NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

Edited By Ramraj Saini | Updated on Nov 16, 2023 11:44 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 Quadratic Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 4.3 class 10 deals concept of finding the roots of quadratic equations by completing squares and applying the quadratic formula. Exercise 4.3 Class 10 Maths, quadratic equations help us to explore the ideas of ways of finding quadratic equations in depth. Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial. The square trinomial can then be solved easily by using algebraic identities and formulas. NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 simple problems with word problems that are easy to solve and also explore the concepts of solving quadratic equations by using quadratic formulas.

NCERT solutions for Class 10 Maths exercise 4.3, focused on the concepts of solving quadratic equations and understanding more about the relationship between the roots of the equation and the nature of the equation. These class 10 maths ex 4.2 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 10 Maths chapter 4 exercise 4.3

Quadratic Equations class 10 chapter 4 Exercise: 4.3

$2x^2 - 7x +3 = 0$

Answer:

Given equation: $2x^2 - 7x +3 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2-\frac{7}{2}x+\frac{3}{2} = 0$

$\Rightarrow (x-\frac{7}{4})^2 + \frac{3}{2} - \frac{49}{16} = 0$

$\Rightarrow (x-\frac{7}{4})^2 = \frac{49}{16} - \frac{3}{2}$

$\Rightarrow (x-\frac{7}{4})^2 =\frac{25}{16}$

$\Rightarrow (x-\frac{7}{4}) =\pm \frac{5}{4}$

$\Rightarrow x =\frac{7}{4}\pm \frac{5}{4}$

$\Rightarrow x = \frac{7}{4}+\frac{5}{4}\ or\ x = \frac{7}{4} - \frac{5}{4}$

$\Rightarrow x = 3\ or\ \frac{1}{2}$

Answer:

Given equation: $2x^2 + x -4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{1}{2}x-2 = 0$

Adding and subtracting $\frac{1}{16}$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 -2 - \frac{1}{16} = 0$

$\Rightarrow (x+\frac{1}{4})^2 =2+\frac{1}{16}$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}$

$\Rightarrow (x+\frac{1}{4}) =\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x =\pm \frac{\sqrt{33}}{4} -\frac{1}{4}$

$\Rightarrow x = \frac{\pm \sqrt{33} - 1}{4}$

$\Rightarrow x = \frac{ \sqrt{33} - 1}{4}\ or\ x = \frac{ -\sqrt{33} - 1}{4}$

Answer:

Given equation: $4x^2 + 4\sqrt3 + 3 = 0$

On dividing both sides of the equation by 4, we obtain

$\Rightarrow x^2+\sqrt3x+\frac{3}{4} = 0$

Adding and subtracting $(\frac{\sqrt3}{2})^2$ in the equation, we get

$\Rightarrow (x+\frac{\sqrt3}{2})^2 +\frac{3}{4} - (\frac{\sqrt3}{2})^2 = 0$

$\Rightarrow (x+\frac{\sqrt3}{2})^2 = \frac{3}{4} - \frac{3}{4} = 0$

$\Rightarrow (x+\frac{\sqrt3}{2}) = 0\ or\ (x+\frac{\sqrt3}{2}) = 0$

Hence there are the same roots and equal:

$\Rightarrow x = \frac{-\sqrt3}{2}\ or\ \frac{-\sqrt3}{2}$

Answer:

Given equation: $2x^2 + x + 4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{x}{2}+2 = 0$

Adding and subtracting $(\frac{1}{4})^2$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 +2- (\frac{1}{4})^2 = 0$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{1}{16} -2 = \frac{-31}{16}$

$\Rightarrow (x+\frac{1}{4}) = \pm \frac{\sqrt{-31}}{4}$

$\Rightarrow x = \pm \frac{\sqrt{-31}}{4} - \frac{1}{4}$

$\Rightarrow x = \frac{\sqrt{-31}-1}{4} \ or\ x = \frac{-\sqrt{-31}-1}{4}$

Here the real roots do not exist (in the higher studies we will study how to find the root of such equations).

Answer:

(i) $2x^2-7x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = -7\ c = 3$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{7 \pm \sqrt{49-24}}{4}$

$\Rightarrow x= \frac{7 \pm 5}{4}$

$\Rightarrow x= \frac{7 + 5}{4} = 3\ or\ x= \frac{7 - 5}{4} = \frac{1}{2}$

Therefore, the real roots are: $x =3,\ \frac{1}{2}$

(ii) $2x^2+x-4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1\ c =-4$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1+32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{33}}{4}$

$\Rightarrow x= \frac{-1 + \sqrt{33}}{4} \ or\ x= \frac{-1 - \sqrt{33}}{4}$

Therefore, the real roots are: $x = \frac{-1+\sqrt{33}}{4}\ or\ \frac{-1-\sqrt{33}}{4}$

(iii) $4x^2+4\sqrt3x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 4,\ b = 4\sqrt{3}\ c =3$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}$

$\Rightarrow x= \frac{-4\sqrt{3} \pm 0}{8}$

Therefore, the real roots are: $x = \frac{-\sqrt{3}}{2}\ or\ \frac{-\sqrt{3}}{2}$

(iv) $2x^2+x+4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1,\ c =4$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1-32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{-31}}{4}$

Here the term inside the root is negative

Therefore there are no real roots for the given equation.

Answer:

Given equation: $x - \frac{1}{x} = 3, x\neq 0$

So, simplifying it,

$\Rightarrow \frac{x^2-1}{x} = 3$

$\Rightarrow x^2-3x-1 = 0$

Comparing with the general form of the quadratic equation: $ax^2+bx+c = 0$ , we get

$a=1,\ b=-3,\ c=-1$

Now, applying the quadratic formula to find the roots:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x= \frac{3 \pm \sqrt{9+4}}{2}$

$\Rightarrow x= \frac{3 \pm \sqrt{13}}{2}$

Therefore, the roots are

$\Rightarrow x = \frac{3+\sqrt{13}}{2}\ or\ \frac{3 - \sqrt{13}}{2}$

Answer:

Given equation: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$

So, simplifying it,

$\Rightarrow \frac{x-7-x-4}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow \frac{-11}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow (x+4)(x-7) = -30$

$\Rightarrow x^2-3x-28 = -30$ or $\Rightarrow x^2-3x+2 = 0$

Can be written as:

$\Rightarrow x^2-x-2x+2 = 0$

$\Rightarrow x(x-1) -2(x-1) = 0$

$\Rightarrow (x-2)(x-1) = 0$

Hence the roots of the given equation are:

$\Rightarrow x = 1\ or\ 2$

Answer:

Let the present age of Rehman be $'x'$ years.

Then, 3 years ago, his age was $(x-3)$ years.

and 5 years later, his age will be $(x+5)$ years.

Then according to the question we have,

$\frac{1}{(x-3)}+\frac{1}{(x+5)} = \frac{1}{3}$

Simplifying it to get the quadratic equation:

$\Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow \frac{2x+2}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow 3(2x+2)= (x-3)(x+5)$

$\Rightarrow 6x+6 = x^2+2x-15$

$\Rightarrow x^2-4x-21 = 0$

$\Rightarrow x^2-7x+3x-21 = 0$

$\Rightarrow x(x-7)+3(x-7) = 0$

$\Rightarrow (x-7)(x+3) = 0$

Hence the roots are: $\Rightarrow x = 7,\ -3$

However, age cannot be negative

Therefore, Rehman is 7 years old in the present.

Answer:

Let the marks obtained in Mathematics be 'm' then, the marks obtain in English will be '30-m'.

Then according to the question:

$(m+2)(30-m-3) = 210$

Simplifying to get the quadratic equation:

$\Rightarrow m^2-25m+156 = 0$

Solving by the factorizing method:

$\Rightarrow m^2-12m-13m+156 = 0$

$\Rightarrow m(m-12)-13(m-12) = 0$

$\Rightarrow (m-12)(m-13) = 0$

$\Rightarrow m = 12,\ 13$

We have two situations when,

The marks obtained in Mathematics is 12, then marks in English will be 30-12 = 18.

Or,

The marks obtained in Mathematics is 13, then marks in English will be 30-13 = 17.

Answer:

Let the shorter side of the rectangle be x m.

Then, the larger side of the rectangle wil be $= (x+30)\ m$ .

Diagonal of the rectangle:

$= \sqrt{x^2+(x+30)^2}\ m$

It is given that the diagonal of the rectangle is 60m more than the shorter side.

Therefore,

$\sqrt{x^2+(x+30)^2} = x+60$

$\Rightarrow x^2+(x+30)^2 = (x+60)^2$

$\Rightarrow x^2+x^2+900+60x = x^2+3600+120x$

$\Rightarrow x^2-60x-2700 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-90x+30x-2700 = 0$

$\Rightarrow x(x-90)+30(x-90)= 0$

$\Rightarrow (x+30)(x-90) = 0$

Hence, the roots are: $x = 90,\ -30$

But the side cannot be negative.

Hence the length of the shorter side will be: 90 m

and the length of the larger side will be $(90+30)\ m =120\ m$

Answer:

Given the difference of squares of two numbers is 180.

Let the larger number be 'x' and the smaller number be 'y'.

Then, according to the question:

$x^2-y^2 = 180$ and $y^2 = 8x$

On solving these two equations:

$\Rightarrow x^2-8x =180$

$\Rightarrow x^2-8x -180 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-18x+10x -180 = 0$

$\Rightarrow x(x-18)+10(x-18) = 0$

$\Rightarrow (x-18)(x+10) = 0$

$\Rightarrow x=18,\ -10$

As the negative value of x is not satisfied in the equation: $y^2 = 8x$

Hence, the larger number will be 18 and a smaller number can be found by,

$y^2 = 8x$ putting x = 18, we obtain

$y^2 = 144\ or\ y = \pm 12$ .

Therefore, the numbers are $18\ and\ 12$ or $18\ and\ -12$ .

Answer:

Let the speed of the train be $x\ km/hr.$

Then, time taken to cover $360km$ will be:

$=\frac{360}{x}\ hr$

According to the question,

$\Rightarrow (x+5)\left ( \frac{360}{x}-1 \right ) = 360$

$\Rightarrow 360-x+\frac{1800}{x} - 5 = 360$

Making it a quadratic equation.

$\Rightarrow x^2+5x-1800 = 0$

Now, solving by the factorizing method:

$\Rightarrow x^2+45x-40x-1800 = 0$

$\Rightarrow x(x+45)-40(x+45) = 0$

$\Rightarrow (x-40)(x+45) = 0$

$\Rightarrow x = 40,\ -45$

However, the speed cannot be negative hence,

The speed of the train is $40\ km/hr$ .

Answer:

Let the time taken by the smaller pipe to fill the tank be $x\ hr.$

Then, the time taken by the larger pipe will be: $(x-10)\ hr$ .

The fraction of the tank filled by a smaller pipe in 1 hour:

$= \frac{1}{x}$

The fraction of the tank filled by the larger pipe in 1 hour.

$= \frac{1}{x-10}$
Given that two water taps together can fill a tank in $9\frac{3}{8} = \frac{75}{8}$ hours.

Therefore,

$\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}$

$\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}$

$\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}$

Making it a quadratic equation:

$\Rightarrow 150x-750 = 8x^2-80x$

$\Rightarrow 8x^2-230x+750 = 0$

$\Rightarrow 8x^2-200x-30x+750 = 0$

$\Rightarrow 8x(x-25) - 30(x-25) = 0$

$\Rightarrow (x-25)(8x+30) = 0$

Hence the roots are $\Rightarrow x = 25,\ \frac{-30}{8}$

As time is taken cannot be negative:

Therefore, time is taken individually by the smaller pipe and the larger pipe will be $25$ and $25-10 =15$ hours respectively.

Answer:

Let the average speed of the passenger train be $x\ km/hr$ .

Given the average speed of the express train $= (x+11)\ km/hr$

also given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Therefore,

$\Rightarrow \frac{132}{x} - \frac{132}{x+11} = 1$

$\Rightarrow 132\left [ \frac{x+11-x}{x(x+11)} \right ] = 1$

$\Rightarrow \frac{132\times11}{x(x+11)} = 1$

Can be written as quadratic form:

$\Rightarrow x^2+11x-1452 = 0$

$\Rightarrow x^2+44x-33x-1452 = 0$

$\Rightarrow x(x+44)-33(x+44)= 0$

$\Rightarrow (x+44)(x-33) = 0$

Roots are: $\Rightarrow x = -44,\ 33$

As the speed cannot be negative.

Therefore, the speed of the passenger train will be $33\ km/hr$ and

The speed of the express train will be $33+11 = 44\ km/hr$ .

Answer:

Let the sides of the squares be $'x'\ and\ 'y'$ . (NOTE: length are in meters)

And the perimeters will be: $4x\ and\ 4y$ respectively.

Areas $x^2\ and\ y^2$ respectively.

It is given that,

$x^2 + y^2 = 468\ m^2$ .................................(1)

$4x-4y = 24\ m$ .................................(2)

Solving both equations:

$x-y = 6$ or $x= y+6$ putting in equation (1), we obtain

$(y+6)^2 +y^2 = 468$

$\Rightarrow 2y^2+36+12y = 468$

$\Rightarrow y^2+6y - 216 = 0$

Solving by the factorizing method:

$\Rightarrow y^2+18y -12y-216 = 0$

$\Rightarrow y(y+18) -12(y+18) = 0$

$\Rightarrow (y+18)(y-12)= 0$

Here the roots are: $\Rightarrow y = -18,\ 12$

As the sides of a square cannot be negative.

Therefore, the sides of the squares are $12m$ and $(12\ m+6\ m) = 18\ m$ .

More About NCERT Solutions for Class 10 Maths Exercise 4.3:

Exercise 4.3 Class 10 Maths consists of a question based on finding the roots of the Quadratic Equation by Completing the Square. In solving the Quadratic Equation by Completing the Square, first, we need to write the quadratic equation in general form that is ax2+bx+c=0 . For that we need to divide both sides of the equation by the coefficient of x2 if it not equal to 1 . Then we need to shift the constant term to the right-hand side. Then we need to add a square of one half on both sides. We need to write the left-hand side as the square and simplify the right-hand side. Take the square root of both sides to find the value of x. The NCERT solutions for Class 10 Maths exercise 4.3 also focused on the concepts of solving quadratic equations by using quadratic formulas. Few questions related to solving quadratic equations by using quadratic formulas are given also in exercise 4.3 Class 10 Maths. Also students can get access of Quadratic Equations Class 10 Notes to revise all the concepts quickly.

Benefits of NCERT Solutions for Class 10 Maths Exercise 4.3 :

• NCERT solutions for Class 10 Maths exercise 4.3 is recommended for students who want to do well in the CBSE Term exams.
• In NCERT book exercise 4.3 Class 10 Maths, all the minute concepts of solution of a Quadratic Equation by Completing the Square are also covered to help students face other competitive exams more confidently.
• By solving the NCERT syllabus Class 10th Maths chapter 4 exercise 4.3 exercises, students are able to answer all the questions based on quadratic equations and also students can attain a good score not only on the final answer but also on each step.

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Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. State true /false : A quadratic equation can be solved using the method of completing square.

Yes, the given statement is true. A quadratic equation can be solved using the method of completing squares.

2. What is the easiest and the fastest method to calculate the roots of a quadratic equation?

One of the fastest methods to calculate the roots of a quadratic equation is using the quadratic formula.

3. What is completing the square according to NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 ?

Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial

4. In NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 , how many questions and what types of questions are covered?

The questions are centred on the concept of finding the roots of quadratic equations by completing squares, and the NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 problems with subsections

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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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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

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 Option 1) Option 2) Option 3) Option 4)

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

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In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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

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With increase of temperature, which of these changes?

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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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