Shorts Ans of 2nd Physics

SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Q#1

The potential is constant throughout a given region of space .is the electric field is zero or non zero in this region? Ans#1:

Electr Electric ic field field E and potent potential ial differe difference nce are related related by equati equation, on, E =-∆ v/∆ r When the potential is taken as constant constant which means v= constant constant then the potential difference becomes zero ∆ v=0 so, E =0/r ∆ → E=0 This This rela relati tion on tells tells that that electr electric ic field field is zero in a regio region n of where where the the potential is constant Q#2:

Suppose that you follow an electric field line due to a positive point charge .do electric field and the potential increases or decreases? Ans#2:

Relation for the electric field and and electric potential are 2 and v=1/4π ε0 x q/ r E= 1/4π 1/4π ε0 x q/r this gives for same charge

E α 1/ r2 Vα 1/ r When we follow electric field line due to positive charge, distance from pos posit itiv ivee charg chargee incr increa ease sess whic which h case casess to decr decrea ease se elec electri tricc field field and and potential. Q#3: How can you identify that which plate of a capacitor is positively charged? Ans#3: When a freely suspended positively charged gold leaf of electroscope is brought near to one of the plates of capacitor the divergence to gold leaf show showss that that the the plat platee of the the capa capaci cito torr is posi positi tive vely ly charg chargee d othe otherw rwis isee negatively charged.. Q#4: Desc Descri ribe be the the forc forcee or forc forces es on a posi positi tive ve poin pointt ch char arge ge when when placed between parallel plates, (a) With similar and equal charges. (b) With opposite opposite and equal charges.

Prepared By; Asif Rasheed BS(Hons) Physics Contact#0344-7846394

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2 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Ans#4: The electric force F on a charge q due to eclectic eclectic field E is given by F= qE (A) When a positive point charge is placed between parallel plates with similar similar and equal charges. charges. it will experience experience no electric electric force force because because electric field between the plate is zero . F=qE

F=q(0) =0 → F=0 (B) When When a posi positi tive ve pint pint char charge ge is plac placed ed betw betwee een n the the plat plates es of the the F=qE S Sca Scapacitor it will experie rience electric ric force rce ,

Due to electric field set up from positive to negative . Q#5: Electric lines of forces never cross each other. Why? Ans#5: No, two electric lines of forces never cross each other because “E” only one direction at a given point .if two lines cross each other then there will be more directions at a point which is impossible. Q#6: If appoint charge q of mass m is released in a non uniform electric field with field lines pointing in the same direction, will it make a rectilinear motion? Ans#6: Motion Motion along a straigh straightt line line is called called rectil rectiline inear ar motion motion when a charge “q” of mass “m” is released in a non uniform electric field. It will not move in a straight line and adopts irregular path due to variable electric field. Q#7: Is E necessarily zero inside a charged rubber balloon if balloon is sphe spheri rica cal? l? Assu Assume me that that ch char arge ge is dist distri ribu bute ted d un unif ifor orml mly y over over the the surface ?


3 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Ans#7: Yes E is necessarily zero inside a charged rubber balloon if balloon is spherical and enclosed no charge but chage is distributed uniformly over its surface. According to Gauss’s law.

→ Фe = 0 --------1 Фe =q/ε0=0/ ε0=0 From definition, Фe = E.A--------2 E.A--------2 0= E.A A # 0 so, E=0 Putting equation 1 in eq 2 Q#8: Is it true that Gauss’s law states that the total number of lines of forc forces es cros crosssing ing any any clos closed ed surf surfac acee in the the outwa utward rd dire direct ctio ion n is proportional to the net positive charge enclosed within surface? Ans#8: Yes, it is true that total no of lines of forces crossing any closed surfa surface ce in outw outwar ard d direc directi tion on prop proport ortio iona nall to the the net net posi positi tive ve char charge ge according with surface in the outward direction proportional to the net positive charge enclosed with in the surface . According to Gauss’s law, Фe =q/ε0 = constant x q →

Фe α q here q = net positive charge . Фe =no of electric lines of force Hence no of electric lines of forces are proportional to to net positive charge Q#9: Do electrons tend to go to region of high potential or of low potential? Ans#9: Electrons tend to go to a region of high potential because high potential is more positive than the lower potential as the charge on electron is negative.

CHAPTER # 13 QUESTIONS AND ANSWERS:Q#1:


4 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS A potential difference difference is applied across across the ends of a copper wire .what is the effect of drift velocity (1) Increasing the potential difference. (2) decreasing the length and the temperature of the wire. Ans#1 The uniform velocity that the free electrons acqure opposite to the electric field of battery is called the drift velocity. (1) the increase increase of potent potential ial differen difference ce makes the electr electric ic field strong which cases to increase the drift velocity length an temperature of of the wire which causes to to (2) the decrease in length increase the drift velocity of the free electron in the wire. wire. Q#2: Do bends in wire effects it’s electrical resistance? Ans#3 We know that R=ρ l/A with bends bends in a wire , its length (L) and and area of cross section (A) remains the same .hence bends in a wire do not affect the electrical resistance (R) of the wire . Q#4: Why does the resistance of a conductor rise with temperature? Ans#4 As temperature of the conductor rises the irrational amplitude of the atoms of the conductor increases and hence the probability of collisions of free electrons with them increases .at high temperature , the atoms offer a bigger target-area to free electrons to collide with them and resistance of conducer

increases. Q5#: What Wh at are are the the diff diffic icul ulti ties es in test testin ing g whet whethe herr the the fila filame ment nt of lighting bulb obeys ohm law? Ans#5 At the beginning, when bulb is turned ON its filament is at low low temrature and it obeys Ohm’s law. I α V


5 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Later on on when when current current rises rises to its maximum maximum value .the maximum maximum power power 2 P=I R dissipated across the filament which increases resistance but current increases at lower rate and does not obey ohm’s law Q#6: Is the filament resistance lower or higher in a 500W, 220V light bulb than in 100W, 220V bulb? Ans#6 Resistance of 1 st bulb= R 1=? Resistance of 2 nd bulb =R 2=? Power of first bulb =P1=500 watts Power of 2nd bulb=P2=100 watts Voltage =V=220 volts By applying the formula, P=V 2/R → R=V2/P For1st bulb =1=V R 2/P1 (220)2/500 = 48400/500 = 96.8Ω For 2nd bulb =R 2=V2/P2 (220)2/500 = 48400/100 = 484Ω Q#7: Desc Descri ribe be a circ circui uitt whic which h will will giva giva a cont contin inuo uous usly ly vary varyin ing g potential?

Ans#7 Potentiome Potentiometer ter is an instrument instrument which gives continuously continuously varying potential. Its circuit diagram is given as. When the sliding contact C is ,r moved from A to B

varies from O to R and potential varies from O to E the current following through is given by ; I=E/R …………1) The potential drops across ‘r” is given by V=Ir as I=E/R so V=E/R xr V= r/R E ………………2) This gives us continuously varying potential


SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Q#8: Expl Explai ain n why why the the term termin inal al pote potent ntia iall diff differ eren ence ce of a batt batter ery y decreases when a current drawn from it is increased? Ans#8 The terminal potential difference is given by / Vt= E –Ir ………1) The emf emf E of cell is cons constan tantt .when .when current current I throug through h is increa increases ses ,the ,the produc productt Ir increa increases ses which which cases cases to decrea decrease se the the termin terminal al poten potentia tiall difference Vt. Q#9: What is the wheat stone bridge? How can it be used to determine an unknown resistance? Ans#9 An electrical circuit devised by Professor Charles Wheatstone used to determine unknown resistance is called wheat stone bridge. It consists of four resistances R 1, R2, R3, and R 4. Connected in the form of mesh ABCDA ABCDA . a battery battery of emf E is connected connected between between points points A and C thro throug ugh h a swit switch ch S .a sens sensit itiv ivee galv galvan anom omet eter er is of resi resist stan ance ce Rg is connected between points B and D. Its equation is , R 1/R 2 = R 3/R 4 ……….1)

Determination of unknown resistance: To determine unknown resistance in the arm containing R 4 (R 4 =R x Then known value of R 1, R2. And R 3are so adjusted that galvanometer no deflection .then from equation no i R 1/R 2 = R 3/R x x = R 2/R 2 x R 3

Chapter #14 Q#1 A plan conducting loop is located in a uniform magnetic field that is directed along the x-axis .for what orientation of the loop is the flux a maximum?. For what orientation is the flux a minimum? Ans#1: When plane of loop is placed perpendicular to the magnetic field the flux will be maximum Ф e = EAcos0 = EA (max)


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7 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS When plane of loop is placed parallel to the magnetic field, the flux will be minimum as, Фe = EAcos90 = 0 (min) Q#2 A current in a conductor produces a magnetic field, which can be calculated using Ampere’s law. Since current is defined as the rat of flow of charges, what can you conclude about the magnetic field due to stationary charges? What about moving charges? Ans#2: Magnetic field due to stationary charges is zero. Moving charges produce magnetic field around them. Q#3 Describe the magnetic field inside a solenoid carrying a steady current I” if (a) the length is solenoid is doubled but the number of turns remains the same and (b)the number of turns doubled but the length remains same Ans#3:. The magnetic field due to solenoid is given by , B = μ0 n I As n = N /l so.B = μ 0 (N/l )I Case #1: If the length is double and N remains the same . B = μ0 (N/2l )I = 1/2 μ0 (N/l )I = 1/2 B Which means magnetic field reduce to half Case #2: If the number of turns (N) is double and l remains the same. B = μ0 (2N/l )I = 2 μ0 (N/l )I = 2 B

Which means magnetic field increases to two times? Q#4 At a given instant the proton moves in the x direction in a region wher wheree ther theree is magn magnet etic ic fiel field d in the the ne nega gati tive ve z dire direct ctio ion. n. Wh What at is direction of the magnetic force? Will the proton continue to move the positive x direction ? Explain Ans# 4: If proton moves along the positive x –axis in a region of magnetic field is along negative z –axis .the magnetic force is given by R.H. rule r ule , F b = q(v x B) , this force(F b) is along along the positive positive Y- axis axis and it it will be be start to move in circular path in xy plane around z –axis and it will not


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SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS continue to move in the positive x – direction.

Q#5 Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the charges are reflected in opposite directions, what can you say about them? Ans#5: Two charge particles are projected in a magnetic field perpendicular to its their velocity and they are deflected in opposite directions. As we know that in magnetic field particles are deflected so they are oppositely charge charged d .So, .So, one partic particle le is positi positivel vely y charge charged d and other other is negati negativel vely y charged. Q#6 Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no work done by the magnetic force that acts on the charge q? -Ans#6: A charge q is moving in uniform magnetic field B, with a velocity V in a Circular path due to magnetic force. The angle between magnetic force and velocity will be 90 0 .work done is given by , 90o 0 0 W = F.d = Fd cos 90 =0 (as cos of 90 is zero). So work is being done . F

Q#7 If a ch char arge ged d part partic icle le move movess in a stra straig ight ht line line thro throug ugh h some some region of space, can you say that the magnetic field in the region is zero? Ans#7: No, No, when when a part partic icle le move movess in stra straig ight ht line line in the the dire direct ctio ion n of magnetic field. B

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q

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SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS The force acting on charge will be zero as angle is 0 0 charge q and magnetic field is parallel so, F= qvBsin00 = qvB(0) = 0 Therefore we can not say magnetic field in this zero. Q#8 Why does the picture on TV screen become distorted when a magnet is brought near the screen? Ans#:8 As the picture is formed on the TV screen due to the electrons , when a magnetic is brought is near the TV screen ,the beam of electron is deflected deflected due the magnetic magnetic force force which is given given by, F= evBsinθ evBsinθ 0 .on the each electron. Q#9 Is it possible to orient a current loop in a uniform magnetic field cush that the loop will tend to rotate? Explain. Ans#9 The torque acting on a current carrying loop in the uniform magnetic field is given by, τ(torqu τ(torquee )=NIAB )=NIABcos cosα α ,Where ,Where α is is the angl anglee betwee between n B and plan planee of 0 0 loop .if α= 90 then cos of 90 = 0 so torque will be zero here this shows that if plane of loop is oreintted at 90 0 with B .Then loop will not tend to rotate. r otate. Q#10 How can a current loop be used to determine the presence of a magnetic field in a given region of space? Ans#10: When current carrying loop is placed in the presence of a magnetic field with its plane makes an angle α with B .The torque acting on it will be τ(torque τ(torque )=NIBAcosα )=NIBAcosα if current current loop deflects deflects , filed is present other other wise not.

Q#11 How How can can you you use use a magn magnet etic ic fiel field d to sepa separa rate te isot isotop opes es of chemical element? Ans#11:


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10 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS When isotopes of chemical element are projected in a magnetic field at right angle. then magnetic force , Fb= evBsin90 = evB Provide centripetal force , Fc = mv2/r Bev = mv2/r r= mv/Be Isotopes has same charge , but due to different masses they adopt different paths due to which fall at different places and are separated . Q 12: What should be the orientation of a current carrying coil in a magnetic field so that torque acting upon the coil (a)maximum (b) minimum Ans#12: When When the curren currentt carryi carrying ng wire wire placed placed in the uniform uniform magnet magnetic ic field . the torque acting on it is given by, τ(torque )=NIBAcosα where α is the angle between and plane of the coil (a) when plane of the coil is oriented parallel to to to B .then α=0 and cos0 = 1 so, τ(torque )=NIBA(1) = NIAB(max) α=90 and cos 90 = 0 τ(torque )=NIBA(0) =0 at this the torque will be minimum. Torque will be maximum (b) when plane of coil is oriented at right angle to B then angle will be 90 so, Q#13 A loop of wire is suspended between the poles of a magnet with its plane parallel to the pole faces. What happens if a direct current uis put through the coil? What happens if an alternating current is used instead? Ans#13 : Torque on loop loop wire is given by , τ(torque )=IBAcos α When the loop of wire is suspended between the poles of a magnet with its plan parallel to pole faces then α =90 and cos 90 = 0 then we get the result τ(torque )=IBAcos90 =0 if direct current or alternating current is passed through the coil .these current have no effect on the loop

Q#14


11 SHORT QUESTIONS AND ANSWERS OF SECOND YEAR PHYSICS Why the resistance of an ammeter should be very low? Ans#14: Ammeter is always connected in series the magnitude of current decreases through through the circuit with the presence of ammeter .to measure

I R maxim maximum um curre current nt the the resi resist stan ance ce of amme ammete terr shou should ld be very very small small as compared to the resistance of circuit. Ans#15: Why the voltmeter should have a very high resistance? Ans#15: Voltmeter is always connected I in parallel with two points where potential is to be measured. the potential across the points decreases with presences of voltmeter because it draws some current Iv of the circuit . to reduce the current current through voltmeter voltmeter ,the resistance resistance of voltmeter should be large as compared to the resistance of circuit .

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IR R

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Shorts Ans of 2nd Physics

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