2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 20 InterviewSolution

951.	Whether electric field inside potentiometer wire is constant or variable ?
Answer» Electric field intensity `E= V//l = a` constant.

Discussion

952.	Why do we prefer a potentiometer to measure emf of a cell rather than a voltmeter ?
Answer» At null point, a potentiometer does not draw any current from the cell whose emf is to be determined, whereas a voltmeter, always draws some little current. Therefore, emf measured by voltmeter is slightly less than actual value of emf of the cell.

Discussion

953.	Two adiabatic containers have volumes `V_(1)` and `V_(2)` respectively. The first container has monoatomic gas at pressure `p_(1)` and temperature `T_(1)`. The second container has another monoatomic gas at pressure `p_(2)` and temperature `T_(2)`. When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. ThenA. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(1)+p_(2)V_(2))`B. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(2)+p_(2)V_(1))`C. `p=(p_(1)V_(2)+p_(2)V_(1))/(V_(1)+V_(2))`D. `p=(p_(1)V_(1)+p_(2)V_(2))/(V_(1)+V_(2))`
Answer» Correct Answer - D

Discussion

954.	A circuit is arranged as shown. Then, the current from `A` to `B` is A. `+ 500 mA`B. `+ 1.5 mA`C. `- 250 mA`D. `- 500 mA`
Answer» Correct Answer - A (a) `I = (sum (E // r))/(1 + R sum (1//r))` `= ((5)/(15) + (10)/(10))/(1 + 10 [ (1)/(15) + (1)/(10)]) = (1)/(2) A = 500 mA`

Discussion

955.	In the circuit of adjoining figure the current through `12 Omega` resister will be A. `1 A`B. `(1)/(5) A`C. `(2)/(5) A`D. `0 A`
Answer» Correct Answer - D (d) Let the current in `12 Omega` resistance is `i` Applying loop theorem in closed mesh `AEFCA` `12i = - E + E = 0 :. i= 0`

Discussion

956.	In the circuit of adjoining figure the current through `12 Omega` resister will be A. `1A`B. `(1)/(5)A`C. `(2)/(5)A`D. 0A
Answer» Correct Answer - D

Discussion

957.	A battery of emf `2.0 V` and internal resistance `0.10 Omega` is being charged with a current of `5.0 A`. Find the potential difference between the terminals of the battery?
Answer» Potential drop across internal resistance `=Ir=0.1xx5=0.5V` Hence potential difference across terminals `=E+Ir=2+0.5=2.5` volt.

Discussion

958.	The reading of the ideal voltmeter in the adjoining diagram will be A. 4VB. 8VC. 12VD. 14V
Answer» Correct Answer - B

Discussion

959.	A current of 7 A flows through the circuit as shown in the figure the potential difference across points B and D isA. `5V`B. `3V`C. `10V`D. `7V`
Answer» Correct Answer - A `i_(1)=(ixxR_(2))/(R_(1)+R_(2))` `i_(2)=(ixxR_(1))/(R_(1)+R_(2))`

Discussion

960.	An ammeter A is connected as shown in the diagram. Ammeter reading isA. `(E)/(r)`B. `(2E)/(r)`C. `(r)/(2E)`D. `(E)/(2r)`
Answer» Correct Answer - B `i=(5E)/(r)-(3E)/(r)=` ….

Discussion

961.	If a rod has resistance `4 Omega` and if rod is turned as half cycle then the resistance along diameterA. `1.56 Omega`B. `2.44 Omega`C. `4 Omega`D. `2 Omega`
Answer» Correct Answer - C

Discussion

962.	The equivalent resistance across XY in fig.A. `r`B. `2r`C. `4r`D. `(r)/(2)`
Answer» Correct Answer - D combination of resistors

Discussion

963.	The equivalent resistance across A and B isA. `2Omega`B. `4Omega`C. `8Omega`D. `12Omega`
Answer» Correct Answer - B combination of resistors

Discussion

964.	If the resistance of a circuit having 12 V source is increased by `4Omega` the current drops by 0.5 A. What is the original resistance of circuitA. `4Omega`B. `8Omega`C. `16Omega`D. `(1)/(16)Omega`
Answer» Correct Answer - B `V=iR` `12=(i-0.5)(R+4)`

Discussion

965.	When `n` wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by `(X)/(Y)Omega`. Then the resistance of each wire isA. `(xn)/(y(n^(2)-1))`B. `(yn)/(x(n^(2)-1))`C. `(xn)/(y(n-1))`D. `(yn)/(x(n-1))`
Answer» Correct Answer - A `R_(S)=R_(p)+((X)/(Y))`

Discussion

966.	An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wire are in the ratio `(4)/(3)` and `(2)/(3)`, then the ratio of the currents passing through the wires will beA. `(1)/(3)`B. `(3)/(1)`C. `(4)/(3)`D. `(3)/(4)`
Answer» Correct Answer - A `V-iR` `V=` constant `i_(1)R_(1)=i_(2)R_(2)` `(i_(1))/(i_(2))=(R_(2))/(R_(1))` `thereforeRprop(l)/(A)`

Discussion

967.	An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wire are in the ratio `(4)/(3)` and `(2)/(3)`, then the ratio of the currents passing through the wires will beA. 3B. `1//3`C. `8//9`D. 2
Answer» Correct Answer - B

Discussion

968.	A student records `Delta Q, Delta U` and `Delta W` for a thermodynamic cycle `A rarr B rarr C rarrA`. Certain entries are missing. Find correct entry in following options. A. `W_(BC)=-70J`B. `Delta Q_(CA)=130J`C. `Delta U_(AB)=190 J`D. `Delta U_(CA)=-160J`
Answer» Correct Answer - D

Discussion

969.	In parallel combination of n cells, we obtainA. more voltageB. more currentC. less voltageD. less current
Answer» Correct Answer - B In parallel combination of cells the voltage across the terminals is same and resistance is minimum. Therefore from V=IR. The current drawn from cell combination will be more.

Discussion

970.	If n cells each of emf E and internal resistance rare connected in parallel, then the total emf and internal resistances will beA. `epsi, (r)/(n)`B. `epsi,nr`C. `"n"epsi,(r)/(n)`D. `"n"epsi,nr`
Answer» Correct Answer - A In the parallel combination, `(epsi_("eq"))/(r_("eq"))=(epsi_(1))/(r_(1))+(epsi_(2))/(r_(2))+,.....+(epsi_(n))/(r_(n))` `(1)/(r_("eq"))=(1)/(r_(1))+(1)/(r_(2))+...+(1)/(r_n)` `(therefore epsi_(1)=epsi_(2)=epsi_(3)=.....=epsi_(n)=epsi and r_(1)=r_(2)=r_(3)=r...r_(n)=r)` `therefore (epsi_("eq"))/(r_("eq))=(epsi)/(r)+(epsi)/(r)+......+(epsi)/(r)=n(epsi)/(r)..(i)` `(1)/(r_("eq"))=(1)/(r)+(1)/(r)+....+(1)/(r)=(n)/(r)` `r_("eq")=r//n...(ii)` From (i) and (ii) `e_("eq")=n(epsi)/(r)xxr_("eq")=nxx(epsi)/(r)xx(epsi)/(n)=epsi`

Discussion

971.	A potential difference of 120 V is applied to a coil at temperature of `12^@C` and the current is 6A. What will be the mean temperature of the coil when the curreny has fallen to 3 A, the applied voltage being the same as before ? Given temperature coefficient of resistance coil is `.00427^@C^(-1)` at `0^@C`.
Answer» In second case the current decreases due to incerese in resistance on heating. Here `R_(12)=V/I=120/6=20 Omega` If `t^@C` is the temperature at which current falls to 3A, then `= R_(1)=120/3=40 Omega` As `R_(t)=R_(0)(1+alpha t)` So `R_(12)=R_(0)(1+alpha xx12)` or `20 =R_(0)(1+.00427 xx 12)` and `R_(t) = R_(0)(1+alpha t)or 40=R_(0)(1+.00427 t)` `:. 40/20 =(1+0.00427 t)/(1+0.00727 xx12)` On solving, `t=258^@C`.

Discussion

972.	If 2 mol of an ideal monatomic gas at temperature `T_(0)` are mixed with 4 mol of another ideal monatoic gas at temperature `2 T_(0)` then the temperature of the mixture isA. `(5)/(3)T_(0)`B. `(3)/(2)T_(0)`C. `(4)/(3)T_(0)`D. `(5)/(4)T_(0)`
Answer» Correct Answer - A

Discussion

973.	The resistance of a silver wire at `0^(@)C " is " 1.25 Omega`. Upto what temperature it must be heated so that its resistance is doubled ? The temperature coefficient of resistance of silver is `0.00375^(@)C^(-1)`. Will the temperature be same for all silver conductors of all shapes ?
Answer» Correct Answer - `267^(@)`, Yes

Discussion

974.	A few straings cells in series are to be charge from a `30V` d.c. supply The end of each cutt is `1.35V` and internal resistance is `0.1 Omega` The charging current is `3.0A` in this arrangement how many calls can be charge and what extra resistance is required be the connected in the circuit ?
Answer» Let `n` be the maximum number of cells in series which can be change and `R` is the extra resistance required if `f` is the currentin the circuit then `(nr + R) 1 = 30 - n epsilon` Hence `epsilon = 1.35 V, r= 0.1 Omega and t = 3.0A` Thus `(n xx 0.1 + R) xx 3= 30 - n xx 1.35 n (1.35 + 0.3) + 3R =m 30 or 1.65 n + 3R = 50`...(i) or `n + (3)/(1.65) R = (30)/(1.65) = 18 + (30)/(165) :. n = 3n and (3R)/(1.65) = (30)/(165) or R = 0.1 Omega`

Discussion

975.	A standard coil marked `3 Omega` is found to have a true resistance of `3.115 Omega` at 300 K. Calculate the temperature at which marking is correct. Temperature coefficient of resistance of the material of the coil is `4.2xx10^(-3) .^(@)C^(-1)`.
Answer» Correct Answer - `290.2 K`

Discussion

976.	Four wires made of same material have different lengths and radii, the wire having more resistance in the following case isA. `l=100cm, r=1mm`B. `l=50cm, r=2mm`C. `l=100cm, r=(1)/(2)mm`D. `l=50cm, r=(1)/(2)mm`
Answer» Correct Answer - C `Ralpha(l)/(r^(2))`. Chech the options

Discussion

977.	Determine the potentials of point A and B with respect to earth
Answer» Current in circute `I = (6)/(1 + 2+ 3) = I A` potential different across `2 Omega = 1 xx 2 = 2V` potential different across `3 Omega = 1 xx 3 = 3V` As earth is at zero potential , so `V_(A) = V_(C) = 2V` or `V_(A) - 0 = 2V` or `V_(A) = 2V` and `V_(C) - V_(B) = 3V` or `0 - V_(B) = 3V` or `V_(B) =- 3V`

Discussion

978.	A wire has a resistance of `6 Omega`. It is cut into two parts and both half values are connected in parallel. The new resistance is ....A. `12 Omega`B. `1.5 Omega`C. `3 Omega`D. `6 Omega`
Answer» Correct Answer - C

Discussion

979.	Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum betweenA. P and QB. Q and RC. P and RD. Any two points
Answer» Correct Answer - A

Discussion

980.	The three resistances of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipationA. `IIltIltIII`B. `IgtIIIgtII`C. `IltIIIltII`D. `IltIIltIII`
Answer» Correct Answer - A `P=i^(2)R` Current is same `P=R` In the first case it is `3r` in second case it is `(2)/(3)r` in third case it is `(r)/(3)` and in fourth case the net resistance is `(3r)/(2)` `R_(III)ltR_(II)ltR_(IV)ltR_(I)becauseP_(III)ltP_(II)ltP_(IV)ltP_(I)`

Discussion

981.	Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum betweenA. P and QB. Q and RC. P and RD. any two points
Answer» Correct Answer - A `R_(PO)=(5)/(11)r,R_(QR)=(4)/(11)r` and `R_(PR)=(3)/(11)r` `thereforeR_(PQ)` is maximum Therefore the correct option is (a)

Discussion

982.	Calculate the radius of the wire of conductance `10 Omega^(-1)` and length `10 cm` whose electrical conductivity is `10^(3) Sm^(-1)`
Answer» Correct Answer - `1.78 mm` Here,`G = 10 Omega^(-1), 1 = 10 cm = 0.10 m` `sigma= 10^(3) Omega m^(-1)` `sigma = (1)/(rho) = (l)/(pir^(2)R) = (lG)/(pir^(2)) or r = sqrt((lG)/(pi sigma))` `:. r = sqrt((0.10 xx 10)/(314 xx 10^(3))) = 1.78 xx 10^(-2)m = 0.0178 "m"`

Discussion

983.	Calculate the potential difference between points B and D of the network of resistance shown in figure
Answer» Here, potential differences across ABC = potential differences across ADC = 12 V Current in arm ABC, `I_(1)=12/6+4 =1.2A` pot. Diff. between A and B, `V_(A) - V_(B)=I_(1) xx 6 =1.2 xx 6=7.2 V` Current in arm ADC, `I_(2)=12/8+12=0.6A` pot. diff. across A and D, `V_(A) - V_(D) = I_(2) xx 8 = 0.6 xx 8 =4.8 V` `:. (V_(A) - V_(B)) - (V_(A) - V_(D))= 7.2 - 4.8 = 2.4 V` or `V_(D) - V_(B) = 2.4 V` `:. V_(B) - V_(A)= -2.4 V` It shows that the point D is at higher potential than that of point B.

Discussion

984.	Six equal resistances each of 4 ohm are connected to form a net work as shown in figure . What is the resistance between A and B ?
Answer» The upper five resistances between A and B will form a balanced wheatstone bridge. Hence there will be no current in arm CO. The equivalent circuit will be as shown in figure. Here, resistance of arm ACB, AOB and Ab are all in parallel. The effective resistance R is given by `1/R =(1)/((4+4)) + 1/(4+4) + 1/4 = (1+1+2)/8 = 4/8 = 1/2` `R = 2 Omega`

Discussion

985.	Resistance of tungsten wire at `150^(@)C` is `133 Omega`. Its resistance temperature coefficient is `0.0045//^(@)C`. The resistance of this wire at `500^(@)C` will beA. `180 Omega`B. `225 Omega`C. `258 Omega`D. `317 Omega`
Answer» Correct Answer - C (c ) `(R_(150))/(R_(500)) = ([1 + alpha (150)])/([1 + alpha (500)])`. Putting `R_(150) = 133 Omega` and `alpha = 0.0045//^(@)C`, we get `R_(500) = 250 Omega`

Discussion

986.	In the circuit shown in figure Find the potential differnece across capacitor.
Answer» When capacitor is fully charged, it fraws no currnet, and hence provides infinite resistance, then no current flows in arm EF. The potential difference across C and F. The resistance of arm AD and they together are in series with resistance of arm DF. The effective resistance between A and F of circuit is, `R = (3xx6/3+6) + 3 = 2 +3 =5 Omega` Main in the circuit, `i = 15/5=3 A` Current through arm BCD ` = 1 xx 3/6+3 = 3xx3/9 = 1 A` Potential difference across C and D `= 1 xx 3 = 3V` Current through arm DF, i =3A Potential differences across D and F `= 3 xx 3 = 9 V` So, potential difference across C and F `= 3+ 9=12 V` Thus, potential difference across the capacitor is 12 V

Discussion

987.	Resistance of tungsten wire at `150^(@)C` is `133 Omega`. Its resistance temperature coefficient is `0.0045//^(@)C`. The resistance of this wire at `500^(@)C` will be
Answer» Correct Answer - `215 Omega` Here, `R_(160) = 132 Omega R_(400) = ?` `alpha = 4.5 xx 10^(-3).^(@)C^(-1)` Now `R_(t) = R_(0)(1 + alpha t)` `:. R_(160) = R_(0) [1 + (4.5 xx 10^(-3))xx 100]` or `132 = R_(0)[1+4.5 xx 10^(-3)xx 100]`……(i) and `R_(400) = R_(0) [1 + 4.5 xx 10^(-3)xx400]`.....(ii) Dividing (ii) by (i) we get `(R_(400))/(132) = (1+ 4.5 xx 10^(-3) xx 400)/(1+ 4.5 xx 10^(-3) xx 400)` `= (1+ 1.80)/(1+ 0.72) = (2.80)/(1.72) = 1.63` `R_(400) = 132 xx 1.63 = 215 Omega`

Discussion

988.	Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is 8 ohm, the resistance between any two adjacent corners isA. 32 ohmB. 8 ohmC. `(1)/(6)` ohmD. 6 ohm
Answer» Correct Answer - D combination of resistors

Discussion

989.	The electrical resistance of a mercury column in a cylindrical container is `R` the mercury is poured into another cylindrical container with half the radius of cross-section. The resistance of the mercury column isA. `R`B. `2R`C. `16R`D. `5R`
Answer» Correct Answer - C `R=(rhol)/(A),V=Al`

Discussion

990.	The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` atA. 1154 KB. 1100 KC. 1400 KD. 1127 K
Answer» Correct Answer - B `R_theta = R_(theta_0) [1 +alpha(theta - theta_0)]` `2 = 1[1 + 0.00125 (theta - 300)]` `theta = (1)/(0.00125) + 300 = 1100 K`.

Discussion

991.	The specific resistance of all metals is most affected byA. TemperatureB. PressureC. Degree of illuminationD. Applied magnetic field
Answer» Correct Answer - A

Discussion

992.	The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` atA. `1154 K`B. `1100 K`C. `1500 K`D. `1127 K`
Answer» Correct Answer - D (d) `(rho_(1))/(rho_(2)) = ((1 + alpha t_(1)))/((1 + alpha t_(2))) implies (1)/(2) = ((1 + 00125 xx 27))/((1 + 0.00125 xx t))` `implies t = 854^(@)C implies T = 1127 K`

Discussion

993.	The positive temperature coefficient of resistance is forA. CarbonB. GermaniumC. CopperD. An electrolyte
Answer» Correct Answer - C

Discussion

994.	A resistance coil market is found to have a true resistance of `3.115 Omega` at `300 K`. Calculate the temperature at which marking is corrent .Temperature coefficient of resistance of the material of is `4.2xx 10^(-3).^(@)C^(-1)`
Answer» Correct Answer - `290.2K` `t = 300 K = 300- 273 = 27^(@)C` `R_(27) = R_(0)(1+ alpha xx 27)` `R_(27) = R_(0)(1+ alpha xx 27)`……(i) `:. 3.115 = R_(0) (1+ 4.3 xx 10^(-3) xx27)` ....(i) and `3 = R_(0)(1+ 4.2 xx 10^(-3) xx t)`…..(ii) `:. (3)/(3.115) = (1 + 4.2 xx 10^(-3) xx t)/(1+ 4.2 xx 10^(-3) xx 27)` On solving we get `t = 17.21^(@)C` `= 17.21 xx 275 = 290.2K`

Discussion

995.	The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` at
Answer» Correct Answer - `1127 K` Here `alpha = 0.00125^(@)C^(-1)` `t_(1) = 300 K= 300 - 273 = 27^(@)C` `R_(27) = R_(0) (1 + alpha xx 27)` and `R_(t) = R_(0) ( 1+alpha xx t)` or `1 = R_(0) (1+ alpha xx 27) and 2 = R_(0) (1+ alpha t)` `:. (2)/(1) (1+alpha t)/(1+alpha xx 27) or 1+ alpha t = 2+ 54 alpha` or `t = (1+ 54 alpha)/(alpha) = (1+ 54 xx 0.00125)/(0.00125) = 854 ^(@)C` `= 854 + 273 = 1127 K`

Discussion

996.	A wire carries a current of 0.5A, when a potential differnece of 1.5 V is applied across it. What is its conductance ? If the wire is of length 3 m and area of cross-section `5.4 mm^(2)`, calculate its conductivity.
Answer» Here, `I=0.5 A , V=1.5 V, l=3 m`, `A = 5.4 mm^(2) = 5.4 xx 10^(-6) m^(2)` Now, resistance, `R=V/I=1.5/0.5 =3 Omega` Conduction, `G=1/R=1/3=0.33 S` Electrical conductivity, `sigma=1/rho =(l)/(RA)` `=3/( 3xx 5.4 xx 10^(-6))=1.85 xx 10^(5) Sm^(-1)`

Discussion

997.	A standard coil marked `5 Omega` is found to have a resistance of `5.128 Omega` at `30^@C`. Calculate the temperature at which the marking is correct. The temperature coefficient of resistance of the material of the coil is `.0042^@C^(-1)`.
Answer» `R_(t_1)=5Omega, R_(t_2)5.128 Omega`, `t_(1)=?, t_(2)=30^@C`. `R_(t_1)/R_(t_2)=(R_(0)(1+alpha t_(1)))/(R_(0)(1+alpha t_(2)))=(1 +alpha t_(1))/(1+alpha t_(2))` or `5/5.128= (1+0.0042 xx t_(1))/(1+0.0042 xx30)` On solving, `t_(1)=23.3^@C`

Discussion

998.	A copper coil has resistance of `20.0 Omega` at `0^@C` and a resistance of `26.4 Omega` at `80^@C`. Find the temperature coefficient of resistance of copper.
Answer» Correct Answer - (i)` 6.45 xx 10^(-3)^(@)C^(-1)`(ii)` 0.267 Omega `(ii)`1.948 xx 10^(-4) Omega m` Here,`r =1 mm = 10^(-3) , l = 50 cm = 0.50 m` , `R_(1)=0.31 Omega,t_(1)=25^(@)C,R_(2)=0.51 cm=0.50 m` , (i) Temperature coefficient of resistance `alpha=(R_(2)-R_(1))/(R_(1)(t_(2)-t_(1)))=(0.51-0.31)/(0.31(125-25))` `=(0.20)/(0.31xx100)=6.45xx10^(3).^(@)C^(-1)` (ii) Resistance at `0^(@)C` is `R_(0)=(R_(1))/(1+alphat_(1))=(0.31)/(1+(6.4xx10^(-3))xx25)=0.267^(@)Omega` (iii) Resistivity at `0^(@)C` `rho_(0)=(R_(0)A)/(l)=(R_(0)xxpir^(2))/(l)=(0.267xx3.142xx(10^(-3))^(2))/(0.50)` `=1.68 xx 10^(-6)m` Resistivity at `25^(@)C,rho_(25)=rho_(0)[1+alpha xx 25]` `=1.68xx10^(-6)[1+6.4xx10^(-3)xx25]` `=1.948xx10^(-6)Omegam`.

Discussion

999.	The fact that the conductance of some metals rises to infinity at some temperature below a few Kelvin is calledA. Thermal conductivityB. Optical conductivityC. Magnetic conductivityD. Superconductivity
Answer» Correct Answer - D

Discussion

1000.	At what temperature(in kelvin) would the resistance of a copper wire be half its resistance at `0^@C`? Temperature coefficient of resistance of copper is `3.9 xx 10^(-3).^(@)C^(-1)`.
Answer» Let at temperature `t^@C,R_(t)=1/2R_(0)` Then `alpha=(R_(t) -R_(0))/(R_(0)xxt)=((R_(0)//2)-R_(0))/(R_()xxt) = -(1)/(2t)` or `t = -(1)/(2 alpha) = - 1/( 2xx 3.9xx10^(-3))= -128.2^@C` `=144.8K`

Discussion

Explore topic-wise InterviewSolutions in .

Whether electric field inside potentiometer wire is constant or variable ?

Why do we prefer a potentiometer to measure emf of a cell rather than a voltmeter ?

A circuit is arranged as shown. Then, the current from `A` to `B` is A. `+ 500 mA`B. `+ 1.5 mA`C. `- 250 mA`D. `- 500 mA`

In the circuit of adjoining figure the current through `12 Omega` resister will be A. `1 A`B. `(1)/(5) A`C. `(2)/(5) A`D. `0 A`

In the circuit of adjoining figure the current through `12 Omega` resister will be A. `1A`B. `(1)/(5)A`C. `(2)/(5)A`D. 0A

A battery of emf `2.0 V` and internal resistance `0.10 Omega` is being charged with a current of `5.0 A`. Find the potential difference between the terminals of the battery?

The reading of the ideal voltmeter in the adjoining diagram will be A. 4VB. 8VC. 12VD. 14V

A current of 7 A flows through the circuit as shown in the figure the potential difference across points B and D isA. `5V`B. `3V`C. `10V`D. `7V`

An ammeter A is connected as shown in the diagram. Ammeter reading isA. `(E)/(r)`B. `(2E)/(r)`C. `(r)/(2E)`D. `(E)/(2r)`

If a rod has resistance `4 Omega` and if rod is turned as half cycle then the resistance along diameterA. `1.56 Omega`B. `2.44 Omega`C. `4 Omega`D. `2 Omega`

The equivalent resistance across XY in fig.A. `r`B. `2r`C. `4r`D. `(r)/(2)`

The equivalent resistance across A and B isA. `2Omega`B. `4Omega`C. `8Omega`D. `12Omega`

If the resistance of a circuit having 12 V source is increased by `4Omega` the current drops by 0.5 A. What is the original resistance of circuitA. `4Omega`B. `8Omega`C. `16Omega`D. `(1)/(16)Omega`

When `n` wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by `(X)/(Y)Omega`. Then the resistance of each wire isA. `(xn)/(y(n^(2)-1))`B. `(yn)/(x(n^(2)-1))`C. `(xn)/(y(n-1))`D. `(yn)/(x(n-1))`

An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wire are in the ratio `(4)/(3)` and `(2)/(3)`, then the ratio of the currents passing through the wires will beA. 3B. `1//3`C. `8//9`D. 2

A student records `Delta Q, Delta U` and `Delta W` for a thermodynamic cycle `A rarr B rarr C rarrA`. Certain entries are missing. Find correct entry in following options. A. `W_(BC)=-70J`B. `Delta Q_(CA)=130J`C. `Delta U_(AB)=190 J`D. `Delta U_(CA)=-160J`

In parallel combination of n cells, we obtainA. more voltageB. more currentC. less voltageD. less current

If n cells each of emf E and internal resistance rare connected in parallel, then the total emf and internal resistances will beA. `epsi, (r)/(n)`B. `epsi,nr`C. `"n"epsi,(r)/(n)`D. `"n"epsi,nr`

If 2 mol of an ideal monatomic gas at temperature `T_(0)` are mixed with 4 mol of another ideal monatoic gas at temperature `2 T_(0)` then the temperature of the mixture isA. `(5)/(3)T_(0)`B. `(3)/(2)T_(0)`C. `(4)/(3)T_(0)`D. `(5)/(4)T_(0)`

The resistance of a silver wire at `0^(@)C " is " 1.25 Omega`. Upto what temperature it must be heated so that its resistance is doubled ? The temperature coefficient of resistance of silver is `0.00375^(@)C^(-1)`. Will the temperature be same for all silver conductors of all shapes ?

A few straings cells in series are to be charge from a `30V` d.c. supply The end of each cutt is `1.35V` and internal resistance is `0.1 Omega` The charging current is `3.0A` in this arrangement how many calls can be charge and what extra resistance is required be the connected in the circuit ?

A standard coil marked `3 Omega` is found to have a true resistance of `3.115 Omega` at 300 K. Calculate the temperature at which marking is correct. Temperature coefficient of resistance of the material of the coil is `4.2xx10^(-3) .^(@)C^(-1)`.

Four wires made of same material have different lengths and radii, the wire having more resistance in the following case isA. `l=100cm, r=1mm`B. `l=50cm, r=2mm`C. `l=100cm, r=(1)/(2)mm`D. `l=50cm, r=(1)/(2)mm`

Determine the potentials of point A and B with respect to earth

A wire has a resistance of `6 Omega`. It is cut into two parts and both half values are connected in parallel. The new resistance is ....A. `12 Omega`B. `1.5 Omega`C. `3 Omega`D. `6 Omega`

Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum betweenA. P and QB. Q and RC. P and RD. Any two points

The three resistances of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipationA. `IIltIltIII`B. `IgtIIIgtII`C. `IltIIIltII`D. `IltIIltIII`

Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum betweenA. P and QB. Q and RC. P and RD. any two points

Calculate the radius of the wire of conductance `10 Omega^(-1)` and length `10 cm` whose electrical conductivity is `10^(3) Sm^(-1)`

Calculate the potential difference between points B and D of the network of resistance shown in figure

Six equal resistances each of 4 ohm are connected to form a net work as shown in figure . What is the resistance between A and B ?

Resistance of tungsten wire at `150^(@)C` is `133 Omega`. Its resistance temperature coefficient is `0.0045//^(@)C`. The resistance of this wire at `500^(@)C` will beA. `180 Omega`B. `225 Omega`C. `258 Omega`D. `317 Omega`

In the circuit shown in figure Find the potential differnece across capacitor.

Resistance of tungsten wire at `150^(@)C` is `133 Omega`. Its resistance temperature coefficient is `0.0045//^(@)C`. The resistance of this wire at `500^(@)C` will be

Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is 8 ohm, the resistance between any two adjacent corners isA. 32 ohmB. 8 ohmC. `(1)/(6)` ohmD. 6 ohm

The electrical resistance of a mercury column in a cylindrical container is `R` the mercury is poured into another cylindrical container with half the radius of cross-section. The resistance of the mercury column isA. `R`B. `2R`C. `16R`D. `5R`

The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` atA. 1154 KB. 1100 KC. 1400 KD. 1127 K

The specific resistance of all metals is most affected byA. TemperatureB. PressureC. Degree of illuminationD. Applied magnetic field

The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` atA. `1154 K`B. `1100 K`C. `1500 K`D. `1127 K`

The positive temperature coefficient of resistance is forA. CarbonB. GermaniumC. CopperD. An electrolyte

A resistance coil market is found to have a true resistance of `3.115 Omega` at `300 K`. Calculate the temperature at which marking is corrent .Temperature coefficient of resistance of the material of is `4.2xx 10^(-3).^(@)C^(-1)`

The temperature coefficient of resistance of a wire is 0.00125 per `.^oC`. At 300K, its resistance is 1 `Omega`. The resistance of the wire will be 2 `Omega` at

A wire carries a current of 0.5A, when a potential differnece of 1.5 V is applied across it. What is its conductance ? If the wire is of length 3 m and area of cross-section `5.4 mm^(2)`, calculate its conductivity.

A standard coil marked `5 Omega` is found to have a resistance of `5.128 Omega` at `30^@C`. Calculate the temperature at which the marking is correct. The temperature coefficient of resistance of the material of the coil is `.0042^@C^(-1)`.

A copper coil has resistance of `20.0 Omega` at `0^@C` and a resistance of `26.4 Omega` at `80^@C`. Find the temperature coefficient of resistance of copper.

The fact that the conductance of some metals rises to infinity at some temperature below a few Kelvin is calledA. Thermal conductivityB. Optical conductivityC. Magnetic conductivityD. Superconductivity

At what temperature(in kelvin) would the resistance of a copper wire be half its resistance at `0^@C`? Temperature coefficient of resistance of copper is `3.9 xx 10^(-3).^(@)C^(-1)`.