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2151.

Two identical calls send the same current in `2 Omega` resistance, whether connected in series of in parallel. The internal resistance of the cell should beA. `1 Omega`B. `2 Omega`C. `(1)/(2) Omega`D. `2.5 Omega`

Answer» Correct Answer - B
(b) In series, `i_(1) = (2E)/(2 + 2r)`
In parallel, `I_(2) = (E)/(2 + (r )/(2)) = (2E)/(4 + r)`
Since `i_(1) = i_(2) implies (2E)/(4 + r) = (2E)/(2 + 2r) implies r = 2 Omega`
2152.

`n` identical calls, each of internal resistance `(r)` are first connected in parallel and then connected in series across a resistance (R). If the current through R is the same in both cases, thenA. `R=(r)/(2)`B. `r=(R)/(2)`C. `R=r`D. `r=0`

Answer» Correct Answer - C
2153.

A cell is to convertA. Chemical energ into electrical energyB. electrical energy into chemical energyC. heat energy into potential energyD. potential energy into heat energy

Answer» Correct Answer - A
2154.

From the following the standard cell isA. Daniel cellB. Cadmium cellC. Leclanche cellD. Lead accumulator

Answer» Correct Answer - B
2155.

What is constant in a battery (also called a source of emf)?A. current supplied by itB. terminal potential differenceC. internal resistanceD. emf

Answer» Correct Answer - D
2156.

In above question, if the bulbs are connected in parallel, total power consumed isA. `(P_(1)+P_(2))/(2)`B. `sqrt(P_(1).P_(2))`C. `(P_(1).P_(2))/(P_(1)+P_(2))`D. `(P_(1)+P_(2))`

Answer» Correct Answer - D
2157.

A cell drives a current through a circuit. The emf of the cell of equal to the work done in moving unit charge (Choose the incorrect option)A. from the positive to the negative plate of the cellB. from the positive plate, back to the positive plateC. from the negative plate, back to the negative plate, back to the negative plateD. from any point in the circuit back to the same point.

Answer» Correct Answer - A
2158.

The terminal voltage of a cell is greater tha its emf. When it isA. being chargedB. an open circuitC. being dischargedD. it never happens

Answer» Correct Answer - A
2159.

Two cells of emfs `E_(1)` and `(E_(2) (E_(1) gt E_(2))` are connected as shows in Fig. `6.45`. When a potentiometer is connected between `A` and `B`, the balancing length of the potentiometer wire is `300 cm`. On connecting the same potentiometer between `A` and `C`, the balancing length is `100 cm`. The ratio `E_(1)//E_(2)` is

Answer» Let K be the potential gradient of the given potentiometer wire.
When potentiometer is connected between A and B, then `epsilon_(1) = K xx 300`
When potentiometer is connected between A and C, then `epsilon_(1) - epsilon_(2) = K xx 100`
`:. (epsilon_(1)-epsilon_(2))/epsilon_(1) = 100/300 = 1/3 or 1 - epsilon_(2)/epsilon_(1) = 1/3`
or `epsilon_(2)/epsilon_(1) = 1 - 1/3 = 2/3 or epsilon_(1)/epsilon_(2)=3/2`
2160.

Which of the following causes production of heat, when current is set up in a wireA. fall of electron from higher orbits to lower orbitsB. inter atomic collisionsC. inter electron collisionsD. collisions of conductions electrons with atoms

Answer» Correct Answer - D
2161.

When an electric cell drives current through load resistance, its back emf,A. Supports the original emfB. Opposes the original emfC. Supports if internal resistance is lowD. Opposes if load resistance is large

Answer» Correct Answer - B
2162.

For the network shown in figure Determine the value of R and the current through it, if the current through the branch AO is zero.

Answer» Since no current flows through arm Ao, so the bridge fomerd by resistance `2 Oemga, 2.5 Omega, 4 Omega` and R will form a balanced Wheatstone bridge.
So, `2/2.5 = 4/R or R =( 4xx2.5)/2= 5 Omega`
The equivalent resistance `R_(eq)` of the network between B and C is
`R_(eq) = ((2+2.5 ) xx (4+5))/((2 +2.5) + (4+5)) + 2 = 3 +2 = 5 Omega`
Current in the circuit, `I=10/5 = 2A`
2163.

Consider four circuits shown in the figure below. In which circuit power dissipated is greatest (Neglect the internal resistance of the power supply)?A. B. C. D.

Answer» Correct Answer - A
(a) `P = (V^(2))/(R_(eq))` , for `P` to be maximum `R_(eq)` should be less.
2164.

In the circuit shown in fig. 5.120 find the current through the branch BD. A. 5AB. 0AC. 3AD. 4A

Answer» Correct Answer - A
2165.

What is the ratio of heat generated in `R` and `2 R`? A. `2:1`B. `1:2`C. `4:1`D. `1:4`

Answer» Correct Answer - A
(a) Both `R` and `2R` in parallel `(V -` constant)
So using `P = (V^(2))/(R ) implies (P_(1))/(P_(2)) = (R_(2))/(R_(1)) implies (H_(1))/(H_(2)) = (R_(2))/(E_(1)) = (2)/(1)`
2166.

A constant voltage is applied between the two ends of a metallic wire . If both the length and the radius of the wire are doubled , the rate of heat developed in the wire willA. will be doubledB. will be halvedC. will remain the sameD. will be quadrupled

Answer» Correct Answer - A
2167.

A battery of four cells in series, each having an emf of `1.14` V and an internal resistance of `2Omega` is to be used to charge a small 2 V accumulator of negligible internal resistance. What is the charging current?A. `0.1A`B. `0.2A`C. `0.3A`D. `0.45A`

Answer» Correct Answer - D
2168.

A battery of 24 cells each of emf 1.5 V and internal resistnace `2Omega` is to be connected in order to send the maximum current through a `12Omega` resistor. The correct arrangement of cells will beA. 2 rows of 13 cells connected in parallelB. 3 rows of 8 cells connected in parallelC. 4 rows of 6 cells connected in parallelD. All of these

Answer» Correct Answer - A
2169.

A resistor `R_(1)` dissipates the power `P` when connected to a certain generator. If the resistor `R_(2)` is put in series with `R_(1)`, the power dissipated by `R_(1)`A. DecreasesB. IncreasesC. Remains the sameD. Any of the above depending upon the relative values of `R_(1)` and `R_(2)`

Answer» Correct Answer - A
(a) Equivalent resistance in the second case` = R_(1) + R_(2) = R`
Now, we know that `P prop (1)/(R )`
Since in the second case the resistance `(R_(1) + R_(2))` is higher than that in the first case `(R_(1))`.
Therefore power dissipation in the second case will be decreased.
2170.

A wire of length 1.6m has a resistance `8 Omega` is connected to a battery of 2 volts and internal resistance `2Omega`. What is the potential gradient ?A. `1.5V//m`B. `0.5V//m`C. `1V//m`D. `2V//m`

Answer» Correct Answer - C
`l=1.6 m, R=8Omega, E=2V, r=2 Omega, I rho=?`
`I rho =((E)/(R+r))(R)/(l)`
`=((2)/(8+2))(8)/(1.6)=(16)/(16)=1 V//m`
2171.

Sensitivity of potentiometer can be increased byA. Incresing the e.m.f. of the cellB. Increasing the length of the potentiometer wireC. Decreasing the length of the potentiometer wireD. None of the above

Answer» Correct Answer - B
(b) The sensitivity of potentiometer can be increased by decreasing the potential gradient i.e., by increasing the length of potentiometer wire.
(Sensitivity `prop (1)/(P.G.)` and Length)
2172.

A battery of 6 volts is connected ot the termainals of a three meter long wire of uniform thickness and resistance of the order of `100 Omega`. The difference of potential between two points separated by `50 cm` on the wire willA. `1 V`B. `1.5 V`C. `2 V`D. `3 V`

Answer» Correct Answer - A
(a) Here same current is passing throughout the length of the wire , hence `V prop R prop l`
` implies (V_(1))/(V_(2)) = (l_(1))/(l_(2)) implies (300)/(50) implies V_(2) = 1V`
2173.

A battery of 6 volts is connected ot the termainals of a three meter long wire of uniform thickness and resistance of the order of `100 Omega`. The difference of potential between two points separated by `50 cm` on the wire willA. 2VB. 3VC. 1VD. 1.5V

Answer» Correct Answer - C
2174.

A resistor `R_(1)` dissipates the power `P` when connected to a certain generator. If the resistor `R_(2)` is put in series with `R_(1)`, the power dissipated by `R_(1)`A. decreasesB. increasesC. Remains the sameD. Any of the above depending upon the relative value of `R_(1)` and `R_(2)`

Answer» Correct Answer - A
2175.

In the circuit shown in figure find a) the rate of conversion of internal (chemical) energy to electrical energy within the battery(b) the rate of dissipation of electrical energy in the battery (c) the rate of dissipation of electrical energy in the external resistor.

Answer» Correct Answer - A::B::C::D
a. `i=12/(5+1)=2A`
` implies P_1=Ei=24W`
b. `P_2=i^2r=(2)^2(1)=4W`
`c. P=P_1-P_2=20W`
2176.

A plantinum wire has resistance of `10 Omega` at `0^(@)C` and `20 Omega` at `273^(@)C`. Find the value of coefficient of resistance.A. 273 per degree CB. 273 per degree KC. `(1)/(273)` per degree CD. `(1)/(273)` per degree K

Answer» Correct Answer - C
`R_(t)=R_(0)(1+alpha t)`
`:.alpha=(R_(1)-R_(0))/(R_(0)t)=(20-10)/(10xx273)`
`=(1)/(273)` per degree C
2177.

A copper disc and a carbon disc of same radius are assembled alternately and co-axially to make a cylindrical conductor whose temperature coefficient conductor whose temperature coefficient of resistance is almost equal to zero. Ratio of thickness of the copper and the carbon disc is (neglect change in length. `alpha_(CU)` and `-alpha_(C)` represent the temperature coefficients of resistivities and `p_(cu)` carbon at room temperature respectively)A. `p_(C)alpha_(C)//p_(CU)alpha_(CU)`B. `p_(CU)alpha_(CU)//p_(C)alpha_(C)`C. `p_(C)alpha_(CU)//p_(CU)alpha_(C)`D. `p_(CU)alpha_(C)//p_(C)alpha_(CU)`

Answer» Correct Answer - A
2178.

In the circuit shown, what is the potential different `V_(PQ)` ? A. `+3V`B. `+2V`C. `-2V`D. None of these

Answer» Correct Answer - D
2179.

In the circuit shown in figure, reading of voltmeter is `V_1` when only `S_1` is closed, reading of voltmeter is `V_2` when only `S_2` is closed, and reading of voltmeter is `V_3` when both `S_1 and S_2` are closed. Then . A. `V_(3) gt V_(2) gt V_(1)`B. `V_(2) gt V_(1) gt V_(3)`C. `V_(3) gt V_(1) gt V_(2)`D. `V_(1) gt V_(2) gt V_(3)`

Answer» Correct Answer - B
2180.

In the circuit shown in figure, reading of voltmeter is `V_1` when only `S_1` is closed, reading of voltmeter is `V_2` when only `S_2` is closed, and reading of voltmeter is `V_3` when both `S_1 and S_2` are closed. Then . A. `V_(2)gtV_(1)gtV_(3)`B. `V_(3)gtV_(2)gtV_(1)`C. `V_(3)gtV_(1)gtV_(2)`D. `V_(1)gtV_(2)gtV_(3)`

Answer» Correct Answer - A
`V_(2)gtV_(1)gtV_(3)`
2181.

In the circuit shown in figure, reading of voltmeter is `V_1` when only `S_1` is closed, reading of voltmeter is `V_2` when only `S_2` is closed, and reading of voltmeter is `V_3` when both `S_1 and S_2` are closed. Then . A. `V gt V gt V`B. `V gt V gt V`C. `V gt V gt V`D. `V gt V gt V`

Answer» Correct Answer - B
2182.

Three equal resistace, each of `R` ohm, are connectedd as shown in the figure. A battery of `2V` and of internal resistance `0.1` ohm is connected across the circuit. The value of `R` for which the heat generated in the circuit maximum will be A. `1.0 Omega`B. `0.2 Omega`C. `0.3 Omega`D. `0.4 Omega`

Answer» Correct Answer - C
2183.

Consider a rectangular slab of length L, and area of cross-section A. A current I is passed through it if the length is doubled the potential drop across the end facesA. Bacomes half of the initial valueB. Becomes one-fotrh of the initial valueC. Becomes double the initial valueD. remains same

Answer» Correct Answer - C
2184.

Figure `7.44` shows three similar lamps `L_(1) , L_(2), and L_(3)` connectged across a power supply. If the lamp ` L_(3)` fuses, how will the light emitted by `L_(1) and L_(2)` change? A. no changeB. brilliance of `L_(1)` decreases and that of `L_(2)` increasesC. brilliance of both `L_(1) ` and `L_(2)` increasesD. brilliance of both `L_(1)` and `L_(2)` decreases

Answer» Correct Answer - B
2185.

Three identical bulbs P, Q and R are connected to a battery as shown in the figure. When the circuit is closed.A. Q and R will be brighter than PB. Q and R will be dimmer than PC. all the bulbs will be equally brightD. Q and R will not shine at all

Answer» Correct Answer - B
2186.

Assertion : Two identical bulbs when connected across a battery, produce a total power P. When they are connected across the same battery in series total power consumed will be `(P)/(4)`. Reason : In parallel, `P=P_(1)+P_(2)` and in series `P=(P_(1)P_(2))/(P_(1)+P_(2))`A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.

Answer» Correct Answer - A
2187.

In the following circuit, bulb rated as 1.5 V , 0.45 W . If bulbs glows with full intensity then what will be the equivalent resistance between X and Y A. `0.45 Omega`B. `1 Omega`C. `3 Omega`D. `5 Omega`

Answer» Correct Answer - B
2188.

Rated power of a bulb at V voltage is P. Now, same voltage V is applied in all conditions mentioned in Column I. Match this Column I with Column II in which actual total power consumed is given.

Answer» Correct Answer - `(Ararrq;Brarrr;,Crarrs;Drarrp)`
2189.

In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m , a resistance R 1 and an accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection in galvanometer is zero. The current supplied by the accumulator will be A. `4 xx 10^(–4) A`B. `8 xx 10^(–4)A`C. `4 xx 10^(-3) A`D. `8 xx10^(-3) A`

Answer» Correct Answer - A
2190.

In the part of circuit shown in figure, match the following two columns for `V_(AB)`

Answer» Correct Answer - `(Ararrp;Brarrr;,Crarrq;Drarrp)`
2191.

As the switch `S` is closed in the circuit shown in figure, current passed through it is. A. `4.5 A`B. `6.0A`C. `3.0A`D. Zero

Answer» Correct Answer - A
2192.

In the circuit diagram shown in figure match the following two columns when switch S is closed.

Answer» Correct Answer - `(Ararrp;Brarrq;,Crarrq;Drarrq)`
2193.

As the switch `S` is closed in the circuit shown in figure, current passed through it is. A. `4.5A`B. `6.0A`C. `3.0A`D. zero

Answer» Correct Answer - A
2194.

In the circuit shown, the reading of ammeter is: A. `0.2 A`B. `0.6 A`C. `2 A`D. `1 A`

Answer» Correct Answer - D
(d) All three resistors are connected in parallel to a supply voltage of 24 volt. Ammeter is connected in the branch containing `24 Omega` resistor. Hence, current throuhg `24 Omega` resistor is correct.
2195.

An ammeter `A` of finite resistance, and a resistor `R` are joined in series to an ideal cell `C`. A potentiometer `P` is joined in parallel to `R`. The ammeter reading is `I_(0)` and the potentiometer reading is `V_(0)`. `P` is now replaced by a voltmeter of finite resistance. The ammeter reading now is `I` and the voltmeter reading is `V`. A. `IgtI_(0),VltV_(0)`B. `IgtI_(0),V=V_(0)`C. `I=I_(0),VltV_(0)`D. `IltI_(0),V=V_(0)`

Answer» Correct Answer - A
Since voltmeter have finite resistance, so resistance of circuit decreases so `IgtI_(0),VltV_(0)`.
2196.

The potentiometer wire `AB` shown in figure is `40 cm` long. Where the free end of the galvanometer should be connected on `AB` so that the galvanometer may show zero deflection?

Answer» Let the free end of galvanometer be connected on `AB` at distance `x` from `A`
`(8)/(12) = (AP)/(PB) = (x)/(40 - x)`
`(2)/(3) = (x)/(40 - x)`
`80 - 2x = 3 x rArr 5x = 80 rArr x = 16 cm`.
2197.

A battery of emf `E_(0) = 12 V` is connected across a `4 m` long uniform wire having resistance `(4 Omega)/(m)`. The cells of small emfs `epsilon_(1) = 2 V` and `epsilon_(2) - 4 V` having internal resistance `2 Omega` and `6 Omega` respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point `N`, the distance of point `N` from teh point `A` is equal to A. `1/6m`B. `1/3 m`C. `25 cm`D. `50 cm`

Answer» Correct Answer - C
`E_(eq)=(E_(2)r_(2)-E_(2)r_(1))/(r_(1)+r_(2))=(2xx6-4xx2)/(6+2)=1/2` volt
At balancing length `l`
`E_(eq)=(E_(0))/(R+16)xx4limplies1/2=12/(8+16)xx4xxl`
`l=25 cm`
2198.

A battery of emf `E_(0) = 12 V` is connected across a `4 m` long uniform wire having resistance `(4 Omega)/(m)`. The cells of small emfs `epsilon_(1) = 2 V` and `epsilon_(2) - 4 V` having internal resistance `2 Omega` and `6 Omega` respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point `N`, the distance of point `N` from teh point `A` is equal to A. `(1)/(6) m`B. `(1)/(3) m`C. `25 cm`D. `50 cm`

Answer» Correct Answer - C
(c ) Potential gradient
`x = ((12)/(8 + 16)) xx 4 = 2 Vm^(-1)`
Effective emf of `E_(1)` and `E_(2)`
`ul(E_(2)) - ul(E_(2))`
`E = (r_(2) r_(1))/(1 // r_(1) + 1 // r_(2)) = (1)/(2)` volt
Balancing length `= ((1)/(2))((1)/(2)) = (1)/(4) = m = 25 cm`
2199.

A battery of emf `E_(0) = 12 V` is connected across a `4 m` long uniform wire having resistance `(4 Omega)/(m)`. The cells of small emfs `epsilon_(1) = 2 V` and `epsilon_(2) - 4 V` having internal resistance `2 Omega` and `6 Omega` respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point `N`, the distance of point `N` from teh point `A` is equal to A. 2.5 mB. 3 mC. 1.25 mD. 0.75 m

Answer» Correct Answer - C
2200.

A parallel combination of two cells of emfs `epsilon_(1)` and `epsilon_(2)`, and internal resistance `r_(1)` and `r_(2)` is used to supply current to a load of resistance R. Write the expression for the current through the load in terms of `epsilon_(1), epsilon_(2),r_(1)` and `r_(2)`.

Answer» Equivalent resistance of two cells in parallel
`r_(eq)=(r_(1) r_(2))/(r_(1) +r_(2))`
Equivalent emf of two cells in parallel
`r_(eq)=(epsilon_(1)r_(1) +epsilon_(2)r_(1))/(r_(1)+r_(2))`
Current in circuit,
`I=(epsilon_(eq))/(R+r_(eq))=((epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)+r_(2)))/(R+(r_(1)r_(2))/((r_(1)+r_(2))))=(epsilon_(1)r_(2)+epsilon_(2)r_(1))/(R(r_(1)+r_(2))+r_(1)r_(2))`