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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.

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

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.

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NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics : Materials, Devices and Simple Circuits

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

Q. 14.1 In an n-type silicon, which of the following statement is true:

(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.

Answer:

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.

Q. 14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

(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

Answer:

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).

Q. 14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy bandgap respectively equal to (E_{g})_{C},(E_{g})_{Si} and (E_{g})_{Ge} . Which of the following statements is true?

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

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

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

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

Answer:

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.

(E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}

Hence correct option would be (c)

Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region beca

(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

Answer:

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)

Q. 14.5 When a forward bias is applied to a p-n junction, it

(a) raises the potential barrier

(b) reduces the majority carrier current to zero.

(c) lowers the potential barrier.

(d) none of the above.

Answer:

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)

Q. 14.6 In half-wave rectification, what is the output frequency if the input frequency is 50 \; Hz. What is the output frequency of a full-wave rectifier for the same input frequency

Answer:

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.

Q. 14.7 A p-n photodiode is fabricated from a semiconductor with bandgap of 2.8\; eV. Can it detect a wavelength of 6000 \; nm\; ?

Answer:

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

Q. 14.8 The number of silicon atoms per m 3 is 5\times 10^{28}. This is doped simultaneously with 5\times 10^{22}. atoms per m^{3} of Arsenic and 5\times 10^{20} per m^{3} atoms of Indium. Calculate the number of electrons and holes. Given that n_{i}=1.5\times 10^{16}\; m^{-3}. Is the material n-type or p-type ?

Answer:

Given:

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

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

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

number of thermally generated electrons 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.

Q. 14.9 In an intrinsic semiconductor the energy gap E_{g} is 1.2\; eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K Assume that the temperature dependence of intrinsic carrier concentration n_{i} is given by

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

Where, n_{0} is constant.

Answer:

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

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

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} .

Q. 14.10 (a) In a p-n junction diode, the current I can be expressed as

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

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}eV/K) and T is the absolute temperature. If for a given diode I_{0}=5\times 10^{-12}A and T=300\; K, then

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

Answer:

As we have

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

Here, I_{0}=5\times 10^{-12}A , T=300\; K, and , k_{B} = Boltzmann constant = (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

Q.14.10 (b) In a p-n junction diode, the current I can be expressed as

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

where I0 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}\; eV/K) and T is the absolute temperature. If for a given diode I_{0}=5\times 10^{12}A and T=300\; K, then

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

Answer:

As we have

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

Here, I_{0}=5\times 10^{-12}A , T=300\; K, and , k_{B} = Boltzmann constant = (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

Q. 14.10 (c) In a p-n junction diode, the current I can be expressed as

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

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^{-10}\; eV/K) and T is the absolute temperature. If for a given diode I_{0}=5\times 10^{-12}A and T=300\; K, then

(c) What is the dynamic resistance?

Answer:

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

Q.14.10 (d) In a p-n junction diode, the current I can be expressed as

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

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?

Answer:

As we have

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

Here, I_{0}=5\times 10^{-12}A , T=300\; K, and , k_{B} = Boltzmann constant = (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.

Q. 14.11 (a) You are given the two circuits as shown in Fig. 14.36. Show that circuit

(a) acts as OR gate while the circuit

1645428242590

Fig. 14.36

Answer:

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

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

Q. 14.11 (b) You are given the two circuits as shown in Fig. 14.36. Show that circuit

(b) acts as AND gate.

1645428384709

Answer:

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

Q. 14.12 Write the truth table for a NAND gate connected as given in Fig. 14.37

1645428401276

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

Answer:

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

Q 14.13 You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

1645428416591

Answer:

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

Q. 14.14 Write the truth table for the circuit given in Fig. 14.39 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

14ncrt

(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.)

Answer:

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

Q. 14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

1645428438468

Figure 14.40

Answer:

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

Semiconductor Class 12 NCERT Solutions: Important Formulas and Diagrams

  • Intrinsic Semiconductors: ne=ni=nn

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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:

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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.

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Question (FAQs)

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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  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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