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NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Edited By Komal Miglani | Updated on Jun 23, 2025 08:51 AM IST

Have you ever faced a situation where solving an equation led to the square root of a negative number? Do you know how engineers, physicists, and mathematicians deal with this situation? Yes, this is where Complex Numbers come in! This chapter discusses the idea of Complex numbers and quadratic equations. A complex number has both real and imaginary parts. The imaginary parts are represented by i(iota). Students will carry out basic algebraic operations on complex numbers in the same way as on real numbers, such as addition, subtraction, multiplication, and division. The quadratic equations, or second-degree order equations, educate one on how to determine the roots of these equations through the discriminant and quadratic formulas. This article on NCERT solutions for class 11 maths will help students to understand all the concepts more thoroughly.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. NCERT Solution for Class 11 Maths Chapter 4 Solutions: Download PDF
  2. NCERT Solutions for Class 11 Maths Chapter 4: Exercise Questions
  3. Class 11 Maths NCERT Chapter 4: Extra Question
  4. Complex Numbers and Quadratic Equations Class 11 Chapter 4: Topics
  5. Complex Numbers and Quadratic Equations Class 11 Solutions: Important Formulae
  6. Approach to Solve Questions of Complex Numbers and Quadratic Equations Class 11
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. NCERT Solutions for Class 11 Maths: Chapter Wise
NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations
NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

The NCERT solutions for Class 11 complex numbers and quadratic equations deliver step-by-step processes to solve exercises and simplify complicated matters. In addition to textbook exercises, students are recommended to solve Class 11 Maths Chapter 4 Question Answer to reinforce learning and ensure they are well prepared for exams. Through the use of resources such as the NCERT Exemplar solutions for Class 11 Maths, students can be able to solve more complex problems and enhance their knowledge of both complex numbers and quadratic equations. Students need to refer to the NCERT Class 10 Maths books to gain more knowledge. With consistent practice and a clear understanding of these concepts, students can approach both theoretical and applied mathematics confidently. For syllabus, notes, and PDF, refer to this link: NCERT.

NCERT Solution for Class 11 Maths Chapter 4 Solutions: Download PDF

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NCERT Solutions for Class 11 Maths Chapter 4: Exercise Questions

Class 10 Maths chapter 4 solutions Exercise: 4.1
Page number: 82-83
Total questions: 14


Question 1: Express the given complex number in the form a+ib: (5i)(-3i/5)?
Answer:

Given a+ib:(5i)×(35i)

(5i)×(35i)=(5×35)(i2)

=3(i2)=3.

Question 2: Express each of the complex number in the form a+ib .

(5i)(3i/5)?

Answer:

We know that i4=1
Now, we will reduce i9+i19 into

i9+i19=(i4)2i+(i4)3i3
=(1)2.i+(1)3.(i) (i4=1,i3=i and i2=1)
=ii=0
Now, in the form of a+ib we can write it as
o+io
Therefore, the answer is o+io

Question 3: Express each of the complex number in the form a+ib.

i39

Answer:

We know that i4=1
Now, we will reduce i39 into

i39 =(i4)9.i3
=(1)9.(i)1 (i4=1,i3=i)
=1i
=1i×ii
=ii2 (i2=1)
=i(1)
=i
Now, in the form of a+ib we can write it as
o+i1
Therefore, the answer is o+i1

Question 4: Express each of the complex number in the form a+ib.

3(7+7i)+i(7+7i)

Answer:

Given problem is
3(7+7i)+i(7+7i)
Now, we will reduce it into

3(7+7i)+i(7+7i) =21+21i+7i+7i2
=21+21i+7i+7(1) (i2=1)
=21+21i+7i7
=14+28i

Therefore, the answer is 14+i28

Question 5: Express each of the complex number in the form a+ib .

(1i)(1+6i)

Answer:

Given problem is
(1i)(1+6i)
Now, we will reduce it into

(1i)(1+6i)=1i+16i
=27i

Therefore, the answer is 27i

Question 6: Express each of the complex number in the form a+ib .

(15+i25)(4+i52)

Answer:

Given problem is
(15+i25)(4+i52)
Now, we will reduce it into

(15+i25)(4+i52)=15+i254i52
=1205+i(425)10
=195i2110

Therefore, the answer is 195i2110

Question 7: Express each of the complex number in the form a+ib .

[(13+i73)+(4+i13)](43+i)

Answer:

Given problem is
[(13+i73)+(4+i13)](43+i)
Now, we will reduce it into

[(13+i73)+(4+i13)](43+i)=13+i73+4+i13+43i
=1+4+123+i(7+13)3
=173+i53

Therefore, the answer is 173+i53

Question 8: Express each of the complex number in the form a+ib .

(1i)4

Answer:

The given problem is
(1i)4
Now, we will reduce it into

(1i)4=((1i)2)2
=(12+i22.1.i)2 (using (ab)2=a2+b22ab)

=(112i)2 (i2=1)
=(2i)2
=4i2
=4

Therefore, the answer is 4+i0

Question 9: Express each of the complex number in the form a+ib .

(13+3i)3

Answer:

Given problem is
(13+3i)3
Now, we will reduce it into

(13+3i)3=(13)3+(3i)3+3.(13)2.3i+3.13.(3i)2 (using (a+b)3=a3+b3+3a2b+3ab2)
=127+27i3+i+9i2

=127+27(i)+i+9(1) (i3=i and i2=1)
=12727i+i9
=12432726i
=2422726i

Therefore, the answer is

2422726i

Question 10: Express each of the complex number in the form a+ib .

(213i)3

Answer:

Given problem is
(213i)3
Now, we will reduce it into

(213i)3=((2)3+(13i)3+3.(2)213i+3.(13i)2.2) (using (a+b)3=a3+b3+3a2b+3ab2)
=(8+127i3+3.4.13i+3.19i2.2)

=(8+127(i)+4i+23(1)) (i3=i and i2=1)
=(8127i+4i23)
=((1+108)27i+2423)
=223i10727

Therefore, the answer is 223i10727

Question 11: Find the multiplicative inverse of each of the complex numbers.

43i

Answer:

Let z=43i
Then,
z¯=4+3i
And
|z|2=42+(3)2=16+9=25
Now, the multiplicative inverse is given by
z1=z¯|z|2=4+3i25=425+i325

Therefore, the multiplicative inverse is

425+i325

Question 12: Find the multiplicative inverse of each of the complex numbers.

5+3i

Answer:

Let z=5+3i
Then,
z¯=53i
And
|z|2=(5)2+(3)2=5+9=14
Now, the multiplicative inverse is given by
z1=z¯|z|2=53i14=514i314

Therefore, the multiplicative inverse is 514i314

Question 13: Find the multiplicative inverse of each of the complex numbers.

i

Answer:

Let z=i
Then,
z¯=i
And
|z|2=(0)2+(1)2=0+1=1
Now, the multiplicative inverse is given by
z1=z¯|z|2=i1=0+i

Therefore, the multiplicative inverse is 0+i1

Question 14: Express the following expression in the form of a+ib:

(3+i5)(3i5)(3+2i)(3i2)

Answer:

Given problem is
(3+i5)(3i5)(3+2i)(3i2)
Now, we will reduce it into

(3+i5)(3i5)(3+2i)(3i2)=32(5i)2(3+2i)(3i2) (using (ab)(a+b)=a2b2)
=95i23+2i3+2i
=95(1)22i (i2=1)
=1422i×2i2i
=72i2i2
=72i2
Therefore, the answer is (0i722)

Class 10 Maths chapter 4 solutions Miscellaneous Exercise
Page number: 85-86
Total questions: 14

Question 1: Evaluate [i18+(1i)25]3 .

Answer:

The given problem is
[i18+(1i)25]3
Now, we will reduce it into
[i18+(1i)25]3=[(i4)4i2+1(i4)6i]3
=[14.(1)+116.i]3 (i4=1,i2=1)
=[1+1i]3
=[1+1i×ii]3
=[1+ii2]3
=[1+i1]3=[1i]3
Now,
(1+i)3=(13+i3+3.12.i+3.1.i2) (using (a+b)3=a3+b3+3.a2.b+3.a.b2)
=(1i+3i+3(1)) (i3=i,i2=1)
=(1i+3i3)=(2+2i)
=22i
Therefore, answer is 22i

Question 2: For any two complex numbers z1 and z2 , prove that Re(z1z2)=Rez1Rez2Imz1Imz2

Answer:

Let two complex numbers are
z1=x1+iy1
z2=x2+iy2
Now,
z1.z2=(x1+iy1).(x2+iy2)
=x1x2+ix1y2+iy1x2+i2y1y2
=x1x2+ix1y2+iy1x2y1y2 (i2=1)
=x1x2y1y2+i(x1y2+y1x2)
Re(z1z2)=x1x2y1y2
=Re(z1z2)Im(z1z2)

Hence proved

Question 3: Reduce (114i21+i)(34i5+i) to the standard form.

Answer:

Given problem is
(114i21+i)(34i5+i)
Now, we will reduce it into

(114i21+i)(34i5+i)=((1+i)2(14i)(1+i)(14i))(34i5+i)
=(1+i2+8i14i+i4i2)(34i5+i)
=(1+9i13i4(1))(34i5+i)
=(1+9i53i)(34i5+i)
=(3+4i+27i36i225+5i15i3i2)=(3+31i+362510i+3)=33+31i2810i=33+31i2(145i)

Now, multiply numerator an denominator by (14+5i)
33+31i2(145i)×14+5i14+5i
462+165i+434i+155i22(142(5i)2) (using (ab)(a+b)=a2b2)
462+599i1552(19625i2)
307+599i2(196+25)=307+599i2×221=307+599i442=307442+i599442

Therefore, answer is 307442+i599442

Question 4: If xiy=aibcid , prove that (x2+y2)2=a2+b2c2+d2.

Answer:

the given problem is

xiy=aibcid
Now, multiply the numerator and denominator by

c+id
xiy=aibcid×c+idc+id
=(ac+bd)+i(adbc)c2i2d2=(ac+bd)+i(adbc)c2+d2
Now, square both the sides
(xiy)2=((ac+bd)+i(adbc)c2+d2)2
=(ac+bd)+i(adbc)c2+d2
x2y22ixy=(ac+bd)+i(adbc)c2+d2
On comparing the real and imaginary part, we obtain

x2y2=ac+bdc2+d2  and  2xy=adbcc2+d2   (i)

Now,
(x2+y2)2=(x2y2)2+4x2y2
=(ac+bdc2+d2)2+(adbcc2+d2)2    (using (i))
=a2c2+b2d2+2acbd+a2d2+b2c22adbc(c2+d2)2
=a2c2+b2d2+a2d2+b2c2(c2+d2)2
=a2(c2+d2)+b2(c2+d2)(c2+d2)2
=(a2+b2)(c2+d2)(c2+d2)2
=(a2+b2)(c2+d2)

Hence proved.

Question 5: If z1=2i,z2=1+i , find |z1+z2+1z1z2+1| .

Answer:

It is given that
z1=2i,z2=1+i
Then,
|z1+z2+1z1z2+1|=|2i+1+i+12i1i+1|=|42(1i)|=|2(1i)|
Now, multiply the numerator and denominator by 1+i
|2(1i)×1+i1+i|=|2(1+i)12i2|=|2(1+i)1+1|=|1+i|
Now,
|1+i|=12+12=1+1=2
Therefore, the value of

|z1+z2+1z1z2+1| is 2

Question 6: If a+ib=(x+i)22x2+1 , prove that a2+b2=(x2+1)2(2x2+1)2 .

Answer:

It is given that
a+ib=(x+i)22x2+1
Now, we will reduce it into

a+ib=(x+i)22x2+1=x2+i2+2xi2x2+1=x21+2xi2x2+1=x212x2+1+i2x2x2+1
On comparing real and imaginary part. we will get
a=x212x2+1 and b=2x2x2+1
Now,
a2+b2=(x212x2+1)2+(2x2x2+1)2
=x4+12x2+4x2(2x2+1)2
=x4+1+2x2(2x2+1)2
=(x2+1)2(2x2+1)2
Hence proved

Question 7 (i): Let z1=2i,z2=2+i. Find

Re(z1z2z1¯)

Answer:

It is given that
z1=2i and z2=2+i
Now,
z1z2=(2i)(2+i)=4+2i+2ii2=4+4i+1=3+4i
And
z¯1=2+i
Now,
z1z2z¯1=3+4i2+i=3+4i2+i×2i2i=6+3i+8i4i222i2=6+11i+44+1 =2+11i5=25+i115
Now,
Re(z1z2z1)=25
Therefore, the answer is

25

Question 7 (ii): Let z1=2i,z2=2+i. Find

Im(1z1z1¯)

Answer:

It is given that
z1=2i
Therefore,
z¯1=2+i
NOw,
z1z¯1=(2i)(2+i)=22i2=4+1=5 (using (ab)(a+b)=a2b2)
Now,
1z1z¯1=15
Therefore,
Im(1z1z¯1)=0
Therefore, the answer is 0.

Question 8: Find the real numbers x and y if (xiy)(3+5i) is the conjugate of 624i.

Answer:

Let
z=(xiy)(3+5i)=3x+5xi3yi5yi2=3x+5y+i(5x3y)
Therefore,
z¯=(3x+5y)i(5x3y)               (i)
Now, it is given that
z¯=624i                    (ii)
Comparing (i) and (ii), we will get
(3x+5y)i(5x3y)=624i
On comparing the real and imaginary part, we will get
3x+5y=6   and   5x3y=24
On solving these we will get
x=3   and   y=3

Therefore, the values of x and y are 3 and -3 respectively

Question 9: Find the modulus of 1+i1i1i1+i .

Answer:

Let
z=1+i1i1i1+i
Now, we will reduce it into
z=1+i1i1i1+i=(1+i)2(1i)2(1+i)(1i)=12+i2+2i12i2+2i12i2 =4i1+1=4i2=2i
Now,
rcosθ=0  and  rsinθ=2
square and add both the sides. we will get,
r2cos2θ+r2sin2θ=02+22
r2(cos2θ+sin2θ)=4
r2=4                  (cos2θ+sin2θ=1)
r=2                  (r>0)

Therefore, the modulus of

1+i1i1i1+i is 2

Question 10: If (x+iy)3=u+iv , then show that ux+vy=4(x2y2).

Answer:

it is given that
(x+iy)3=u+iv
Now, expand the Left-hand side
x3+(iy)3+3.(x)2.iy+3.x.(iy)2=u+iv
x3+i3y3+3x2iy+3xi2y2=u+iv
x3iy3+3x2iy3xy2=u+iv (i3=i  and  i2=1)
x33xy2+i(3x2yy3)=u+iv
On comparing real and imaginary part. we will get,
u=x33xy2   and   v=3x2yy3
Now,
ux+vy=x(x23y2)x+y(3x2y2)y
=x23y2+3x2y2
=4x24y2
=4(x2y2)
Hence proved

Question 11: If α and β are different complex numbers with |β|=1 , then find |βα1α¯β| .

Answer:

Let
α=a+ib and β=x+iy
It is given that
|β|=1x2+y2=1x2+y2=1
and
α¯=aib
Now,
|βα1α¯β|=|(x+iy)(a+ib)1(aib)(x+iy)|=|(xa)+i(yb)1(ax+iayibxi2yb)|
=|(xa)+i(yb)(1axyb)i(bxay)|
=(xa)2+(yb)2(1axyb)2+(bxay)2
=x2+a22xa+y2+b2yb1+a2x2+b2y22ax+2abxyby+b2x2+a2y22abxy
=(x2+y2)+a22xa+b2yb1+a2(x2+y2)+b2(x2+y2)2ax+2abxyby2abxy
=1+a22xa+b2yb1+a2+b22axby (x2+y2=1 given)
=1

Therefore, value of |βα1α¯β| is 1

Question 12: Find the number of non-zero integral solutions of the equation |1i|x=2x.

Answer:

Given problem is
|1i|x=2x
Now,
(12+(1)2)x=2x
(1+1)x=2x
(2)x=2x
2x2=2x
x2=x
x2=0
x = 0 is the only possible solution to the given problem

Therefore, there are 0 number of non-zero integral solutions of the equation |1i|x=2x

Question 13: If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2

Answer:

It is given that
(a+ib)(c+id)(e+if)(g+ih)=A+iB,
Now, take mod on both sides
|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|
|(a+ib)||(c+id)||(e+if)||(g+ih)|=|A+iB| (|z1z2|=|z1||z2|)
(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)
Square both sides, we will get

(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)

Hence proved

Question 14: If (1+i1i)m=1, then find the least positive integral value of m .

Answer:

Let
z=(1+i1i)m
Now, multiply both numerator and denominator by (1+i)
We will get,
z=(1+i1i×1+i1+i)m
=((1+i)212i2)m
=(12+i2+2i1+1)m
=(11+2i2)m (i2=1)
=(2i2)m
=im
We know that i4=1
Therefore, the least positive integral value of m is 4.

Also, read,

Class 11 Maths NCERT Chapter 4: Extra Question

Question:
If α is a root of the equation x2+x+1=0 and k=1n(αk+1αk)2=20, then n is equal to ______

Solution:
We have,
α=ω(ωk+1ωk)2=ω2k+1ω2k+2=ω2k+ωk+2ω3k=1

k=1n(ω2k+ωk+2)=20(ω2+ω4+ω6++ω2n)+(ω+ω2+ω3++ωn)+2n=20
Now if n=3m,mI
Then 0+0+2n=20n=10 (not satisfy) if n=3m+1, then

ω2+ω+2n=201+2n=20n=212 (not possible)  if n=3m+2,(ω8+ω10)+(ω4+ω5)+2n=20(ω2+ω)+(ω+ω2)+2n=202n=22

n=11 satisfy n=3 m+2n=11

Hence, the correct answer is 11.

Complex Numbers and Quadratic Equations Class 11 Chapter 4: Topics

The topics discussed in the NCERT Solutions for class 11, chapter 4, Complex Numbers and Quadratic Equations are:

Complex Numbers and Quadratic Equations Class 11 Solutions: Important Formulae

Complex Numbers:

A complex number is expressed as a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit.

Imaginary Numbers: The square root of a negative real number is called an imaginary number, represented as √ − 1 = i.

Equality of Complex Number: Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if x1 = x2 and y1 = y2.

Algebra of Complex Numbers:

Addition: (z1 + z2) = (x1 + x2) + i(y1 + y2)

Subtraction: (z1 - z2) = (x1 - x2) + i(y1 - y2)

Multiplication: (z1 * z2) = (x1x2 - y1y2) + i(x1y2 + x2y1)

Division: (z1 / z2) = [(x1x2 + y1y2) + i(x2y1 - x1y2)] / (x22 + y22), where z2 ≠ 0.

Conjugate of Complex Number: The conjugate of a complex number z = x + iy is represented as z¯ = x - iy.

Modulus of a Complex Number: |z| = √(x2 + y2)

Argument of a Complex Number: The angle made by the line joining the point z to the origin, with the positive X-axis in an anti-clockwise sense is called the argument (arg) of the complex number.

Principal Value of Argument:

  • When x > 0 and y > 0 ⇒ arg(z) = θ

  • When x < 0 and y > 0 ⇒ arg(z) = π - θ

  • When x < 0 and y < 0 ⇒ arg(z) = -(π - θ)

  • When x > 0 and y < 0 ⇒ arg(z) = -θ

Polar Form of a Complex Number: z = |z| (cosθ + isinθ), where θ = arg(z).

The general polar form of z is z = |z| [cos(2nπ + θ) + isin(2nπ + θ)], where n is an integer.

Approach to Solve Questions of Complex Numbers and Quadratic Equations Class 11

Here are some approaches that students can use to solve complex numbers and quadratic equation problems.

  • Familiarise yourself with the standard form of complex numbers, i.e., a + bi, where ‘a’ and ‘b’ are real numbers and ‘i’ is the imaginary unit.
  • For quadratic equations, the standard form is ax2+bx+c. The discriminant, D = b24ac can be used to determine the nature of the roots.
  • Practice formulas like addition, subtraction, multiplication, division, conjugate, and modulus of a complex number.
    Grasp the concepts of roots and their properties for finding the sum and product of roots. This would be extremely helpful while attempting MCQ questions.
    - Sum =α+β=ba
    - Product =αβ=ca
  • Develop a strong understanding of plotting complex numbers as points and vectors.
    Also, realise that a quadratic graph is a parabola, and you must analyse the roots depending on where it crosses the x-axis.
  • Also, realise the Euler and polar form of complex numbers employed in multiplication and division. De Moivre's theorem is also of great use.

What Extra Should Students Study Beyond NCERT for JEE?

NCERT Solutions for Class 11 Maths: Chapter Wise


The links below allow students to access all the Maths solutions from the NCERT book.

Also, read,

NCERT Solutions For Class 11 - Subject-wise

Given below are the subject-wise NCERT solutions of class 11 NCERT:

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

Frequently Asked Questions (FAQs)

1. What are the properties of complex numbers in Chapter 4?

Addition/Subtraction: Combine real & imaginary parts separately.
Multiplication: Distribute like algebra, using i2=1.
Division: Multiply by the conjugate to simplify.
Modulus: Distance from origin, |z|=a2+b2.
Conjugate: Flip the sign of i, useful in division & simplifying expressions.

2. How to solve quadratic equations using complex numbers?

If the discriminant b24ac<0, use i to express the square root:

x=b±b24ac2a

3. What is the conjugate of a complex number, and how is it useful?

Definition: If z=a+bi, its conjugate is z¯=abi.

 Its Uses:

  • Simplifies the division of complex numbers.
  • Helps find modulus: |z|2=zz¯.
  • Appears in electrical engineering & signal processing.
4. How to express a complex number in polar form?

Expressed as:

z=r(cosθ+isinθ) or z=reiθ
Where:
- r=|z|=a2+b2 (modulus)
- θ=tan1(b/a) (argument)

Useful for multiplication, division, and powers of complex numbers.

5. What are the applications of complex numbers in real life?
  • Engineering: AC circuits & signal processing.
  • Physics: Quantum mechanics & electromagnetism.
  • Control Systems: Stability analysis.
  • Computer Graphics: Rotations & transformations.
  • Aerospace & Robotics: Navigation & control algorithms.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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0.34\; J

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Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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