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In This Post we are  providing Chapter- 4 ANALYSIS OF FINANCIAL STATEMENT NCERT MOST IMPORTANT QUESTIONS for Class 12 ACCOUNTANCY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ANALYSIS OF FINANCIAL STATEMENT

1. How would you show the following two items in a company‘s Balance Sheet as at 31st March, 2012 as per the requirement of Schedule VI:
   General Reserve (Since 31st March, 2011) Rs. 3,00,000, Statement of Profit and Loss (Debit Balance) for 2011-12 Rs. 2,00,000.Ans.
   Balance Sheet
   As at 31^st march, 2012 Equity and LiablitiesNote No.Rs.Shareholders‘ fund
   Reserve and Surplus 
   1 
   1,00,000Notes to Accounts:Reserve and SurplusGeneral Reserve(1st April, 2011)3,00,000Less: Statement of Profit and Loss(Dr. Balance)2,00,0001,00,000

2. Under Which main headings and sub-headings of Equity and Liabilities of the balance sheet as per the Revised Schedule VI of a company will you classify the following items:
   1. Proposed dividend.
   2. Fixed Deposit from Public
   Ans.Sr. No.ItemsMain-HeadingSub-Headingi.Proposed dividendCurrent-LiabilitiesShort-term provisionii.Fixed deposit from PublicNon-current liabilitiesLong term borrowing
3. State any two items which are shown under the head ’Investment‘ in a company balance sheet.
   Ans.
   (i) Government Securities.
   (ii) Sinking Fund Investment.
4. How is analysis of Financial statements suffered from the limitation of window dressing?
   Ans. Analysis of financial statements is affected from the limitation of window dressing as companies hide Some vital information or show items at incorrect value to portray better profitability and financial Position of the business, for example the company may overvalue closing stock to show higher profits.
5. What is the interest of Shareholders in the analysis of Financial Statements?
   Ans.
   (i) They want to judge the present and future earning capacity of the business.
   (ii) They want to judge the safety of their investment.

6. Name two tools of Financial Analysis?
   Ans. 
   1. Comparative Financial Statements.
   2. Ratio Analysis etc.
7. What is Horizontal Analysis?
   Ans: The analysis which is made to review and compare the financial statements of two or more then two Years is called Horizontal Analysis.
8. Give the example of Horizontal Analysis.
   Ans. Comparative Financial Statement.

9 Give the Main Heading and Sub- Heading of Equity and Liabilities of the Balance sheet of a company as per the Revised Schedule VI of the companies Act.1956.Ans.
2. EQUITY AND LIABILITIES
(5) Shareholders’ Funds
(d) Share Capital
(e) Reserves and Surplus
(f) Money received against share warrants
(6) Share Applications Money Pending Allotment
(7) Non-Current Liabilities
(e) Long-term borrowings
(f) Deferred tax liabilities(Net)
(g) Other Long-term Liabilities
(h) Long-term provisions
(8) Current Liabilities
(e) Short-term borrowings
(f) Trade payables
(g) Other current liabilities
(h) Short-term provisionsTOTAL

2. Give the Main Heading and Sub- Heading of Assets of the Balance sheet of a company as per the Revised Schedule VI of the companies Act.1956.
   Ans. ASSETS
   (1) Non-Current Assets
   (a) Fixed Assets 
   1. Tangible Assets
   2. Intangible assets
   3. Capital work-in progress
   4. Intangible assets under development
   2. Non-current investments
   3. Deferred tax assets (net)
   4. Long-term loans and advances
   5. Other non-current assets(2) Current Assets
   1. Current investments
   2. Inventories
   3. Trade receivables
   4. Cash and cash equivalents
   5. Short-term loans and advances
   6. Other current assets
3. Rearrange the following items under assets according to Revised or New Schedule VI:Ans.
   1. Livestock
   2. Loose Tools.
   3. Goodwill
   4. Trademarks
   5. Bills Receivable
   6. Debtors
   7. Land
   8. Leasehold
   9. Stock-in-Trade
   10. Stores and Spare Parts
   11. Vehicles
   12. Cash at Bank
   13. Work in Progress(Machinery)
   14. Interest accrued on Investment
   15. Furniture
   16. Advance to Subsidiaries
   17. Cash in Hand
   18. Plant
   19. Deposits with electricity supply company.
   20. Fixed Assets(Tangible): Livestock, Land, Leasehold, furniture, vehicles and plant
   21. Capital Work-in-progress: Work in progress(Machinery)
   22. Fixed Assets(Intangible): Goodwill and Trademarks
   23. Inventories: Loose Tools, Stock-in-Trade, Stores and Spare Parts.
   24. Trade Receivables: Bill Receivables, Debtors
   25. Cash and Cash Equivalents: Cash at Bank, Cash in Hand
   26. Long term Loans and Advances: Advance to Subsidiaries, Deposits with Electricity Supply Company.
   27. Other Current Assets: Interest Accrued on Investments.
4. List any three items that can be shown as contingent Liabilities in a company‘s Balance sheet.Ans:
   1. Claims against the Company not acknowledged as debts.
   2. Uncalled Liability on partly paid shares.
   3. Arrears of Dividend on Cumulative preference shares.108
5. How is a Company‘s balance sheet different from that of a Partnership firm? Give Two point only
   Ans. 
   1. For company‘s Balance Sheet there are two standard forms prescribed
      under the companies Act.1956 .Whereas, there is no standard form prescribed under the Indian partnership Act,1932 for a partnership Firms balance sheet.
   2. In case of a company‘s Balance sheet previous year‘s figures are required to be given whereas it is not so in the case of a partnership firms balance sheet.

10 Prepare Comparative and Common Size income statement from the following information for the year‘s ended march 31, 2008 and 2009.
Particulars2008(Rs.)2009(Rs.)1.Net Sales
2.Cost of Goods Sold
3.Indirect Expenses
4.Income Tax rate8,00,000
60% of sales
10% of Gross profit
50%10,00,000
60% of sales
10% of Gross Profit
60%Ans.Comparative Income statement:articular2008
amount2009
amountChange in
amountChange in
PercentageNet Sales
Less: C.O.G.S.8,00,000
4,80,00010,00,000
6,00,0002,00,000
1,20,00025%
25%Gross Profit
Less: Indirect Expenses3,20,000
32,0004,00,000
40,00080,000
8,00025%
25%Operating Profit/ PBT
Less: tax2,88,000
1,44,0003,60,000
2,16,00072,000
72,00025%
50%Profit after tax1,44,.0001,44,000———-————Common Size Income statementParticular2008
amount2009
amountPercentage of
Net sales in
P.Y.Percentage of
Net sales in
C.Y.Net Sales
Less: C.O.G.S.8,00,000
4,80,00010,00,000
6,00,000100%
60%100%
60%Gross Profit
Less: IndirectExpenses3,20,000
32,0004,00,000
40,00040%
4%40%
4%Operating Profit/ PBT
Less: tax2,88,000
1,44,0003,60,000
2,16,00036%
18%36%
21.6%Profit after tax1,44,.0001,44,00018%14.4%

In This Post we are  providing Chapter- 5 ACCOUNTING RATIOS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ACCOUNTING RATIOS

1.OM Ltd has a current ratio of 3.5 : 1 and quick ratio of 2 : 1. If the excess of current assets over quick assets as represented by inventory is Rs 1,50,000, calculate current assets and current liabilities. (Delhi2012)
Ans.

2.X Ltd has a current ratio of 3 : 1 and quick ratio of 2 :1. If the excess of current assets over quick assets as represented by inventory is Rs 40,000, calculate current assets and current liabilities. (All India 2012)
Ans.

3.From the following information, calculate any two of the following ratios
(i)Debt-equity ratio
(ii)Working capital turnover ratio (iii) Return on investment
Information Equity share capital Rs 10,00,000, general reserve Rs 1,00,000, balance of statement of profit and loss after interest and tax Rs 3,00,000, 12% debentures Rs 4,00,000, creditors Rs 3,00,000, land and buildings Rs 13,00,000, furniture Rs 3,00,000, debtors 12,90,000, cash Rs 1,10,000.Revenue from operations i.e. sales for the year ended 31st March, 2011 was Rs 30,00,000. Tax rate is 50%. (All India 2012; Modified)
Ans.

4.From the following information, calculate the following ratios
(i)Liquid ratio
(ii)Proprietary ratio

Ans.

5.From the following information, calculate any two of the following ratios
(i) Operating ratio      (ii) Inventory turnover ratio (iii) Proprietary ratio

Ans.

6.From the following information, calculate any two of the following ratios (i) Liquid ratio (ii) Debt equity ratio

Ans.

7.From the following information, calculate any two of the following ratios (i) Net profit ratio (ii) Debt equity ratio

Ans.

8.From the following calculate the ‘gross profit ratio’ and ‘working capital turnover ratio’:

Ans.

9.(i)From the following information, compute ‘debt equity ratio’

(ii) The current ratio of X Ltd is 2 : 1. State with reason which of the following transaction would increase, decrease or not change the ratio 
(a)Included in the trade payables was a bills payable of Rs 9,000 which was met on maturity.
(b)Company issued 1,00,000 equity shares of Rs 10 each to the vendors of machinery purchased. (Delhi 2014)
Ans.

10.The quick ratio of a company is 1.5 : 1. State with reason which of the following
transactions would (a) increase (b) decrease or (c) not change the ratio
(a)Paid rent Rs 3,000 in advance.
(b)Trade receivables included a debtor Shri Ashok who paid his entire amount due Rs 9,700.
(ii) From the following information compute ‘proprietary ratio’

Ans. (i) (a) Not change the ratio
Reason As there is a simultaneous increase and decrease in current asset, i.e. prepaid expenses and cash, therefore it will not affect the value of current asset.
(b) Not change the ratio
Reason As there is a simultaneous increase and decrease it will not affect the value of current asset.

In This Post we are  providing Chapter- 1 ACCOUNTING FOR A SHARE CAPITAL NCERT MOST IMPORTANT QUESTIONS for Class 12 ACCOUNTANCY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ACCOUNTING FOR A SHARE CAPITAL

Illustration 1: S T L Global Ltd. was formed with a nominal Share Capital of Rs. 40,00,000 divided into 4,00,000 shares of Rs. 10 each. The Company offers 1,30,000 shares to the public payable Rs. 3 per share on Application, Rs. 3 per share on Allotment and the balance on First and Final Call. Applications were received for 1,20,000 shares. All money payable on allotment was duly received, except on 200 shares held by Y. First and Final Call was not made by the Company.
How would you show the relevant items in the Balance Sheet of STL Global Ltd.?
Solution 1:
Balance Sheet (Extract) of S T L Global Ltd. (Relevant Part only)
As at ________________

Illustration 2: On 1st April, 2012, Janta Ltd. was formed with an authorized capital of 50,00,000 divided into 1,00,000 equity shares of 50 each. The company issued prospectus inviting application for 90,000 Shares. The issue price was payable as under:
On Applicant : Rs. 15
On Allotment : Rs. 20
On call : Balance amount
The issue was fully subscribed and the company allotted shares to all he applicants. The company did not make the call during the year.
Show the following :

1. Share capital in the Balance Sheet of the company as per revised schedule – VI, Part-I of the companies Act, 1956.
2. Also prepare Notes to Account’s for the same.

Solution :
Balance Sheet of Janta Ltd.
As at……………………. (As per schedule iii)

Notes to Accounts

Issue of shares against Lump sum payment : When whole amount due on shares is payable in one instalment. The journal entries will be as follow:
Illustration 3 : Vaibhav Ltd. issued 1,00,000 shares of Rs. 10 each at per. The whole amount was payable with application. Pass the necessary journal entries in the books of company.
Solution:
Journal

Illustration 4 : X Ltd. invited application for 10,000 shares of the value of Rs. 10 each. The amount is payable as Rs. 2 on application and Rs. 5 on allotment and balance on First and Final Call. The whole of the above issue was applied and cash duly received. Give Journal entries for the above transaction.

Illustration 5 : V Ltd. Issued 20,000 Equity shares of Rs. 10 each at a premium of Rs. 3 payable as follows:
On Application Rs. 4
On Allotment Rs. 5 (including Securities Premium Reserve)
On First Cell Rs. 2
On Final Call Rs. 2
All shares were duly subscribed and all money duly received. Pass necessary Journal Entries.
Solution :
In the Book of X Ltd.

Issue of shares at discount [Section 53] : A company cannot issue shares at discount other than sweat equity shares.
Shares Issue for Consideration Other than Cash
When a company purchases any fixed asset or business and makes the payment to the vendor in form of issue of shares in place of cash it is called the issue of shares for consideration other than cash.
Share can be issued at par, at premium.
Journal entries for issue of shares to vendors/consideration other than cash

Note: When name of vendor is given then we write the name of vendor
Illustration 6 : Atlas Co. Ltd. Purchased a machine from HMT Co. for Rs 64,000. It was decided to pay Rs. 10,000 in cash and balance will be paid by issue of shares of Rs. 10 each,
Pass journal entries shares

1. Issued at par
2. Issued at premium of 20%

Solution :
Journal

Illustration 7 : A company issued 15,000 fully paid up equity shares of Rs. 100 each for the purchases of the following assets and liabilities from Gupta Bros.
Plant – Rs. 3,50,000; Stock Rs. 4,50,00;
Land and Building Rs. 6,00,000; Sundry Creditors Rs. 1,00,000
Pass necessary Journal entries.
Solution:
Journal

Note : Calculation : Goodwill = Purchases consideration + Liabilities – assets = Rs. 15,00,000 + Rs. 1,00,000 = Rs, 14,00,000 Rs. 2,00,000.
Illustration 8 : A company purchased a running business from Mahesh for a sum of Rs.
1,50,000 payable as Rs. 1,20,000 in fully paid equity shares of Rs. 10 each and balance in cash. The assets and liabilities consisted of the following Plant and Machinery Rs. 40,000; Stock Rs. 50,000; Building Rs. 40,000; Cash Rs, 20,000 Sundry debtors Rs, 30,000; Sundry creditors Rs. 20,000
Pass necessary Journal entries.
Solution
Journal

Note : Calculation; Net assets – liabilities = Rs. 1,800,000- Rs. 20,000 Rs. 1,60,000 Capital reserve = Net Asset – Purchase consideration = Rs. 1,60,000 – Rs. 1,50,000 = Rs. 10,000
Illustration 9 : Pass necessary journal entries for the following transactions in the Books of Rajan Ltd.

1. Rajan Ltd. purchased machinery of Rs. 7,20,000 from Kundan Ltd. The payment was made to Kundan Ltd. by issue of equity shares of Rs. 100 each at 20% Premium.
2. Rajan Ltd. purchased a running business from Vikas Ltd. for a sum of Rs. 2,50,000 payable as Rs. 2,20,000 in fully paid equity shares of Rs.10 each and balance by a bank draft. The assets and liabilities consisted of the following:
   Plant & Machinery Rs, 90,000; Buildings Rs, 90,000;
   Sundry Debtors Rs. 30,000; Stock Rs. 50,000; Cash Rs. 20,000;
   Sundry Creditors Rs. 20,000

Solution
Rajan Ltd.
Journal

Illustration 10 : Ram holding 10 shares of Rs. 10 each of which Rs. 2 on application Rs. 3 on allotment but could not pay Rs. 3 on first call. His shares were forfeited by the Directors. The Final call is not made as yet. Give Journal entries in the book of company.
Solution:
Journal

Forfeiture of Shares Issued at Premium :

1. when the premium has been received;
2. When the premium has not been received.

Case 1: When the premium has been received : In such cases premium received will not be forfeited and will not record anywhere in the forfeiture journal entry.
Journal

In This Post we are  providing Chapter- 10 WAVE OPTICS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON WAVE OPTICS

1.(a) State Huygens’s principle for constructing wavefronts?
(b) Using Huygens’s principle deduce the laws of reflection of light?
(c) What changes in diffraction pattern of a single slit will you observe. when the monochromatic source of light is replaced by a source of white light?
Ans.(a) According to Huygens’s principle
(1) Each source of light spreads waves in all directions.
(2) Each point on the wavefront give rise to new disturbance which produces secondary wavelets which travels with the speed of light.
(3) Only forward envelope which encloses the tangent gives the new position of wavefront.
(4) Rays are always perpendicular is the wavefront.
(b) A plane wave front AB incident at A hence every point on AB give rise to new waves. Time taken by the ray to reach from P to R







Since all the secondary wavelets takes the same time to go form the incident wavefront to the reflected wavefront so it must be independent of Q
i.e sin i– sin r = O
sin i = sin r
or i = r  law of Reflection of light
(c) (1) The diffracted light consists of different colours.
(2) It results in overlapping of different colours.

2.(a) Coloured spectrum is seen, when we look through a muslin cloth. Why?
(b) What changes in diffraction pattern of a single slit will you observe. when the monochromatic source of light is replaced by a source of white light?
Ans. (a) Muslin cloth consist of very fine threads which acts as fine slits and when light pass through it, light gets diffracted giving rise to a coloured spectrum.
(b) (i) Diffracted lights consist of different colours
(ii) It results in overlapping of different colours.

3.A slit of width ‘a’ is illuminated by light of wavelength. For what value
of ‘a’ will the :-
(i) First maximum fall at an angle of diffraction of?
(ii) First minimum fall at an angle of diffraction?
Ans.

(1) For first maximum



(2) For first minimum





 

4.(a) Derive all expression for the fringe width in young’s double slit experiment?
(b) If the two slits in young’s double slit experiment have width ratio 4:1, deduce the ratio of intensity of maxima and minima in the interference pattern?
Ans.Path difference between

——–(A)


——(1)
Using Binomial theorem expand equation. (1) and neglect higher terms

Similarity ——(2)
Substituting equation (1) & (2) in equation (A)





For bright fringes
Path difference = 




For fringe width


(b) 


Using 
 

5.Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index o
f water is 1.33.
Ans.Wavelength of incident monochromatic light,
 = 589 nm = 
Speed of light in air, c = 3 × 108 m/s
Refractive index of water,  = 1.33
(a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,


















Hence, the speed, frequency, and wavelength of the reflected light are  m/s,  Hz, and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency,  = Hz
Speed of light in water is related to the refractive index of water as:


Wavelength of light in water is given by the relation,




Hence, the speed, frequency, and wavelength of refracted light are  respectively.

6.In Young’s double-slit experiment using monochromatic light of wavelength , the intensity of light at a point on the screen where path difference is , is K units. What is the intensity of light at a point where path difference is  /3?
Ans.Let be the intensity of the two light waves. Their resultant intensities can be obtained as:

Where,
= Phase difference between the two waves
For monochromatic light waves,



Phase difference = 
Since path difference =  ,
Phase difference, 

Given,
I‘ = K
 …(1)
When path difference,
Phase difference, 
Hence, resultant intensity, 

Using equation (1), we can write:

Hence, the intensity of light at a point where the path difference is is units.

7. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 
Ans.Wavelength of the light beam, 



Wavelength of another light beam, 
Distance of the slits from the screen = D
Distance between the two slits = d
(a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

For third bright fringe. N=3

(b) Let the nth bright fringe due to wavelength and (n – 1)th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:

520n=650n-650
650=130n

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


Note: The value of d and D are not given in the question.

8.Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
Ans.No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression:
C sin i=v sin r
C sin i=v sin r
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as:

Hence, equation (i) reduces to

But, > 1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v.
The wave picture of light is consistent with the experimental results.

9.Answer the following questions: (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Ans.(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.
(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.
 

10.Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Ans.Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Thus, the radial spread of the radio waves should not exceed 50 km.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:
 = 20 km =  m
Aperture can be taken as:
a = d = 50 m
Fresnel’s distance is given by the relation,

Where,
 = Wavelength of radio waves


Therefore, the wavelength of the radio waves is 12.5 cm.Search

In This Post we are  providing Chapter- 11 DUAL NATURE OF RADIATION AND MATTER NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON DUAL NATURE OF RADIATION AND MATTER

Question 1.
Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. (Delhi 2016)
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.

Question 2.
Sketch the graph showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v_A > v_B.
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain. (All India 2016)
Answer:

Question 3.
The graphs, drawn here, are for the phenomenon of photoelectric effect.


(i) Identify which of the two characteristics (intensity/frequency) of incident light, is being kept constant in each case.

(iii) Justify the existence of a ‘threshold frequency’ for a given photosensitive surface. (Comptt. Delhi 2016)
Answer:
(i) (a) In graph 1 : intensity is being kept constant.
(b) In graph 2 : frequency is being kept constant.
(ii) (a) In graph 1 : Saturation current
(b) In graph 2 : Stopping potential.
(iii) For a given photo-sensitive surface electrons need a minimum energy to be emitted, this is called work function of the surface W.
∴ Photons energy hv should be greater/ equal to the work function.

which is justified to be called as threshold frequency.

Question 4.
Point out two distinct features observed experimentally in photoelectric effect which’ cannot be explained on the basis of wave theory of light. State how the ‘photon picture’ of light provides an explanation of these features. (Comptt. All India 2016)
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.

(b) Basic features of photon picture of electromagnetic radiation :
(i) Radiation behaves as if it is made of particles like photons. Each photon has energy E = hv and momentum p = h/λ.
(ii) Intensity of radiation can be understood in terms of number of photons falling per second on the surface. Photon energy depends only on frequency and is independent of intensity.
(iii) Photoelectric effect can be understood as the result of one to one collision between an electron and a photon.
(iv) When a photon of frequency
(v) is incident on a metal surface, a part of its energy is used in overcoming the work function and other part is used in imparting kinetic energy, so KE = h(v – v₀).

Question 5.
(i) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
(ii) The work function of the following metals is given : Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and Ni 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away? (Delhi 2017)
Answer:

Because the work function of Mo and Ni is more than the energy of the incident photons; so photoelectric emission will not take place from these two metals Mo and Ni. When the laser source is brought nearer and placed 50 cm away, the kinetic energy of photo-electrons will not change, only photoelectric current will change.

Question 6.
In the study of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown here:
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10¹⁴ Hz for this metal.
Answer:

Question 7.
Explain giving reasons for the following:
(a) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation.
(b) The stopping potential (V₀) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces.
(c) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
Answer:
(a) The collision of a photon can cause emission of a photoelectron (above the threshold frequency). As the intensity increases, number of photons increases. Hence, the current increases.

hence, it depends on the frequency and not on the intensity of the incident radiation.

Question 8.
The given graph shows the variation of photocurrent for a photosensitive metal:

(a) Identify the variable X on the horizontal axis.
(b) What does the point A on the horizontal axis represent?
(c) Draw this graph for three different values of frequencies of incident radiation v₁ v₂ and v₃ (v₁ > v₂ > v₃) for same intensity.
(d) Draw this graph for three different values of intensities of incident radiation I₁ I₂ and I₃ (I₁ > I₂ > I₃) having same frequency.
Answer:
(a) ‘X’ is a collector plate potential.
(b) ‘A’ represents the stopping potential.
(c) Graph for different frequencies :

Question 9.
Draw a graph showing the variation of de Broglie wavelength of a particle of charge q and mass m with the accelerating potential. Proton and deuteron have the same de Broglie wavelengths. Explain which has more kinetic energy.
Answer:

Question 10.
(a) Draw the graph showing the variation of de Broglie wavelength of a particle of charge q and mass m with the accelerating potential.
(b) An electron and proton have the same de Broglie wavelengths. Explain, which of the two has more kinetic energy.
Answer:
(a) For Graph :


Since the mass of electron is less than that of proton, hence electron will have more kinetic energy.

In This Post we are  providing Chapter- 15 COMMUNICATION SYSTEM NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON COMMUNICATION SYSTEM

Question 1.
flow will you classify’ communication systems?
Answer:
Communication systems can be classified based on the nature of source. mode of communication. type of modulation and nature of channel used.

Question 2.
What are the lips of channels used for transmission?
Answer:
i. Space communication (Broadcasting, microwave mobile etc.)
ii. Line communication (Two wire, co-axial cables, fiber optical etc.)

Question 3.
What is the length of antenna required to transmit wave of frequency 40 Hz and 40 MHz?
Answer:
The minimum length of antenna required is λ4
Velocity, c = υλ
λ = cυ

Question 4.
Which physical quantity of wave is varied in AM. FM and PM?
Answer:
In AM. the physical quantity of carrier that changes is amplitude.
In FM, the physical quantity of carriers that change is frequency.
In PM, the physical quantity of carrier that changes is a phase.

Question 5.
What are the advantages and limitations of AM and FM?
Answer:

Question 6.
What is ground wave propagation?
Answer:
Ground wave follows curvature of the earth and has carrier frequencies up to 2MHz. e.g. AM radio.
Ground waves progress along the surface of the earth and must be vertically polarized to prevent short circuiting the electric equipments. A wave induces currents in the ground over which it passes and thus loses some energy by absorption. This is made up by energy diffracted downwards from the upper portion of the wavefront.

There is another way also by which the ground waves get attenuated. Because of diffraction, the wavefront gradually tilts over, as shown in the figure. As the wave propagates over the earth, the tilt increases and this tilt causes greater short-circuiting of the electric component of the wave. Hence there is a reduction in the field strength. Eventually, at some distance from the antenna, the wave gets weakened and dies off. The maximum range of such a transmitter depends on its frequency and power. The ground wave propagation is effective only at VLF.

Question 7.
Through which atmospheric layer. does the propagation take place in ground. space and sky communications?
Answer:
Ground wave – Troposphere
Space wave – Troposphere
Sky wave – Ionosphere

Question 8.
For establishing a communication between a transmitting and receiving station, a physical medium is used.
Answer:
(a) Name the two principal classes of communication based on the physical medium used for propagation.
(b) Construct a table showing advantages and one practical application each for the two wire, coaxial cable and optic fiber communication.
(c) in cable TV transmission usually channel in UHF band carries relatively more noise, compared to VHF band. Justify
Answer:
(a) Line communication and space communication.

(c) At higher frequency, radiation loss is high.

Question 9.
Schematic diagram for three types of satellite orbits are shown below and named as A.B.C.
Answer:

(a) Identify the polar orbit and give its approximate height from earth.
(b) Give the criteria for selecting frequency of em wave to be used in photographs from satellites.
(c) A satellite T V company attempts to use 25,000 kHz for up linking signal to a sat ellite. Say whether they have selected apt frequency. Justify.
Answer:
(a) Orbit C. Its height is about 1000 km.
(b) i. Nature of the atmosphere.
ii. Reluctance of the object.
(c) No. Because frequency below 20 MHz will undergo total internal reflection at the ionosphere.

Question 10.
The following diagrams represent some of the modulated signals.



Which among is following correct
a. i only
b. ii only
c. iii only
d. both i and ii
Answer:
d. both i and ii

In This Post we are  providing Chapter- 14 SEMICONDUCTOR ELECTRONIC : MATERIAL, DEVICES AND SIMPLE CIRCUITS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON SEMICONDUCTOR ELECTRONIC : MATERIAL, DEVICES AND SIMPLE CIRCUITS

1.Draw a pn junction with reverse bias? Which biasing will make the resistance of a
p-n-junction high?
Ans.

Reverse biasing will make the resistance high as it will not allow the current to pass.

2.Write the truth table for the following combination of gates?

Ans.

3.Draw the voltage current characteristics of a zener diode?
Ans.

4.For a extrinsic semiconductor, indicate on the energy band diagram the donor and
acceptor levels?
Ans. N-type Extrinsic Semiconductor P-type Extrinsic Semiconductor

5.What do you mean by depletion region and potential barrier in junction diode?
Ans.A layer around the junction between p and n-sections of a junction diode where charge carriers electrons and holes are less in number is called depletion region. The potential difference created across the junction due to the diffusion of charge carriers across the junction is called potential barrier.

6.A transistor has a current gain of 30. If the collector resistance is 6k, input resistance is 1k, calculate its voltage gain?
Ans.Given 


Voltage gain = current gain Rgain
Voltage gain = 306 = 180

7. What are the advantages and disadvantages of semiconductor devices over vacuum tubes?
Ans.Advantages – Semiconductor devices are very small in size as compared to the vacuum tubes. It requires low voltage for their operation
Disadvantage – Due to the rise in temperature and by applying high voltage it can be damaged.

8.The base of a transistor is lightly doped. Explain why?
Ans.In a transistor, the majority carries form emitter region moves towards the collector region through base. If base is made thick and highly doped, majority carriers will combine with the other carriers within the base and only few is collected by the collector which leads to small output collector current. Thus in order to have large output collector current, base is made thin and lightly doped.
 

9. Determine the currents through resistance R of the circuits (i) and (ii) when similar diodes are connected as shown in the figure.

Ans.In figure (i) are forward biased

In figure (ii) is forward biased but is reverse biased due to which  offers infinite resistance

10.What do you mean by hole in a circuit? Write its two characteristics?
Ans.A vacancy created in a covalent bond in a semiconductor due to the release of electron is known as hole in a semiconductor.
Characteristics of hole
(i) Hole is equivalent to a positive electronic charge.
(ii) Mobility of hole is less than that of an electron

In This Post we are  providing Chapter- 12 ATOMS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ATOMS

Question 1.
A hydrogen atom in its excited state emits radiations of wavelengths 1218 Å and 974.3 Å when it finally comes to the ground state. Identify the energy levels from where transitions occur. Given Rydberg constant R = 1.1 × 10⁷ m^-1. Also, specify the spectral series to which these lines belong.
Answer:
We know that


On solving n = 4
Lyman series.

Question 2.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, i.e. an atom where the electron is replaced by a negatively charged muon (μ^–) of mass about 207 me that orbits around a proton. (Given for hydrogen atom, the radius of first orbit and ground state energy are 0.53 × 10^-10m and – 13.6 eV respectively)
Answer:
In Bohr’s model of hydrogen atom the radius of n^th orbit is given by
r_n = n2h24π2e2mek

As n = 1
Therefore, we have 1

Question 3.
The electron in a given Bohr orbit has a total energy of – 1.5 eV. Calculate Its
(a) kinetic energy.
(b) potential energy.
(C) the wavelength of radiation emitted, when this electron makes a transition to the ground state.
(Given Energy in the ground state = – 13.6 eV and Rydberg’s constant = 1.09 × 1o⁷ m^-1)
Answer:
Total energy of the electron In a Bohr’s orbit is – 1.5 eV

We know that kinetic energy of the electron in any orbt is half of the potential energy in magnitude and potential energy is negative
(a) Total energy = kinetic energy + potential energy
– 1.5 = E_k – 2E_k
1.5 = E_k

(b) E_p = – 2 × 1.5 = – 3 eV

(c) Energy released when the transition of this electron takes place from this orbit to the ground state
= – 1.5 – (- 13.6)
= 12.1 eV
= 12.1 × 1.6 × 10¹⁹ = 1.936 × 10^-18 J

Let be the wavelength of the Light emitted then,

Question 4.
In a Geiger—Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an a-particle of 8 MeV energy Impinges on it before It comes momentarily to rest and reverses its direction. How will the distance of the closest approach be affected when the kinetic energy of the a-particle is doubled?
Answer:
Given Z = 80, E = 8 MeV = 8 × 10⁶ × 1.6 × 10¹⁹ J, r_o =?
Using the expression

When the kinetic energy of a-particle has doubled the distance of the closest approach becomes half its previous value, i.e. 1.44 × 10¹⁴ m

Question 5.
The ground state energy of the hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to – 3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
Answer:
Energy released = – 0.85 – (- 3.4) = 2.55 eV = 2.55 × 1.6 × 10^-19 J
Using E = hc/λ we have
λ = hcE=6.62×10−34×3×1082.55×1.6×10−19 = 4.87 × 10^-7 m

It belongs to the Balmer series.

Question 6.
A hydrogen atom in the third excited state de-excites to the first excited state. Obtain the expressions for the frequency of radiation emitted in this process.
Also, determine the ratio of the wavelengths of the emitted radiations when the atom de-excites from the third excited state to the second excited state and from the third excited state to the first excited state.
Answer:
We know that


Question 7.
Obtain the expression for the ratio of the de Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom.
Answer:
We know that
2πr = nλ ….(i)
For the second excited state (n = 3)
r = 0.529(n)² A = 0.529(3)²

Putting in (i) we get 2π(0.529)(3)² = 3 λ₂

For third excited state n = 4
r = 0.529 (4)²
Putting in (i) we get 2π (0.529)(4)² = 4 λ₃

Question 8.
(a) The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-10 m. Calculate its radius in n = 3 orbit.
(b) The total energy of an electron in the first excited state of the hydrogen atom is 3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.
Answer:
(a) The radius is given by
r_n = n2h24π2me2k = n²ao

where n is the number of orbit, hence r3 = 5.3 × 10^-10 × 32 = 4.77 × 10^-9 m

(b) Total kinetic energy = + 3.4 eV
Total potential energy = – 6.8 eV

Question 9.
Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Answer:
The de-Broglie wavelength is given by 2πrn = nλ.
In ground state, n = 1 and r_o = 0.53 Å, therefore, λ_o = 2 × 3.14 × 0.53 = 3.33 Å

In first excited state, n = 2 and r₁ = 4 × 0.53 Å = 2.12 Å, therefore,
r₁ = (2 × 3.14 × 2.12)/2 = 6.66Å

Therefore, λ₁ – λ_o = 6.66 – 3.33 = 3.33 Å.
In other words, the de-Broglie wavelength becomes double.

Question 10.
When is the H_α line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.
Answer:
Hα is obtained when n_i = 3 and n_f = 2

In This Post we are  providing Chapter- 13 NUCLEI NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON NUCLEI

Question 1.
Calculate the binding energy per nucleon of Fe⁵⁶₂₆ Given m_Fe = 55.934939 u, m_n = 1.008665 u and m_p = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z m_p + (A – Z)m_n – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 10²³ nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10^-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 10⁹ years. Calculate the activity of 1 g sample of ₉₂²³⁸U.
Answer:
Given T = 4.5 × 10⁹ years.
Number of nuclei of U in 1 g
= N = 6.023×1023238 = 2.5 × 10²¹

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day^-1
or
T_1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = N_oe^-λt
or
N = N_o e^{-0 3456 × 4}
or
N = N₀ e^{-1 3824} = N_o × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/N_o = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day^-1

Question 7.
A radioactive substance decays to 1/32^th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = N_o / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T₁/2 = 30 s, N = 3N_o / 4, λ = ?, t = ?
(i) Decay constant
λ = 0.693T1/2=0.69330 = 0.0231 s^-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = N_oe^-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given N_ox = N_oY, T_X = 1 h, T_Y = 2 h, therefore
λXλY=21 = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are R_x and R_y respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 10³² kg and it generates energy at the rate of 5 × 10³⁰ watt. How long will it take to convert all the helium to carbon at this rate?

Answer:

As 4 × 10^-3 kg of He consists of 6.023 × 10²³ He nuclei so 5 × 10³² kg He will contain
6.023×1023×5×10324×10−3 = 7.5 × 10⁵⁸ nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10^-13 J of energy
So all nuclei in the star will produce
E = 7.27×1.6×10−133 × 7.5 × 10⁵⁸
= 2.9 × 10⁴⁶ J

As power generated is P = 5 × 10³⁰ W, therefore time taken to convert all He nuclei into carbon is
t = EP=2.9×10465×1030 = 5.84 × 10¹⁵ s
or
1.85 × 10⁸ years

In This Post we are  providing Chapter-4 MOVING CHARGES AND MAGNETISM NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON MOVING CHARGES AND MAGNETISM

Question 1.
Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle.
Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Question 2.
Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.
(i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path.
(ii) What is resonance condition? How is it used to accelerate the charged particles? (All India 2017)
Answer:
(i) Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

(ii) The frequency v_a of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one-half of the revolution. The requirement v_a = v_c is called the resonance condition.
The phase of the supply is adjusted so that when the positive ions arrive at the edge of D₁, D₂ is at a lower potential and the ions are accelerated across the gap.

Question 3.
(a) Two straight long parallel conductors carry currents I₁ and I₂ in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them.
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
(i) What is the direction of the magnetic moment of the current loop?
(ii) When is the torque acting on the loop
(A) maximum,
(B) zero?

Answer:
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(i) Magnetic moment will be out of the plane from the surface HEFG.
(ii) Torque
(A) Torque is maximum when MII B i.e., when it gets rotated by 90°.
(B) Torque is minimum when M and B are at 270° to each other.

Question 4.
(a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? (All India)
Answer:
(a) Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.

(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) For radial magnetic field, sin θ = 1,
so torque τ = NIAB.
Thus when radial magnetic field is used, the deflection of the coil is proportional to the current flowing through it. Hence a linear scale can be used to determine the deflection of the coil.

(c) A high resistance is joined in series with a galvanometer so that when the arrangement (voltmeter) is used in parallel with the selected section of the circuit, it should draw least amount of current. In case voltmeter draws appreciable amount of current, it will disturb the original value of potential difference by a good amount.

To convert a galvanometer into ammeter, a shunt is used in parallel with it so that when the arrangement is joined in series, the maximum current flows through the shunt, and thus the galvanometer is saved from its damage, when the current is passed through ammeter.

Question 5.
(a) Derive an expression for the force between two long parallel current carrying conductors.

(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?
Answer:
(a) For (a) and (b) :
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(c) Force experienced by the proton,

As magnetic field due to the current carrying wire is directed into the plane of the paper (θ = 90°)

Force is directed away from the current carrying wire or in the right direction of observer.

Question 6.
State Biot-Savart law, giving the mathematical expression for it.
Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet? (Delhi 2011)
Answer:
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.


Magnetic field on the axis of circular current loop :

As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain

In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

Question 7.
With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles. Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.

Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Yes, there is an upper limit. The increase in the kinetic energy of particles is qv. Therefore, the radius of their path goes on increasing each time, their kinetic energy increases. The lines are repeatedly accelerated across the dees, untill they have the required energy to have a radius approximately that of the dees. Hence, this is the upper limit on the energy required by the particles due to definite size of dees.

Question 8.
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement
(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. (All India 2011)
Answer:
(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Since vs=IsR increase in current sensitivity may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(c) Conversion of galvanometer into ammeter: By just connecting a low resistance known as shunt in parallel to the galvanometer, it can be converted into an ammeter.

Let G = resistance of the galvanometer.
I_g = the current with which galvanometer gives full scale deflection.
S = shunt resistance
I – I_g = current through the shunt.
As the galvanometer and shunt are connected in parallel,
Potential difference across the galvanometer = Potential difference across the shunt

Question 9.
(a) Write the expression for the force, F→, acting on a charged particle of charge ‘q’, moving with a velocity latex]\overrightarrow{\mathbf{v}}[/latex] in the presence of both electric field E→ and magnetic field B→. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field B→. Prove that the torque τ⃗  acting on the loop is given by τ⃗ =m⃗ ×B→, where m→ is the magnetic moment of the loop. (All India 2011)
Answer:
(a) A charge q in an electric field E→ experiences the electric force, F→e=qE→

This force acts in the direction of field E→ and is independent of the velocity of the charge.

The magnetic force experienced by the charge q moving with velocity v→ in the magnetic field B is given by

This force acts perpendicular to the plane of V→ and B→ and depends on the velocity v→ of the charge.

The total force, or the Lorentz force, experienced by the charge q due to both electric and magnetic field is given by

Hence, A stationary charged particle does not experience any force in a magnetic field. (b) Torque on a current loop in a uniform magnetic field.
Let I = Current flowing through the coil PQRS

Its magnitude is, F3 = IaB sin(90° + 0)

Question 10.
(a) Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter and
(ii) an ammeter.
(b) Two long straight parallel conductors carrying steady currents I₁ and I₂ are separated by a distance’d’ Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
Answer:
(a) (i) Voltmeter is connected in parallel with the circuit element across which the potential difference is intended to be measured.

A galvanometer can be converted into a voltmeter by connecting a higher resistance in series with it. The value of this resistance is so adjusted that only current I which produces full scale deflection in the galvanometer, passes through the galvanometer.

(ii) A galvanometer can be converted into an ammeter by connecting a low value
resistance in parallel with it.


(b)
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A