2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 8 InterviewSolution

351.	An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt isA. `0.09 Omega`B. `0.03 Omega`C. `0.3 Omega`D. `0.9 Omega`
Answer» Correct Answer - A (a) `(i)/(i_(g)) = 1 + (G)/(S) implies (10)/(1) = 1 + (0.81)/(S) implies S = 0.09 Omega`

Discussion

352.	An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt isA. `0.03 Omega`B. `0.3 Omega`C. `0.9 Omega`D. `0.09 Omega`
Answer» Correct Answer - D (d) `i_g xx G = (i-i_g)S` `:. S = (i_g xx G)/(i-i_g) = (1 xx 0.81)/(10-1) = 0.09 Omega`.

Discussion

353.	A galvanometer of resistance `25 Omega` gives full scale deflection for a current of 10 milliampere , is to be changed into a voltmeter of range 100 V by connecting a resistance of ‘ R ’ in series with galvanometer. The value of resistance R in `Omega` isA. 10000B. 10025C. 975D. 9975
Answer» Correct Answer - D

Discussion

354.	A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `900Omega`B. `1800Omega`C. `500Omega`D. `1000Omega`
Answer» Correct Answer - A

Discussion

355.	A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `500 Omega`B. `900 Omega`C. `1000 Omega`D. `1800 Omega`
Answer» Correct Answer - B

Discussion

356.	A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `1000Omega`B. `900Omega`C. `1800Omega`D. `500Omega`
Answer» Correct Answer - 2 Let required resistance b e R then `(R+R_(g))I_(g)=VrArr(R+100)(30xx10^(-3))=30` `rArrR=900Omega`

Discussion

357.	A galvanometer of resistance `25 Omega` gives full scale deflection for a current of 10 milliampere , is to be changed into a voltmeter of range 100 V by connecting a resistance of ‘ R ’ in series with galvanometer. The value of resistance R in `Omega` isA. 10000B. 975C. 10025D. 9975
Answer» Correct Answer - D

Discussion

358.	The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistance. The graph between shunt resistance and range will have the nature A. PB. QC. RD. S
Answer» Correct Answer - B (b) To make range `n` times, the galvanomter resistance should be `G//n`, where `G` is initial resistance.

Discussion

359.	The graph which represents the relation between the total resistance `R` of a multi range moving coil voltmeter and its full scale deflection A. (i) (iii)B. (ii) (iv)C. (iii)D. (iv)
Answer» Correct Answer - D

Discussion

360.	The graph which represents the relation between the total resistance `R` of a multi range moving coil voltmeter and its full scale deflection A. `(i)`B. `(ii)`C. `(iii)`D. `(iv)`
Answer» Correct Answer - D (d) For conversion of a galvanometer into a voltmeter `(V)/(R + G) = i_(g) implies (V)/(R_(V)) = i_(g)` Where `R_(V) = R + G = ` Total resistance `implies R_(V) = (V)/(i_(g)) implies R_(V) prop V`

Discussion

361.	The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistance. The graph between shunt resistance and range will have the nature A. PB. QC. RD. S
Answer» Correct Answer - B

Discussion

362.	A galvanometer of 10 ohm resistance gives full scale deflection with 0.01 ampere of current. It is to be converted into an ammeter for measuring 10 ampere current. The value of shunt resistance required will beA. `(10)/(999)` ohmB. 0.1 ohmC. 0.5 ohmD. 1.0 ohm
Answer» Correct Answer - A

Discussion

363.	In the circuit shown the resistance of voltmeter is `10000 ohm` and that of ammeter is `20 ohm`. The ammeter reading is `0.10 amp` and voltmeter reading is 12 volts. Then `R` is equal to A. `122Omega`B. `100Omega`C. `118Omega`D. `116Omega`
Answer» Correct Answer - 2 `V=I(R+R_(A))rArr12=0.1(R+20)` `rArrR=(12)/( 0.1)-20 R=100Omega`

Discussion

364.	A galvanometer has a resistance of `25 ohm` and a maximum of `0.01 A` current can be passed throught it. In order to change it into an ammeter of range `10 A`, the shunt resistance required isA. `5//999 ohm`B. `10//999 ohm`C. `20//999 ohm`D. `25//999 ohm`
Answer» Correct Answer - D (d) `i_(g) = i(S)/(G + S) implies 0.01 = 10 (S)/(24 + S)` `implies 1000 S = 25 + S implies S = (25)/(999) Omega`

Discussion

365.	A voltmeter having resistance of `50 × 10^(3)` ohm is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required isA. `10^(5)` ohmB. 105 k ohmC. 900 k ohmD. `9 xx 10^(6)` ohm
Answer» Correct Answer - A

Discussion

366.	A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will beA. `0.001`B. `0.01`C. 1D. `0.05`
Answer» Correct Answer - A

Discussion

367.	A part of a circuit is shown in figure. Here reading of ammeter is `5A` and voltmeter is `100 V`. If voltmeter resistance is `2500 ohm`, then the resistance `R` is approximately A. `20Omega`B. `10Omega`C. `100Omega`D. `200Omega`
Answer» Correct Answer - A

Discussion

368.	The reading of voltmeter is A. 50 VB. 60 VC. 40 VD. None of these
Answer» Correct Answer - D

Discussion

369.	A voltmeter has resistance of 2000 ohms and it can measure upto `2 V`. If we want ot increase its range to `10 V` then the, required resistance in series will beA. `2000 Omega`B. `4000 Omega`C. `6000 Omega`D. `8000 Omega`
Answer» Correct Answer - D (d) Here, `n = (10)/(2) = 5` `:. R = (n - 1) G = (5 - 1) 2000 = 8000 Omega`

Discussion

370.	The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is `0.5 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is . where i is the current in the potentiometer wire.A. `(30E)/(100)`B. `(30E)/(100.5)`C. `(30E)/((100-0.5))`D. `(30(E-0.5i))/(100),(30(E-0.5))/(100)` where I is the current in the potentiometer
Answer» Correct Answer - A

Discussion

371.	Three ammeters `A,B`, and `C` of resistances `R_A,R_b` and `R_C` respectively are joined as shown. When some potential difference is appllied across the terminals `T_1` and `T_2` their readings are `I_A,I_B and I_C` respectively Then, A. `I_A = I_B`B. `I_A R_A + I_B R_B = I_C R_C`C. `(I_A)/(I_C) = (R_C)/(R_A)`D. `(I_B)/(I_C) = (R_C)/(R_A + R_B)`
Answer» Correct Answer - C

Discussion

372.	Three ammeters P,Q and R with internal resistances r, 1.5r, 3r respectively. Q and R parallel and this combination is in series with P, the whole combination is in series with P, T the whole combination connected between X and Y. When the battery connected between X and Y, the ratio of the readings of P, Q and R isA. `2:1:1`B. `3:2:1`C. `3:1:2`D. `1:1:1`
Answer» Correct Answer - B `i=(V)/(R)`

Discussion

373.	The ammeter shown in figure consists of a `480 Omega` coil connected in parallel to `20 Omega` shunt. Find the reading of the ammeter A. 0.68 AB. 0.46 AC. 1.32 AD. 0.27 A
Answer» Correct Answer - A

Discussion

374.	The current through the ammeter shown in figure is 1 A. If each of the `4Omega` resistor is replaced by `2Omega` resistor, the current in circuit will become nearly : A. `10/9A`B. `5/4`C. `9/8A`D. `5/8A`
Answer» Correct Answer - A

Discussion

375.	A galvanometer together with an unkonwn resistance in series is connected across two identical cells, each of emf 1.5 V. When the cells are connected inseries, the galvanometer records a current of 1 A and when the cells are connected in parallel, the current =0.6 A. What is the internal resistance of each cell ?A. `r=2/3Omega`B. `r=2/5Omega`C. `r=1/3Omega`D. `r=3/2Omega`
Answer» Correct Answer - C

Discussion

376.	If the reading of ammeter `A_(3)` in figure is 0.75 A. Neglecting the resistances of the ammeters, the reading of ammeter `A_(2)` will be : A. 1.5 AB. 0.125C. 4.5 AD. 0.25
Answer» Correct Answer - B

Discussion

377.	The ammeter shown in the figure consists of a `480Omega` coil conneted in parallel to `20 Omega` shunt. Find the reading of the ammeter. A. `50/73A`B. `40/53A`C. `50/53A`D. `73/50A`
Answer» Correct Answer - A

Discussion

378.	In the circuit shown the reading of ammeter and voltmeter are `4 A` and `20 V` respectively. The meters are non ideal, then `R` ia A. `5Omega`B. less than `5Omega`C. greater than `5Omega`D. between `4Omega" and "5Omega`
Answer» Correct Answer - C

Discussion

379.	A student connects a voltmeters, ammeter and resistance according to the circuit given. If the voltmeter reading is 20V and ammeter reading is 4A, then the resistance will be A. equal to `5Omega`B. more than `5Omega`C. less than `5Omega`D. more of less depending on the material of wire
Answer» Correct Answer - 2

Discussion

380.	Figure shows two flasks connected to each other. The volume of the flask `1` is twice that of flask `2`. The system is filled with an ideal gas at temperature `100 K` and `200 K` respectively. If the mass of the gas in `1` be `m` then what is the mass of the gas in flask `2` A. mB. `(m)/(4)`C. `(m)/(2)`D. 2m
Answer» Correct Answer - C P(2V)=(m/M)R × 100 PV=(m'/M)R×200 from both,we get m'=M/4

Discussion

381.	A charged particle having drift velocity of `7.5xx10^(-4)ms^(-1)` in electric field of `3xx10^(-10)Vm^(-1)` mobility isA. `6.5xx10^(6)m^(2)V^(-1)s^(-1)`B. `2.5xx10^(6)m^(2)V^(-1)s^(-1)`C. `2.5xx10^(4)m^(2)V^(-1)s^(-1)`D. `6.5xx10^(4)m^(2)V^(-1)s^(-1)`
Answer» Correct Answer - B Mobility of charged particle `mu=(\|v_(d)\|)/(E)=(7.5xx10^(-4))/(3xx10^(-10))=2.5xx10^(6)m^(2)V^(-1)s^(-1)`

Discussion

382.	The dimensions of mobility of charge carriers areA. `[M^(-2)T^(2)A]`B. `[M^(-1)T^(2)A]`C. `M^(-2)T^(3)A]`D. `[M^(-1)T^(3)A]`
Answer» Correct Answer - B

Discussion

383.	Write the mathematical relation between mobility and drift velocity of charge carriers in a coductor. Name the mobile charge carriers responsible for conduction of electric current in (a) an electrolyte (b) an ionised gas.
Answer» Mobility, `mu = ("drift velocity")/("electric field") = v_(d)/E` (a) The charge carriers in an electrolyte are positive and negative ions. (b) The charge carriers in an ionised gas are electrons and positively charged ions.

Discussion

384.	Two batteries, one of emf 18V and internal resistance `2Omega` and the other of emf 12 and internal resistance `1Omega` , are connected as shown. The voltmeter V will record a reading of A. 18voltB. 30voltC. 14voltD. 15volt
Answer» Correct Answer - 3 Reading of voltmeter `=(E_(1)/(r_(1))+(E_(2))/(r_(2)))/((1)/(r_(1))+(1)/(r_(2)))=(E_(1)r_(2)+E_(2)r_(1))/(r_(1)+r_(2))` =14volt.

Discussion

385.	A piece of copper and another of germanium are cooled from room temperature to `80 K`. The resistance ofA. each of the them increasesB. each of them decreasesC. copper increases and germanium decreasesD. copper decreases and germanium increases
Answer» Correct Answer - D

Discussion

386.	Two batteries one of the emf3V, internal resistance In and the other of emf l SV, internal resistance `2Omega` are connected in series with a resistanceR as shown. If the potential difference between points a and b is zero, the resistance R is : A. `5Omega`B. `7Omega`C. `3Omega`D. `1Omega`
Answer» Correct Answer - A

Discussion

387.	While connecting 6 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective resistance of load is 24 ohm and internal resistance of each cell is one ohm and emf 1.5 V the current delivered by the battery isA. 0.1 AB. 0.2 AC. 0.3 AD. 0.4 A
Answer» Correct Answer - B `i=((n-2m)E)/((R+nr))`

Discussion

388.	Five cells have been connected in parallel to form a battery. The emf and internal resistances of the cells have been shown in figure. A load resistance R is connected to the battery. (a) Which of the 5 cells will have maximum current flowing through it? (b) find the current through load resistance R.
Answer» Correct Answer - (a) cell with emf 16e (b) `I=(80 epsilon)/(31R+16r)`

Discussion

389.	In the circuit shown, potential difference between X and Y will be A. zeroB. 20 VC. 60 VD. 120 V
Answer» Correct Answer - D

Discussion

390.	A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volts having internal resistance of 0.15 ohm. The potential difference measured, in volts, across both the ends of the cell will beA. `1.35`B. `1.50`C. `1.00`D. `1.20`
Answer» Correct Answer - D

Discussion

391.	In the given circuit current flowing through the resistance `20Omega` is `0.3A,` while the ammeter reads 0.8 A. What is the value of `R_1` A. `30 Omega`B. `40 Omega`C. `50 Omega`D. `60 Omega`
Answer» Correct Answer - D

Discussion

392.	In the given circuit current flowing through the resistance `20Omega` is `0.3A,` while the ammeter reads 0.8 A. What is the value of `R_1` A. `30 Omega`B. `40 Omega`C. `50 Omega`D. `60 Omega`
Answer» Correct Answer - D `V_(20Omega)=V_(Total)` `:. (20)(0.3)=(R_(Total))(0.8)` `: R_(Total)=30/4Omega` `:. 4/30=1/R_1+1/20+1/15` Solving we get `R_1=60Omega`

Discussion

393.	The battery in the diagram is to be charged by the generator `G`. The generator has a terminal voltage of `120` volts when the charging current is `10` amperes. The battery has an emf of `100` volts and an internal resistance of `1` ohm. In order to charge the battery at `10` amperes charging current, the resistance `R` should be set at A. `0.1 Omega`B. `0.5 Omega`C. `1.0 Omega`D. `5.0 Omega`
Answer» Correct Answer - C Applying loop law `120-10R-10xx1-100=0 implies R=1 Omega`

Discussion

394.	In a galvanometer, the deflection becomes one half when the galvanometer is shunted by a `20Omega` resistance is A. `5 Omega`B. `10 Omega`C. `40 Omega`D. `20 Omega`
Answer» Correct Answer - D `I_(g)=20/(20+R_(g))Iimplies 1/2=20/(20+R_(g))` `20+R_(g)=40impliesR_(g)=20 Omega`

Discussion

395.	A galvanometer having a coil resistance of `100 omega` gives a full scale deflection , when a current of `1 mA` is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of `10A` , is :A. `2Omega`B. `0.1Omega`C. `3Omega`D. `0.01Omega`
Answer» Correct Answer - D

Discussion

396.	`50 Omega` and `100 Omega` resistors are connected in series. This connection is connected with a battery of 2.4 volts. Whne a voltmeter, then the reading of the voltmeter will be `100 Omega` resistor, then eh reading of the voltmeter will beA. `1.6 V`B. `1.0 V`C. `1.2 V`D. `2.0 V`
Answer» Correct Answer - C (c ) Equivalent resistance of the circuit `R_(eq) = 100 Omega` current through the circuit `i= (2.4)/(100) A` `P.D.` across combination of voltmeter and `100 Omega` resistance `= (2.4)/(100) xx 50 = 1.2 V` Since the voltmeter and `100 Omega` resistance are in parallel, so the voltmeter reads the same value i.e.,

Discussion

397.	A moving coil galvanometer has a resistance of `9.8 Omega` and gives a full scale deflection when a current of 10 mA is passed through it. The value of the shunt required to convert it into a milliameter to measure currents upto 500 mA isA. `0.02 Omega`B. `0.2 Omega`C. `2 Omega`D. `0.4 Omega`
Answer» Correct Answer - B

Discussion

398.	A galvanometer coil has a resistance `90 Omega` and full scale deflection current `10 mA`. A `910 Omega` resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is `0.1 V` the number of divisions on its scale isA. 90B. 91C. 100D. None
Answer» Correct Answer - C (c ) `R_(v) = (V)/(i_(g)) - G implies 910 = (V)/(10 xx 10^(-3)) = 90` `implies V = 10 implies` No. of division `= (10)/(0.12) = 100`

Discussion

399.	What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate `R_("unknown")` by any other method?
Answer» In a Wheatstone bridge, the null point method involves a balanced Wheatstone bridges, in which the resistance of galvanometer does not affect the balance point and there is no need to determine the currents in resistances and galvanometer. The unknown resistance can also be determined by using unbalanced Wheatstone bridge. But in this method we require the additional accurate measurement of all the currents in resistors and galvanometer as well as the internal resistance of galvanometer.

Discussion

400.	Four resistances of `16 Omega , 12 Omega , 4 Omega and 9 Omega` respectively are connected in cycle order so from a Wheatstone bridge. Calculate the resistance in be connected in parallel with `9 Omega` resistance to balance the bridge.A. `13.5Omega`B. `15.5Omega`C. `27Omega`D. `12Omega`
Answer» Correct Answer - A To balance the bridge equivalent resistance of `R_(4)` and shunt x should be `9 Omega` `:.R_(P)=(R_(4)xx x)/(R_(4)+ x)` `9=(27 xx x)/(27 + x) :.x=13.5Omega`.

Discussion

Explore topic-wise InterviewSolutions in .

An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt isA. `0.09 Omega`B. `0.03 Omega`C. `0.3 Omega`D. `0.9 Omega`

An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt isA. `0.03 Omega`B. `0.3 Omega`C. `0.9 Omega`D. `0.09 Omega`

A galvanometer of resistance `25 Omega` gives full scale deflection for a current of 10 milliampere , is to be changed into a voltmeter of range 100 V by connecting a resistance of ‘ R ’ in series with galvanometer. The value of resistance R in `Omega` isA. 10000B. 10025C. 975D. 9975

A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `900Omega`B. `1800Omega`C. `500Omega`D. `1000Omega`

A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `500 Omega`B. `900 Omega`C. `1000 Omega`D. `1800 Omega`

A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will beA. `1000Omega`B. `900Omega`C. `1800Omega`D. `500Omega`

A galvanometer of resistance `25 Omega` gives full scale deflection for a current of 10 milliampere , is to be changed into a voltmeter of range 100 V by connecting a resistance of ‘ R ’ in series with galvanometer. The value of resistance R in `Omega` isA. 10000B. 975C. 10025D. 9975

The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistance. The graph between shunt resistance and range will have the nature A. PB. QC. RD. S

The graph which represents the relation between the total resistance `R` of a multi range moving coil voltmeter and its full scale deflection A. (i) (iii)B. (ii) (iv)C. (iii)D. (iv)

The graph which represents the relation between the total resistance `R` of a multi range moving coil voltmeter and its full scale deflection A. `(i)`B. `(ii)`C. `(iii)`D. `(iv)`

The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistance. The graph between shunt resistance and range will have the nature A. PB. QC. RD. S

A galvanometer of 10 ohm resistance gives full scale deflection with 0.01 ampere of current. It is to be converted into an ammeter for measuring 10 ampere current. The value of shunt resistance required will beA. `(10)/(999)` ohmB. 0.1 ohmC. 0.5 ohmD. 1.0 ohm

In the circuit shown the resistance of voltmeter is `10000 ohm` and that of ammeter is `20 ohm`. The ammeter reading is `0.10 amp` and voltmeter reading is 12 volts. Then `R` is equal to A. `122Omega`B. `100Omega`C. `118Omega`D. `116Omega`

A galvanometer has a resistance of `25 ohm` and a maximum of `0.01 A` current can be passed throught it. In order to change it into an ammeter of range `10 A`, the shunt resistance required isA. `5//999 ohm`B. `10//999 ohm`C. `20//999 ohm`D. `25//999 ohm`

A voltmeter having resistance of `50 × 10^(3)` ohm is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required isA. `10^(5)` ohmB. 105 k ohmC. 900 k ohmD. `9 xx 10^(6)` ohm

A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will beA. `0.001`B. `0.01`C. 1D. `0.05`

A part of a circuit is shown in figure. Here reading of ammeter is `5A` and voltmeter is `100 V`. If voltmeter resistance is `2500 ohm`, then the resistance `R` is approximately A. `20Omega`B. `10Omega`C. `100Omega`D. `200Omega`

The reading of voltmeter is A. 50 VB. 60 VC. 40 VD. None of these

A voltmeter has resistance of 2000 ohms and it can measure upto `2 V`. If we want ot increase its range to `10 V` then the, required resistance in series will beA. `2000 Omega`B. `4000 Omega`C. `6000 Omega`D. `8000 Omega`

The ammeter shown in figure consists of a `480 Omega` coil connected in parallel to `20 Omega` shunt. Find the reading of the ammeter A. 0.68 AB. 0.46 AC. 1.32 AD. 0.27 A

The current through the ammeter shown in figure is 1 A. If each of the `4Omega` resistor is replaced by `2Omega` resistor, the current in circuit will become nearly : A. `10/9A`B. `5/4`C. `9/8A`D. `5/8A`

If the reading of ammeter `A_(3)` in figure is 0.75 A. Neglecting the resistances of the ammeters, the reading of ammeter `A_(2)` will be : A. 1.5 AB. 0.125C. 4.5 AD. 0.25

The ammeter shown in the figure consists of a `480Omega` coil conneted in parallel to `20 Omega` shunt. Find the reading of the ammeter. A. `50/73A`B. `40/53A`C. `50/53A`D. `73/50A`

In the circuit shown the reading of ammeter and voltmeter are `4 A` and `20 V` respectively. The meters are non ideal, then `R` ia A. `5Omega`B. less than `5Omega`C. greater than `5Omega`D. between `4Omega" and "5Omega`

A student connects a voltmeters, ammeter and resistance according to the circuit given. If the voltmeter reading is 20V and ammeter reading is 4A, then the resistance will be A. equal to `5Omega`B. more than `5Omega`C. less than `5Omega`D. more of less depending on the material of wire

A charged particle having drift velocity of `7.5xx10^(-4)ms^(-1)` in electric field of `3xx10^(-10)Vm^(-1)` mobility isA. `6.5xx10^(6)m^(2)V^(-1)s^(-1)`B. `2.5xx10^(6)m^(2)V^(-1)s^(-1)`C. `2.5xx10^(4)m^(2)V^(-1)s^(-1)`D. `6.5xx10^(4)m^(2)V^(-1)s^(-1)`

The dimensions of mobility of charge carriers areA. `[M^(-2)T^(2)A]`B. `[M^(-1)T^(2)A]`C. `M^(-2)T^(3)A]`D. `[M^(-1)T^(3)A]`

Write the mathematical relation between mobility and drift velocity of charge carriers in a coductor. Name the mobile charge carriers responsible for conduction of electric current in (a) an electrolyte (b) an ionised gas.

Two batteries, one of emf 18V and internal resistance `2Omega` and the other of emf 12 and internal resistance `1Omega` , are connected as shown. The voltmeter V will record a reading of A. 18voltB. 30voltC. 14voltD. 15volt

A piece of copper and another of germanium are cooled from room temperature to `80 K`. The resistance ofA. each of the them increasesB. each of them decreasesC. copper increases and germanium decreasesD. copper decreases and germanium increases

Two batteries one of the emf3V, internal resistance In and the other of emf l SV, internal resistance `2Omega` are connected in series with a resistanceR as shown. If the potential difference between points a and b is zero, the resistance R is : A. `5Omega`B. `7Omega`C. `3Omega`D. `1Omega`

In the circuit shown, potential difference between X and Y will be A. zeroB. 20 VC. 60 VD. 120 V

A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volts having internal resistance of 0.15 ohm. The potential difference measured, in volts, across both the ends of the cell will beA. `1.35`B. `1.50`C. `1.00`D. `1.20`

In the given circuit current flowing through the resistance `20Omega` is `0.3A,` while the ammeter reads 0.8 A. What is the value of `R_1` A. `30 Omega`B. `40 Omega`C. `50 Omega`D. `60 Omega`

In the given circuit current flowing through the resistance `20Omega` is `0.3A,` while the ammeter reads 0.8 A. What is the value of `R_1` A. `30 Omega`B. `40 Omega`C. `50 Omega`D. `60 Omega`

In a galvanometer, the deflection becomes one half when the galvanometer is shunted by a `20Omega` resistance is A. `5 Omega`B. `10 Omega`C. `40 Omega`D. `20 Omega`

`50 Omega` and `100 Omega` resistors are connected in series. This connection is connected with a battery of 2.4 volts. Whne a voltmeter, then the reading of the voltmeter will be `100 Omega` resistor, then eh reading of the voltmeter will beA. `1.6 V`B. `1.0 V`C. `1.2 V`D. `2.0 V`

A moving coil galvanometer has a resistance of `9.8 Omega` and gives a full scale deflection when a current of 10 mA is passed through it. The value of the shunt required to convert it into a milliameter to measure currents upto 500 mA isA. `0.02 Omega`B. `0.2 Omega`C. `2 Omega`D. `0.4 Omega`

A galvanometer coil has a resistance `90 Omega` and full scale deflection current `10 mA`. A `910 Omega` resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is `0.1 V` the number of divisions on its scale isA. 90B. 91C. 100D. None

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate `R_("unknown")` by any other method?

Four resistances of `16 Omega , 12 Omega , 4 Omega and 9 Omega` respectively are connected in cycle order so from a Wheatstone bridge. Calculate the resistance in be connected in parallel with `9 Omega` resistance to balance the bridge.A. `13.5Omega`B. `15.5Omega`C. `27Omega`D. `12Omega`