c)(b-d)}]/ [(a-c) ² + (b-d) ²]
As a²+b² = c²+d² the first term of the numerator is zero.
Hence
(z1+z2)/(z1-z2)
= i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]
So this is a pure imaginary number.
If b+d as well as a+c are equal zero, the expression can be zero also.
4. If two compounds have the same empirical formula but different molecular formulae they must have ------
a. different percentage composition
b. different molecular weights
c. same viscosity
d. same vapour density
answer (b)
Different molecular formulae but same empirical formula imply same empirical mass but different molecular masses
Saturday, May 31, 2008
IIT JEE June 1 st Weekly Cumulative Practice Test
1. State whether the statement is true or false
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is
a. three
b. two
c. one
d. infinite
e. none of these
3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be
a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these
4. If two compounds have the same empirical formula but different molecular formulae they must have ------
a. different percentage composition
b. different molecular weight
c. same viscosity
d. same vapour density
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is
a. three
b. two
c. one
d. infinite
e. none of these
3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be
a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these
4. If two compounds have the same empirical formula but different molecular formulae they must have ------
a. different percentage composition
b. different molecular weight
c. same viscosity
d. same vapour density
Tuesday, May 27, 2008
Solutions for May 25 Practice set
covers topics
Complex Numbers
Vectors (common to physics and mathematics)
Mole concept
1. The inequality |z-4|<|z-2| represents the region given by
a. Re(z)>0
b. Re(z)<2
c. Re(z)<0
d. none of these
Answer (d)
Answer
Let z = x+iy then
|z-4|<|z-2|
=> |z-4²|<|z-2|²
=> (x-4) ²+y²<(x-2) ²+y²
=> -8x+16<-4x+4
=>12<4x
=> x>3
Means Re(z)>3
2. The points with the position vectors 60i +3j, 40i-8j, ai-52j are collinear if
a. a = -40
b. a = 40
c. a = 20
d. none of these
If the vectors x and y are collinear,
x = py
If we take given position vectors m.n,o, it implies
o-m = p(n-m)
[(ai-52j)-( 60i +3j)] = k[(40i-8j)-( 60i +3j)]
=> (a-60)i – 55j = k(-20i-11j)
=> a-60 = -20k and -55 = -11k
=> k = 55/11
=> a = -20*5 + 60
=> a = -40
3. One molal solution is one that contains one mole of a solute in :
a. 1000 g of the solvent
b. one litre of the solvent
c. one litre of the solution
d. 22.4 litres of the solution
Answer (a)
As per the definition, molality is amount of solute per kg of solvent.
Complex Numbers
Vectors (common to physics and mathematics)
Mole concept
1. The inequality |z-4|<|z-2| represents the region given by
a. Re(z)>0
b. Re(z)<2
c. Re(z)<0
d. none of these
Answer (d)
Answer
Let z = x+iy then
|z-4|<|z-2|
=> |z-4²|<|z-2|²
=> (x-4) ²+y²<(x-2) ²+y²
=> -8x+16<-4x+4
=>12<4x
=> x>3
Means Re(z)>3
2. The points with the position vectors 60i +3j, 40i-8j, ai-52j are collinear if
a. a = -40
b. a = 40
c. a = 20
d. none of these
If the vectors x and y are collinear,
x = py
If we take given position vectors m.n,o, it implies
o-m = p(n-m)
[(ai-52j)-( 60i +3j)] = k[(40i-8j)-( 60i +3j)]
=> (a-60)i – 55j = k(-20i-11j)
=> a-60 = -20k and -55 = -11k
=> k = 55/11
=> a = -20*5 + 60
=> a = -40
3. One molal solution is one that contains one mole of a solute in :
a. 1000 g of the solvent
b. one litre of the solvent
c. one litre of the solution
d. 22.4 litres of the solution
Answer (a)
As per the definition, molality is amount of solute per kg of solvent.
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