2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 45 InterviewSolution

2201.	A house writing, supplied with a `220V` supply line is protected by a 9 ampere fuse. Find the maximum number of 60 W in parallel that can be turned on.
Answer» Correct Answer - `33` Current in each bulb `I = (P)/(V) = (60)/(220) = (3)/(11)A` No of bulb used `n = (9)/(3//11) = 33`

Discussion

2202.	When a potential difference is applied across, the current passing throughA. an insulator at 0 K is zeroB. a semiconductor at 0 K is zeroC. a metal at 0 K is finiteD. a p-n diode at 300K is finite, if it is reverse biased
Answer» Correct Answer - A::B::D (a,b,d) At 0K an insulator does not permit any current to flow through it. Option (a) is correct. At 0K a semiconductor behaves as an insulator. Option (b) is correct. In reverse biasing at 300 K, a very small current. Flows through a p-n junction diode. Option (d) is correct. In case of metal, the current flowing will be very-very high because a metal becomes super conductor at 0K. Option (c ) is incorrect.

Discussion

2203.	Currents of the order of 0.1A through the human body are fatal. What causes the death: heating of the body due to electric current or something else?
Answer» The causes of death is not heating due to current passing through a person. Though he may receive burns if the currents are too large. The cause of death is the interfernce of the external currents with our highly sensitive nervous system which is basically electrical in nature, which is basically electrical in nature, which inturn affect the heat beating. Beyond a certain point, this interference becomes fatal.

Discussion

2204.	How is the current conducted in metals ? Expalin.
Answer» Every metal coductor has large number of free electrons which move at random at room temperature. Their average thermal velocity at any instant is zero. When a pot. Diff. is applied across the ends of the conductor, an electic field is set up in the conductor. Due to it, the free electrons of the condutor experience force due to electic field and drift towards the positive end of the coductor, causing the electric current (i.e., conduction current) in the conductor. The direction of conventional current is opposite to the direction of motion of the free electrons in the conductor.

Discussion

2205.	A house is fitted with `20` length of `60` watt each `10` fans consuming `0.5` ampere each an electric kettle of resistance `110 Omega`. If the energy is supplied at `120 V` and costs `150` paise `kWh`, calculate monthly bill for running these appliances for 6 hours a day (1 length = 30 days).
Answer» Correct Answer - `739.80 Rs` Total energy connected in a all appliance ltnrgt `= (n_(1)P_(1) + n_(2)//V = (V^(2))/(R )) t xx 30` `= ( 20 xx 600 + 10 xx 0.5 xx 220 + (220^(2))/(110)) xx 6 xx 30` `= (1200 + 1100 + 440 xx 6 xx 30 `watt hour `= (2740 xx 6 xx 30)/(1000) kWh` Cost of electricity consumption `= (1.50 xx 2740 xx 6 xx 30)/(1000) = 739.80 Rs`

Discussion

2206.	There is an impression among many people that a person touching a high power line gets stuck with the line. Is that true ? Expalin.
Answer» This impression is misleading. Infact there is no special attractive force that keeps a person stuck with a high power line while touching that wire, whereas a current of few milliampere is enough to disorganize our nervous system. As a result of it, the affected person loses temporarily his ability to exercise his nervous control to get himself free the high power line.

Discussion

2207.	Assertion : A person touching a high power line gas stuck with the line. Reason : The current carrying wires attract the man toward it.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false
Answer» Correct Answer - D

Discussion

2208.	Assertion : A person touching a high power line gas stuck with the line. Reason : The current carrying wires attract the man toward it.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false.
Answer» Correct Answer - D (d) Because there is no special attractive force that keeps a person stuck with a high power line. The actaul reason is that a current of the order of `0.05 A` or even less is enough to bring disorder in our nervous system. As a result of it, the affected person may lose temporaily his ability to exercise his nervous control to get himself free from the high power line.

Discussion

2209.	An electric kettle has two coils of same power . When one coil is switched on , it takes `15 min` to boil water , and when the second coil is switched on , it takes `30 min`. How long will it take to boil water when both the coils are used in `i`. Series and `ii`. parallel?A. `9:2`B. `2:9`C. `4:5`D. `5:4`
Answer» Correct Answer - A `t_(S)=t_(1)+t_(2),t_(P)=(t_(1)t_(2))/(t_(1)+t_(2))`

Discussion

2210.	Two uniform wires of same material , each weighing `1 g` but one having double the length of the other, are connected in series , carrying a current of `10 A`. The length of the longer wire is `20 cm`. Calculate the rate of consumption of energy in each of the two wires. which wire gets hotter ? The resistivity of the material of the wire is `20 xx 10^(-5) Omega cm`. (Density of material =11 units )
Answer» Let `a_(1)` and `a_(2)` be the area of cross-section of shorter and longer wire respectively. As, = mass = volume `xx` density `= a l rho` `:. 1 = a_(1) xx 10 xx11 = a_(2) xx 20 xx 11` or `a_(1) = 1/10xx11 cm^(2) and a_(2) = 1/20xx11 cm^(2)` `:. R_(1) = 20 xx 10^(-5) xx 10/(1//(10 xx 11))` `= 20 xx 10^(-5) xx 10 xx 10 xx 11 = 22 xx 10^(-2) Omega` `:. R_(2) = 20 xx 10^(-5) xx 20/(1//(20 xx 11)) = 88 xx 10^(-2) Omega` And rate of heat produced, `H_(1) I^(2)R_(1) = (10)^(2) xx 22 xx 10^(-2) = 22 W` and `H_(2) = I^(2) R_(2) = (10)^(2) xx 88 xx 10^(-2) = 88 W`

Discussion

2211.	A copper electrical kettle weighing 1 kg contians 0.5 kg of water at `[email protected]`. It takes 10 minutes to raise the temperature to `100^@C`. If the electric energy is supplied at 220V, calculate the strength of the current,assuming that 20% heat is washed. Specific heat of copper is 0.1.
Answer» Here, mass of kettle, `m_(k) = 1000 g`, mass of water, `m_(w) = 500 g` , sp. Heat of copper, `S_(Cu) = 0.1 cal g^(-1) C^(-1)`, sp. Heat of water, `S_(w)= 1 cal g^(-1) .^(@)C^(-1)` Rise in temperature `= theta_(2) - theta_(1) = 100 -20 = 80^@C` Heat required = `(m_(k) S_(Cu) + m_(w) S_(w)) (theta_(2) - theta_(1))` `= (1000 xx 0.1 + 500 xx 1) xx 80 = 48000 cal` Heat produced, `H = (VIt)/J = (220 xx I xx (10 xx 60))/4.2 cal`. Useful heat produced = (100 -20)= 80% `= 80/100 xx (220 xx I xx 10 xx 60)/4.2 cal`. `:. 80/100 = (220 xx I xx 10 xx 60)/4.2 = 48000` On solving, `I= 1.9A`

Discussion

2212.	Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?A. 4 if wires are in parallelB. 2 if wires are in seriesC. 1 if wires are in seriesD. 0.5 if wires are in parallel
Answer» Correct Answer - B::D (b,d) `H = V^2/R xx 4 ` ………(i) where `R = (4rhol)/(pid^2)` When resistances are connected in series Total resistance `=R_1 + R_2 = 2[(4rhol)/(4pid^2)] = 2xx R/4 = R/2` `:. H = (V^2/R//2) xx t^2` ………(ii) From (i) and (ii) `t_2 = 2min.` Therefore (b) is correct. When resistance are connected in parallel `Total resistance `= (R_1R_2)/(R_1+R_2) = (R_(1)^2)/(2R_1) = (R//4)/2 = R/8` `:. H = v^2/ (R//8) xx t^2` From (i) and (iii) `t_2 = 0.5 min` `:. (d) is correct.

Discussion

2213.	An electric kettle has two coils. When one coil is switched an it takes `5` minutes to boil water and when second coil is switched on it takes `10` minutes .How long will it take to boil water, when both the coil are used in series ?
Answer» Correct Answer - `15 min` Let `R_(1) ,R_(2)` be the resistance of the two coils , `V` be the supply voltage and `H` be the heat required to boil the water For the first coil `H = (V^(2)t_(1))/(JR_(1)) = (V^(2) xx (5 xx 60))/(4.2 xx R_(1)) cel` For the second coil `H = (V^(2)t_(2))/(JR_(2))= (V^(2)xx 10 xx 60)/(4.2 xx R_(2)) cal` `:. (V^(2) xx 5 xx 60)/(4.2 R_(1)) = (V^(2) xx 10 xx 60 )/(4.2 xx R_(2) or (R_(2))/(R_(1)) = (10)/(5) = 2` when the coil are connected in series the effective resistance `R = R_(1) + R_(2)` Let `r` be the time in min boiling water , then `H = (V^(2)t)/(J(R_(1) + R_(2))) = (V^(2) xx t_(1))/(JR_(1))` `or t = (t_(1)(R_(1) + R_(2)))/(R_(1)) = t_(1) ( 1 + (R_(2))/(R_(1))) = 5 ( 1+ 2)` = 15 min

Discussion

2214.	A copper electrical kettle weighing 1 kg contians 0.5 kg of water at `[email protected]`. It takes 10 minutes to raise the temperature to `100^@C`. If the electric energy is supplied at 220V, calculate the strength of the current,assuming that 20% heat is washed. Specific heat of copper is 0.1.
Answer» Here, mass of kettle, `m_(k) = 1000 g`, mass of water, `m_(w) = 500 g` , sp. Heat of copper, `S_(Cu) = 0.1 cal g^(-1) C^(-1)`, sp. Heat of water, `S_(w)= 1 cal g^(-1) .^(@)C^(-1)` Rise in temperature `= theta_(2) - theta_(1) = 100 -20 = 80^@C` Heat required = `(m_(k) S_(Cu) + m_(w) S_(w)) (theta_(2) - theta_(1))` `= (1000 xx 0.1 + 500 xx 1) xx 80 = 48000 cal` Heat produced, `H = (VIt)/J = (220 xx I xx (10 xx 60))/4.2 cal`. Useful heat produced = (100 -20)= 80% `= 80/100 xx (220 xx I xx 10 xx 60)/4.2 cal`. `:. 80/100 = (220 xx I xx 10 xx 60)/4.2 = 48000` On solving, `I= 1.9A`

Discussion

2215.	Find the resistance of `240 V - 200` watt electric bulb when glowing. If this resistance is `10` times the resistance at `0^(@)C` and the temperature of the glowing filament is `2000^(@)C`, then find the temperature coefficient of resistance of the filament.
Answer» Resistance of hot bulb, `R_(t) = V^(2)/P = (200 xx 200)/100 = 400 Omega` Resistance of bulb at room temperature, `R_(0) = 400/10 = 40 Omega` Now, `R_(t) = R_(0) ( 1 +alpha t)` `:. 400 = 40 (1 +alpha xx 2000)` or `10 = 1 + alpha xx 2000` On solving, we get `alpha = 9/2000 = 4.5 xx 10^(-3).^(@)C^(-1)`

Discussion

2216.	A thin metallic wire of resistance `100 Omega` is immersed in a calorimeter containing 250 g of water at `10^@C` and a current of 0.5 ampere is passed through it for half an hour. If the water equivalent of the calorimeter is 10 kg, find the final temperature of water.
Answer» Here, `R = 100 Omega, m = 250g`, `theta_(1) = 10^@C, I = 0.5 A, t = 30 xx 60 s, w= 10g`, `theta_(2)` = final temperature of water (in `.^(@)C`) Heat produced, `H = I^(2) Rt` `= (0.5)^(2) xx 100 xx (30 xx 60)= 45000 J` Heat gained by water and caloimeter `= (m +w) c(theta_(2)- theta_(1))` `=(250 + 10) xx 1 xx (theta_(2) - theta_(1)) cal` `= 260 xx 4.2 xx (theta_(2)- theta_(1))= J` `:. 260 xx 4.2 xx (theta_(2)- theta_(1))=45000` or `theta_(2) - theta_(1) = 45000/(260 xx 4.2) = 41.2^@C` `theta_(2) = 41.2 + theta_(1) = 41.2 + 10 = 51.2^(@)C`

Discussion

2217.	The resistance of a thin silver wire is `1.0 Omega` at `20^@C`. The wire is placed in a liquid bath and its resistance rises to `1.2 Omega`. What is the temperature of the bath? `alpha` for silver is `3.8xx10^03//(.^(@) C)`
Answer» `R(T)R_0[1+alpha(T-T_0)]` Here, `R(T)=1.2Omega, R_0=1.0Omega, alpha=3.8xx10^-3//(.^@C) "and" T_0=2^@C` substituting the values we have `1.2=1.0[1+3.8x10^-3(T - 20)]` or `3.8xx10^-3(T-20)=0.2` solving this we get `T=72.6^@C`

Discussion

2218.	Compare the drift speed obtained above with (i) Thermal speed of copper atoms at ordinary temperatures. (ii) Speed of propagation of electric field along the conductor which causes the drift motion
Answer» (i) At a temperature T, the thermal speed of a copper atom of mass is obtained from `(1)/(2) M_(e) V_(rms)^(2) = [(3)/(2)] K_(B) T` `implies V_(rms) = sqrt((3K_(B) T)/(M_(e)))` Here `K_(B) = 1.38 xx 10^(-23)` `T = 300 K, M_(e)) = 9.1 xx 10^(-31) kg` `implies V_("rms") = sqrt((3 xx 1.38 xx 10^(-23) xx 300)/(9.1 xx 10^(-31)))` `= sqrt((1.38 xx 9 xx 10^(10))/(9.1))` ltBrgt `= sqrt(1.36) xx 10^(5) m//s` `:.` Thermal speed, `V_("rms") = 1.17 xx 10^(5) m//s` `(V_(d))/(V_(rms)) = (1.225 xx 10^(-3))/(1.17 xx 10^(5)) = 1.047 xx 10^(-8)` `:.` drift speed of electron`(V_(d)) = 1.047 xx 10^(-8)` `= 10^(-8)` times of thermal speed at ordinary temperature. (ii) The electric field travels along conductor with speed of EMW. `C = 3 xx 10^(8) m//s` `V_(d) = 1.25 xx 10^(-3) m//s` `(V_(d))/(C ) = (1.225 xx 10^(-3))/(3 xx 10^(8))` `V_(d) =0.408 xx 10^(-11) C` `:.` Drift speed is, in comparision of C, extermely smallar by a factor of `10^(-11)`.

Discussion

2219.	The number of free electrons per 100 mm of ordinary copper wire is `2xx10^(21)`. The average drift speed of electorn is `0.25 mm//s`. What is the current flowing?
Answer» Let A be the area of cross-section of wire. Therefore, number of electrons per unit volume of copper wire is, `n=2xx10^(21)/(A xx (100xx10^(-3)))` Now, `I=n A e v_(d)` `=(2xx 10^(21))/(A xx (100xx10^(-3))) xxA xx (1.6 xx10^(-19))` `xx (0.25 xx 10^(-3))` `= 0.8A`

Discussion

2220.	A potential difference V is applied to a conductor of length L, diameter D. How are electric field E, the drift velocity `v_(d)` and the resistance R affected when (i) V is doubled (ii)L is doubled (iii) D is doubled ?
Answer» We know, Electic field, `E = V/L`, Drift velocity, `v_(d) = (eE)/(m) tau = (eV)/(mL) tau` Resistance, `R = rho L/A = (4 rho L)/(pi D^(2))` (i) When V is doubled, E becomes double, `v_(d)` becomes double R remains unchanged. (ii) When L is doubled, E becomes half `v_(d)` becomes half and R becomes double. (iii) When D is doubled, E remains unchanged, `v_(d)` is also unchanged and R becomes one-fourth.

Discussion

2221.	For ohmic conductor, the drift velocity `v_(d)` and the electric field applied across it are related asA. `v_(d) prop sqrt(E)`B. `v_(d) prop E`C. `v_(d) prop E^(3//2)`D. `v_(d) prop E^(2)`
Answer» Correct Answer - B

Discussion

2222.	Unit of currenguish between Ct density isA. `A/m^2`B. `A/m^3`C. `m^2/A`D. `C/m^2`
Answer» Correct Answer - A

Discussion

2223.	Electric cume, temperature, luminous intensity, electric rrent flows fromA. higher potential to lower potentialB. lower potential to higher potentialC. can flow in any directionD. none of these
Answer» Correct Answer - A

Discussion

2224.	The equivalent resistance of two resistor connected in series is `6 Omega` and their equivalent resistance is `(4)/(3)Omega`. What are the values of resistances ?A. `4 Omega, 6 Omega`B. `8 Omega, 1 Omega`C. `4 Omega, 2 Omega`D. `6 Omega, 2 Omega`
Answer» Correct Answer - C

Discussion

2225.	Four cells, each of emf E and internal resistance r, are connected in series across an external resistance reverse. Then, the current in the external circuit isA. `(2E)/(4r+R)`B. `(3E)/(4r+R)`C. `(3E)/(3r+R)`D. `(2E)/(3r+R)`
Answer» Correct Answer - A

Discussion

2226.	The resistance of a wire is 5 ohm at `50^@C` and 6 ohm at `100^@C`. The resistance of the wire at `0^@C` will beA. 1 ohmB. 2 ohmC. 3 ohmD. 4 ohm
Answer» Correct Answer - D

Discussion

2227.	The net resistance of an ammeter should be small to ensure thatA. It does not get overheatedB. it does not draw excessive currentC. it can measure large potential differencesD. it does not appreciably change the potential difference to be measured.
Answer» Correct Answer - B

Discussion

2228.	the nonideal batteries are connected in series. Consider the following statements : (A)The equivalent emf is larger than either of the two emfs. (B) the equivalent internal resistances is smaller than either of the two internal resistances.A. Each of A and B is correctB. A is correct but B is wrongC. B is correct but A is wrongD. Each of A and B is wrong
Answer» Correct Answer - B

Discussion

2229.	In an atom electrons revolves around the nucleus along a path of radius 0.72Ã… making `9.4xx 10^(18)` revolution per second. The equivalent current is `(e=1.6xx10^(-19) C) `A. 1.2AB. 1.5AC. 1.4AD. 1.8A
Answer» Correct Answer - B Radius fo electron orbit `r=0.72Ã…=0.72xx10^(-10)m` Frequency of revolution of electron in orbit of given atom `u=9.4xx10^(18)rev//s` (where T is the time period of revolution of elctron in orbit) Then equivalent current is `I=(e)/(T)=ev=1.6xx10^(-19)xx9.4xx10^(18)=1.504A`

Discussion

2230.	If E is the emf of a cell of internal resistance r and external resistance R, then potential difference across R is given asA. `V=E//(R+r)`B. `V = E`C. `V=E//(1+r//R)`D. `V=E//(1+R//r)`
Answer» Correct Answer - C

Discussion

2231.	If an electron revolves in the circular path of radius `0.5A^(@)` at a frequency of `5xx10^(15)` cycles/sec. The equivalent electric current isA. `0.4 mA`B. `0.8mA`C. `1.2 mA`D. `1.6 mA`
Answer» Correct Answer - B

Discussion

2232.	Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. `R = r//2`B. `R = r`C. `R = 2r`D. `R = 0`
Answer» Correct Answer - A

Discussion

2233.	Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. `R=r/2`B. `R=r`C. `R=2r`D. `R=0`
Answer» Correct Answer - A Power delivered to the resistor is R or current through R is maximum when total internal resistance of the circuit is equal to the external resistance i.e. `R=r//2`.

Discussion

2234.	At what temperature will the resistance of a copper wire become three times its value at `0^(@)C` (Temperature coefficient of resistance for copper `= 4 xx 10^(3)` per C)A. `400^(@) C`B. `450^(@)C`C. `500^(@)C`D. `550^(@)C`
Answer» Correct Answer - C

Discussion

2235.	At what temperature will the resistannce of a copper wire become three times its value at `0^(@)C` (Temperature coefficient of resistance for copper `=4xx10^(-3)"per".^(@)C`:-A. `400^(@)C`B. `450^(@)C`C. `500^(@)C`D. `500^(@)C`
Answer» Correct Answer - 3

Discussion

2236.	The potential drop across the `3 Omega` resistor is A. 1VB. 1.5VC. 2VD. 3V
Answer» Correct Answer - A

Discussion

2237.	10,000 electrons are passing per minute through a tube of radius 1 cm. the resulting current is:A. 10000AB. `0.25xx10^(-16)A`C. `10^(-9)A`D. `0.5xx10^(-19)A`
Answer» Correct Answer - 2

Discussion

2238.	In the given figure, potential difference between A and B is
Answer» Correct Answer - C

Discussion

2239.	If i=0.25 amp. In fig then value of R is:- A. `48Omega`B. `12 Omega`C. `120Omega`D. `42Omega`
Answer» Correct Answer - 4

Discussion

2240.	If each resistance in the figure is of `9Omega` then reading of ammeter is A. 5AB. 8AC. 2AD. 9A
Answer» Correct Answer - A

Discussion

2241.	If each resistance in the figure is of `9Omega` then reading of ammeter is A. 5AB. 8AC. 2AD. 9A
Answer» Correct Answer - 1

Discussion

2242.	The resistance of a copper wire and an iron at `20^@C` are `4.1 Omega` and `3.9Omega` respectively. Neglecting any thermal expansion, find the temperature at which resistane of both are equal. `alpha_(Cu)=4.0xx10^-3K^-1` an `alpha_(Fe)=5.0xx10^-3K^-1`
Answer» Correct Answer - C `4.1[1+4.0x10^-3(theta-20)]` `=3.9[1+5.0xx10^-3(theta-20)]` solving we get `theta~~85^@C`

Discussion

2243.	Show that the volume thermal expansion coefficient for an ideal gas at constant pressure is `(1)/(T)`.A. TB. `T^(2)`C. `(1)/(T)`D. `(1)/(T^(2))`
Answer» Correct Answer - C

Discussion

2244.	The figure shows part of certain circuit, find, a. power dissipted in `5Omega` resistance, b.Potential difference `V_C-V_B` c.Which battery is being charged?
Answer» Correct Answer - A::B::C a. Current flowing through resistance `5Omega` is `11 A` power dissipted `=i^2R` `=(121)5=605W` `b. V_B+8V+3V=12V-12V-5V=V_C` `V_B+11V-5V=V_C` `6V=V_C-V_B` c. both batteries are being charged.

Discussion

2245.	In the figure shown the power generated by `y` is maximum when `y=5Omega`. Then `R` is A. `2 Omega`B. `6 Omega`C. `5 Omega`D. `3 Omega`
Answer» Correct Answer - D `P=I^(2)y=(10/(2+y+R))^(2)Y=100/((2+y+R)^(2)) y` For `P` to be maximum `(dP)/(dy)=0 implies d/(dy)(100y)/((2+y+R)^(2))=0` `R=y-z` put `y=5implies R=3 Omega`

Discussion

2246.	A steel tape measures that length of a copper rod as `90.0` cm when both are at `10^(@)C`, the calibration temperature, for the tape. What would the tape read for the length of the rod when both are at `30^(@)C`. Given `alpha_("steel")=1.2xx10^(-5)" per".^(@)Cand alpha_(Cu)=1.7xx10^(-5)per .^(@)C`A. `89.00cm`B. `90.21cm`C. `89.80cm`D. `90.01cm`
Answer» Correct Answer - D

Discussion

2247.	Power generated across a uniform wire connected across a supply is `H`. If the wire is cut into `n` equal parts and all the parts are connected in parallel across the same supply, the total power generated in the wire isA. `(H)/(n^(2))`B. `n^(2)H`C. `nh`D. `(H)/(n)`
Answer» Correct Answer - B

Discussion

2248.	The variation of current `(I)` and voltage `(V)` is as shown in figure `A`. The variation of power `P` with current `I` is best shown by which of the following graph A. B. C. D.
Answer» Correct Answer - B

Discussion

2249.	The variation of current `(I)` and voltage `(V)` is as shown in figure `A`. The variation of power `P` with current `I` is best shown by which of the following graph B. C. D.
Answer» Correct Answer - B From graph `V=KI^(2)` We know `P=VI implies P=KI^(3)`

Discussion

2250.	If tha ammter in the given circuit reads `2 A`, the resistance `R` is A. 1 ohmB. 2 ohmC. 3 ohmD. 4 ohm
Answer» Correct Answer - A

Discussion

Explore topic-wise InterviewSolutions in .

A house writing, supplied with a `220V` supply line is protected by a 9 ampere fuse. Find the maximum number of 60 W in parallel that can be turned on.

When a potential difference is applied across, the current passing throughA. an insulator at 0 K is zeroB. a semiconductor at 0 K is zeroC. a metal at 0 K is finiteD. a p-n diode at 300K is finite, if it is reverse biased

Currents of the order of 0.1A through the human body are fatal. What causes the death: heating of the body due to electric current or something else?

How is the current conducted in metals ? Expalin.

A house is fitted with `20` length of `60` watt each `10` fans consuming `0.5` ampere each an electric kettle of resistance `110 Omega`. If the energy is supplied at `120 V` and costs `150` paise `kWh`, calculate monthly bill for running these appliances for 6 hours a day (1 length = 30 days).

There is an impression among many people that a person touching a high power line gets stuck with the line. Is that true ? Expalin.

A copper electrical kettle weighing 1 kg contians 0.5 kg of water at `[email protected]`. It takes 10 minutes to raise the temperature to `100^@C`. If the electric energy is supplied at 220V, calculate the strength of the current,assuming that 20% heat is washed. Specific heat of copper is 0.1.

An electric kettle has two coils. When one coil is switched an it takes `5` minutes to boil water and when second coil is switched on it takes `10` minutes .How long will it take to boil water, when both the coil are used in series ?

A copper electrical kettle weighing 1 kg contians 0.5 kg of water at `[email protected]`. It takes 10 minutes to raise the temperature to `100^@C`. If the electric energy is supplied at 220V, calculate the strength of the current,assuming that 20% heat is washed. Specific heat of copper is 0.1.

Find the resistance of `240 V - 200` watt electric bulb when glowing. If this resistance is `10` times the resistance at `0^(@)C` and the temperature of the glowing filament is `2000^(@)C`, then find the temperature coefficient of resistance of the filament.

A thin metallic wire of resistance `100 Omega` is immersed in a calorimeter containing 250 g of water at `10^@C` and a current of 0.5 ampere is passed through it for half an hour. If the water equivalent of the calorimeter is 10 kg, find the final temperature of water.

The resistance of a thin silver wire is `1.0 Omega` at `20^@C`. The wire is placed in a liquid bath and its resistance rises to `1.2 Omega`. What is the temperature of the bath? `alpha` for silver is `3.8xx10^03//(.^(@) C)`

Compare the drift speed obtained above with (i) Thermal speed of copper atoms at ordinary temperatures. (ii) Speed of propagation of electric field along the conductor which causes the drift motion

The number of free electrons per 100 mm of ordinary copper wire is `2xx10^(21)`. The average drift speed of electorn is `0.25 mm//s`. What is the current flowing?

A potential difference V is applied to a conductor of length L, diameter D. How are electric field E, the drift velocity `v_(d)` and the resistance R affected when (i) V is doubled (ii)L is doubled (iii) D is doubled ?

For ohmic conductor, the drift velocity `v_(d)` and the electric field applied across it are related asA. `v_(d) prop sqrt(E)`B. `v_(d) prop E`C. `v_(d) prop E^(3//2)`D. `v_(d) prop E^(2)`

Unit of currenguish between Ct density isA. `A/m^2`B. `A/m^3`C. `m^2/A`D. `C/m^2`

Electric cume, temperature, luminous intensity, electric rrent flows fromA. higher potential to lower potentialB. lower potential to higher potentialC. can flow in any directionD. none of these

The equivalent resistance of two resistor connected in series is `6 Omega` and their equivalent resistance is `(4)/(3)Omega`. What are the values of resistances ?A. `4 Omega, 6 Omega`B. `8 Omega, 1 Omega`C. `4 Omega, 2 Omega`D. `6 Omega, 2 Omega`

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance reverse. Then, the current in the external circuit isA. `(2E)/(4r+R)`B. `(3E)/(4r+R)`C. `(3E)/(3r+R)`D. `(2E)/(3r+R)`

The resistance of a wire is 5 ohm at `50^@C` and 6 ohm at `100^@C`. The resistance of the wire at `0^@C` will beA. 1 ohmB. 2 ohmC. 3 ohmD. 4 ohm

The net resistance of an ammeter should be small to ensure thatA. It does not get overheatedB. it does not draw excessive currentC. it can measure large potential differencesD. it does not appreciably change the potential difference to be measured.

In an atom electrons revolves around the nucleus along a path of radius 0.72Ã… making `9.4xx 10^(18)` revolution per second. The equivalent current is `(e=1.6xx10^(-19) C) `A. 1.2AB. 1.5AC. 1.4AD. 1.8A

If E is the emf of a cell of internal resistance r and external resistance R, then potential difference across R is given asA. `V=E//(R+r)`B. `V = E`C. `V=E//(1+r//R)`D. `V=E//(1+R//r)`

If an electron revolves in the circular path of radius `0.5A^(@)` at a frequency of `5xx10^(15)` cycles/sec. The equivalent electric current isA. `0.4 mA`B. `0.8mA`C. `1.2 mA`D. `1.6 mA`

Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. `R = r//2`B. `R = r`C. `R = 2r`D. `R = 0`

Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. `R=r/2`B. `R=r`C. `R=2r`D. `R=0`

At what temperature will the resistance of a copper wire become three times its value at `0^(@)C` (Temperature coefficient of resistance for copper `= 4 xx 10^(3)` per C)A. `400^(@) C`B. `450^(@)C`C. `500^(@)C`D. `550^(@)C`

At what temperature will the resistannce of a copper wire become three times its value at `0^(@)C` (Temperature coefficient of resistance for copper `=4xx10^(-3)"per".^(@)C`:-A. `400^(@)C`B. `450^(@)C`C. `500^(@)C`D. `500^(@)C`

The potential drop across the `3 Omega` resistor is A. 1VB. 1.5VC. 2VD. 3V

10,000 electrons are passing per minute through a tube of radius 1 cm. the resulting current is:A. 10000AB. `0.25xx10^(-16)A`C. `10^(-9)A`D. `0.5xx10^(-19)A`

In the given figure, potential difference between A and B is

If i=0.25 amp. In fig then value of R is:- A. `48Omega`B. `12 Omega`C. `120Omega`D. `42Omega`

If each resistance in the figure is of `9Omega` then reading of ammeter is A. 5AB. 8AC. 2AD. 9A

If each resistance in the figure is of `9Omega` then reading of ammeter is A. 5AB. 8AC. 2AD. 9A

The resistance of a copper wire and an iron at `20^@C` are `4.1 Omega` and `3.9Omega` respectively. Neglecting any thermal expansion, find the temperature at which resistane of both are equal. `alpha_(Cu)=4.0xx10^-3K^-1` an `alpha_(Fe)=5.0xx10^-3K^-1`

Show that the volume thermal expansion coefficient for an ideal gas at constant pressure is `(1)/(T)`.A. TB. `T^(2)`C. `(1)/(T)`D. `(1)/(T^(2))`

The figure shows part of certain circuit, find, a. power dissipted in `5Omega` resistance, b.Potential difference `V_C-V_B` c.Which battery is being charged?

In the figure shown the power generated by `y` is maximum when `y=5Omega`. Then `R` is A. `2 Omega`B. `6 Omega`C. `5 Omega`D. `3 Omega`

Power generated across a uniform wire connected across a supply is `H`. If the wire is cut into `n` equal parts and all the parts are connected in parallel across the same supply, the total power generated in the wire isA. `(H)/(n^(2))`B. `n^(2)H`C. `nh`D. `(H)/(n)`

The variation of current `(I)` and voltage `(V)` is as shown in figure `A`. The variation of power `P` with current `I` is best shown by which of the following graph A. B. C. D.

The variation of current `(I)` and voltage `(V)` is as shown in figure `A`. The variation of power `P` with current `I` is best shown by which of the following graph B. C. D.

If tha ammter in the given circuit reads `2 A`, the resistance `R` is A. 1 ohmB. 2 ohmC. 3 ohmD. 4 ohm