NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

# NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

Edited By Vishal kumar | Updated on Sep 11, 2023 09:08 AM IST | #CBSE Class 12th

## NCERT Solutions for Class 12 Physics Chapter 8 –Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves- If you're a Class 12 student concerned about NCERT solutions electromagnetic waves, you're in the right place on Careers360. On this NCERT solutions page, you'll find answers to all the exercises and additional exercise questions from one to fifteen. These class 12 chapter 8 physics NCERT solutions, authored by subject experts, are detailed and presented in easy-to-understand language

The Electromagnetic Waves NCERT solutions cover the question related to the concept of displacement current. Problems related to the Electromagnetic Spectrum are also discussed in NCERT Solutions for Class 12 Physics chapter 8 Electromagnetic Waves. The class 12 physics ch 8 NCERT solutions help in testing the knowledge about the concepts studied in a chapter. Students must refer to these solutions of NCERT to prepare for the examination effectively. The NCERT Solutions for Class 12 Physics Chapter 8 PDF download also helps in preparing for various entrance exams like JEE, NEET, etc.

The NCERT textbook is usually explicitly referenced in the questions that are asked in the CBSE EMW class 12 exams. One of the subjects that comes up frequently on the board test is electromagnetism. Therefore, it is advised that students use the electromagnetic waves NCERT solutions to get a firm understanding of both the chapter and the subject.

##### PTE Registrations 2024

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

Free download electromagnetic waves class 12 ncert solutions PDF for CBSE exam.

## NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves: Exercise Question and Answer

Calculate the capacitance and the rate of change of potential
difference between the plates.

Radius of the discs(r) = 12cm

Area of the discs(A) = $\pi r^{2}=\pi (0.12)^{2}=.045m^{2}$

Permittivity, $\epsilon _{0}$ = $8.85\times 10^{-12}C^{2}N^{-1}m^{2}$

Distance between the two discs = 5cm=0.05m

$Capacitance= \frac{\varepsilon _{0}A}{d}$

= $\frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}$

= $8.003\times 10^{-12}F$

=8.003 pF

$Rate\ of \ change\ of \ potential =\frac{dv}{dt}$

But

$V =\frac{Q}{C}$

Therefore,

$Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C}$

$=\frac{0.15A}{8.003pF}$

= $1.87\times 10^{10}Vs^{-1}$

Obtain the displacement current across the plates.

The value of the displacement current will be the same as that of the conduction current which is given to be 0.15A. Therefore displacement current is also 0.15A.

Is Kirchhoff’s first rule (junction rule) valid at each plate of the
capacitor? Explain.

Yes, Kirchoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first but once we take into consideration both the conduction and displacement current the Kirchoff's first rule will hold good.

What is the rms value of the conduction current?

Capacitance(C) of the parallel plate capacitor = 100 pF

Voltage(V) = 230 V

Angular Frequency (ω) = $300\ rad\ s^{-1}$

$Rms\ Current(I) =\frac{Voltage}{Capacitive\ Reactance}$

$Capacitive \ Reactance(X_{c})= \frac{1}{C\omega }=\frac{1}{100\ pF\times 300\ rad\ s^{-1}}$

$X_{c}= 3.33\times 10^{7} \Omega$

$I=\frac{V}{X_{c}}=\frac{230V}{3.33\times 10^{7}\Omega }= 6.9\times 10^{-6}A$

RMS value of conduction current is $6.9\mu A$

Is the conduction current equal to the displacement current?

Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere – Maxwell Law.

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

We know the Ampere - Maxwells Law,

$\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})$

Between the plates conduction current $i_{c}=0$ .

For a loop of radius r smaller than the radius of the discs,

$\mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}$

$B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}$

$B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}$

Since we have to find the amplitude of the magnetic field we won't use the RMS value but the maximum value of current.

$\\i_{max}=\sqrt{2}\times i_{rms}\\ i_{max}=\sqrt{2}\times6.9\mu A\\ i_{max}=9.76\mu A$

$\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\ B_{amp}=1.63\times 10^{-11}T$

The amplitude of B at a point 3.0 cm from the axis between the plates is $1.63\times 10^{-11}T$ .

The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to $3\times 10^{8}ms^{-1}$ (speed of light in vacuum).

Since the electromagnetic wave travels along the z-direction its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.

Frequency of wave = 30 MHz

Wavelength=

$\frac{Speed\ of\ light}{Frequency}=\frac{3\times 10^{8}}{30\times 10^{6}}$

$=10m$ .

Frequency range = 7.5 MHz to 12 MHz

Speed of light = $3\times 10^{8}ms^{-1}$

Wavelength corresponding to the frequency of 7.5 MHz =

$\frac{3\times 10^{8}ms^{-1}}{7.5\times 10^{6}Hz}$

$=40m$

Wavelength corresponding to the frequency of 12 MHz =

$\frac{3\times 10^{8}ms^{-1}}{12\times 10^{6}Hz}$

$=25m$

The corresponding wavelength band is 25m to 40m.

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore electromagnetic waves produced will have a frequency of 10 9 Hz.

Magnetic Field (B 0 )=510 nT = 510 $\times$ 10 -9 T

Speed of light(c) = 3 $\times$ 10 8 ms -1

Electric Field = B 0 $\times$ c

= 510 $\times$ 10 -9 T $\times$ 3 $\times$ 10 8 ms -1

= 153 NC -1

E 0 = 120 NC -1

$\nu =50.0\ MHz$

(a)

$Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}$

= $\frac{120}{3\times 10^{8}}$

$=400 nT$

Angular frequency ( $\omega$ ) = 2 $\pi \nu$

$=2$ $\times \pi \times 50$ $\times$ $10^{6}$

=3.14 $\times$ 10 8 rad s -1

Propagation constant(k)

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

= $\frac{2\pi \nu }{\lambda\nu }$

= $\frac{\omega }{c}$

= $\frac{3.14\times 10^{8} }{3\times 10^{8}}$

Wavelength( $\lambda$ ) = $\frac{c}{\nu }$

= $\frac{3\times 10^{8}}{50\times 10^{6}}$ $= 6 m$

Assuming the electromagnetic wave propagates in the positive z-direction the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction as they are mutually perpendicular and $E_{0}\times B_{0}$ gives the direction of propagation of the wave.

$\vec{E}=E_{0}sin(kx-\omega t)\hat{i}$

$= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}$

$\vec{B}=B_{0}sin(kx-\omega t)\hat{j}$

$= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}$

$E=h\nu=\frac{hc}{\lambda}$

$E=h\nu=\frac{6.6\times 10^{-34}\times3\times10^8}{\lambda \times1.6\times10^{-19}}eV=\frac{12.375\times 10^{-7}}{\lambda}eV$

Now substitute the different range of wavelength in the electromagnetic spectrum to obtain the energy

 EM wave one wavelength is taken from the range Energy in eV Radio 1 m $1.2375\times10^{-6}$ Microwave 1 mm $1.2375\times10^{-3}$ Infra-red 1000 nm 1.2375 Light 500 nm 2.475 Ultraviolet 1nm 1237.5 X-rays 0.01 nm 123750 Gamma rays 0.0001 nm 12375000

Frequency( $\nu$ ) =20 $\times$ 10 10 Hz

E 0 = 48 Vm -1

$Wavelength(\lambda) =\frac{Speed\ of\ light(c)}{Frequency (\nu )}$

= $\frac{3\times 10^{8}}{20\times 10^{10}}$

$=1.5 mm$

The amplitude of the oscillating magnetic field(B 0 ) = $\frac{E_{0}}{c}$

$=$ $\frac{48}{3\times 10^{8}}$

$=160 nT$

The average energy density of the Electric field(U E )

= $\frac{1}{2}\epsilon E^{2}$

= $\frac{1}{2}\epsilon (Bc)^{2}$ (as E=Bc)

= $\frac{1}{2}\epsilon \frac{B^{2}}{\mu \epsilon }$ $(c=\frac{1}{\sqrt{\mu \epsilon }})$

=U B

Therefore the average energy density of the electric field is equal to the average energy density of the Magnetic field.

$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$

$E = \left \{ ( 3.1 N/C )\sin [ (\frac{\pi }{2}-(( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t ))] \right \} \hat i$

The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.

From the equation of the wave given we can infer k = 1.8 rad m -1

$Wavelength(\lambda )=\frac{2\pi }{k}$

$=\frac{2\pi }{1.8}$

$=3.49 m$

From the given equation of the electric field we can infer angular frequency( $\omega$ ) = 5.4 $\times$ 10 8 rad s -1

Frequency( $\nu$ ) = $\frac{\omega }{2\pi }$

= $\frac{5.4\times10^{8}}{2\pi }$

=8.6 $\times$ 10 7 Hz

=86 MHz

From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1 NC -1

$Magnetic field amplitude (B_0) =\frac{E_{0}}{c}$

$=\frac{3.1}{3\times 10^{8}}$

=1.03 $\times$ 10 -7 T

As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction the magnetic field vector must be directed along the negative z-direction. ( $-\hat{i}\times -\hat{k}=-\hat{j}$ )

Therefore, $\vec{B}= \left \{ B_{0}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat k$

$\vec{B}= \left \{ 1.03\times 10^{-7}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \}T \hat k$

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 1m

$=\frac{5}{4\pi (1)^{2}}$

=0.398 Wm -2

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 10 m

$=\frac{5}{4\pi (10)^{2}}$

=3.98 $\times$ 10 -3 Wm -2

 EM wave one wavelength is taken from the range Temperature $T=\frac{0.29}{\lambda}$ Radio 100 cm $2.9\times 10^{-3}K$ Microwave 0.1cm 2.9 K Infra-red 100000ncm 2900K Light 50000 ncm 5800K Ultraviolet 100ncm $2.9\times10^{6}K$ X-rays 1 ncm $2.9\times10^8K$ Gamma rays 0.01 nm $2.9\times10^{10}K$

These numbers indicate the temperature ranges required for obtaining radiations in different parts of the spectrum

Frequency( $\nu$ )=1057 MHz

Wavelength( $\lambda$ ) $=\frac{c}{\nu }$

$=\frac{3\times 10^{8}}{1057\times 10^{6} }$

=0.283 m

=28.3 cm

Using formula, $\lambda _{m}T=0.29\ cmK$

$\lambda =\frac{0.29}{2.7}$

= 0.107cm

Microwaves.

E=14.4 keV

Wavelength( $\lambda$ ) = $\frac{hc}{E}$

= $\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{14.4\times 10^{3}\times 1.6\times 10^{-19}}$

= 0.85 A 0

X-rays

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

As TV signals are of high frequencies they are not reflected by the ionosphere and therefore satellites are to be used to reflect them.

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiations coming from the sun which are very harmful to humans.

If the earth did not have an atmosphere its average surface temperature be lower than what it is now as in the absence of atmosphere there would be no greenhouse effect.

The use of nuclear weapons would cause the formation of smoke clouds preventing the light from the sun reaching earth surface and it will also deplete the atmosphere and therefore stopping the greenhouse effect and thus doubling the cooling effect.

Class 12 physics ch 8 NCERT solutions is an essential resource for students preparing for both board exams and competitive exams like JEE. This chapter can be considered moderately challenging but also offers scoring potential. These solutions provide a clear understanding of electromagnetic waves, making complex concepts manageable, and play a crucial role in helping students excel in both board exams and JEE, contributing significantly to their success. There are 15 questions in the Electromagnetic Waves NCERT solutions. The NCERT solution for Class 12 Physics chapter 8 covers theoretical as well as numerical problems.

NCERT solutions for class 12 physics chapter-wise

 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

### NCERT Solutions for Class 12 Physics Chapter 8: Important Formulas and Diagrams

• Gauss’ law of electricity:

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

• Gauss’s law of magnetism:

• Faraday’s laws of Electromagnetic induction:

Since emf can be defined as the line integral of the electric field, the above relation can be expressed as

• Ampere-Maxwell’s Circuital law:

Where

### Electromagnetic Spectrum

• Gamma Rays:

Wavelength range: 1 x 10¹⁴ to 1 x 10⁻¹⁰m

Frequency range: 3 x 10²² to 3 x 10¹⁸Hz

Production: By transition of atomic nuclei and decay of certain elementary particles.

• X- rays:

Wavelength range: 1 x 10⁻¹¹ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁹ to 1 x 10¹⁶Hz

Production: By sudden deceleration of high-speed electrons at a high-atomic number target, and also by the electronic transition among the innermost orbits of atoms.

• Ultraviolet- Rays:

Wavelength range: 1 x 10⁻⁸ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁶ to 1 x 10¹⁴Hz

Production: By sun, arc, vacuum spark, and ionized gases.

• Visible Light:

Wavelength range: 1 x 10⁻⁷ to 1 x 10⁻⁷m

Frequency range: 7 x 10¹⁴ to 4 x 10¹⁴Hz

Production: Radiated by excited atoms in gases and incandescent bodies.

Wavelength range: 8 x 10⁻⁷ to 5 x 10⁻³m

Frequency range: 4x 10¹⁴ to 6 x 10¹⁰ Hz

Production: From hot bodies, and by rotational and vibrational transitions in molecules

Wavelength range: 1 x 10⁻¹ to 1 x 10⁴m

Frequency range: 3 x 10⁹ to 3 x 10⁴ Hz

Production: By oscillating electric circuits.

Also, check

Electromagnetic Wave Class 12 Physics NCERT: Topics

The following are the main headings covered in the Chapter 8 Physics Class 12-

• Concepts of displacement current and Maxwell's equation
• Basics of electromagnetic waves
• Electromagnetic Spectrum

### Significance of NCERT solutions for class 12 physics chapter 8 electromagnetic induction:

• For the CBSE board exam, 4% of questions are expected from chapter 8 physics class 12.

• NCERT solutions for class 12 will give a clear idea about how to use the formulas studied in the NCERT chapter 8 Physics Class 12.

• For exams like NEET and JEE Main one or two questions are asked from the Electromagnetic Waves NCERT.

• Questions based on the wave equation are expected from the chapter for the CBSE board exam. The relation between electric field, magnetic field and speed of light is also important.

• The electromagnetic Waves Class 12 pdf download button is available for the ease of students to prepare offline using Electromagnetic Waves Class 12 NCERT solutions.

### Key Features of Electromagnetic Waves Class 12 NCERT SolutionsPDF

1. Comprehensive Coverage: These ncert solutions electromagnetic waves comprehensively cover all the topics and questions presented in the Class 12 Physics chapter on electromagnetic waves.

2. Detailed Explanations: Each class 12 chapter 8 physics ncert solutions provides detailed step-by-step explanations, simplifying complex concepts for students.

3. Clarity and Simplicity: The solutions are presented in clear and simple language, ensuring ease of understanding.

4. Exam Preparation: These solutions are crucial for board exam preparation and provide valuable support for competitive exams like JEE.

5. Balanced Complexity: The chapter balances complexity with scoring potential, making it accessible for students aiming to score well.

6. Free Access: These solutions are available for free, ensuring equal access for all students.

Excluded Content:

Certain portions have been omitted, which include:

• Example 8.1.
• Section 8.3.2 discusses the nature of electromagnetic waves (excluding references to ether and content on page 277).
• Examples 8.4 and 8.5.
• Exercises 8.11 to 8.15.

### NCERT solutions subject wise

1. What is the prerequisite of the NCERT chapter electromagnetic waves?

Before starting the preparation of NCERT class 12 chapter 8 it is better to have an idea of chapters 1 to 7. Chapter 1 to 7 is interconnected.

2. How many questions are expected from Class 12 Physics Chapter 8 NCERT solutions for NEET?

One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.

3. What is the weightage of the NCERT class 12 chapter electromagnetic wave for JEE Main?

From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.

4. How many marks are allotted to electromagnetic wave for CBSE board exams?

3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.

5. How can students complete electromagnetic waves ncert solutions for chapter 8 physics class 12 with a perfect score?

Experts at Careers360 have created the chapter 8 physics class 12 ncert solutions after completing extensive research on each concept. To assist students perform well on class assignments and board exams, every small element is well addressed. Additionally, it enables students to successfully complete their tasks on time.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

#### National Institute of Open Schooling 10th examination

Exam Date:20 September,2024 - 07 October,2024

#### National Institute of Open Schooling 12th Examination

Exam Date:20 September,2024 - 07 October,2024

Edx
1113 courses
Coursera
804 courses
Udemy
394 courses
Futurelearn
222 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD

### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

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.

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