2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 35 InterviewSolution

1701.	The e.m.f. of a cell is independent ofA. quantity of electrolyteB. distance between the electrodesC. area of the electrodesD. all of these
Answer» Correct Answer - D

Discussion

1702.	Find the minimum number of cells required to produce an electric current of 1.5 A through a resistance of `30 Omega`. Given that the emf of each cell is 1.5 V and internal resistance `1.0 Omega`.
Answer» Correct Answer - 120 cells

Discussion

1703.	Electromotive force is the force, which is able to maintain a constantA. CurrentB. ResistanceC. PowerD. Potential difference
Answer» Correct Answer - D

Discussion

1704.	A 10 m long wire ofresistance `20 Omega` is connected in series with battery of EMF `3V` and negligible internal resistance and a resistance of `10 Omega`. The potential gradient along the wire is :A. 0.02B. 0.1C. 0.2D. 1.2
Answer» Correct Answer - C As resistances are in series, `R_("total")=R_(1)+R_(2)=20+10=30 Omega` `i=(4)/(R_("total"))=(3)/(30)=(1)/(10)A` So, `V_("wire")=iR_("wire")=(1)/(10)xx20=2V` Hence, potential gradient is `(V_("wire"))/(l)=(2)/(10)=0.2 Vm^(-1)`

Discussion

1705.	Two cells connected in series have electromotive force of 1.5 V each. Their internal resistances are `0.5 Omega` and `0.25 Omega` respectively. This combination is connected to a resistance of `2.25 Omega`. Calculate the current flowing in the circuit and the potential differnce across the terminals of each cell.
Answer» Correct Answer - `1.0 A, 1.0 V, 1.25 V`

Discussion

1706.	Electromotive force of a cell is basically aA. forceB. powerC. workD. current capacity
Answer» Correct Answer - 3

Discussion

1707.	In S.I. system, unit of electromotive force of a cell isA. newtonB. dyneC. voltD. ampere
Answer» Correct Answer - C

Discussion

1708.	For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit isA. 1.38 VB. 1.68 VC. 2.03 VD. 3.10 V
Answer» Correct Answer - A If i is the current in a wire then the charge flowing through the wire in t sec is `q=it=2xx6xx60=720 C` The work done in taking q coulomb (720 C) of charge from one end of wire to other end under a potential difference of V volt (in this case emf) is W = V q `rArr " " V=(W)/(q)=(1000)/(720)=1.38 V`

Discussion

1709.	Four identical cells each having an electromotive force (e.m.f.) of 12 V , are connected in parallel. The resultant electromotive force (e.m.f.) of the combination isA. `48V`B. `12V`C. `4V`D. `3V`
Answer» Correct Answer - B

Discussion

1710.	In the shown circuit, what is the potential difference across A and B A. `50V`B. `45V`C. `30V`D. `20V`
Answer» Correct Answer - D

Discussion

1711.	Two batteries of e.m.f. `4V` and `8V` with internal resistances `1 Omega` and `2 Omega` are connected in a circuit with a resistance of `9 Omega` as shown in figure. The current and potential difference between the points `P` and `Q` A. `(1)/(3)A` and `3V`B. `(1)/(6)A` and `4V`C. `(1)/(9) A` and `9V`D. `(1)/(2)A` and `12 V`
Answer» Correct Answer - A

Discussion

1712.	Consider the circuit shown in the figure. The current `I_(3)` is equal to A. 5 ampB. 3 ampC. `-3` ampD. `-5//6` amp
Answer» Correct Answer - D

Discussion

1713.	If `V_(AB) = 4V` in the given fig are, then resistance `chi` will be A. 5B. 10C. 15D. 20
Answer» Correct Answer - 4

Discussion

1714.	Two batteries of e.m.f. `4V` and `8V` with internal resistances `1 Omega` and `2 Omega` are connected in a circuit with a resistance of `9 Omega` as shown in figure. The current and potential difference between the points `P` and `Q` A. `(1)/(3)A " and " 3V`B. `(1)/(6)A " and " 4V`C. `(1)/(9)A " and " 9V`D. `(1)/(12)A " and " 12V`
Answer» Correct Answer - A

Discussion

1715.	If `V_(AB) = 4V` in the given fig are, then resistance `chi` will be A. `5 Omega`B. `10 Omega`C. `15 Omega`D. `20 Omega`
Answer» Correct Answer - D

Discussion

1716.	Two resistance `R_(1)` and `R_(2)` are joined as shown in figure to two batteries of emf `E_(1)` and `E_(2)`. If `E_(2)` is short circuited, what is the current through `R_(1)` ? A. `E_(1)//R_(1)`B. `E_(2)//R_(1)`C. `E_(2)//R_(2)`D. `E_(1)//(R_(2)+R_(1))`
Answer» Correct Answer - A

Discussion

1717.	A galvanometer whose resistance is `120 Omega` gives full scale deflection with a curretn of `0.5 A` so that it can read a maximum current of `10 A`. A shunt resistance is added in parallel with it. The resistance of the ammeter so formed isA. `0.06 Omega`B. `0.006 Omega`C. `0.6 Omega`D. `6 Omega`
Answer» Correct Answer - C (c ) Resistance of shunted ammeter `= (GS)/(G + S)` Also `(i)/(i_(g)) = 1 + (G)/(S) implies (GS)/(G + S) = (i_(g). G)/(i)` `implies (GS)/(G + S) = (0.05 xx 120)/(10) = 0.6 Omega`

Discussion

1718.	A milliammeter of range of `10 mA` gives full-scale deflection for a current of `100 mA`, when a shunt of `0.1 Omega` is connected in parallel to it. The coil of the milliammeter has a resistance of.A. `0.9 Omega`B. `1 Omega`C. `1.1 Omega`D. `0.11 Omega
Answer» Correct Answer - A `(i-i_g) S = i_g G` `(100 - 10)(0.1) = 10 G rArr G = 0.9 Omega`.

Discussion

1719.	A voltmeter of resistance `1000 Omega` gives full scale deflection when a current of `100 mA` flow through it. The shunt resistance required across it to enable it to be used as an ammter reading `1 A` at full scale deflection isA. `10000 Omega`B. `9000 Omega`C. `222 Omega`D. `111Omega`
Answer» Correct Answer - D

Discussion

1720.	We have a galvanometer of resistance `25 Omega`. It is shunted by a `2.5 Omega` wire. The part of total current that flows through the galvanometer is given asA. `(I)/(I_(0)) = (1)/(11)`B. `(I)/(I_(0)) = (1)/(10)`C. `(I)/(I_(0)) = (3)/(11)`D. `(I)/(I_(0)) = (4)/(11)`
Answer» Correct Answer - A

Discussion

1721.	The resistance of 10 metre long potentiometer wire is 1 ohm/meter . A cell of e.m.f. 2.2 volts and a high resistance box are connected in series to this wire. The value of resistance taken from resistance box for getting potential gradient of 2.2 millivolt/metre will beA. `790 Omega`B. `810 Omega`C. `990 Omega`D. `1000 Omega`
Answer» Correct Answer - C

Discussion

1722.	An ammeter of 5 ohm resistance can read 5 mA . If it is to be used to read 100 volts, how much resistance is to be connected in seriesA. `19.9995 Omega`B. `199.995 Omega`C. `1999.95 Omega`D. `19995 Omega`
Answer» Correct Answer - D

Discussion

1723.	The potential gradient long the length of a unifrom wire is `10 "volt"//meter`. `B` and `C` are the two points at `30 cm` and `60 cm` point on a meter scale fitted along the wire. The potential diffenence between `B` and `C` will beA. 3 voltB. 0.4 voltC. 7 voltD. 4 volt
Answer» Correct Answer - A

Discussion

1724.	The potential gradient long the length of a unifrom wire is `10 "volt"//meter`. `B` and `C` are the two points at `30 cm` and `60 cm` point on a meter scale fitted along the wire. The potential diffenence between `B` and `C` will beA. 3 VB. 0.4 VC. 7 VD. 4 V
Answer» Correct Answer - B,D

Discussion

1725.	A wire 1m long has a resistance of `1 Omega`. If it is uniformly stretched, so that its length increases by 25% then its resistance will increase byA. 0.25B. 0.5C. 0.5625D. 0.7733
Answer» Correct Answer - C `R_(2)=n^(2)R_(1)=(1.25)^(2)R_(1)=1.5625xx1` `=1.5625 Omega` Thus, increase in resistance is , `R_(2)-R_(1)=1.5625-1=0.5625Omega=56.25 %`

Discussion

1726.	In an electric circuit, potential difference across a lamp is 20 V and current through the lamp is 0.5A. The resistance of the lamp isA. `10Omega`B. `20Omega`C. `30Omega`D. `40Omega`
Answer» Correct Answer - D `R=(V)/(I)=(20)/(0.5)= 40 Omega`

Discussion

1727.	When a wire carries a current of `1.20 A`, the drift velocity is `1.20xx10^-4 m//s`. What is the velocity when the current is `6.00 A`?
Answer» Correct Answer - A::D `i="ne"Av_d`ltbr. `i.e. v_d propi` when current has increased from `9=1.2A` to `i=6.0A`, i.e five times, then drift velocity will increase to five times.

Discussion

1728.	A metal wire of cross-sectional area `1m m^(2)` contains `5xx10^(22)` electrons per `cm^(3)` . If the electrons move along the wire with average drift velocity `1m m//s`, then the current in the wire is (`e=1.6xx10^(-19)C`)A. 8AB. 4AC. 2AD. 1A
Answer» Correct Answer - A `v_(d)=(I)/(n A e)` `:.I=v_(d) nA e` `=1xx10^(-3)xx5xx10^(22)xx10^(6)xx10^(-6)xx1.6xx10^(-19)` `= 8A`

Discussion

1729.	Calculate the drift velocity for the electrons in a silver wire which has a radius of 0.1 cm and carries a current of 2 A. Atomic mass of silver=108 and density of silver `=10.5xx10^(3) kg//m^(3)`.
Answer» Correct Answer - `7xx10^(-4) ms^(-1)`

Discussion

1730.	What is the drift velocity of electrons in a copper conductor having a cross-sectional area of `5xx10^(-6) m^(2)` if the current is 10 A. Assume there are `8.0xx10^(28) "electrons"//m^(3)`.
Answer» Correct Answer - `1.56 ms^(-1)`

Discussion

1731.	An electorn moves in a circle of radius 10cm with a constant speed of `4.0xx10^(6)ms^(-1)`.Find the electric current at a point on the circle.A. `2xx10^(-12)A`B. `1.019xx10^(-12)A`C. `1xx10^(-13)A`D. `1xx10^(-14)A`
Answer» Correct Answer - B `I=(ev)/(2pi r)=(1.6xx10^(-19)xx4xx10^(6))/(6.28xx10^(-1))` `=1.019xx10^(-12)A`

Discussion

1732.	A wire of 50 cm long, `1mm^(2)` in cross-section carries a current of 4 A, when connected to a 2 V battery, the resistivity of wire isA. `5xx10^(-6)Omega m`B. `2xx10^(-6)Omega m`C. `4xx10^(-6)Omega m`D. `1xx10^(-6)Omega m`
Answer» Correct Answer - D `rho=(RA)/(l)=(V)/(I)(A)/(l)=(2xx10^(-6))/(4xx0.5)=1xx10^(-6)Omega m`

Discussion

1733.	A certain piece of copper is to be shaped into a conductor of minimum resistance, its length and cross sectional area should beA. L and AB. 2L and `A/2`C. `L/2` and AD. `L/3` and 4A
Answer» Correct Answer - D For minimum resistance length should be minimum and area should be maximum.

Discussion

1734.	The potential gradient along the length of a uniform wire is 5 V/m. There are two points on the same wire at a distance of 20 cm and 40 cm from initial end of the wire. The potential difference between these points isA. 1VB. 2VC. 3VD. 4V
Answer» Correct Answer - A `V_(1)-V_(1)=Irho(l_(2)-l_(1))` `=5(40-20)xx10^(-2)` `=5xx20xx10^(-2)` `=1V`

Discussion

1735.	A current of 2 A flows in an electric circuit as shown in figure. The potential difference `(V_(R )- V_(S))`, in volts `(V_(E ) " and " V_(S)` are potenitals at R and S respectively) is A. `-4`B. `+2`C. `+4`D. `-2`
Answer» Correct Answer - C Current through each arm PQR and PSQ = 1A `V_(p)-V_(R)=3V` `V_(P)-V_(S)=7V` From Eqs. (i) and (ii), we get `V_(R)-V_(S)= +4V`

Discussion

1736.	Four resistances are connected in a circuit in the given figure. The electric cirrent flowing through `4 ohm` and `6 ohm` resistance is respectively A. `2 amp` and `3 amp`B. `1 amp` and `2 amp`C. `1 amp` and `1 amp`D. `2 amp` and `2 amp`
Answer» Correct Answer - D (d) Equivalent resistance `=(4 xx 4)/(4 + 4) + (6 xx 6)/(6 + 6) = 5 ohm` So the current in the circuit `= (20)/(5) = 4` ampere. Hence, the current flowing through each resistance `= 2` ampere.

Discussion

1737.	Five resistors are connected as shown in figure. Find the equivalent resistance between the points B and C. A. `(70)/(19)Omega`B. `(19)/(70)Omega`C. `(16)/(5)Omega`D. `(15)/(8)Omega`
Answer» Correct Answer - A Resistance in the branch ADC. `R_(1)=3Omega+7Omega = 10 Omega` Since, arms `ADC(10 Omega)` and `AC (10 Omega)` are in parallel their equivalent resistance, `R_(2)=(10xx10)/(10+10)Omega = 5 Omega` Since, `R_(2)` is in series with `9 Omega` in arm AB, equivalent resistance between B and C, i.e., `R_(BC)=(14xx5)/(14+5)=(70)/(19)Omega`

Discussion

1738.	If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolerance of the combinationsA. 0.2B. 0.05C. 0.1D. 0.15
Answer» Correct Answer - B Resistance of combination, `R_(e)=4R` `(Delta R_(e))/(R_(e))=(Delta R)/(R )=(5xx100)/(100)=5%`

Discussion

1739.	If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolereance of the combinationsA. `5%`B. `10%`C. `15%`D. `20%`
Answer» Correct Answer - A

Discussion

1740.	If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolerance of the combinationsA. `5%`B. `10%`C. `15%`D. `20%`
Answer» Correct Answer - A `4R=400+-20`

Discussion

1741.	When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will beA. `(R)/(4)`B. `(R)/(8)`C. `4R`D. `(R)/(2)`
Answer» Correct Answer - A

Discussion

1742.	A wire of resistance `12 Omega m^(-1)` is bent to from a complete circle of radius `10 cm`. The resistance between its two diametrically opposite points, `A` and `B` as shown in the figure, is A. `6Omega`B. `0.6piOmega`C. `3Omega`D. `6piOmega`
Answer» Correct Answer - 2 `R_(AB)=(R)/(2)=((1)/(2))(pirp)` `=((1)/(2))(pixx10xx10^(-2)xx12)=0.6piOmega`

Discussion

1743.	`n` conducting wires of same dimensions but having resistivities 1,2,3,….n are connected is series. The equivalent resistivity of the combination isA. `(n(n+1))/(2)`B. `(n+1)/(2)`C. `(n+1)/(2n)`D. `(2n)/(n+1)`
Answer» Correct Answer - B `R=R_(1)+R_(2)+…R_(n)` `(rho(nl))/(A)=l(l)/(A)+2(l)/(A)+….+n(l)/(A)` `rhon=1+2+3+…+n` `rhon=(n(n+1))/(2)=(n+1)/(2)`

Discussion

1744.	Two wires X,Y have the same resistivity, but their cross-sectional areas are in the ratio `2 : 3` and lengths in the ratio `1 : 2`. They are first connected in series and then in parallel to a.d.c. source. Find out the ratio of drift speeds of the electrons in the two wires for the two cases.
Answer» Correct Answer - `3;2;2;1` Given, `(A_(1))/(A_(2)) = (2)/(3) and (l_(1))/(l_(2)) = (1)/(2)` When two wire are in series , the current through two wires is the same As ` I = nAev_(d)` so , `upsilon_(d) = (1)/(nAe) or upsilon_(d) prop (1)/(A)` `:. (upsilon_(d_1))/(upsilon_(d_2)) = (A_(2))/(A_(1)) = (3)/(2)` When two wire are in parallel , the pot, diff `V` is same across each wire .Then `1 = (V)/(R ) = nAe upsilon_(d) or upsilon_(d) = (V)/(nAeR) = (V)/(nAe rhol//A)` or `upsilon_(d) = (V)/(n e rhol) or upsilon_(d) prop (1)/(l)` `:. (upsilon_(d_1))/(upsilon_(d_2)) = (l_(2))/(l_(1)) = (2)/(1)`

Discussion

1745.	A hollow cylinder of specific resistance `rho`, inner radius `R`, outer radius `2R` and length l is as shown in figure. What is the net resistance between the inner and outer surfaces?A. `(3piR^(2)p)/(l)`B. `(pR)/(2pi)`C. `(pIn(2))/(2pil)`D. `(pln(2))/(l)`
Answer» Correct Answer - C

Discussion

1746.	A cylindrical tube of uniform cross-sectional area A is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially the pressure of the gas is `P_0` and temperature is `T_0`, atmospheric pressure is also `P_0`. Now the temperature of the gas is increased to `2T_0`, the tension in the wire will beA. `2p_(0)A`B. `p_(0)A`C. `(p_(0)A)/(2)`D. `4p_(0)A`
Answer» Correct Answer - B

Discussion

1747.	The ends of a copper rod of length 1m and area of cross-section `1cm^2` are maintained at `0^@C` and `100^@C`. At the centre of the rod there is a source of heat of power 25 W. Calculate the temperature gradient in the two halves of the rod in steady state. Thermal conductivity of copper is `400 Wm^-1 K^-1`.A. `150.50^(@)C//m`B. `325.75^(@)C//m`C. `212.5^(@)C//m`D. `126.25^(@)C//m`
Answer» Correct Answer - C

Discussion

1748.	A rod AB of uniform cross-section consists of 4 sections AC, CD, DE and EB of different metal with thermal conductivities K, `0.8,1.2K` and `1.5` respectively. Their lengths are respectively `L,1.2L, 1.5L` and `0.6L`. They are jointed rigidly in Succession at C, D and E to from the rod AB The end A is mainained at `100^(@)C` and the end B is the maintained at `0^(@)C`. The steady state temperatures p the joints C, D and E are respectively `T_(C),T_(D)` and `T_(E)` Match the following two columns
Answer» Correct Answer - (A)R, (B) S, (C) Q, (D)P

Discussion

1749.	The heat capacity at constant volume of a monoatomic gas is `35 j//K`. Find (a) the number of moles (b) the internal energy at `0^@ C`. (c ) the molar heat capacity at constant pressure.A. TTTB. TFTC. FTTD. FFT
Answer» Correct Answer - C

Discussion

1750.	An ideal monoatomic gas is taken through one of the following reversible processes expressed by the equation in Table-1. Match the molar heat capacity of the gas, expressed in multiplies of R, in Table-2 with the appropriate process:
Answer» Correct Answer - (A) S, (B) R, (C) Q, (D)P

Discussion

Explore topic-wise InterviewSolutions in .

The e.m.f. of a cell is independent ofA. quantity of electrolyteB. distance between the electrodesC. area of the electrodesD. all of these

Find the minimum number of cells required to produce an electric current of 1.5 A through a resistance of `30 Omega`. Given that the emf of each cell is 1.5 V and internal resistance `1.0 Omega`.

Electromotive force is the force, which is able to maintain a constantA. CurrentB. ResistanceC. PowerD. Potential difference

A 10 m long wire ofresistance `20 Omega` is connected in series with battery of EMF `3V` and negligible internal resistance and a resistance of `10 Omega`. The potential gradient along the wire is :A. 0.02B. 0.1C. 0.2D. 1.2

Electromotive force of a cell is basically aA. forceB. powerC. workD. current capacity

In S.I. system, unit of electromotive force of a cell isA. newtonB. dyneC. voltD. ampere

For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit isA. 1.38 VB. 1.68 VC. 2.03 VD. 3.10 V

Four identical cells each having an electromotive force (e.m.f.) of 12 V , are connected in parallel. The resultant electromotive force (e.m.f.) of the combination isA. `48V`B. `12V`C. `4V`D. `3V`

In the shown circuit, what is the potential difference across A and B A. `50V`B. `45V`C. `30V`D. `20V`

Consider the circuit shown in the figure. The current `I_(3)` is equal to A. 5 ampB. 3 ampC. `-3` ampD. `-5//6` amp

If `V_(AB) = 4V` in the given fig are, then resistance `chi` will be A. 5B. 10C. 15D. 20

If `V_(AB) = 4V` in the given fig are, then resistance `chi` will be A. `5 Omega`B. `10 Omega`C. `15 Omega`D. `20 Omega`

Two resistance `R_(1)` and `R_(2)` are joined as shown in figure to two batteries of emf `E_(1)` and `E_(2)`. If `E_(2)` is short circuited, what is the current through `R_(1)` ? A. `E_(1)//R_(1)`B. `E_(2)//R_(1)`C. `E_(2)//R_(2)`D. `E_(1)//(R_(2)+R_(1))`

A galvanometer whose resistance is `120 Omega` gives full scale deflection with a curretn of `0.5 A` so that it can read a maximum current of `10 A`. A shunt resistance is added in parallel with it. The resistance of the ammeter so formed isA. `0.06 Omega`B. `0.006 Omega`C. `0.6 Omega`D. `6 Omega`

A milliammeter of range of `10 mA` gives full-scale deflection for a current of `100 mA`, when a shunt of `0.1 Omega` is connected in parallel to it. The coil of the milliammeter has a resistance of.A. `0.9 Omega`B. `1 Omega`C. `1.1 Omega`D. `0.11 Omega

A voltmeter of resistance `1000 Omega` gives full scale deflection when a current of `100 mA` flow through it. The shunt resistance required across it to enable it to be used as an ammter reading `1 A` at full scale deflection isA. `10000 Omega`B. `9000 Omega`C. `222 Omega`D. `111Omega`

We have a galvanometer of resistance `25 Omega`. It is shunted by a `2.5 Omega` wire. The part of total current that flows through the galvanometer is given asA. `(I)/(I_(0)) = (1)/(11)`B. `(I)/(I_(0)) = (1)/(10)`C. `(I)/(I_(0)) = (3)/(11)`D. `(I)/(I_(0)) = (4)/(11)`

An ammeter of 5 ohm resistance can read 5 mA . If it is to be used to read 100 volts, how much resistance is to be connected in seriesA. `19.9995 Omega`B. `199.995 Omega`C. `1999.95 Omega`D. `19995 Omega`

The potential gradient long the length of a unifrom wire is `10 "volt"//meter`. `B` and `C` are the two points at `30 cm` and `60 cm` point on a meter scale fitted along the wire. The potential diffenence between `B` and `C` will beA. 3 voltB. 0.4 voltC. 7 voltD. 4 volt

The potential gradient long the length of a unifrom wire is `10 "volt"//meter`. `B` and `C` are the two points at `30 cm` and `60 cm` point on a meter scale fitted along the wire. The potential diffenence between `B` and `C` will beA. 3 VB. 0.4 VC. 7 VD. 4 V

A wire 1m long has a resistance of `1 Omega`. If it is uniformly stretched, so that its length increases by 25% then its resistance will increase byA. 0.25B. 0.5C. 0.5625D. 0.7733

In an electric circuit, potential difference across a lamp is 20 V and current through the lamp is 0.5A. The resistance of the lamp isA. `10Omega`B. `20Omega`C. `30Omega`D. `40Omega`

When a wire carries a current of `1.20 A`, the drift velocity is `1.20xx10^-4 m//s`. What is the velocity when the current is `6.00 A`?

A metal wire of cross-sectional area `1m m^(2)` contains `5xx10^(22)` electrons per `cm^(3)` . If the electrons move along the wire with average drift velocity `1m m//s`, then the current in the wire is (`e=1.6xx10^(-19)C`)A. 8AB. 4AC. 2AD. 1A

Calculate the drift velocity for the electrons in a silver wire which has a radius of 0.1 cm and carries a current of 2 A. Atomic mass of silver=108 and density of silver `=10.5xx10^(3) kg//m^(3)`.

What is the drift velocity of electrons in a copper conductor having a cross-sectional area of `5xx10^(-6) m^(2)` if the current is 10 A. Assume there are `8.0xx10^(28) "electrons"//m^(3)`.

An electorn moves in a circle of radius 10cm with a constant speed of `4.0xx10^(6)ms^(-1)`.Find the electric current at a point on the circle.A. `2xx10^(-12)A`B. `1.019xx10^(-12)A`C. `1xx10^(-13)A`D. `1xx10^(-14)A`

A wire of 50 cm long, `1mm^(2)` in cross-section carries a current of 4 A, when connected to a 2 V battery, the resistivity of wire isA. `5xx10^(-6)Omega m`B. `2xx10^(-6)Omega m`C. `4xx10^(-6)Omega m`D. `1xx10^(-6)Omega m`

A certain piece of copper is to be shaped into a conductor of minimum resistance, its length and cross sectional area should beA. L and AB. 2L and `A/2`C. `L/2` and AD. `L/3` and 4A

The potential gradient along the length of a uniform wire is 5 V/m. There are two points on the same wire at a distance of 20 cm and 40 cm from initial end of the wire. The potential difference between these points isA. 1VB. 2VC. 3VD. 4V

A current of 2 A flows in an electric circuit as shown in figure. The potential difference `(V_(R )- V_(S))`, in volts `(V_(E ) " and " V_(S)` are potenitals at R and S respectively) is A. `-4`B. `+2`C. `+4`D. `-2`

Four resistances are connected in a circuit in the given figure. The electric cirrent flowing through `4 ohm` and `6 ohm` resistance is respectively A. `2 amp` and `3 amp`B. `1 amp` and `2 amp`C. `1 amp` and `1 amp`D. `2 amp` and `2 amp`

Five resistors are connected as shown in figure. Find the equivalent resistance between the points B and C. A. `(70)/(19)Omega`B. `(19)/(70)Omega`C. `(16)/(5)Omega`D. `(15)/(8)Omega`

If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolerance of the combinationsA. 0.2B. 0.05C. 0.1D. 0.15

If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolereance of the combinationsA. `5%`B. `10%`C. `15%`D. `20%`

If `400 Omega` of resistance is made by adding four `100 Omega` resistance of tolerance `5%`, then the tolerance of the combinationsA. `5%`B. `10%`C. `15%`D. `20%`

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will beA. `(R)/(4)`B. `(R)/(8)`C. `4R`D. `(R)/(2)`

A wire of resistance `12 Omega m^(-1)` is bent to from a complete circle of radius `10 cm`. The resistance between its two diametrically opposite points, `A` and `B` as shown in the figure, is A. `6Omega`B. `0.6piOmega`C. `3Omega`D. `6piOmega`

`n` conducting wires of same dimensions but having resistivities 1,2,3,….n are connected is series. The equivalent resistivity of the combination isA. `(n(n+1))/(2)`B. `(n+1)/(2)`C. `(n+1)/(2n)`D. `(2n)/(n+1)`

Two wires X,Y have the same resistivity, but their cross-sectional areas are in the ratio `2 : 3` and lengths in the ratio `1 : 2`. They are first connected in series and then in parallel to a.d.c. source. Find out the ratio of drift speeds of the electrons in the two wires for the two cases.

A hollow cylinder of specific resistance `rho`, inner radius `R`, outer radius `2R` and length l is as shown in figure. What is the net resistance between the inner and outer surfaces?A. `(3piR^(2)p)/(l)`B. `(pR)/(2pi)`C. `(pIn(2))/(2pil)`D. `(pln(2))/(l)`

The heat capacity at constant volume of a monoatomic gas is `35 j//K`. Find (a) the number of moles (b) the internal energy at `0^@ C`. (c ) the molar heat capacity at constant pressure.A. TTTB. TFTC. FTTD. FFT

An ideal monoatomic gas is taken through one of the following reversible processes expressed by the equation in Table-1. Match the molar heat capacity of the gas, expressed in multiplies of R, in Table-2 with the appropriate process: