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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The effective between A&B in the given circuit isA. `7Omega`B. `2Omega`C. `6Omega`D. `5Omega` |
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Answer» Correct Answer - C using combination of resistor |
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| 2. |
A source of e.m.f. `E = 15V` and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as `i = 1.2 t +3` Then, the total charge that will flow in first five second will beA. `10C`B. `20C`C. `30C`D. `40C` |
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Answer» Correct Answer - C |
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| 3. |
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. The battery that runs the potentiometer should have voltage of 8 VB. The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 VC. The first portion of 50 cm of wire itself should have a potential drop of 10 VD. Potentiometer is usually used for comparing resistance and not voltages |
| Answer» In a potentiometer experiment, the emf of a cell can be measured, if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V | |
| 4. |
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. The battery that runs the potentionmeter should have voltage of 8 VB. The battery of potentionmeter can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10 VC. The first portion of 50cm of wire itself should have a potential drop of 10VD. Potentiometer is usually used for comparing resistances and not voltages |
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Answer» Correct Answer - B In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. As values of emfs of two cells are approximately 5V and 10V, therefore, the potential drop along the potentiometer wire must be more than 10V. Hence, option (b) is correct. |
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| 5. |
Two cells of e.m.f. 10V & 15V are connected in parallel to each other between points A&B. The cell of e.m.f. 10V is ideal but the cell of e.m.f. 15V has internal resistance `1Omega`. The equivalent e.m.f. between A and B is: A. `25/2V`B. not definedC. 15 VD. 10 V |
| Answer» Correct Answer - D | |
| 6. |
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. Toe battery that runs the potentiometer should have voltage of 8 V.B. Toe battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.C. The first portion of 50 cm of wire itself should have a potential drop of 10 VD. Potentiometer is usually used for comparing resistances and not voltages. |
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Answer» Correct Answer - B As `R=rho(l)/(A)` resistance is maximum when l is large and A is least. For the give dimensions of wire, resistane will be maximum for l=10cm and `A=1cmxx(1)/(2)cm` |
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| 7. |
In the `R_(1) = 10 Omega ,R_(2) = 20 Omega, R_(3) = 40 Omega R_(4) = 80 Omega` and `V_(A) = 5V,V_(B) = 10V,V_(c) = 20V,V_(D) = 15V` The current in the resistance `R_(1)` will be A. `0.4 A` toward OB. `0.4A` away from OC. `0.8A` toward OD. `0.8A` toward O |
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Answer» Correct Answer - b Let `V_(0)` be the potential at `O` and `I_(1),I_(2),I_(3) and I_(4)` are the current through `R_(1),R_(2),R_(3)and R_(4)` respectively Then `I_(1),I_(2),I_(3),I_(4)= 0` `(V_(0) - V_(A))/(R_(1)) + (V_(0) - V_(B))/(R_(2)) +(V_(0) - V_(C))/(R_(3)) +(V_(0) - V_(D))/(R_(4)) = 0` `(V_(0) - 5)/(10) + (V_(0) - 10)/(20) +(V_(0) - 20)/(40) +(V_(0) - 15)/(80) = 0` On solving we shall get `V_(0) = 9V` Current in the resistance `R_(1)` is `I_(1) =(V_(0) - V_(A))/(R_(1)) = (9-5)/(10) = 0.4A` As `V_(0) gt V_(A)`, so the currect is flowing from `O` toward A i.e. away from O. |
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| 8. |
The amount of charge Q passed in time t through a cross-section of a wire is `Q=5t^(2)+3t+1`. The value of current at time t=5 s isA. 9AB. 49AC. 53AD. None of these |
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Answer» Correct Answer - C Current, `i=(dQ)/(dt)=(d)/(dt)(5t^(2)+3t+1)` `= 10t+3` When `t = 5 s i=10xx5+3=53 A` |
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| 9. |
The charge flowing through a conductor varies with time as `q= 8 t - 3t^(2)+ 5t^(3)` Find (i) the initial current (ii) time after which the current reaches a maximum value of current |
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Answer» Correct Answer - (i)` l = 8A`,(ii)` 0.2 s`,(iii)`7.4A` Given `q = 8 t - 3t^(2) + 5t^(3)` (i) current `I = (dq)/(dt) = 8 - 6t + 15t^(2)`…..(i) When `t = 0 I = 8A` (ii) `(dI)/(dt) = - 6+30t` For `l` to the maximum or minimum `(dI)/(dt) =0 or - 6 + 30 t = 0 or t = 0.2 s` (iii) putting this value of `t` in (i) we get `I = 8 - 6 xx 0.2 + 15(0.2)^(2) = 8 - 1.2 + 0.6 = 7.4 A` As this value of `l` is less than at `t = 0`, it will be minimum value , so minimum value of current `= 7.4A` |
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| 10. |
The current in a conductor varies with time `t` as `I=3t+4t^(2)` Where I in amp and t in sec. The electric charge flows through the section of the conductor between `t=1s` and `t=3s`A. `(100)/(3)C`B. `(127)/(4)C`C. `(140)/(3)C`D. `(150)/(3)C` |
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Answer» Correct Answer - C `q=int_(t_(1))^(t_(2))I.dt` |
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| 11. |
A silver wire has a resistance of `2.1 Omega` at `27.5^(@)C`, and a resistance of `2.7 Omega` at `100^(@)C`. Determine the temperature coefficient of resistivity of silver. |
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Answer» Here `R_(27.5) = 2.1 Omega, R_(100) = 2.7 Omega, alpha = ?` `alpha = (R_(100) - R_(27.5))/(R_(27.5) xx(100 - 27.5)) = (2.7 - 2.1)/(2.1 xx (100 - 27.5)) = 0.0039^(@)C^(-1)` |
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| 12. |
In the circuit shown in figure-3.348 when the switch is closed, the initial current through the `1 Omega` resistor just after closing the switch is A. 2AB. 4AC. 3AD. 6A |
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Answer» Correct Answer - B,C,D |
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| 13. |
A monoatomic ideal gas is used in a carnot engine as the working substance. If during the adiabatic expansion the volume of the gas increases from `(V)/(8)` to V, the efficiency of the engine isA. `0.50`B. `0.25`C. `0.75`D. `0.40` |
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Answer» Correct Answer - C |
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| 14. |
For a monoatomic ideal gas undergoing an adiabatic change, the relation between temperature and volume `TV^(x)` = constant, where x isA. `(7)/(5)`B. `(2)/(5)`C. `(2)/(3)`D. `(1)/(3)` |
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Answer» Correct Answer - C |
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| 15. |
A certain amount of ideal monoatomic gas undergoes a process given by `TV^(1//2)`= constant. The molar specific heat of the gas for the process will be given byA. R/2B. `-3R//2`C. `3R//2`D. `-R//2` |
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Answer» Correct Answer - D |
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| 16. |
If `R_(1)` and `R_(2)` are respectively the filament resistances of a 200 watt bulb and 100 watt bulb designed to operate on the same voltage, thenA. `R_(1)` is two times `R_(2)`B. `R_(2)` is two times `R_(1)`C. `R_(2)` is four times `R_(1)`D. `R_(1)` four times `R_(2)` |
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Answer» Correct Answer - B |
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| 17. |
If `R_(1)` and `R_(2)` are respectively the filament resistances of a 200 watt bulb and 100 watt bulb designed to operate on the same voltage, thenA. `R_(1) " is two times "R_(2)`B. `R_(2)"is two times"R_(1)`C. `R_(2)"is four times "R_(1)`D. `R_(1)"is four times"R_(2)` |
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Answer» Correct Answer - D |
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| 18. |
If `R_(1)` and `R_(2)` are respectively the filament resistances of a 200 watt bulb and 100 watt bulb designed to operate on the same voltage, thenA. `R_(1)=2R_(2)`B. `R_(2)2R_(1)`C. `R_(2)=4R_(1)`D. `R_(1)=4R_(2)` |
| Answer» Correct Answer - B | |
| 19. |
The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown in figure. What is the emf and internal resistance of each cell? |
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Answer» Total emf of three cells in series (`3 epsilon`) = pot. Diff. corresponding to zero current `:. 3 epsilon = 6 V or epsilon = 6/3 = 2V` The internal resistance of each cell `r= epsilon/I_(max) = 2/1 = 2 Omega` |
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| 20. |
Two wires of equal length one of copper and other of manganin have the same resistance. Which wire is thicker? |
| Answer» `R=rho l//A` or `A prop rho` if l and R are constant. Since `rho` is grater for manganin than for copper, hence manganin wire is thicker than copper wire. | |
| 21. |
When a current is established in a wire, the free electrons drift in the direction opposite to the current, Does the number of free electrons in the wire continuously decrease? |
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Answer» Correct Answer - B |
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| 22. |
If the current in the primary circuit is decreased, then balancing length is obtained atA. lower lengthB. higher lengthC. same lengthD. 1/3rd length |
| Answer» Correct Answer - B | |
| 23. |
For the circuit would the balancing length increase, decrease or remain the same if (i) `R_(1)` is decreased (ii) `R_(2)` is increased, without any change (in each case) in the rest of the circuit ? Justify your answer in each case. |
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Answer» (i) When `R_(1)` is decreased, the balancing length decreases. It is because as `R_(1)` decreases, the current in the potentiometer circuit increases. Due to it, the potential gradient of potentiometer wire increases. As a result of it the same emf will be balanced on smaller length of potentiometer wire. (ii) When `R_(2)` is increased, the balancing length increases. It is because as `R_(2)` increases, the current `I( = epsilon/r+R_(2))` decreases. This increases the terminal pot. diff. `V (epsilon - Ir)` across the unknown cell. As a result of it, the terminal potential difference of cell `epsilon` will be balanced on larger length of potentiometer wire. |
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| 24. |
As temperature increases, the vicosity of liquids decrease considerably. Will this decrease the resistance of an electrolyte as the temperature increases? |
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Answer» Correct Answer - C |
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| 25. |
A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass`=9xx10^(-6)//^(@)C` and the coefficient of volume expansion of `Hg=1.8xx10^(-4)//^(@)C` ?A. 50 `cm^(3)`B. 100 `cm^(3)`C. 150 `cm^(3)`D. 200 `cm^(3)` |
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Answer» Correct Answer - C |
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| 26. |
Two moles of helium are mixed with n moles of hydrogen. The root mean spure (rms) speed of the gas molecules in the mexture is `sqrt2` times the speed of sound in the mixture. Then value of n isA. 1B. 3C. 2D. `3//2` |
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Answer» Correct Answer - C |
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| 27. |
The relation between U, p and V for an ideal gas in an adiabatic process is given by relation `U=a+bpV`. Find the value of adiabatic exponent `(gamma)` of this gas.A. `(b+1)/(b)`B. `(b+1)/(a)`C. `(a+1)/(b)`D. `(a)/(a+b)` |
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Answer» Correct Answer - A |
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| 28. |
Which of the following quantities is independent of the nature of the gas at same temperature ?A. The number of molecules in 1 moleB. The number of molecules in equal volumeC. The translational kinetic energy of 1 moleD. The kinetic energy of unit mass |
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Answer» Correct Answer - A::C |
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| 29. |
Which of the following quantities depend on temperature only for a given ideal gas ?A. Total internal energy of the gasB. Product pV of the gasC. The ratio of pressure and density of the gasD. Root mean square speed of the gas |
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Answer» Correct Answer - C::D |
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| 30. |
When the temprature of a gas filled in a closed vessel is increased by `1^(@)C`, its pressure increases by 0.4 percent. The initial temperature of gas wasA. 250 KB. 500 KC. `250^(@)C`D. `25^(@)C` |
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Answer» Correct Answer - A |
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| 31. |
Three identical rods `AB`, `CD` and `PQ` are joined as shown. `P` and `Q` are mid points of `AB` and `CD` respectively. Ends `A`, `B`, `C` and `D` are maintained at `0^(@)C`, `100^(@)C`, `30^(@)C` and `60^(@)C` respectively. The direction of heat flow in `PQ` is A. from P to QB. from Q to PC. heat does not flow in PQD. Data is insufficient |
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Answer» Correct Answer - A |
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| 32. |
n moles of an ideal gas undergo a process in which the temperature changes with volume as `T=kv^(2)`. The work done by the gas as the temperature changes from `T_(0)` to `4T_(0)` isA. `3nRT_(0)`B. `(5//2)nRT_(0)`C. `(3//2)nRT_(0)`D. zero |
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Answer» Correct Answer - C |
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| 33. |
If two rods of length L and 2L having coefficients of linear expansion `alpha` and `2alpha` respectively are connected so that total length becomes 3L, the average coefficient of linear expansion of the composite rod equalsA. `(3)/(2)alpha`B. `(5)/(2)alpha`C. `(5)/(3)alpha`D. None of these |
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Answer» Correct Answer - C |
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| 34. |
A vessel contains 1 mole of `O_2` gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (relative molar mass 4) at temperature 2T has a pressure ofA. `p//8`B. `p`C. `2p`D. `8p` |
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Answer» Correct Answer - C |
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| 35. |
`rho`-T equation of a gas in adiabatic process is given byA. `T^(gamma-1)rho="constant"`B. `rho^(gammaT)="constant"`C. `Trho^(1-gamma)="constant"`D. `T^(gamma)rho^(gamma-1)="constant"` |
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Answer» Correct Answer - C |
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| 36. |
What must be the lengths of steel and copper rods at `0^(@)C` for the difference in their lengths to be 10 cm at any common temperature? `(alpha_(steel)=1.2xx10^(-5).^(@)K^(-1)and alpha_("copper")=1.8xx10^(-5).^(@)K^(-1))`A. 30 cm for steel and 20 cm for copperB. 20 cm for steel and 40 cm for copperC. 40 cm for steel and 30 cm for copperD. 30 cm for steel and 40 cm for copper |
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Answer» Correct Answer - A |
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| 37. |
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled ifA. Both the length and radius of the wire are halvedB. Both the length and radius of the wire are doubledC. the radius of the wire is doubledD. the length of the wire are doubled |
| Answer» Correct Answer - b | |
| 38. |
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled ifA. both the length and the radius of the wire we halvedB. both the length and the radius of the wire are doulbedC. The radius of the is doubledD. the length of the wire is doulbed |
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Answer» Correct Answer - B (b) `P = (V^(2))/(R ) = (V^(2) A)/(rho l) prop (r^(2))/(l) [V rarr same]` |
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| 39. |
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled ifA. both the length and the radius of the wire are halved.B. both the length and the radius of the wire are doubled.C. the radius of the wire is doubled.D. the length of the wire is doubled . |
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Answer» Correct Answer - B (b) KEY CONCEPT : The heat produced is given by `H = V^2/R and R=(rhol)/(pir^2) :. H= V^2 ((pir^2)/(rhol))` or `H = (piV^2)/rho (r^2/l)` Thus heat (H) is doubled if both length (l) and radius (r ) are doubled. |
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| 40. |
Temperature dependence of resistivity p(T) of semiconductors, insulators and metals is significantly based on the following factors:A. Numbr of charge carriers can change with temperature TB. time interval between two successie collisions can depend of TC. length of material can be a function of TD. mass of carriers is a function of T. |
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Answer» Correct Answer - A::B The resistivity of a metallic conductor is given by `e=(m)/(m e^(2)tau)` When n is number of charge carriers per unit volume which can change with temperature T and `tau` is time interval between two successive collisions which decreases with the increase of temperature. |
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| 41. |
The electrostatic field due to a point charge depends on the distance r as `1/r^2`. Indicate which of the following quantities shows same dependence on r.A. Intensity of light from a point sourceB. Electrostatic potential due to a point charge.C. Electrostatic potential at a distance r from the centre of a charged metallic spher. Given rlt radius of the sphere.D. None of these |
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Answer» Correct Answer - A (a) `I prop 1/(r^2), V prop 1/r , V prop r^0` . |
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| 42. |
The current I in the circuit is A. 1/45 AB. 1/15 AC. 1/10 AD. 1/5 A |
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Answer» Correct Answer - C (c) BC and AC are in series `:. R_(BCA) = 30 + 30 = 60 Omega` Now BA and DC are in parallel. `1/R_(eq) = 1/30 + 1/60 = 90/(30 xx 60)` `R_(eq) = 20Omega , V = IR` `rArr I = 2/20 = 0.1 Amp. ` |
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| 43. |
In the circuit shown in fig the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor isA. `1cal//sec`B. `2cal//sec`C. `3cal//sec`D. `4cal//sec` |
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Answer» Correct Answer - B (b) Since `R_(AB) = 2R_(CD)` therefore, current in AB will be half as compared to current in CD. `P_4/P_5 = ((i//2)^2 4)/(i^2 xx 5) = 1/5 rArr P_4 = 10/5 = 2cal//s` Here `P_4 = power dissipation in 4Omega` `P_5 = Power dissipation in 5Omega`. |
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| 44. |
In the circuit shown in fig the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor is |
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Answer» Correct Answer - `80 cal` Let `l, l_(1) ` be the current through upper and lower area of the circuit .resistance of upper arm of circuit `= 2 + 3 = 5 Omega` resistance of lower arm `= 6 + 4 = 10 Omega` Heat prodeced in `4 Omega` resistance `= l_(1)^(2) xx 4 =40 xx 4.2 J` `or l_(1)^(2) = 10 xx 4.2 or l_(1) sqrt(42)A` pot diff across lower arm `= l_(1) xx 10 = sqrt(42) xx 10 V` pot diff across arm `= 10 sqrt(42) V` Cyrrent through upper arm `l = 10 sqrt(42)//5 = 2 sqrt(42)A` Here produced per second in `2 Omega` resistance `= ((2 sqrt(42))^(2) xx 2 xx 1)/(4.2) = 80 cal` |
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| 45. |
Two identical slabs of given metal are joined together in two different ways as shown in figure. What is the ratio of the resistance of these two combinations? |
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Answer» Let `l` and A be the length and area of cross-section of the each of the two slabs. Then `R= rho l//A` `R_(1) = (rho 2l)/A = 2R` `R_(2) = (rho l)/(2A)= R/2` `R_(1)/R_(2)=(2R)/(R//2) = 4 or R_(1) : R_(2)=4:1` |
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| 46. |
A `100W, 200V` bulb is operated on a `110V` line. The power consumed isA. `50W`B. `75 W`C. `90 W`D. `25 W` |
| Answer» Correct Answer - d | |
| 47. |
Under what condition, the current passing through the resistance R can be increased by short - circuiting the battery of emf `E_2` The internal resistances of the two batteries are `r_1 and r_2,` respectively. A. `E_(2)r_(1) gt E_(1)(R + r_(2))`B. `E_(1)r_(2) gt E_(2)(R + r_(1))`C. `E_(2)r_(2) gt E_(1)(R + r_(2))`D. `E_(1)r_(1) gt E_(2)(R + r_(1))` |
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Answer» Correct Answer - B |
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| 48. |
If a copper wire is stretched to make its radius decrease by `0.1%`, then the percentage increase in resistance is approximately -A. `-0.4%`B. `+0.8%`C. `+0.4%`D. `+0.2%` |
| Answer» Correct Answer - C | |
| 49. |
If a copper wire is stretched to make it `0.1%` longer wha is the percentage change in its resistance?A. increases by 0.05%B. increases by 0.2%C. decreases by 0.2%D. decrease by 0.05% |
| Answer» Correct Answer - B | |
| 50. |
Consider the two insulating sheets with thermal resistance `R_(1)` and `R_(2)` as shown in figure. The temperature `theta` isA. `(theta_(1)theta_(2)R_(1)R_(2))/((theta_(1)+theta_(2))(R_(1)+R_(2)))`B. `(theta_(1)R_(1)+theta_(2)R_(2))/(R_(1)+R_(2))`C. `((theta_(1)+theta_(2))R_(1)R_(2))/(R_(1)^(2)+R_(2)^(2))`D. `(theta_(1)R_(2)+theta_(2)R_(1))/(R_(1)+R_(2))` |
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Answer» Correct Answer - D |
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