InterviewSolution
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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. |
Current `i` is being driven through a cell of emf `epsilon` and internal resistance3 `r`, as shown (i) the cell absorbs energy at rate of `epsilon i` (ii) The cell stores chemical energy at the rate of `(epsilon i - i^(2)r)` (iii) The potential differnece across the cell is `epsilon+ir` (iv) some heat is produced in the cellA. (i) , (ii)B. (ii), (iii)C. (i),(iii),(iv)D. all |
| Answer» Correct Answer - D | |
| 2. |
The production of emf by maintaining a difference of tempreture between the two junctions of two different metals is known asA. Joule effectB. Seebeck effectC. Peltirer effectD. Thomson effect |
| Answer» Correct Answer - B | |
| 3. |
In a thermocouple, which of the following statements is not true?A. Neutral temperature depends upon the nature of materials in the thermocoupleB. Temperature of inversion depends upon the tenperature of cold junctionC. When the temperature of the hot junction is equal to the temperature of inversion, the thermo emf becomes zero.D. when the temperature of cold junction increases, the temperature of inversion also increases. |
| Answer» Correct Answer - D | |
| 4. |
Thermoelectric constant of a thermocouple are `alpha` and `beta`. Thermoelectric power at inversion tempreture isA. `alpha`B. `- alpha`C. `(alpha)/(beta)`D. `-(alpha)/(beta)` |
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Answer» Correct Answer - B `E = alpha T + 1/3 beta T^(2)` For inversion temperature `E = 0, T_(i) = -(2alpha)/(beta)` Thermoelectric power ltbr. `P = (dE)/(dT) = alpha + 1/2 beta.(2T) = alpha+beta T_(i)` `alpha+beta(-(2alpha)/(beta))=-alpha`. |
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| 5. |
The expression for thermo emf in a thermocouple given by the relation `E=40theta-(theta^(2))/(20)`, where `theta` is the temperatue difference of two junctons. For this, the neutral temperature will beA. `100^@C`B. `200^@C`C. `300^@C`D. `400^@C` |
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Answer» Correct Answer - D `E = 40 theta -(theta^2)/(20)` For neutral temperature. `(dE)/(d theta) = 40-(theta)/(10)=0 implies theta = 400^(@)C`. |
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| 6. |
If for a thermocouple `T_(n))` is the neutral temperature, `T_(c )` is the tempreture of the cold junction and `T_(i))` is the temperature of inversion, thenA. `T_i = 2T_(n)-T_(c)`B. `T_i = T_(n)-2T_(c)`C. `T_i = T_(n)-T_(c)`D. None of these |
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Answer» Correct Answer - A `T_(n) = (T_i-T_c)/(2) implies T_(i) = 2T_(n) - T_(c)`. |
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| 7. |
The cold junction of a thermocouple is maintained at `15^(@)C`. No thermo-emf is developed when the hot junction is mainted at `525^(@)C` . Find neutral temperature. |
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Answer» Since `E=0` then `T_(i) = 525^(@)C` `T_(c) = 15^@C` `T_(i) - T_(n) = T_(n)-T_(c)` `T_(n) = (T_(i)+T_(c))/(2) = (525+15)/(2) = 270^@C`. |
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| 8. |
A `100W` bulb and a 25W bulb are desigened for the same voltage. They have filaments of the same length and material. The ratio of the diameter of `100W` bulb to that of the `25W` bulb isA. `4:1`B. `2:1`C. `sqrt(2):1`D. `1:2` |
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Answer» Correct Answer - B `R_(1) = (V^2)/(P_1) = (V^2)/(100) =R` `R_(2) = (V^2)/(P_2) = (V^2)/(25) = 4R` `R = (rhol)/(A) = (rhol)/(pir^2)=(rhol)/(pid^(2)//4), d: diameter` `R prop 1//d^(2)` `(R_1)/(R_2) = ((d_2)/(d_1))^(2) implies R/(4R) = ((d_2)/(d_1))^(2) implies (d_2)/(d_1) = 1/2` `d_(1)/d_(2) = 2//1`. |
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| 9. |
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 is doubled. |
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Answer» Correct Answer - A `H prop (V^2)/R implies H prop 1/R` `R prop 1/(pi r^2)`. |
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| 10. |
The walls of a closed cubical box of edge `50cm` are made of a material of thickness `1mm` and thermal conductivity `4xx10^(-4) cm^(-1)(.^@C)^(-1)`. The interior of the box is maintained `100^(@)C` above the outside temperature by a heater placed inside the box and connected across `420V` d.c. calculate the resistance of copper. |
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Answer» `A` = total surface area of box `= 6xx` area of each face `= 6xx50xx50xx10^(-4)m^(2) = 1.5m^(2)` `K = 4xx4.2xx10^(-2)J//sm.^(@)C` `theta_(2)-theta_(1) = 100^(@)C, L = 1 mm = 10^(-3)m` The rate of heat transfer through walls `P = Q/t = (KA(theta_(2)-theta_(1)))/(L) = (4xx4.2xx10^(-2)xx1.5xx100)/(10^(-3))` `=25200W` Heat produced by `"heater"//"sec" = (V^2)/(R)` `P = (V^2)/(R)` `R = (V^2)/(P) = (420xx420)/((4xx4.2xx10^(-2)xx1.5xx100)//10^(-3))` `=7Omega`. |
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| 11. |
The relation between faraday constant `(F)`, chemical equicalent `(E)` and electrochemical equivalent `(Z)` isA. `F = EZ`B. `F = Z/E`C. `F=E/Z`D. `F = E/(Z^2)` |
| Answer» Correct Answer - C | |
| 12. |
In producing chlorine through electrolysis `100 W` power at `125 V` is being consumed. How much chlorine per min is liberated ? `ECE` of chlorine is `0.367 xx 10^(6) kg//C`A. 24.3 mgB. 16.6 mgC. 17.6 mgD. 21.3 mg |
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Answer» Correct Answer - C `P = Vi implies 100=125 I implies I = 4//5 = 0.8A` `m = Zi t = 0.367xx10^(-6) xx 0/8xx60` `= 0.0176xx10^(-3)kg = 17.6 mg`. |
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| 13. |
An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage. If the voltage is `V`, it drawas power. ThenA. `P = ((V_0)/(V))P_(0)`B. `P = ((V)/(V_0))P_(0)`C. `P = ((V)/(V_0))^(2)P_(0)`D. `P = ((V_0)/(V))^(2)P_(0)` |
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Answer» Correct Answer - C `R_(B) = (V_0^2)/(P_0)` `P = (V^2)/(R_B) = ((V)/(V_0))^(2)P_(0)`. |
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| 14. |
The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` isA. `(a^2R)/(6b)`B. `(a^3R)/(3b)`C. `(a^3R)/(2b)`D. `(a^3R)/(b)` |
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Answer» Correct Answer - A `Q = a t - bt^(2)` `I = (dQ)/(dt) = a-2bt, i=0 at t = a/(2b)` `H = int_(0)^(t=(a)/(ab)) i^(2)Rdt = R int_(0)^(a/2b) (a-2bt)^(2) dt` `= R int_(0)^(a/2b) (a^(2)-4abt+4b^(2)t^(2))dt` `R|a^(2)t - 4ab(t^2)/(2)+(4b^2)t^3))/(3)|_(0)^(a/2b)` `=R[a^(2) . a/(2b) -(4ab)/(2) * (a^2)/(4b^2) +(4b^2)/(3) * (a^3)/(8b^3)]` `=(a^3R)/(b) [1/2 - 1/2 + 1/6] = (a^3R)/(6b)`. |
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| 15. |
a capacitor of capacitance `C` is connected to two voltmeter `A` and `B`. `A` is ideal , having infinite resistance, while `B` has resistance `R`. The capcitor is charged and then switch `S` is closed. The reading of `A` and `B` will be equal A. at all timesB. after time RCC. after time `RC 1n 2`D. only after a very long time |
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Answer» Correct Answer - A In discharging `RC` circuit `V_(R)=V_(C)` (always) |
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| 16. |
You are given a resistance coil and a battery. In which of the following cases is largest amount of heat generated?A. When the coil is connected to the battery directlyB. when the coil is divided into two equal parts and both the parts are connected to the battery in parallelC. when the coil is divided into four equal parts and all the four parts are connected to the battery in parallelD. When only half the coil is connected to the battery . |
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Answer» Correct Answer - C `H prop (V^2)/R` (a) `H prop (V^2)/R` (b) `H prop (V^2)/(R//4)` (c) `H prop (V^2)/(R//16)` (d) `H prop (V^2)/(R//2)` |
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| 17. |
Two indential heated produce heat `H_(1)` in time `t` when connected in parallel across the main supply. They produce heat `H_(2)` in time `t` when connected in series. Then `H_(1)//H_(2)` isA. `1/4`B. 4C. `1/2`D. 2 |
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Answer» Correct Answer - B `H_(1) = (V^2)/(R//2) t , H_(2) = (V^2)/(2R) t` `(H_1)/(H_2) = 4` . |
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| 18. |
The energy supplied by the cell during charging is equal toA. `H_(1)`B. `H_(2)`C. `3H_(2)`D. `2H_(1)` |
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Answer» Correct Answer - D Energy supplied by battery `=QE=CE^(2)=2H_(1)` |
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| 19. |
In the following diagram A. `i_(1) = 1/3 mA`B. `i_(2) = 1/3 mA`C. `i_(3) =0`D. all |
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Answer» Correct Answer - D `i_(3) = 0` `i_(1) = i_(2) = (9)/(12+15) = 1/3 mA` . |
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| 20. |
Find the time requried to liberate 1.0 litre of hydrogen at STP in an electrolytic cell by a current of `5.0 A`. |
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Answer» Equivlent volume of `H_2 = 11.2 "litre"` Number of equivalents of `H_(2) = 1/11.2` `1` equivalent `-= 1F` charge `1/11.2` equivalent `-= 1/11.2 F =Q` `Q="it"` `96500/11.2 = 5t implies t = 17235 ~= 29 min`. |
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| 21. |
A `100 W` bulb `B_1`, and two` 60 W` bulbs `B_2 `and `B_3` are connected to a `250 V` source as shown in the figure. Now `W_1,W_2` and `W_3` are the output powers of the bulbs `B_1, B_2` and `B_3` respectively. Then A. `W_(1) gt W_(2) = W_(3)`B. `W_(1) gt W_(2) gt W_(3)`C. `W_(1)lt W_(2) = W_(3)`D. `W_(1) lt W_(2) lt W_(3)` |
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Answer» Correct Answer - D `R_(1) = ((250)^(2))/(100) , R_(2) = R_(3) = ((20)^(2))/(60)` `(R_(2)=R_(3)) gt R_(1)` In upper branch containing `B_(1)` and `B_(2)` current is same `W_(1) = i^(2) R_(1), W_(2) = i^(2) R_(2)` `W_(2) gt W_(1)` Total power consumed in upper branch = `((250)^2)/(R_1+R_2)` Total power consumed in lower branch = `(250^2)/(R_2) = W_(3)` `W_(1) lt W_(2) lt W_(3)`. |
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| 22. |
Two electric bulbs, one of `200V, 40W`, and the othere `220V`, `100W` are connected in a house wiring circuitA. they have equal currents through themB. the resistance of the filaments in both the bulbs is sameC. The resistance of the filament in `40 watt` bulb is more than the resistance in `100watt` bulbD. The resistance of the filament in `100 watt` bulb is more than the resistance in `400watt` bulb |
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Answer» Correct Answer - C `R_(1) = ((200)^(2))/(40), R_(2) = ((200)^(2))/(100)` `R_(1) gt R_(2)`. |
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| 23. |
A capacitor charges from a cell through a resistance. The time constant is `tau`. In what time will the capacitor collect `10%` of the final charge?A. `tau 1n (0.1)`B. `tau 1n (0.9)`C. `tau 1n 10/9`D. `tau 1n 11/10` |
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Answer» Correct Answer - C `q = q_(0) (1-e^(-t//tau))` `1/10 q_(0)=q_(0) (1-e^(-t//tau))implies e^(-t//tau) = 9/10` `e^(-t//tau)=10//9` `t/(tau) 1n e = 1n(10//9)` `t=tau 1n(10//9)`. |
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| 24. |
A copper wire having area of cross section `0.4mm^(2)` and a length of 10 cm is initially at `75^(@)C` and is thermally insulated from the surroundings. If a current of `10A` is set up in this wire. ltbr. (a) Find the time in which the wire will strat melting. ignore variation of resistance with temperature. (b) If length of wire is doubled, find time. Given for copper : density =`10^(4) kg//m^(3)`, specific heat = `9xx10^(-2) kcal kg^(-1)(.^@C)^(-1)`, melting point =`1075^(@)C`, resispectively `=1.6xx10^(-8)Omega * m`. |
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Answer» (a) Resistace of wire `R = (rhol)/(A)` Heat required to melt the copper wire `H = i^2 Rt` ..(i) also `H = msDelta theta` ..(ii) `i^(2)Rt = ms Delta theta` `i^2 (rhol)/(A)t = (dAl)s Delta theta` `t = (dA^(2)s Delta theta)/(i^2rho)` `=(10^4xx(0.4xx10^(-6))^(2)(9xx10^(-2))xx4.2xx10^3xx(1075-75))/((10)^(2)xx1.6xx10^(-8))` `=378 sec` (b) `t` is independent of length of wire. Time of melting remains same. |
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| 25. |
In an `RC` circuit while charging, the graph of `1n i` versis time is as shown by the dotted line in the diagram figure. Where `i` is the current. When the value of the resistance is doubled, which of the solid curve best represents the variation of `1n i` versus time A. PB. QC. RD. S |
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Answer» Correct Answer - B `q= CE(1-e^(-t//RC))` `I = (dq)/(dt) = E/R e^(-t//RC)` `1n I = 1n(E)/R = t/(RC)` `1n I V//s t` will be straight line with -ve slope and positive intercept `Slope = -(1)/(RC)`, intercept = `1n(E/R)` When resistance is doubled, intercept decrease and slope increase. |
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| 26. |
Figure, shows an electrolyte of `AgCI` through which a current is passed. It is observed that `2.68g` of silver is deposited in 10 minutes on the cathode. Find the heat developed in the `20 Omega` resistor during this period. Atomic weight of silver is `107.9 g mol^(-1)`. (Figure) |
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Answer» `m = (Eit)/(F) ` ltbr. `i=(mF)/(Et) = (2.68xx96500)/(107.9xx10xx60)=4A` `H = i^2RT = (4)^2 xx 20 xx 10 xx 60` `=192xx10^3 J` `=192kJ`. |
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| 27. |
Silver and copper voltmeters are connected in parallel with a battery of emf `12V`. In `30` minutes, `1g` of silver and `1.8g` of copper are liberated. The power supplied by the battery is `(Z_(Cu) = 6.6xx10^(-4) g//C and Z_(Ag) = 11.2xx10^(-4)g//C`)A. `24.13 J//sec`B. `2.413 J//sec`C. `0.2416 J//sec`D. `2413 J//sec` |
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Answer» Correct Answer - A Current (silver voltmeter) `i_(1) = (m)/(Zt) = (1)/(11.2xx10^(-4)xx30xx60) = 0.496A` Copper voltmeter `i_(2) = (1.8)/(6.6xx10^(-4)xx30xx60) = 1.515A` `I = i_(1) +i_(2) = 2.011 A` `P = Vi = 12 xx 2.011 = 24.132 J//sec`. |
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| 28. |
The potential difference across the terminals of a battery of `emf 12 V` and internal resistance `2 Omega` drops to `10 V` when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is `107.9 g mol^(-1)`. |
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Answer» `V = E-ir` `10=12-ixx2 implies i=1A` `m = (Eit)/(F) = (107.9xx1xx30xx60)/(96500) = 2g`. |
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| 29. |
A capacitor charged to `100V` is discharged by connecting the two ptates at `t=0` . If the potential difference across the plates drops to `1.0V` at `10 ms`, what will be the potential difference at `t= 20 ms`. |
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Answer» In discharging `RC` circuit charge on capacitor `q = q_(0)r^(-1//RC) = CV_(0)e^(-t//RC), V_(0)=100V` p.d. across capacitor `V_(C) = q/C = V_(0)e^(-t//RC)` `1 = 100 e^(-10xx10^(-3))//RC) implies e^((-10^(-2))/(RC)) = 1/100` At `t = 20 ms` `V_(C) = V_(0)e^(-t//RC)= 100e^((-20xx10^(-3))/(RC) = 100e^((-2xx10^(-2))/(RC))` `=100(e^((-10^(-2))/(RC))^(2) = 100 xx 1/((100)^(2)) = 0.01 V`. |
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| 30. |
At `t=0` switch `S` is closed . (a) After how much time , charge on capacitor is 50% of maximum charge stored on capacitor. (b) After how much time, energy stored in capacitor is halt of maximum energy stored in capacitor. |
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Answer» (a) `q = CE (1-e^(-t//RC))` `q_(max) = CE, q = (q_(max))/(2) = 1/2 CE` `1/1 CE = CE(1-e^(-t//RC))` `e^(-t//RC)=2` `t/(RC)`log_(e) e = log_(e) 2 = 0.693RC` (b) `U =(q^2)/(2C) = 1/2 CE^(2)(1-e^(-t//RC))^(2)` `U_(max) = 1/2 CE^2, U = (U_(max))/(2) = (CE^2)/(4)` `(CE^2)/(4) = (CE^2)/() (1-e^(-t//RC))^(2)` `(1-e^(-t//RC))^(2) = 1/2 implies (1-e^(-t//RC))=1/(sqrt(2))` `e^(-t//RC)=1-1/(sqrt(2)) = (sqrt(2)-1)/(sqrt(2))` `e^(-t//RC) = (sqrt(2))/(sqrt(2)-1) = (sqrt(2))/(sqrt(2)-1)* ((sqrt(2)-1))/((sqrt(2)+1)) = (2+sqrt(2))` `t/(RC ) log_(e) e = log_(e) (2+sqrt(2))` ltbr .`t = RC log_(e) (2+sqrt(2))`. |
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| 31. |
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, `100W, 60W and 40W` bulbs have filament resistances `R_(100), R_(60) and R_(40)`, respectively, the relation between these resistances isA. `(1)/(R_(100))=(1)/(R_(40))+(1)/(E_(60))`B. `R_(100)=R_(40)+R_(60)`C. `R_(100)gtR_(60)gtR_(40)`D. `(1)/(R_(100))gt(1)/(R_(60))gt(1)/(R_(40))` |
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Answer» Correct Answer - D `R_(100) = (V^2)/(100), R_(60) = (V^2)/(60), R_(40) = (V^2)/(40)` `(1)/(R_(100)) gt (1)/(R_60) gt 1/(R_40)`. |
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| 32. |
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. `0.1 Omega`B. `0.2 Omega`C. `0.3 Omega`D. `0.4 Omega` |
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Answer» Correct Answer - C `(R_(eq))_(e3xt) = R//3` `r = R//3 implies R = 3r = 3 xx 0.1 = 0.3 Omega`. |
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| 33. |
The three resistance of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation.(I) (II) (IV) A. III lt II ltIV lt IB. II lt III lt IV lt IC. I lt IV lt III lt IID. I lt III lt II ltIV |
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Answer» Correct Answer - C `P_(1) = (V^2)/(3R) , P_(II) = (V^2)/(R//3), P_(IV) = (V^2)/(3R//2)` `I lt IV lt II lt III`. |
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| 34. |
An electrically heating coil was placed in a calorimeter containing `360g` of water at `10^(@)C`. The coil consumes energy at the rate of `70W`. The water equivalent of calorimeter and coil is `40g` . Find temperature of the water after `10min`. Given specific heat of water = `1 cal//g.^(@)C, 1 cal = 4.2J`. |
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Answer» The energy consumed by the coil in `10 min` `W =Pt = 70xx(10xx60) = 4.2xx10^(4)J` ..(i) Let final temperature of water = `10^(@)C` (given) Heat absorbed by (water+calorimeter + coil) `Delta Q = msDelta theta = (360+40)xx 11 xx(theta_(2)-10)cal` `=40(theta_(2)-10)xx4.2J` ...(ii) `Delta Q = W` `400(theta_(2)-10)xx4.2 = 4.2xx10^(4)` `theta_(2)-10=25` `theta_(2) = 35^(@)C`. |
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| 35. |
Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are `Z_1` and `Z_2` respectively the charge which flows through the silver voltameter isA. `q(Z_1)/(Z_2)`B. `q(Z_2)/(Z_1)`C. `(q)/(1+(Z_1)/(Z_2))`D. `(q)/(1+(Z_2)/(Z_1))` |
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Answer» Correct Answer - D `m = Zq implies Z prop 1//q` `(Z_1)/(Z_2) = (q_2)/(q_1)` `q = q_(1)+q_(2) implies q/(q_2)= (q_1)/(q_2)+1` `q_(2)=q//(1+q_(1)//q_(2))` `q_(2)=q//(1+Z_(1)//Z_(2))`. |
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| 36. |
The capacitor C is initially without charge. X is now joined to Y for a long time, during which `H_1` heat is produced in the resistance R. X is now joined to Z for a long time, during which `H_2` heat is produced in R A. `H_(1)=H_(2)`B. `H_(1)=H_(2)//2`C. `H_(1)=2H_(2)`D. the maximum energy stored in `C` at any time is `H_(2)` |
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Answer» Correct Answer - A `H_(1)=1/2CE^(2)` `H_(2)=1/2CE^(2)` |
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