Careers360 Logo
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 Apr 30, 2025 02:22 PM IST | #CBSE Class 12th

From mobile phones to PCS, LEDS to solar cells — semiconductors are the fundamental building materials of contemporary electronic devices. The knowledge of their operation is necessary for every student who is attempting to excel at physics and has a career linked with technology.

Chapter 14 – Semiconductor Electronics: Materials, Devices and Simple Circuits is a crucial chapter in NCERT Class 12 Physics, especially for those students who are preparing for board-level exams or competitive exams like the JEE and NEET. In order to have a better grasp of the topic, it is highly recommended to study the NCERT Solutions of the chapter under study.

This Story also Contains
  1. NCERT Solutions for Class 12 Physics Chapter 14: Download PDF
  2. Additional Questions
  3. Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions
  4. Approach to Solve Questions of Semiconductor Electronics:Materials, Devices and Simple Circuits
  5. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  6. NCERT solutions for class 12 physics chapter-wise
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

These responses are the correct and proper answers to each question that are located in the NCERT Class 12 Physics book. Authored by subject experts, the content is written in an understandable form that is apt for students. The descriptions are meant to demystify complex concepts, such as semiconductor materials, p-n junction diodes, rectifiers, transistors, and applications. For their optimal achievement, the students must supplement their textbook study with these NCERT solutions for Class 12 and other study materials for better understanding and effective revision.

Background wave

Also read :

NCERT Solutions for Class 12 Physics Chapter 14: Download PDF

Download PDF

class 12 physics chapter 14: 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 (Eg)C,(Eg)Si and (Eg)Ge . Which of the following statements is true?

(a) (Eg)Si<(Eg)Ge<(Eg)C

(b) (Eg)C<(Eg)Ge>(Eg)Si

(c) (Eg)C>(Eg)Si>(Eg)Ge

(d) (Eg)C=(Eg)Si=(Eg)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.

(Eg)C>(Eg)Si>(Eg)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 50Hz. 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.

Additional Questions

Q.1 A p-n photodiode is fabricated from a semiconductor with bandgap of 2.8eV. Can it detect a wavelength of 6000nm?

Answer:

Given

the energy band gap of photodiode is 2.8eV.

wavelength = λ = 6000nm = 6000109

The energy of signal will be hcλ

where c is speed of light(300000000m/s) , h is planks constant ( = 6.6261034Js )

putting the corresponding value

The energy of signal = (6.62610343108)6000109

= 3.3131020J

= 0.207eV(since1.61020=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.

Q.2 The number of silicon atoms per m 3 is 5×1028. This is doped simultaneously with 5×1022. atoms per m3 of Arsenic and 5×1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni=1.5×1016m3. Is the material ntype or ptype ?

Answer:

Given:

number of Silicon atoms per m3 = 5×1028.

number of Arsenic atoms per m3 = 5×1022.

number of Indium atoms per m3 = 5×1020

number of thermally generated electrons ni=1.5×1016m3.

Now,

Number of electrons

ne= 510221.51016 = 4.991022(approx)

number of holes is nh

in thermal equilibrium

nhne=ni2

nh=ni2/ne

nh=(1.51016)2/4.991022

nh=4.51109

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

Q.3 In an intrinsic semiconductor the energy gap Eg is 1.2eV. 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 ni is given by

ni=n0exp[Eg2KBT]

Where, n0 is constant.

Answer:

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

temperature dependence of intrinsic carrier concentration ni is given by

ni=n0exp[Eg2KBT]

Where is constant, KB is Boltzmann constant = 8.862105eV/K ,

T is temperature

Initial temperature = T1 = 300K

the intrinsic carrier concentration at this temperature :

ni1=n0exp[Eg2KB300]

Final temperature = T2 = 600K

the intrinsic carrier concentration at this temperature :

ni2=n0exp[Eg2KB600]

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

ni2ni2=n0exp[Eg2KB600]n0exp[Eg2KB300]

=expEg2KB[13001600]=exp[1.228.6210521600]

=exp[11.6]=1.09105

Therefore the ratio between the conductivities is 1.09105 .

Q.4 In a p-n junction diode, the current I can be expressed as

I=I0[expeVKBT1]

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, kB is the Boltzmann constant (8.6×105eV/K) and T is the absolute temperature. If for a given diode I0=5×1012A and T=300K, then

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

Answer:

As we have

I=I0[expeVKBT1]

Here, I0=5×1012A , T=300K, and , kB = Boltzmann constant = (8.6×105eV/K) =(1.3761023J/K)

When the forward voltage is 0.6V:

I=51012[exp1.610190.61.37610233001]=0.0625A

Hence forward current is 0.0625A

Q.5 In a p-n junction diode, the current I can be expressed as

I=I0[expeVKBT1]

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, kB is the Boltzmann constant (8.6×105eV/K) and T is the absolute temperature. If for a given diode I0=5×1012A and T=300K, then

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

Answer:

As we have

I=I0[expeVKBT1]

Here, I0=5×1012A , T=300K, and , kB = Boltzmann constant = (8.6×105eV/K) =(1.3761023J/K)

When the forward voltage is 0.7V:

I=51012[exp1.610190.71.37610233001]=3.029A

When the forward voltage is 0.6V:

I=51012[exp1.610190.61.37610233001]=0.0625A

Hence the increase in the forward current is

I(whenv=0.7)I(whenv=.6) =3.0290.0625=2.967A

Q.6 In a p-n junction diode, the current I can be expressed as

I=I0[expeVKBT1]

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, kB is the Boltzmann constant (8.6×1010eV/K) and T is the absolute temperature. If for a given diode I0=5×1012A and T=300K, then

(c) What is the dynamic resistance?

Answer:

Dynamic Resistance = voltagechangecurrentchange

Resistance change = 0.7 - 0.6 = 0.1

Current change = 2.967(calculated in prev question)

Therefore

, DynamicResistance=0.12.967=0.0337Ω

Q.7 In a p-n junction diode, the current I can be expressed as

I=I0[expeVKBT1]

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, kB is the Boltzmann constant (8.6×105V/K) and T is the absolute temperature. If for a given diode I0=5×1012A and T=300K, then

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

Answer:

As we have

I=I0[expeVKBT1]

Here, I0=5×1012A , T=300K, and , kB = Boltzmann constant = (8.6×105eV/K) =(1.3761023J/K)

When reverse voltage is 1V, V= -1

I=51012[exp1.61019(1)1.37610233001]5×1012

When the reverse voltage is -2V:

I=51012[exp1.61019(2)1.37610233001]5×1012

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

(a) acts as OR gate while the circuit

1645428242590

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 = A+B

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

Y=A+B=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 BY
0 00
0 11
1 01
1 11

Q. 8(b) You are given the two circuits as shown in Fig. 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 = (A+B)

= A.B A+B=A.B

= AB

Hence the circuit functions as AND gate.

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


INPUTS
OUTPUT
A BY
0 00
0 10
1 00
1 11

The truth table is the same as that of AND gate

Q.9 Write the truth table for a NAND gate connected as given in the figure.

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=AA

Y=A+A

Y=A

Hence circuit functions as a NOT gate.

The truth table for the given figure:

InputOutput
AY
01
10

Q 10 You are given two circuits as shown in Fig, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

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 = (A.B)(A.B)

Y= (A.B) + (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 = A.B

Y = A. + 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

INPUTINTERMEDIATE OUTPUTOUTPUT
0010
0110
1010
1101

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

fig. b

INPUTSOUTPUT
000
011
101
111

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

Q.11 Write the truth table for the circuit given in Fig 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 = (A+B+A+B)

Y = A+B . A+B

Y = A + B

Hence Circuit behaves as OR gate.

Truth table

INPUTSOUTPUT
000
011
101
111

Q.12 Write the truth table for the circuits given in Fig consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

1645428438468

Answer:

a)

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

Y = A+A

Y = A

Hence circuit functions as a NOT gate.

TRUTH TABLE:

INPUTOUTPUT
01
10

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

Y = A+B

Y = A . B

Y = A.B

Hence it functions as AND gate.

TRUTH TABLE:

INPUTSOUTPUT
000
010
100
111

Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions

Q1:

For an amplifier (NPN), VBC = 0, β=50, IC = 2.475 mA. Then the value of IB = ?

Answer:

As we learn
Relation between emitter current, Base current, collector current -
IE=IB+IC
- wherein
IE= Emitter Current
IB= Base Current
IC= Collector Current
IB=ICβ=2.47550=49.5 mA


Q2:

A Si diode has a saturation current of 10-7 A. Calculate the junction current for a forward bias of 0.7 V and 300 k (y = 2 for Si and V = 26 mV)

Answer:

As we learn,

Relation between current I & Voltage V -

I=I0(ecvKT1)

- wherein

K = Boltzmann constant

I0 = reverse saturation current

In forward bias

ervKT>>1

Then, the forward biasing current is

I=I0ecvKT

I=I0(ecvKT1)=I0(evy+VT1)=107(0.72×261)=70 mA


Q3:

Find the current through the circuit for Si diode`

Given that -

Knee voltage for Ge is 0.3 V

Knee voltage for Si is 0.7 V

Answer:

Knee voltage of P-N junction -

It is defined as that forward voltage at which the current through the junction starts rising rapidly with increase in voltage .

Knee voltage for Ge is 0.3 V

Knee voltage for Si is 0.7 V

I=3.40.7600=2.7600=4.5 mA


Q4:

The truth table for the above logic circuit is the same as that of :

Answer:

As we learn

NOR Gate -

NOT + OR Gate

- wherein

Y=A+B

A and B are input

Y is output

The output (y) of two input (A, B) NOR gates is :

Y=A+B


Q5:

For the circuit shown current through 1.5KΩ is:

Answer:

As we learn

Zener diode can operate continuously without being damaged in the region of reverse bias

- wherein

1) It acts as a voltage regulator

2) In forward biasing it acts as an ordinary diode.

IL=VZ1.5 K=61.5×103=4 mA


Approach to Solve Questions of Semiconductor Electronics:Materials, Devices and Simple Circuits

  • Master the basic fundamentals of semiconductors.
NEET/JEE Coaching Scholarship

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

  1. Know what conductors, insulators, and semiconductors are.
  2. There are two categories of semiconductors: intrinsic (pure) and extrinsic (doped).
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook
  • Know the types of doping
JEE Main Important Mathematics Formulas

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

  1. n-type: Introduction of pentavalent impurity → Additional electrons
  2. p-type: Introduction of trivalent impurity → Increased holes
  • Learn important parts:
  1. Diode: Recognize forward and reverse bias.
  2. Zener diode: Used for voltage control
  3. Transistors (n-p-n, p-n-p): Current amplification and switching
  • Learn circuit behavior
  1. Explain how voltage and current act in forward/reverse-biased diodes
  2. Study rectifier circuits: Half-wave and full-wave.
  • Apply proper formulas-
  1. Current gain in a transistor: β=ICIB

  2. IE=IB+IC (Emitter current = Base + Collector )

  • Study I-V characteristics- Draw and label graphs for diodes and transistors.
  • Learn logic gates (brief introduction)- AND, OR, NOT (if included) – Know the truth tables and how they apply.
  • Practice circuit numericals- Recognize biasing, compute currents, and perceive role of power supply
  • Know applications-
  1. Diodes: Rectification, switching
  2. Transistors: Logic circuits, amplifiers
  • Study the theory and practice NCERT questions- Most questions are questions about numbers or ideas, so put both.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT solutions for class 12 physics chapter-wise

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Questions (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.

Articles

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 ?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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.

View All

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

Back to top