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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.
| 201. |
The `80 Omega` galvanometer deflects full scale for a potential of 20 mV. A voltmeter deflecting full scale of 5 V is to made using this galvanometer. We must connectA. a resistance of `19.92 k Omega` parallel to the galvanometerB. a resistance of `19.92 k Omega` in series with the galvanometerC. a resistance of `20 k Omega` parallel to the galvanometerD. a resistance of `20 k Omega` in series with galvanometer |
| Answer» Correct Answer - B | |
| 202. |
A electric current passes through non uniform cross-section wire made of homogeneous and isotropic meterial. If the `j_(A)` and `J_(B)` be the current densities and `E_(A)` and `E_(B)` be the electric field intensities at `A` and `B` respectively, then A. `j_(a) gt j_(B) , E_(A) gt E_(B)`B. `j_(a) gt j_(B) , E_(A) lt E_(B)`C. `j_(a) lt j_(B) , E_(A) gt E_(B)`D. `j_(a) lt j_(B) , E_(A) lt E_(B)` |
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Answer» Correct Answer - A (a) `(I)/(A) gt J_(B)` and `j = (E)/(rho)` `:. J_(A) gt j_(B)` and `E_(A) gt E_(B)`. |
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| 203. |
The size of a carbon block, having specific resistance. `3.5xx10^(-3)Omega-cm` is `2cmxx2cmxx2cm`. The resistance of the block between two square end faces and two opposite rectangular faces are respectivelyA. `17.5xx10^(-4)` and `1.75xx10^(-4)Omega`B. `1.75xx10^(-4)` and `175xx10^(-4)Omega`C. `175xx10^(-4)` and `1.75xx10^(-4)Omega`D. `1.75xx10^(-4)` and `17.5xx 10^(-4)Omega` |
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Answer» Correct Answer - C Specific resistance `rho = 3.5xx10^(-3)Omega -cm` `rho =R(A)/(l)rArr R = (rho l)/(A)` Resistance R between square faces `= (3.5xx10^(-3)xx20)/(4)=17.5xx10^(-3)=175xx10^(-4)Omega` Resistance between opposite rectangular faces `R = (3.5xx10^(-3)xx2)/(2xx20)=1.75xx10^(-4)Omega` |
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| 204. |
A resistance of `1980Omega` is connected in series with a voltmeter, after which the scale division becomes 100 times larger. Find the resistance of voltmeter.A. `10Omega`B. `20Omega`C. `30Omega`D. `40Omega` |
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Answer» Correct Answer - B |
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| 205. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_(1)R_(2)R_(5)=R_(3)R_(4)R_(6)`B. `(1)/(R_(5))+(1)/(R_(6))=(1)/(R_(1)+R_(2))+(1)/(R_(3)+R_(4))`C. `R_(1)R_(4)=R_(2)R_(3)`D. `R_(1)R_(3)=R_(2)R_(4)=R_(5)R_(6)` |
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Answer» Correct Answer - C Since current i is independent of the value of `R_(6)`, it is clear that the circuit is of a balanced Wheatstone bridge. As per condition of balance, we have `(R_(1))/(R_(3))=(R_(2))/(R_(4))rArr R_(1)R_(4)=R_(2)R_(3)` |
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| 206. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_(1)R_(2)R_(3)=R_(3)R_(4)R_(5)`B. `(1)/(R_(5))+(1)/(R_(6))=(1)/(R_(1)+R_(2))+(1)/(R_(3)+R_(4))`C. `R_(1)R_(4)=R_(2)R_(3)`D. `R_(1)R_(3)=R_(2)R_(4)=R_(5)R_(6)` |
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Answer» Correct Answer - C |
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| 207. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_(1)R_(2)R_(5)=R_(3)R_(4)R_(6)`B. `(1)/(R_(5))+(1)/(R_(6))=(1)/(R_(1)+R_(2))+(1)/(R_(3)+R_(4))`C. `R_(1)R_(4)=R_(2)R_(3)`D. `R_(1)R_(3)=R_(2)R_(4)` |
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Answer» Correct Answer - C Current `I` can be independent of `R_(6)` only whhen `R_(1),R_(2),R_(3),R_(4)` ad `R_(6)` from a balanced wheatstone bridge Therefore `(R_(1))/(R_(2))=(R_(3))/(R_(4))` |
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| 208. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_1 R_2 R_5 = R_3 R_4 R_6`B. `(1)/(R_5) + (1)/(R_6) = (1)/(R_1 + R_2) + (1)/(R_3 + R_4)`C. `R_1 R_4 = R_2 R_3`D. `R_1 R_3 = R_2 R_4 = R_5 R_6` |
| Answer» Correct Answer - C | |
| 209. |
As the temperature of a metallic resistor is increased,the product of its resistivity and conductivityA. increaseB. decreaseC. remain constantD. may increase or decrease |
| Answer» Correct Answer - C | |
| 210. |
The current density in a wire is `10A//cm^(2)` and the electric field in the wire is 5 V/cm. If p = resistivity of material, `sigma`= conductivity of the material then (in SI unit)A. `rho=5xx10^(-3)`B. `rho=200`C. `sigma=5xx10^(-3)`D. `sigma=200` |
| Answer» Correct Answer - A::D | |
| 211. |
As the temperature of a metallic resistor is increased,the product of its resistivity and conductivityA. increasesB. decreasesC. may increase or decreaseD. remains constant |
| Answer» Correct Answer - 4 | |
| 212. |
One mole of an ideal gas undergoes a process in which `T = T_(0) + aV^(3)`, where `T_(0)` and `a` are positive constants and V is molar volume. The volume for which pressure with be minimum isA. `((T_(0))/(2a))^(1//3)`B. `((T_(0))/(3a))^(1//3)`C. `((a)/(2T_(0)))^(2//3)`D. `((a)/(3T_(0)))^(2//3)` |
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Answer» Correct Answer - A |
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| 213. |
In electrolysis the mass deposited on an electrode is directly proportional to:A. CurrentB. Square of currentC. Concentration of solutionD. inverse of current |
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Answer» Correct Answer - A In electrolysis, according to first law of Faraday, the mass of a substance deposited at an electrode is directly proportional to the charge passed through the electrolyte i.e, `m prop q` If a current I passes for a time t, then as we know, q=it Hence, `m prop it ` or `m prop i` Thus, mass deposited at an electrode is directly proportional to current |
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| 214. |
Three rods of equal length of same material are joined to form an equilateral triangle ABC as shown in figure. Area of cross-section of rod AB is S, of rod BC is 2S and that of AC is S. Then match the following. |
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Answer» Correct Answer - (A) Q, (B) R,S (C) S |
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| 215. |
Two cylindrical rods of uniform cross-section area `A` and `2A`, having free electrons per unit volume `2n` and `n` respectively are joined in series. A current `I` flows through them in steady state. Then the ratio of drift velocity of free electron in left rod to drift velocity of electron in the right rod is `((v_(L))/(v_(R)))` is A. `(1)/(2)`B. 1C. 2D. 4 |
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Answer» Correct Answer - B (b) Since current `I = n e Avd` through both rods is same `2 (n) eAv_(L) = n e (2A) v_(R)` or `(v_(L))/(v_(R)) = 1` |
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| 216. |
In the bohr model, the electron of a hydrogen atom moves in circular orbit of radis `5.3xx10^-11` m with speed of `2.2xx10^6 m//s`. Determine its frequency f and the current l in the orbit. |
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Answer» Correct Answer - A `f=v/(2pir)=(2.2xx10^6)/((2pi)(5.3xx10^-11))` `=6.6xx10^15Hz` `I=qf` `=(1.6xx10^-19)(6.6xx10^15)` `=1.06 xx 10^-3A` `=1.06mA` |
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| 217. |
In your city electricity cost 40 paise per kWh. You pay forA. electric chargeB. electric powerC. electric energyD. electric current |
| Answer» Correct Answer - C | |
| 218. |
A negligibly small current is passed through a wire of length 15 m and uniform cross-section `6.0 xx 10^(-7) m^(2)`, and its resistance is measured to be `5.0 Omega`. What is the resistivity of the material at the temperature of the experiment ? |
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Answer» Length of the wire, `l = 15` m Area of the cross-section of the wire, `a = 6.0 xx 10^(-7) m^(2)` Resistance of the material of the wire, `R = 5.0 Omega` Resistivity of the material of the wire `= rho` Resistance is related with the resistivity as `R = rho (l)/(A)` `rho = (RA)/(l)` `= (5 xx 6 xx 10^(-7))/(15) = 2 xx 10^(-7) Omega` m Therefore, the resistivity of the material is `2 xx 10^(-7) Omega` m. |
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| 219. |
A negligibly small current is passed through a wire of length 15 m and uniform cross-section `6.0 xx 10^(-7)Omega m^(2)`, and its resistance is measured to be `5.0 Omega`. What is the resistivity of the material at the temperature of the experiment ?A. `1xx10^(-7)Omega-m`B. `2xx10^(-7)Omega-m`C. `3xx10^(-7)Omega-m`D. `4xx10^(-7)Omega-m` |
| Answer» Correct Answer - B | |
| 220. |
A negligibly small current is passed through a wire of length 15 m and unifrom cross-section `6.0 xx 10^(-7) m^(2)` and its resistance is measured to be `5 Omega`. What is the resistivity of the material at the temperature of the experiment? |
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Answer» Here `L = 15 m, A = 6.0 xx 10^(-7) m^(2), R = 5.0 Omega, rho = ?` `rho = (R_(A))/(1) = (5.0610)/(15) = 2.0 xx 10^(-7) Omega m` |
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| 221. |
A negligibly small current is passed through a wire of length 15 m and uniform cross-section `6.0 xx 10^(-7)Omega m^(2)`, and its resistance is measured to be `5.0 Omega`. What is the resistivity of the material at the temperature of the experiment ? |
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Answer» Here, `l = 15 m , A = 6.0 xx 10^(-7) m^(2) , R = 5.0 Omega , rho = ?` `rho = (RA)/(l) = (5.0 xx 6 xx 10^(-7))/15 = 2.0 xx 10^(-7) Omega m` |
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| 222. |
In house wiring, copper wire `2.05 mm` in diameter is often used. Find the resistane of `35.0 m` long wire. Specific resistance of copper is `1.72xx10^-8Omega-m`. |
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Answer» Correct Answer - A `R=rhol/A=((1.72xx10^-8)(35))/((pi//4)(2.05xx10^-3)^2)=0.18Omega` |
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| 223. |
A cell of negligible internal resistance is connected to a potentiometer wire and potential gradient is found. Keeping the length as constant, if the radius of potentiometer wire is increased four times, the potential gradient will become (no series resistance in primary)A. 4 timesB. 2 timesC. halfD. constant |
| Answer» Correct Answer - D | |
| 224. |
At room temperature `(27.0^@C)` the resistance of a heating element is `100 Omega`. What is the temperature of the element if the resistane is found to be `117 Omega`, given that the temperature co-efficicent of the material of the resistor is `1.70 xx 10^(-4) .^@C^(-1)`. |
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Answer» Here, `R_(27) = 100 Omega , R_(t) = 117 Omega , t = ? , alpha = 1.70 xx 10^(-4) .^(@)C^(-1)` We know that, `alpha = R_(t)-R_(27)/(R_(27) (t - 27)) or t- 27 = (R_(t)-R_(27))/(R_(27)xx alpha)` `:. t = (R_(t)-R_(27))/( R_(27) xx alpha) + 27 = ( 117 - 100)/(100 xx 1.7 xx 10^(-4)) + 27 = 1000 + 27 = 1027^@C` |
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| 225. |
A uniform wire of resistance `18 Omega` is bent in the form of a circle. The effective resistance across the points a and b is A. `3Omega`B. `2 Omega`C. `2.5Omega`D. `6Omega` |
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Answer» Correct Answer - C `R_(ab)=((3xx15))/(3+15)=2.5Omega` As `R_(60^@)=(18/(360^@))(60^@)=3Omega` |
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| 226. |
On increasing the resistance of the primary circuit of potentiometer, its potential gradient willA. become moreB. become lessC. not changeD. become infinite |
| Answer» Correct Answer - B | |
| 227. |
(a) Three resistors `2 Omega, 4 Omega and 5 Omega` are combined in series. What is the total resistance of the combination ? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery. |
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Answer» (a) Here, `R_(1) = 2 Omega , R_(2) 4Omega , R_(3) = 5 Omega , V = 20 V` In parallel combination,total resistane `R_(P)` is given by `1/R_(P) = 1/R_(1) + 1/R_(2) + 1/R_(3) = 1/2 + 1/4 + 1/5 = (10 + 5 + 4)/20 = 19/20 or R_(P) = 20//19 Omega`. (b) Current through `R_(1) = V//R_(1) = 20//2 = 10A`, Current through `R_(2) = 20//4 = 5A` Current through `R_(3) = 20//5 = 4A` , Total current `= 20//(20//19)=19A` |
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| 228. |
(a) Three resistors `1 Omega, 2Omega and 3 Omega` are combined in series. What is the total resistance of the combination ? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor. |
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Answer» (a) Here, `R_(1) = 1 Omega , R_(2)2 Omega , R_(3) = 3 Omega , V = 12 V` `{:(,"In series, total resistance",,R_(S)=R_(1)+R_(2)+R_(3)=1+2+3=6 Omega),((b),"Current through the circuit",,I = V//R_(s)=12//6=2A),( :.,"Potential drop across",,R_(1)=I R_(1)=2xx1 =2V),(,"Potential drop across",,R_(2)=I R_(2)=2xx2=4V),(,"Potential drop across",,R_(3)=IR_(3)=2xx3=6V),(,,,):}` |
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| 229. |
(a) Three resistors `2 Omega, 4 Omega and 5 Omega` are combined in parallel. What is the total resistance of the combination ? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery. |
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Answer» (a) There are the resistiors of resistances, `R_(1)=2 Omega,R_(2) = 4 Omega, and R_(3) = 5 Omega` They are connected in parallel. Hence, total resistance (R) of the combination is given by, `(1)/(R)=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))` `:. R = (20)/(19) Omega` Therefore, total resistance of the combination is `(20)/(19) Omega` (b) Emf of the battery, `V = 20 V` Current `(I_(1))` flowing through resistor `R_(1)` is given by, `I_(1)=(V)/(R_(1))` `=(20)/(2)=10 A` Current `(I_(2))` flowing through resistor `R_(2)` is given by, `I_(2)=(V)/(R_(2))` `= (20)/(4)=5A` Current `(I_(2))` flowing through resistor `R_(3)` is given by, `I_(3)=(V)/(R_(3))` `=(20)/(5) = 4A` The current, `I = I_(1)+I_(2)+I_(3)=10+5+4 = 19 A` Therefore, the current through each resistor is 10 A, 5 A and 4 A respectively and the total current is 19 A. |
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| 230. |
A resistor of `24 Omega` resistance is bent in the form of a circle as . What is the effective resistance between points A and B? |
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Answer» Resistance of the portion of resistor forming an angle of `60^@` at the centre O of the circle is `R_(1)=60^@/360^@ xx 24=4 Omega` Resistance of remaining portion of resistor, `R_(2)=24-4=20 Omega` Here, resistance, `R_(1) and R_(2)` are in parallel, the effective resistance is `R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(4xx20)/(4+20)=80/24=10/3 Omega` |
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| 231. |
A wire of uniform cross-section and length `l` has a resistance of `16 Omega ` is cut into four equal parts. Each part is stretched uniform to length `l` and all the four stretched parts are connected in parallel calcuate the total resistance of the combination so formed. Assume that stretching of wire does not cause any change in the density of its material |
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Answer» Correct Answer - `16 Omega` Resistance of the each of the four parts of length `l//4` is `R_(1) = (16)/(4) = 4 Omega` When each part is stretched to length `l`, its volume remains constant so `V = A_(1) ,l_(1) = A_(2) l_(2)or (A_(2))/(A_(1)) = (l_(1))/(l_(2)) = (1//4)/(l) = (1)/(4)` Now `R_(2) = rho 1//A or R prop l//A` `:.(R_(2))/(R_(1)) = (l_(2))/(l_(1)) xx (A_(1))/(A_(2)) = (l)/(1//4) xx 4 = 16` or `R_(1) = 16 R_(1) = 10xx 4 = 64 Omega` These resistance part `= 64 Omega` when these four are connected in parallel the effective resistance `R_("eff")` of the combination is given by `(1)/(R_(eff)) = (1)/(64) + (1)/(64) + (1)/(64) + (1)/(64) = (4)/(64) = (1)/(16)` or `R_(eff) = 16 Omega` |
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| 232. |
Two coils have a combined resistance of `9 Omega` when connected in series and `2 Omega` when connected in parallel. Find the resistance of each coil. |
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Answer» Let `R_(1)` and `R_(2)` be the resistance of two coils When coils are in series, then `R_(s)=R_(1) +R_(2)=9` When coils are in paralle, then `R_(p) =(R_(1)R_(2))/(R_(1)+R_(2))= 2` or `R_(1) R_(2)=2(R_(1) +R_(2))=2 xx 9 =18` From (i), `R_(1) =9 -R_(2)` Putting this value in (ii), we get `(9 -R_(2))R_(2)=18` or `9 R_(2) - R_(2)^(2) =18` or `R_(2)^(2)-9R_(2) +18=0` On solving, we get `R_(2)=3Omega` or `6 Omega` and `R_(1)=6 Omega` or `3 Omega` |
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| 233. |
What is the colour of the third band of a coded resistor of resistance `2.3 xx 10^(2)Omega`? |
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Answer» Resistance of resistor `=2.3 xx10^(2)Omega = 23 xx10^(2)Omega` Therefore, the colour of third band of a colour coded resistor will be related to number 1, i.e., brown. |
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| 234. |
What is the colour of the third band of a coded resistor of resistance `0.34 Omega` ? |
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Answer» Correct Answer - silver Resistence of resistor `=0.34 Omega = 34 xx 10^(-2) Omega`. The colour of the third hand of a coded resistor is due to multiplier `10^(-2)` which is for silver. |
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| 235. |
A voltage of 5 V is applied across a colour coded carbon resistor. A current of 5 mA flows through it. What are the colours of second and third band of coded colour resistor? |
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Answer» Here, `V=5V, I=5 mA = 5 xx 10^(-3)A` Resistance of carbon resistor, `R=V/I=5/(5xx10^(-3)) =10^(3)Omega =10 xx 10^(2)Omega` Therefore, the colours of second and third bands will be related to numbers 0 and 2,i.e., black and red. |
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| 236. |
A voltage of 30 V is applied across a colour coded carbon resistor with first, second and third rings of blue, black and yellow colurs. What is the current flowing through the resistor ? |
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Answer» Correct Answer - `5xx10^(-5)A` |
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| 237. |
A potentiometer wire of length 1 m has a resistance of `100Omega`. It is connected to a 6V battery in series with a resistance of `5Omega`. Determine the emf of the primary cell which gives a balance point at 40cm.A. 1.2 VB. 1.8 VC. 1.6 VD. 1.9 V |
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Answer» Correct Answer - C Given, length of wire, l = 1m = 100 cm Resistance, `R = 10 Omega` The emf of a battery, `E_(1)=6V` `R_(1)=5 Omega` `X = 40 cm` `therefore " "` Current, `I=(E_(1))/(R+R_(1))=(6)/(10+5)=(6)/(15)=(22)/(5)A` `V_(AB)=IR=(2)/(5)xx10=4 V` `therefore` The emf of the primary cell `= (V_(AB))/(l)xx x=(4)/(100)xx40 = 1.6 V` |
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| 238. |
There are two wires A and B of same mass and of the same material. The diameter of wire A is one-third the diameter of wire B. If the resistance of wire A is `30 Omega`, find the resistance of wire B. |
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Answer» Here, `R_(A) =30 Omega`. Let `l^(A), D^(A)` be the length and diameter of wire A and `l_(B),D^(B)` be the length and diameter of wire B. Let d be the density of the material of wires A and B. Mass of the wire = volume `xx` density = area of cross-section `xx` lenght `xx` density As mass of two wires is same to `m=pi D_(A)^(2)/4 xx l_(A) xx d = pi D_(B)^(2)/4 xx l_(B) xx d` or `D_(A)^(2)l_(A)=D_(B)^(2)l_(B)` or `l_(B)/l_(A) =D_(A)^(2)/D_(B)^(2)=(D_(B//3))^(2)/D_(B)^(2)=1/9` `R_(B)/R_(A) = (4 rho l_(B)//(pi D_(B)^(2)))/(4 rho l_(A)//(pi D_(A)^(2))) = l_(B)/l_(A) xx D_(A)^(2)/D_(B)^(2) = 1/9 xx 1/9 =1/81` `R_(B)=R_(A)/81 =30/81=0.37 Omega` |
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| 239. |
A wire of resistance `5.0 Omega` is used to wind a coil of radius 5 cm. The wire has a diameter 2.0 mm and the specific resistance of its material is `2.0xx10^(-7)Omega m`. Find the number of turns in the coil. |
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Answer» Here, `R=5.0 Omega , r_(1) =5 xx10^(-2)m`, `D=2.0xx10^(-3) m, rho =2.0xx10^(-7)Omega m`. `R=rho l/(pi D^(2)//4)` or `l=(R pi D^(2))/(4 rho)` Let n be the number of turns in the coil. Then total length of the wire used, `l= 2 pi r_(1) n` or `n=l/2pi r_(1) = (R pi D^(2))/( 4 rho xx 2pi r_(1))=(R D^(2))/ (8 rho r_(1))` `=(5.0 xx (2.0xx10^(- 3))^(2))/(8xx(2.0xx10^(-7)) xx(5xx10^(-2)))=250` |
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| 240. |
An electron of hydrogen atoms is considered to be revolving round the proton in the circuit orbit of radius `(h^(2))/(2pi^(2)me^(2)) `with velocity` (2pie^(2))/(h)` the equivalent current due to circuiting charge isA. `(4pi^(2)me^(4))/(h^(3))`B. `(4pi^(2)me^(5))/(h^(3))`C. `(4pi^(2)m^(2)e^(4))/(h^(3))`D. None of above |
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Answer» Correct Answer - b `T = (2pi r)/(upsilon) = 2pi xx (h^(2))/(4pi^(2)e^(2))xx(h)/(2pi e^(2)) = (h^(2))/(4pi^(2)me^(4))` Current `i = (e)/(T) = (4pi^(2)m e^(5))/(h^(3))` |
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| 241. |
Find the current flowing through a copper wire of length 0.2m, area of cross-section `1 mm^(2)`, when connected to a battery of 4 V. Given that electron mobility `=4.5xx10^(-6) m^(2)V^(-1) s^(-1)` and charge on electron `=1.6xx10^(-19)C`. The number density of electron in copper is `8.5 xx10^(28) m^(-3)`. |
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Answer» Correct Answer - `1.22 A` `I = n A e upsilon_(d) = n A e mu E = n= n A e mu (V)/(l)` `= (8.5 xx 10^(28)) xx (10^(-3))^(2) xx (1.6 xx 10^(-19)) xx (4.5 xx 10^(-6)) xx (4)/(0.2)` `= 1.22 A` |
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| 242. |
A rheostat has 100 turns of a wire of radius 0.4 mm having resistivity `4.2xx10^(-7) Omega`m. The diameter of each turn is 3 cm. What is the maximum value of resistance that it can introduce ? |
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Answer» Correct Answer - `7.875 Omega` |
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| 243. |
A current of 3 A is flowing through a wire of length 2 m and cross- sectional area `1mm^(2)`. If wire contains `10^(29) "electrons"//m^(3)`, calculate the average time taken by an electron to cross the length of the wire. |
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Answer» Correct Answer - `3.36 xx 10^(4)s` Here, `I = 2A ,l = 1 m, A = 2 mm^(2) = 2 xx 10^(-6) m^(2)`, `n = 8.5 xx 10^(28) m^(-1)` Driff speed `upsilon_(d) = (I)/(n A e)` `= (2)/((8.5 xx 10^(28)) xx (2 xx 10^(-6)) xx (1.6 xx 10^(-19))` ` = 7.35 xx 10^(-5)ms^(-1)` Time taken, `t = (l)/(upsilon_(d)) = (1)/(7.35 xx 10^(-5)) = 3.36 xx 10^(4)s` |
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| 244. |
Calculate the resistivity of the material of a wire 1.0 m long, 0.4 mm diameter and having a resistance of `2.0 Omega`. |
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Answer» Correct Answer - ` 2.53 xx 10^(-6) Omega m` `rho = (RA)/(l) = (R pi D^(2))/(4l) = (2.0 xx 3.142 xx (0.4 xx 10^(-3))^(2))/( 4xx 10)` `= 2.51 xx 10^(-8) Omega m` |
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| 245. |
A current of `4.4A` is flowing in a copper wire of radius `1mm` density of copper is `9 xx 10^(3)kg m^(-3)`and its atoms is `63.5u` If every atoms of copper contributes one condition electron , then the drift velocity of electrons is nearly [density of copper `9 xx 10^(3) kg m^(3)]`A. `0.1mm s^(-1)`B. `0.5mm s^(-1)`C. `1mm s^(-1)`D. `1.5mm s^(-1)` |
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Answer» Correct Answer - a Atomic mass of copper `M = 63.5g = 0.0635 kg` No of copper atoms in `0.0635kg` coppere `= 6.02 xx 10^(23)` Density of copper `= (M)/(d) = (0.0635)/(9 xx 10^(3)) m^(3)` Total no of atoms in unit volume of copper `= (9 xx 10^(3))/(0.0635) xx 6.02 xx 10^(23) = 8.5 xx 10^(28) "atoms"//m^(-3)` As one atom of copper contributes one conduction electron to so the electron density of copper `n = 8.5 xx 10^(28)m^(-3)` Drift velocity`v_(d) = (1)/(n Ae)` `= (44)/((6.5 xx 10^(28)) xx pi(0.001)^(2) xx (1.6 xx 10^(-19))` `0.1 xx 10^(-3)m//s = 0.1mm//s` |
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| 246. |
A current of `4.4A` is flowing in a copper wire of radius `1mm` density of copper is `9 xx 10^(3)kg m^(-3)`and its atoms is `63.5u` If every atoms of copper contributes one condition electron , then th drift velocity of electrons is nearly [density of copper `9 xx 10^(3) kg m^(3)]` |
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Answer» Correct Answer - `10^(-4) ms^(-1)` Here, density, `d = 8.9 xx 10^(3) kg m^(-3), M = 63.5 kg`, `N = 6.02 xx 10^(26)` per kg atom , `I = 1.5 A` `D = 1.2 mm = 1.2 xx 10^(-3) m` Mass of `1m^(3)` of copper `= 8.9 xx 10^(3) kg` Number of atoms per `m^(3)` of copper `= (N)/(M//d) = (Nd)/(M)` As 1 atoms contributes one conduction electron, therefore number of conbination electron per `m^(3)` of copper is `n = (Nd)/(M) = ((6.20 xx 10^(26)) xx (8.9 xx 10^(3)))/(63.5)` `= 8.44 xx 10^(28)m^(-3)` `upsilon_(d) = (I)/(n A e)` `=(1.5)/((8.44xx10^(28))xx[(22)/(7)xx(1.2 xx10^(-3))^(2)/(4)]xx(1.6 xx10^(-19)))` `= 0.98 xx 10^(-4) = 10^(-4) ms^(-1)` |
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| 247. |
A fan with copper winding in its motor consumes less power as compared to an otherwise similar fan having aluminium winding.Explain. |
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Answer» Correct Answer - B |
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| 248. |
The energy density `u/V` of an ideal gas is related to its pressure P asA. `(U)/(V)=3p`B. `(U)/(V)=(3)/(2)p`C. `(U)/(V)=(p)/(3)`D. `(U)/(V)=(5)/(2)p` |
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Answer» Correct Answer - B |
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| 249. |
Consider the adjacent figure. A piston divides a cylindrical container into two equal parts. The lets part contains 1 mole of helium gas and the right part contains two moles or oxygen gas. The initial temperatures and pressures of the gases in the left chamber are `T_(0)` and `p_(0)` and the right chamber are `(T_(0))/(2)` and `p_(0)` respectively as shown in the figure. The piston as well as the walls of the container are adiabatic. After removal of the piston, gases mix homoganeously and the final pressure becomes `p=(4n)/(13)p_(0)`. Find the value of n. |
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Answer» Correct Answer - C |
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| 250. |
A black body emits maximum radiation of wavelength `lambda_(1)` at a certain temperature `T_(1)`. On increasing the temperature, the total energy of radiation emitted is increased 16 times at temperature `T_(2)`. If `lambda_(2)` is the wavelength corresponding to which maximum radiation is emitted at temperature `T_(2)`. Calculate the value of `((lambda_(1))/(lambda_(2)))`. |
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Answer» Correct Answer - B |
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