NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

# NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

Edited By Vishal kumar | Updated on Sep 13, 2023 09:18 AM IST | #CBSE Class 12th

## NCERT Solutions for Class 12 Physics Chapter 14 – Free PDF Download

NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit - This chapter is one of the most important chapter in Class 12 Physics. Students preparing for the upcoming board examination must refer to the NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit. These NCERT Solutions for Class 12 Physics Chapter 14 comprise of the solutions to the questions aksed in the NCERT books for Class 12 Physics.The Semiconductor Class 12 solutions are given in an easy-to-understand language. These NCERT Solutions will be beneficial for academics as well as competitive examinations.

Semiconductor Electronics is one of the important topics that comes under the Unit Modern Physics of Class 12. Semiconductor Class 12 discuss the basics of semiconductors electronics and devices like diode, rectifiers and transistors. Students can use the following materials along with the Semiconductor NCERT solutions. The NCERT Solutions for Class 12 contain solutions provided by subject matter experts, and students can use these solutions to prepare for their board exams or another competitive exam like JEE or NEET.

Free download chapter 14 physics class 12 ncert solutions pdf for CBSE exam.

## NCERT Solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits: Exercise Solution

(a) Electrons are majority carriers and trivalent atoms are the dopants.

(b) Electrons are minority carriers and pentavalent atoms are the dopants.

(c) Holes are minority carriers and pentavalent atoms are the dopants.

(d) Holes are majority carriers and trivalent atoms are the dopants.

An N-type semiconductor has electron as majority carriers and holes as minority carriers. It is formed when we dope pentavalent impurity in Silicon atom. Some pentavalent dopants are phosphorus, arsenic, and bismuth.

Hence the correct option is C.

(a) Electrons are majority carriers and trivalent atoms are the dopants.

(b) Electrons are minority carriers and pentavalent atoms are the dopants.

(c) Holes are minority carriers and pentavalent atoms are the dopants.

(d) Holes are majority carriers and trivalent atoms are the dopants

In a p-type semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atom-like aluminium is doped in a silicon atom. Hence correct option for p-type conductor would be (d).

(a) $\inline (E_{g})_{Si} < (E_{g})_{Ge}< (E_{g})_{C}$

(b) $\inline (E_{g})_{C} < (E_{g})_{Ge}> (E_{g})_{Si}$

(c) $\inline (E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$

(d) $\inline (E_{g})_{C} = (E_{g})_{Si}= (E_{g})_{Ge}$

Since carbon is a non-metal, its energy band gap would be highest and energy band gap of Ge would be least as it is a metalloid.

$\inline (E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$

Hence correct option would be (c)

(a) free electrons in the n-region attract them

(b) they move across the junction by the potential difference.

(c) hole concentration in p-region is more as compared to n-region.

d) All the above

Charge flows from higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the p-region to n-region and hence the concentration of hole is greater in p region.

and hence correct option would be (c)

(a) raises the potential barrier

(b) reduces the majority carrier current to zero.

(c) lowers the potential barrier.

(d) none of the above.

When a p-n junction is forward biased, the negative voltage repels the electron toward junction and give them the energy to cross the junction and combine with the hole which is also being pushed by a positive voltage. This leads to a reduction in the depletion layer which means a reduction in potential barrier across the junction.

Hence correct option would be (c)

As we know :

output frequency for half-wave rectifier = input frequency, and hence output frequency in half-wave rectifier will be 50Hz.

also, output frequency for full-wave rectifier = 2*(input frequency) and Hence output frequency in full-wave rectifier will be 2*50 = 100 Hz.

Given

the energy band gap of photodiode is 2.8eV.

wavelength = $\lambda$ = 6000nm = $6000*10^{-9}$

The energy of signal will be $\frac{hc}{\lambda }$

where c is speed of light(300000000m/s) , h is planks constant ( = $6.626 * 10^{-34}Js$ )

putting the corresponding value

The energy of signal = $\frac{(6.626 * 10^{-34} * 3*10^8)}{6000*10^{-9}}$

= $3.313*10^{-20}J$

= $0.207eV (since 1.6*10^{-20}= 1eV)$

The energy of the signal is 0.207eV which is less than 2.8eV ( the energy and gap of photodiode). Hence signal can not be detected by the photodiode.

## NCERT Solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits: Additional Exercise Solution

Given:

number of Silicon atoms per $\inline m^{3}$ = $\inline 5\times 10^{28}.$

number of Arsenic atoms per $\inline m^{3}$ = $\inline 5\times 10^{22}.$

number of Indium atoms per $\inline m^{3}$ = $\inline 5\times 10^{20}$

number of thermally generated electrons $\inline n_{i}=1.5\times 10^{16}\; m^{-3}.$

Now,

Number of electrons

$n_e =$ $5 * 10 ^{22}-1.5*10^{16}$ = $4.99*10^{22}(approx)$

number of holes is $n_h$

in thermal equilibrium

$n_h*n_e=n_i^2$

$n_h=n_i^2/n_e$

$n_h= (1.5*10^{16})^2/4.99*10^{22}$

$n_h= 4.51 * 10^9$

Now, since the number of electrons is higher than number of holes, it is an n-type semiconductor.

Where, $\inline n_{0}$ is constant.

Energy gap of given intrinsic semiconductor = E g = 1.2eV

temperature dependence of intrinsic carrier concentration $\inline n_{i}$ is given by

$\inline n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$

Where is constant, $K_B$ is Boltzmann constant = $8.862 * 10^{-5}eV/K$ ,

T is temperature

Initial temperature = T1 = 300K

the intrinsic carrier concentration at this temperature :

$n_{i1} = n_0exp[\frac{-E_g}{2K_B*300}]$

Final temperature = T2 = 600K

the intrinsic carrier concentration at this temperature :

$n_{i2} = n_0exp[\frac{-E_g}{2K_B*600}]$

the ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures

$\frac{n_{i2}}{n_{i2}} = \frac{n_0exp[\frac{-E_g}{2K_B*600}]}{n_0exp[\frac{-E_g}{2K_B*300}]}$

$= exp\frac{E_g}{2K_B}[\frac{1}{300}-\frac{1}{600}]=exp[\frac{1.2}{2*8.62*10^{-5}}* \frac{2-1}{600}]$

$= exp[11.6] = 1.09 * 10^{5}$

Therefore the ratio between the conductivities is $1.09 * 10^{5}$ .

where $I_{0}$ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $\inline k_{B}$ is the Boltzmann constant $\inline (8.6\times 10^{-5}eV/K)$ and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{-12}A$ and $\inline T=300\; K,$ then

(a) What will be the forward current at a forward voltage of $\inline 0.6\; V\; ?$

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ , $\inline T=300\; K,$ and , $\inline k_{B}$ = Boltzmann constant = $\inline (8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence forward current is 0.0625A

where I0 is called the reverse saturation current, $\inline V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $\inline I$ is the current through the diode, $\inline k_{B}$ is the Boltzmann constant $\inline (8.6\times 10^{-5}\; eV/K)$ and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{12}A$ and $\inline T=300\; K,$ then

(b) What will be the increase in the current if the voltage across the diode is increased to $\inline 0.7 \; V?$

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ , $\inline T=300\; K,$ and , $\inline k_{B}$ = Boltzmann constant = $\inline (8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.7V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.7}{1.376*10^{-23}*300}-1 ]=3.029A$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence the increase in the forward current is

$I(whenv=0.7) - I(whenv=.6)$ $= 3.029- 0.0625 = 2.967A$

where $\inline I_{0}$ is called the reverse saturation current, $\inline V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $\inline I$ is the current through the diode, $\inline k_{B}$ is the Boltzmann constant $\inline (8.6\times 10^{-10}\; eV/K)$ and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{-12}A$ and $\inline T=300\; K,$ then

(c) What is the dynamic resistance?

Dynamic Resistance = $\frac{voltage-change}{ current-change}$

Resistance change = 0.7 - 0.6 = 0.1

Current change = 2.967(calculated in prev question)

Therefore

, $Dynamic Resistance = \frac{0.1}{2.967} = 0.0337\Omega$

where $I_{0}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}V/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ , $\inline T=300\; K,$ and , $\inline k_{B}$ = Boltzmann constant = $\inline (8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When reverse voltage is 1V, V= -1

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-1)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

When the reverse voltage is -2V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-2)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

In both case current is very small and approximately equal to the reverse saturation current, hence their difference is negligible which causes dynamic resistance of infinity.

(a) acts as OR gate while the circuit

Fig. 14.36

Here, THE Input = A and B

Output = Y

The left part of the figure acts as a NOR and right part acts as NOT Gate.

The output of NOR gate = $\overline{A+B}$

the output of the NOR gate would be the input of NOT Gate and hence

Y = =

Hence the figure functions like an OR Gate.

or compare the truth table by giving different input and observing the output

INPUTS

OUTPUT
A B Y
0 0 0
0 1 1
1 0 1
1 1 1

The output of NOT gate ( left part of the circuit) is the input of the NOR gate

Hence the output of total circuit Y = $\over(\overline A + \overline B)$

= $\overline{\overline A}.\overline{\overline B}$ $\overline{A+B}=\overline A. \overline B$

= $A*B$

Hence the circuit functions as AND gate.

or give the inputs 00,01,10,11 and observe the truth table

INPUTS

OUTPUT
A B Y
0 0 0
0 1 0
1 0 0
1 1 1

The truth table is the same as that of AND gate

Hence identify the exact logic operation carried out by this circuit.

Here A is both input of the NAND gate and hence Output Y will be

$Y = \overline {A*A}$

$Y = \overline {A} + \overline A$

$Y = \overline {A}$

Hence circuit functions as a NOT gate.

The truth table for the given figure:

 Input Output A Y 0 1 1 0

a)

A and B are inputs of a NAND gate and output of this gate is the input of another NAND gate so,

Y = $\over(\overline {A.B})(\overline {A.B})$

Y= $\over(\overline {A.B})$ $+$ $\over(\overline {A.B})$

Y= $AB$

Hence this circuit functions as AND gate.

b)

A is input to the NAND gate output of whose goes to the rightmost NAND gate. Also, B is input to the NAND gate whose output goes to the rightmost NAND gate.

Y = $\over \overline A .\overline B$

Y = $\over\overline A .$ + $\over\overline B.$

Y = A + B

Hence the circuit functions as an OR gate .

Alternative method

fig. a

construct the truth table by giving various input and observe the output

 INPUT INTERMEDIATE OUTPUT OUTPUT 00 1 0 01 1 0 10 1 0 11 0 1

The above truth table is the same as that of an AND gate

fig. b

 INPUTS OUTPUT 00 0 01 1 10 1 11 1

The above truth table is the same as that of an OR gate

(Hint: $A=0,B=1$ then $A$ and $B$ inputs of second NOR gate will be $0$ and hence $Y=1.$ Similarly work out the values of $Y$ for other combinations of $A$ and $B.$ Compare with the truth table of OR, AND, NOT gates and find the correct one.)

A and B are the input od a NOR gate and Output of this NOR gate is the Input of Another NOR gate whose Output is Y. Hence,

Y = $\over(\overline{A+B} + \overline{A+B})$

Y = $\over\overline {A+B}$ . $\over\overline {A+B}$

Y = A + B

Hence Circuit behaves as OR gate.

Truth table

 INPUTS OUTPUT 00 0 01 1 10 1 11 1

Figure 14.40

a)

A is the two input of the NOR gate and Hence Output Y is:

Y = $\overline {A+A}$

Y = $\overline {A}$

Hence circuit functions as a NOT gate.

TRUTH TABLE:

 INPUT OUTPUT 0 1 1 0

b) A is the two input of a NOR gate whose output(which is $\overline {A}$ ) is the one input of another NOR gate. B is the two input of NOR gate whose output (which is $\overline {B}$ ) is the input of another NOR gate. Hence,

Y = $\over\overline {A} + \overline {B}$

Y = $\over\overline {A}$ . $\over\overline {B}$

Y = A.B

Hence it functions as AND gate.

TRUTH TABLE:

 INPUTS OUTPUT 00 0 01 0 10 0 11 1

Semiconductors ncert solutions holds significant importance for students preparing for board exams, as well as competitive ones like JEE and NEET. Its relevance lies in the practical applications of semiconductor devices in modern technology. Scoring in this ncert solutions class 12 physics chapter 14 can be relatively manageable, provided students have a solid grasp of semiconductor concepts. Therefore, mastering this chapter not only ensures success in board exams but also aids in tackling competitive exams with confidence, making it a valuable part of the syllabus.

## 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

### Semiconductor Class 12 NCERT Solutions: Important Formulas and Diagrams

• Intrinsic Semiconductors: ne=ni=nn

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

Where:

ne represents the free electron density in the conduction band of a semiconductor, ni stands for the intrinsic carrier concentration and nn represents the hole density in the valence band of a semiconductor.

• Extrinsic Semiconductors:

N-type: ne≅ Nd>nn

P- type: nn ≅ Nd>ne

Where: Nd is the number density

• Mobility

=Vd/E

Where: Vd = the drift velocity and E is the electric field

• Action of Transistor

IE=IC+IB

Where:

IE is the emitter current, IC is the collector current and IB is the base current

## Important Topic Covered In Semiconductor Class 12

Some of the important topic of ncert solutions of chapter 14 physics class 12 are listed below:

• Basics of Semiconductors and Valance Band Theory
• Types of Semiconductors
• Diode and its Chara
• Rectifiers, Photo Diodes, LED and Solar Cell
• Logic Gates

### Significance of NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

• As far as the CBSE board exam is considered you can expect more theoretical questions from Class 12 Physics Chapter 14 NCERT solutions.

• With the help of NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits, you can prepare well and it is easy to score in this chapter.

• NCERT Solutions for class 12 help in preparations of exams like NEET and JEE Mains. For NEET exams one or two questions and for JEE Mains one question is expected from this chapter

### How to use NCERT class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits:

• Go through the NCERT Syllabus for Class 12 Physics first. Therein, check all the important topics and subtopics.

• Try to solve all the questions by your own first. Afterwards check the NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit.

• Along with the NCERT solutions, also practice from the CBSE previous year question papers and sample papers to score well in the examination.

• NCERT Solutions for Class 12 Physics Chapter 14 PDF Download will also be available soon.

## Key features of NCERT Solutions Class 12 Physics Chapter 14

1. Comprehensive Coverage: These semiconductors ncert solutions encompass all the topics and questions found in Chapter 14, ensuring a thorough understanding of semiconductors and electronic devices.

2. Detailed Explanations: Each solution offers comprehensive, step-by-step explanations, making complex semiconductor concepts accessible to students.

3. Clarity and Simplicity: The chapter 14 physics class 12 ncert solutions pdf are presented in clear and straightforward language, ensuring ease of understanding.

4. Practice Questions: Exercise questions are included for practice and self-assessment, enhancing students' problem-solving skills.

5. Exam Preparation: These ncert solutions class 12 physics chapter 14 are essential for board exam preparation and provide valuable support for competitive exams.

6. Real-Life Applications: Concepts covered in this semiconductor class 12 ncert solutions have practical applications in the world of electronics and technology, enhancing students' real-world knowledge.

7. Free Access: These ncert solutions of chapter 14 physics class 12 are available for free, ensuring accessibility to all students.

### NCERT solutions subject wise

1. How many questions are asked for JEE main from the chapter semiconductor electronics?

1 or maximum 2 questions can be expected for JEE Main from NCERT chapter Semiconductor electronics. The questions may be from any topics of either analog electronics or digital electronics(Logic Gates). Cover the topics from JEE Main syllabus to get good score. Students should practice enough questions to crack an exam like JEE Main or NEET. To pratice students can refer to NCERT questions, NCERT Exemplar questions and JEE Main previous year papers.

2. What is the weightage of class 12 chapter semiconductor electronics materials devices and simple circuits for NEET exam?

Two or three questions can be expected from the chapter. Questions from both analogue and digital electronics part can be expected.

3. Is the chapter semiconductor devices useful for higher studies?

Yes, this chapter is the basics for higher studies in electronics-related branches and for VLSI and Nano Technology.

4. What types of questions are asked from the NCERT chapter semiconductor electronics materials devices and simple circuits for board exams?

Mostly theoretical questions are asked from the chapter semiconductor electronics for board exams. These covers mainly the analog electronics part that has topics like diode and its working, intrinsic and extrinsic semiconductors, how pn junction is formed, concepts of rectifiers, photo diodes, leds and solar cells etc.

5. Explain intrinsic and extrinsic semiconductors according to the semiconductors class 12 ncert solutions?

According to semiconductor electronics class 12 Intrinsic semiconductors are pure semiconductors such as pure silicon or germanium, with a small number of free electrons and holes, and low electrical conductivity. Extrinsic semiconductors are doped with impurities, increasing the number of free electrons or holes, which increases its electrical conductivity. Extrinsic semiconductors are further divided into p-type and n-type, depending on the type of impurity added.

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### 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