NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

# NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Edited By Vishal kumar | Updated on Sep 11, 2023 01:26 PM IST | #CBSE Class 12th

## NCERT Solutions for Class 12 Physics Chapter 10 – Download Free PDF

NCERT solutions for class 12 Physics chapter 10 Wave Optics - Students appearing in the Class 12 board exams must go through this Solutions for Class 12. These NCERT solutions for Class 12 Physics chapter 10 Wave Optics gives you an insight into how to deal with the questions asked in the NCERT books for Class 12 Physics. NCERT Class 12 Physics solutions chapter 10 Wave Optics comprise of the step by step solutions in an easy-to-understand language. The NCERT solutions for Class 12 Physics chapter 10 Wave Optics are helpful in academics as well as in competitive exams. Read further to know the Class 12 Physics Chapter 10 NCERT solutions.

Free download ncert class 12 physics chapter 10 pdf for CBSE exam.

## NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics: Exercises Solution

Given a monochromatic light has a wavelength of $589 nm\ or\ 589\times 10^{-9}m$ .

And as we know the speed of the light in air is, $c = 3\times 10^{8} m/s$ .

Refractive index of water, $\mu = 1.33$

Therefore we have now,

The ray is incident on the medium, and then it gets reflected back to the same medium therefore, the wavelength, speed, and frequency will be the same as that of the incident ray.

We have the frequency of light is given by the relation,

$v = \frac{c}{\lambda}$

$= \frac{3\times 10^8 m/s}{589\times 10^{-9}}$

$= 5.09\times 10^{14}Hertz$ $= 5.09\times 10^{14}Hertz$

Therefore the speed, frequency, and the wavelength of the reflected light are $3\times 10^{8} m/s$ , $5.09\times 10^{14}Hertz$ , and $589 nm$ respectively.

Given,

The wavelength of the monochromatic light $\lambda =589nm=589*10^{-9}m$

Refractive index of the water $\mu_{water}=1.33$

b)In the case of refraction, speed and wavelength will change as the medium is changing however, the frequency will remain the same because it doesn't change when the medium is changed,

so, frequency $f=5.09 * 10^{14}Hz$

Speed of rays: speed of rays in water with refractive index 1.33 is

$v=\frac{c}{\mu _{water}}=\frac{3*10^8}{1.33}=2.26*10^8m/s$

Now, the Wavelength of light

$\lambda =\frac{v}{f}=\frac{2.26*10^8}{5.09*10^{14}}=444.01*10^{-9}m$

Hence the wavelength of light is 444.01nm, frequency is $5.09 * 10^{14}Hz$ and speed is $2.26*10^8m/s$ .

(a) Light diverging from a point source.

The shape of wavefront when light is diverging from a point source is Spherical since light travels in all direction.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

The shape of light emerging out of a convex lens when a point source is placed at its focus is Parallel .when light rays come from infinity parallelly, they intersect at focus of convex lens and hence when light is emerging from the focus, the rays will get parallel to each other after coming out of the convex lens, because path of light rays are reversible.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth

The portion of the wavefront of the light from the distant star which is seen from earth is plane since a small area of a large sphere will nearly look like a plane.

Given,

Refractive index of the glass $\mu _{glass}$ = 1.5

Speed of light in vaccum $c=3*10^8m/s$

Now,

As we know,

Refractive index of a medium

$\mu _{medium }=\frac{c}{v_{medium}}$ where ${v_{medium}}$ is the speed of light in that medium.

so from here,

${v_{glass}}=\frac{c}{\mu _{glass}}=\frac{3*10^8}{1.5}=2*10^8m/s$

Hence the speed of light in water is $2*10^8m/s$

No, the speed of light in glass is not independent of the colour. The colour of the light does influence the refractive index and speed of light in the medium. The refractive index of the violet light is greater than the refractive index of red light and hence red component of the white light travels faster in the glass than the red component of the light.

as

$v=\frac{c}{\mu }$

the more the refractive index, the lesser the speed.

Given,

Distance between screen and slit $D=1.4m$

Distance between slits $d=0.28mm=0.28*10^{-3}m$

Distance between central and fourth bright fringe $u=1.2cm=1.2*10^{-2}m$

Now,

as we know, the distance between two fringes in a constructive interference is given by

$u=n\lambda \frac{D}{d}$

where $n=$ order of fringe (which is 4 here) and $\lambda$ is the wavelength of light we are using.

so from here,

$\lambda = \frac{ud}{nD}=\frac{1.2*10^{-2}*0.28*10^{-3}}{4*1.4}=6*10^{-7}m$

Hence wavelength os the light is 600nm

Given, in youngs double-slit experiment.

the wavelength of monochromatic light = $\lambda$

The intensity of light when the path difference is $\lambda$ = K

Now,

As we know,

The phase difference $\phi$ is given by

$\phi =\frac{2\pi }{\lambda }(PathDiffernce)$

also

Total Intensity

$I=I_1+I_2+2\sqrt{I_1I_2}cos\phi$

Let $I_1=I_2=I_0$

Now, when path difference is $\lambda$

the phase difference angle

$\phi=\frac{2\pi }{\lambda}*\lambda=2\pi$

so,

$I_0+I_0+2\sqrt{I_0I_0}cos2\pi=K$

$K=4I_0$

Now, when path difference is $\frac{\lambda }{3}$

$\phi=\frac{2\pi }{\lambda}*\frac{\lambda}{3}=\frac{2\pi}{3}$

Intensity of light

$K'=I_0+I_0+2\sqrt{I_0I_0}cos\frac{2\pi}{3}$

$K'=I_0$

Now comparing intensity at both cases

$\frac{K'}{K}=\frac{I_0}{4I_0}=\frac{1}{4}$

$K'=\frac{K}{4}$

Hence intensity will reduce to one-fourth of initial when path difference changes from $\lambda$ to $\frac{\lambda}{3}$ .

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

Given,

The wavelength of one light beam :

$\lambda_1=650nm=650*10^{-9}m$

The wavelength of another Light beam

$\lambda_2=520nm=520*10^{-9}m$

Let, the distance between the two-slit be $d$ and distance between slit and screen is $D$

Now,

As we know, the distance $x$ of nth bright fringe from central maxima is given by

$x=n\lambda \frac{D}{d}$

so for 3rd fringe,

$n=3$

$x=n\lambda_1 \frac{D}{d}=3*650*10^{-9}*\frac{D}{d}=1950\frac{D}{d}nm$

Hence distance of 3rd fringe from central maxima is $1950\frac{D}{d}nm$ . Here value D and d are not given in the question.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Let the nth bright fringe due to wavelength, $\lambda_2$ and (n − 1) th bright fringe due to wavelength $\lambda_1$ coincide on the screen. We can equate the conditions for bright fringes as:

$n\lambda_2=(n-1)\lambda _1$

$520n=650n-650$

$650=130n$

$n=5$

Hence, the least distance from the central maximum can be obtained by the relation:

$X=n\frac{\lambda_2D}{d}=5*\frac{520*10^{-9}*D}{d}$

Hence we can find out X if D and d are given.

Given

The angular width of the fringe when the medium is air

$\theta _{air}=0.2^0$

The distance of the screen from the slit $D = 1m$

The wavelength of light we are using $\lambda=600nm=600*10^{-9}m$

Refractive index of water $\mu_{water}=4/3$

let angular width of fringe when the medium is water $\theta _{water}$

Now, as we know the angular width is given by

$\theta =\frac{\lambda }{d}$

so,

$d=\frac{\lambda _{air}}{\theta _{air}}=\frac{\lambda _{water}}{\theta _{water}}$

$d=\frac{\lambda _{air}}{\lambda _{water}}=\frac{\theta _{air}}{\theta _{water}}=\mu$

From here

$\theta _{water}=\frac{\theta _{air}}{\mu _{water}}=\frac{3}{4}0.2^0=0.15^0$

Hence angular width of the fringe in the water is $0.15^0$ .

Given,

Refractive index of glass $\mu_{glass}=1.5$

Now as we know,

$tan\theta =\mu$

where $\theta$ is the polarizing angle, also called the Brewster angle.and $\mu$ is the refractive index.

so from here

$\theta =tan^{-1}(\mu _{glass})=tan^{-1}1.5=56.31^0$

Hence Brewster angle is $56.31^0$ .

Given,

Wavelength of light $\lambda =5000\AA =5000*10^{-10}m$

Speed of light $c=3*10^8$

Now,

Wavelength and frequency will be the same when the ray is reflected.

Frequency of reflected light

$f=\frac{c}{\lambda }=\frac{3*10^{8}}{5000*10^{-10}}=6*10^{14}Hz$

Hence wavelength and frequency of light is $5000*10^{-10}m$ and $6*10^{14}Hz$ respectively.

Now,

as per the law of reflection, angle of incidence $i$ is always equal to angle of reflection $r$ .

$i=r$

Now, when the reflected ray is perpendicular with incidence ray,

$i+r=90$

$i+i=90$

$2i=90$

$i=45$

Hence the angle of incidence is 45 for this condition.

Given

Aperture $a=4mm=4*10^{-3}m$

Wavelength of light $\lambda =400nm=400*10^{-9}m$

Now,

Distance for which ray optics is a good approximation also called Fresnel's distance:

$Z_f=\frac{a^2}{\lambda }=\frac{(4*10^{-3})^2}{400*10^{-9}}=40m$

Hence distance for which ray optics is a good approximation is 40m.

## NCERT solutions for Class 12 Pysics chapter 10 Wave Optics additional exercises:

Given,

wavelength Hα line emitted by hydrogen:

$\lambda=6563*10^{-10}m$

star is red-shifted by

$\lambda'-\lambda=15*10^{-10}m$

let velocity of the star be $v$

Now,

as we know,

$\lambda'-\lambda=\frac{v}{c}*\lambda$

from here

$\lambda'-\lambda\frac{1}{{c}*\lambda }={v}$

$v=\frac{3*10^8*15*10^{-10}}{6563*10^{-10}}=6.87*10^5m/s$

Hence speed at which star is receding away is $6.87*10^5m/s$

According to corpuscular theory, when corpuscle of the light goes from rare medium to denser medium, the component of their velocity along the surface of the interface remains the same.

So we can write

$v_1sini=v_2sinr$

$\frac{v_2}{v_1}=\frac{sini}{sinr}=\mu$

As $\mu>1$ ,

$v_2>v_1$

That is light should be faster in the dense medium than in rare medium.this is the opposite of what we see experimentally.

Huygens wave theory predicts that light is faster in a rare medium which matches with our experiments observation.

Let an object M is placed in front of a plane mirror AB at a distance r .

A circle is drawn from the centre, such that it just touches the plane mirror at point P. according to the Huygens’ principle, AB is the wavefront of the incident light .

If the mirror is absent then a similar wavefront A'B' would form behind M at a distance r.

A'B' can be considered as a virtual reflected ray for the plane mirror. Hence a point object placed in front of the plane mirror produces an image at the same distance as the object.

(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum depend?

The speed of light in a vacuum is constant and independent of anything according to Einstein's theory of relativity.

Question:

The speed of light in any medium depends upon the wavelength of the light and does not depends on the nature of the source, direction of propagation, the motion of the source and/or observer, and intensity of the wave.

The sound wave requires a medium for propagation.so, even though both given situation may relate to the same relative motion, they are not identical physically since, the motion of the observer, relative to the medium is different in two situations. Hence, we cannot expect the Doppler formula to be identical in both given cases.

When light waves are in a vacuum, there is clearly nothing to distinguish between two cases.

for light propagation in a medium, two situations are not identical for the same reason as in the case of sound waves.

Given,

Wavelength of light $\lambda$ = 600nm

Angular fringe width

$B_{\theta}=\frac{\lambda}{d}$

$d=\frac{\lambda}{B_{\theta}}=\frac{600*10^{-9}*180}{0.1*\pi }=3.44*10^{-4}m$

Hence spacing required between the two slits is $3.44*10^{-4}m$ .

Q10.17 (a) Answer the following questions:

In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

As we know,

width of the central diffraction band is given by

$2D\frac{\lambda}{d}$

where d is the width of the slit.

So when we double the width of the slit, the size of the central diffraction band reduces to half of its value. But, the light amplitude becomes double, which increase the intensity 4 times.

10.17 (b) Answer the following questions:

When we have a width in the order of $\lambda$ , the intensity of interference fringes in Young's double-slit experiment is modified by the diffraction pattern of each slit.

Q10.17 (c) Answer the following questions:

A bright spot is seen at the centre of the shadow of the obstacle because wave diffracted from the edge of a circular obstacle interfere constructively at the centre of the shadow producing the bright spot.

Q10.17 (d) Answer the following questions:

The size of obstacle should be comparable to the wavelength for diffraction of waves by obstacles, through a large scale.

This comes from

$sin\theta =\frac{\lambda}{a}$

$sin\theta =\frac{10^{-7}}{10}=10^{-8}$

This implies $\theta\rightarrow 0$

it means the light goes almost unbent and hence student are unable to see each other.

Q10.17 (e) Answer the following questions:

Typical size of the obstacle is much larger than the wavelength of light. Hence the diffraction effect is negligibly small. thus the assumption that light travels in a straight line can be safely used in day to day life.

Given,

Distance between two towers = 40km

size of aperture = $a=50m$

Now,

As we know

Fresnel's distance is equal to half of the distance between towers

$Z_f=\frac{40}{2}=20km$

Also from the formula:

$Z_f=\frac{a^2}{\lambda}=20m$

$\lambda=\frac{a^2}{Z_f}=\frac{50^2}{20*10^3}=12.5cm$

Hence this is the required longest wavelength of the radio wave, which can be sent in between the towers without considerable diffraction effect.

Given

The distance of the screen from the slit, $D=1m$

The distance of the first minimum $X_1=2.5mm=10^{-3}=2.5*10^{-3}mm$

The wavelength of the light $\lambda=500nm=500*10^{-9}m$

Now,

As we know,

$X_n=n\frac{\lambda D}{d}$

$d=n\frac{\lambda D}{X_n}=1*\frac{500*10^{-9}*1}{2.5*10^{-3}}=2*10^{-4}m=0.2mm$

Hence, the width of the slit is 0.2 mm.

When a low flying aircraft passes overhead, we notice slight shaking in pictures of the TV. This is because aircraft interferes with signals and reflects it. So the shaking we see is the interference of direct signal and reflected signal.

Q10.20 (b) Answer the following questions:

The superposition principle comes from the linear character of the differential equation of wavemotion.that is if $x_1$ and $x_2$ are the solution of any wave equation, then linear combination of $x_1$ and $x_2$ is also the solution of the wave equation.

Let the width of the slit $b$ be divided into n equal parts so that

$b'=\frac{b}{n}$

$b=b'n$

Now,

$\theta=\frac{n\lambda}{b}=\frac{n\lambda}{b'n}=\frac{\lambda}{b'}$

At this angle, each slit will make the first diffraction minimum. therefore the resultant intensity for all the slits will be zero at the angle of $\frac{n\lambda}{b}$ .

Wave optics class 12 exercise solutions hold great significance for board exams, as well as competitive exams like JEE and NEET. This chapter's questions range from moderately challenging to complex, offering students the opportunity to earn substantial marks. By thoroughly understanding concepts related to interference, diffraction, and polarization, and practising with these solutions, students can confidently tackle questions and secure good marks in both board and competitive exams.

Chapter wise NCERT Solutions for class 12 physics

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism NCERT solutions for class 12 physics chapter 5 Magnetism and Matter NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current NCERT solutions for class 12 physics chapter8 Electromagnetic Waves NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

• ### Huygens' Principle

Path difference (Δ):

T.D. (Time Difference):

Resultant Amplitude:

Resultant intensity:

• ### Young’s Double Slit Experiment (YDSE)

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Path difference:

Where x be the position of point P from central maxima

For maxima at P:

For minima at P:

• Brewster’s law:

• Resolving power of Microscope:

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Resolving power of Telescope:

## Wave Optics NCERT Topics

The topics covered in the Wave Optics Class 12 NCERT Physics are Huygens Principle, Explanation of Refraction and Reflection using Huygens Principle, Addition of Waves, Interference, Defraction, Polarisation and related subtopics. One important experiment of Wave Optics is Young's Double Slit Experiment.

### Importance of NCERT solutions for class 12 physics chapter 10 wave optics:

• Previously, in the CBSE board exam, 11-mark questions were asked from the unit optics which includes chapters 9 Ray Optics and Wave Optics of NCERT Class 12 Physics.

• For competitive exams also the Wave Optics NCERT solutions are important. Questions from the topic of young's double-slit experiment are frequently asked in competitive exams like NEET and JEE Main.

• NCERT solutions for Class 12 Physics Chapter 10 wave optics are important to score well in exams.

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

### Key features of Wave Optics Class 12 NCERT Solutions

1. Comprehensive Explanations: These wave optics class 12 exercise solutions offer comprehensive explanations, aiding students in understanding intricate wave optics concepts.

2. Problem-Solving Aid: Students can utilize these ncert class 12 physics chapter 10 pdf to practice and enhance their problem-solving skills in wave optics.

3. Clarity in Language: The solutions are presented in straightforward language, ensuring students can easily grasp and apply the principles.

4. Exam Preparation: These wave optics ncert solutions are a valuable resource for preparing for both board exams and competitive exams like JEE and NEET.

5. Performance Enhancement: By leveraging these solutions, students can improve their overall performance in physics.

6. Accessible for All: These wave optics class 12 solutions are freely accessible, making them available to all students seeking assistance with wave optics.

### How to use NCERT solutions for class 12 Physics chapter 10 Wave Optics

• Firstly go through the complete NCERT syllabus for Class 12 Physics and check all the topics. Through this, you will get a clear understanding of what to study.

• Afterwards, try to solve the questions on your own before attempting the solutions.

• If you still have any doubts then check the Wave Optics Class 12 NCERT Solutions.

• Along with the Class 12 Physics Chapter 10 NCERT solutions, solve the previous year’s question papers and sample papers too.

NCERT solutions subject-wise

1. Are NCERT books for Class 12 Physics enough to prepare for CBSE Board exams?

Yes, NCERT books are enough to prepare for the board exams, but you can refer to other reference books and sample papers as well. Try to cover all the concepts based on the NCERT syllabus. To get a good score in the CBSE board exam understand all the topics in the NCERT book and solve all the questions of NCERT exercise. Additionally students can refer NCERT exemplar problems and CBSE previous year question papers.

2. Will the NCERT solutions be helpful for competitive exams as well?

Yes, NCERT solutions will be helpful for competitive exams as well. Solving NCERT problems will give a better idea of concepts studied in a chapter and this in turn helps in competitive exams like JEE Main ans NEET.

3. Will the questions in the CBSE board exams be directly asked from the NCERT books?

Questions will be based on NCERT topics. But may be an application level question. The questions maynot be directly from the NCERT Questions but are related to the NCERT syllabus.

4. How can I quickly solve the wave optics-based problems in Chapter 10 of wave optics class 12 ncert?

To quickly solve ncert wave optics problems in ch 10 physics class 12, make sure you have a clear understanding of the fundamental concepts and practice solving problems. Read the problem statement carefully and use diagrams to visualize the problem. Review your work and seek help if needed.

5. How wave optics ncert solutions are important for NEET?

wave optics class 12 ncert solutions is important for NEET as the exam covers a wide range of topics from physics and wave optics is one of the important topics that is covered. Concepts and problem-solving skills learned in this chapter will be useful for understanding and solving questions related to wave optics that may appear on the NEET exam.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.

Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

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

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

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) 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

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

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

 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