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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.
| 1901. |
Why are constantan and manganin used for making standard resistances ?A. Both (A) and (R) are true and (R) is the correct explanantion of A.B. Both (A) and (R) are true but (R) is not the correct explanation of A.C. (A) is true but (R) is falseD. (A) is false but (R) is true. |
| Answer» Correct Answer - A | |
| 1902. |
Why are constantan and manganin used for making standard resistances ?A. Specific resistance is lowB. Density is highC. Temperature coefficient of resistance is negligibleD. Melting point is high |
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Answer» Correct Answer - C |
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| 1903. |
A `100 V` voltmeter of internal resistance `20 k Omega` in series with a high resistance `R` is connected to a `110 V` line. The voltmeter reads `5 V`, the value of `R` isA. `210kOmega`B. `315kOmega`C. `420kOmega`D. `440kOmega` |
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Answer» Correct Answer - C |
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| 1904. |
A `100 V` voltmeter of internal resistance `20 k Omega` in series with a high resistance `R` is connected to a `110 V` line. The voltmeter reads `5 V`, the value of `R` isA. `210 k Omega`B. `315 k Omega`C. `420 k Omega`D. `440 k Omega` |
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Answer» Correct Answer - C |
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| 1905. |
In the circuit diagram shown in Fig. 4.53, a voltmeter reads 30 V when connected across `400 Omega` resistance. Calculate what the same voltmeter reads when it is connected across `300 Omega` resistance. |
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Answer» Correct Answer - `22.5 V` |
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| 1906. |
A students connects an ammeter `A `and a voltmeter `V` to measure a resistancer as shown in figure. If the voltmeter reads `20 V` and the ammeter reads `4A`, then `R` is A. equal to `5Omega`B. greater than `5Omega`C. less than `5Omega`D. greater or less than `5Omega` depending upon the direction of current |
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Answer» Correct Answer - C `4(R+R_A)=20V` `:. R=5-R_A` where `R_A`= resistane of ammeter |
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| 1907. |
In the circuit shown in the figure, cell is ideal and `R_(2) = 100 Omega`. A voltmeter of internal resistance `200 Omega` reads `V_(12) = 4 V and V_(23) = 6 V` between the pair of points `1-2 and 2-3` respectively. What will be the reading of the voltmeter between the points `1-3`. |
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Answer» Correct Answer - 12 V |
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| 1908. |
A students connects an ammeter `A `and a voltmeter `V` to measure a resistancer as shown in figure. If the voltmeter reads `20 V` and the ammeter reads `4A`, then `R` is A. Equal to `5Omega`B. Greater than `5Omega`C. Less than `5Omega`D. Greater or less than `5Omega` depending,upon the direction of current |
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Answer» Correct Answer - B |
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| 1909. |
A voltmeter of resistance `R_(V)` and an ammeter of resistance `R_(A)` are connected as shown in an attempt to measure the resistance `R`. The measured value of the resistance is `R_(M) = (V)/(I_(0))` where `V` is reading of voltmeter and `I_(0)` is reading of the ammeter. find the true value of the resistance in terms of `R_(M), R_(V) and R_(A)`. |
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Answer» Correct Answer - `R=(R_(M)R_(V))/(R_(V)-R_(M))` |
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| 1910. |
Calculate battery current and equivalent resistance of the network shown in figure. |
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Answer» Correct Answer - A All four resistors are in parallel `:. 1/R=1/8+1/4+1/6+1/12` `R=8//5Omega` `:. i=24/(8/5)` `=15A` |
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| 1911. |
Find the current supplied by the battery in the circuit shown in figure. |
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Answer» Correct Answer - A `8Omega` and `12 Omega` resistors are in parallel `:. R_("net")=(8xx12)/(8+12)=4.8Omega` `:. i=24/4.8` `=5A` |
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| 1912. |
The equivalent resistance resistance between A and B of network shown in figure is A. `(3R)/(4)`B. `(4R)/(3)`C. 6 RD. 2 R |
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Answer» Correct Answer - A |
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| 1913. |
A galvanometer having resistance of `50 Omega` requires a current of `100 Omega A` to given full scale deflection. How much resistance is required to convert it into an ammeter of range of 10 A ?A. `5xx10^(-3)Omega` in seriesB. `5xx10^(-4)Omega` in parallelC. `10^(5)Omega` in seriesD. `10^(5)Omega` in parallel |
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Answer» Correct Answer - B |
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| 1914. |
A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is A. `1.5 V`B. `3.0 V`C. `0.67 V`D. `1.33 V` |
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Answer» Correct Answer - B (b) `E = xl = (V)/(l) = (iR)/(L) xx l implies E = (e)/((R + R_(h) + r)) xx (R )/(L) xx l` `implies E = (10)/((5 + 4 + 1)) xx (5)/(5) xx 3 = 3 V` |
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| 1915. |
The resistance of a galvanometer is `50 ohm` and the current required to give full scale deflection is `100 muA`. In order to convert it into an ammeter, reading upto `10 A`, it is necessary to put a resistance of A. `5 xx 10^(-3) Omega` in parallelB. `5 xx 10^(-4) Omega` in parallelC. `10^(5) Omega` in seriesD. `99,950 Omega` in series |
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Answer» Correct Answer - B (b) Resistance in parallel `S = (Gi_(g))/(i- i_(g)) = (50 xx 100 xx 10^(-6))/((10 - 100 xx 10^(-6)))` `implies S = 5 xx 10^(-4) Omega` |
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| 1916. |
The resistance of a galvanometer is `50 ohm` and the current required to give full scale deflection is `100 muA`. In order to convert it into an ammeter, reading upto `10 A`, it is necessary to put a resistance of A. `5 xx 10^(-3) Omega` in parallelB. `5 xx 10^(-4) Omega` in parallelC. `10^(5) Omega` in seriesD. `99,950 Omega` in series |
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Answer» Correct Answer - B |
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| 1917. |
A potentiometer circuit shown in the figure is set up to measure e.m.f. of a cell E . As the point P moves from X to Y the galvanometer G shows deflection always in one direction, but the deflection decreases continuously until Y is reached. In order to obtain balance point between X and Y it is necessary to A. Decreases the resistance RB. Increase the resistance RC. Reverse the terminals of battery VD. Reverse the terminals of cell E |
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Answer» Correct Answer - A |
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| 1918. |
The circuit shown here is used to compare the e.m.f. of the two cells `E_(2) (E)_(1) gt E_(2)`. The null point is at `C` when the galvanometer is connected to `E_(1)`. When the galvanometer is connected to `E_(2)`, the null point will be A. To the left of CB. To the right of CC. At C itselfD. Nowhere on AB |
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Answer» Correct Answer - A |
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| 1919. |
In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of `2 m` when the cell is shunted by a `5 Omega` resistance and is at a length of `3 m` when the cell is shunted by a `10 Omega` resistance, the internal resistance of the cell is thenA. 0.1ohmB. 1 ohmC. 2 ohmD. 0.5ohm |
| Answer» Correct Answer - 2 | |
| 1920. |
In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of `2 m` when the cell is shunted by a `5 Omega` resistance and is at a length of `3 m` when the cell is shunted by a `10 Omega` resistance, the internal resistance of the cell is thenA. `1.5 Omega`B. `10 Omega`C. `15 Omega`D. `1 Omega` |
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Answer» Correct Answer - B |
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| 1921. |
In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of `2 m` when the cell is shunted by a `5 Omega` resistance and is at a length of `3 m` when the cell is shunted by a `10 Omega` resistance, the internal resistance of the cell is thenA. `12 Omega`B. `8 Omega`C. `16 Omega`D. `1 Omega` |
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Answer» Correct Answer - B As, `((1-2)/(2))4=((1-3)/(3))8 implies l=6` Therefore, `r=((1-2)/(2))4=8 Omega` |
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| 1922. |
In an experiment with a potentiometer to measure the internal resistance of a cell. when the cell in the secondary circuit is by shounted by `5 Omega`, the null point is at `220 cm`. When the cell is shunted by `20 Omega ` the null point is at `300 cm`. Find the internal resistance of the cell.A. `2Omega`B. `4Omega`C. `6Omega`D. `8Omega` |
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Answer» Correct Answer - B `(V_(1))/(V_(2))=(l_(1))/(l_(2))` `V_(1)=[(E)/(R_(1)+r)]R_(1),V_(2)[(E)/(R_(2)+r)]R_(2)` `(R_(1)(R_(2)+r))/(R_(2)(R_(1)+r))=(l_(1))/(l_(2))impliesr=4Omega` |
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| 1923. |
In an experiment with a potentiometer to measure the internal resistance of a cell. when the cell in the secondary circuit is by shounted by `5 Omega`, the null point is at `220 cm`. When the cell is shunted by `20 Omega ` the null point is at `300 cm`. Find the internal resistance of the cell. |
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Answer» Correct Answer - `2.76 Omega` Let `epsilon` and `r` be the emf and internal resistance of the cell used in secondary circuit and `K` be the potential gradent along the potentiometer wire. According to equation `((epsilon)/(r + 5)) 5 = K xx 220 `….(i) and `(epsilon)/(r + 20)) 20 = K xx 300 `….(ii) On solving (i) and (ii) we get `r = 2.76 Omega` |
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| 1924. |
In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of `2 m` when the cell is shunted by a `5 Omega` resistance and is at a length of `3 m` when the cell is shunted by a `10 Omega` resistance, the internal resistance of the cell is thenA. `1 Omega`B. `1.5 Omega`C. `10 Omega`D. `15 Omega` |
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Answer» Correct Answer - c `(Er_(1))/((R_(1)+r)) = kl_(1) and (Er_(2))/((R_(2)+r)) = kl_(1)` `:. (Er_(1)//(R_(1)+r))/(Er_(2)//(R_(2)+r)) = (l_(1))/(l_(2)) or (R_(1)(R_(2)+r))/(R_(2)(R_(1)+r)) = (l_(1))/(l_(2))` `:. (5(10+r))/(10(5+r)) = (2)/(3)` On solving `r = 10 Omega` |
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| 1925. |
Three unequal resistor in parallel are equivalent to a resistance `1` ohm If two of them are in the ratio `1:2` and if no resistance value is fractional the largest of three resistance in ohm isA. 4B. 6C. 8D. 12 |
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Answer» Correct Answer - B `(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))=1` `R_(2):R_(3)=1:2` |
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| 1926. |
Three unequal resistor in parallel are equivalent to a resistance `1Omega` If two of them are in the ratio 1:2 and if no resistance value is fractional the largest of the three resistance in ohm isA. `4`B. `5`C. `6`D. `10` |
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Answer» Correct Answer - c Given, `(R_(1))/(R_(2)) =(1)/(2) or R_(2) = 2R_(1)` The equivalent resistance of parallel combination is `(1)/(R ) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))` `(1)/(R_(1)) + (1)/(2R_(2)) + (1)/(R_(3))= (3)/(2R_(1)) + (1)/(R_(3))` or `(1)/(R_(3)) = (1)/(R )-(3)/(2R_(1)) = (1)/(1)- (3)/(2R_(1))` or `1 = R_(3) = (3R_(3))/(2R_(1))` `R_(3) = (3R_(3))/(2R_(1))` Since no resistance value is fractional, hence minimum value of `(R_(1))/(R_(2)) = (2)/(3)` So `R_(2) = 1+ (3)/(2)xx(2)/(3) = 1+1 = 2 Omega` and `R_(1) = (3R_(3))/(2) = (3xx2)/(2) = 3 Omega` `:.` Maximum resistance value is of `R_(2) = 2R_(1) = 2 xx 3 = 6 Omega` |
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| 1927. |
If gas molecules undergo, inelastic collision with the walls of the containerA. the temperature of the gas will decreaseB. the pressure of the has will increaseC. neither the temperature nor the pressure will changeD. the temperature of the gas will incease |
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Answer» Correct Answer - C |
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| 1928. |
On what factors, does the potential gradient of the potentimeter wire depend? |
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Answer» The fall of potential across a potentiometer wire of length l is given by `V = IR = I rho l //A` `:.` Potential gradient, `K=V/l = (I rho)/(A)` Thus potential gradient of a wire depends upon the following factors: (i) `K prop I` (i.e., current passing through the potentiometer wire) (ii) `K prop rho` (i.e., specific resistance of the material of the potentiometer wire) (iii) `K prop 1/A`, where A is the area of cross-section of the potentiometer wire. |
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| 1929. |
Current passing through `1 Omega` resistance is zero. Then the emf E is A. `8 V`B. `6 V`C. `4 V`D. `12 V` |
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Answer» Correct Answer - B |
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| 1930. |
For the arrangement of the potentimeter shown in the figure, The balance point is obtained at a distance `75 cm` from `A` when the key `k` is open The second balance point is obtained at `60 cm` from `A` when the key `k` is closed. Find the internal resistance of the battery `E_(1)`. |
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Answer» Correct Answer - 6 |
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| 1931. |
A block of ice with mass m falls into a lake. After impact, a mass of ice `(m)/(5)` melts. Both the block of ice and the lake have a temperature of `0^(@)C`. If L represents the latent heat of fusion, the distance the ice falls before striking the surface isA. `(L)/(5g)`B. `(5L)/(g)`C. `(gL)/(5m)`D. `(mL)/(5g)` |
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Answer» Correct Answer - A |
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| 1932. |
A solution of sodium chloride discharge `6.0 xx 10^(16)Na^(+) and 4.5 xx 10^(16)Cl^(-1)` lons in `2` seconds. What is the current passing through the solution ? |
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Answer» Correct Answer - ` 8.4 mA` `I = I_(Na)+ I_(C) = (n_(Na)+ n_(Cl)) e//t` `= ((6.0 xx 10^(16)+ 4.5 xx 10^(16)) xx 1.6 xx 10^(-19))/(2)` `= 8.4 xx 10^(-3)A = 8.4 mA` |
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| 1933. |
In hydrogen atom, the electron makes `6.6 xx 10^(15)` revolutions per second around the nucleus in an orbit of radius `0.5 xx 10^(-10)m`. It is equivalent to a current nearlyA. `1A`B. `1mA`C. `1 muA`D. `1.6 xx 10^(-19)A` |
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Answer» Correct Answer - B |
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| 1934. |
In Bohr model of hydrogen atom , the electron revolves around the nucleus in a circular orbit of radius `5.1 xx 10^(-11)m` at a frequency of `6.8 xx 10^(15)` revolutions per second. Find the equivalent current at may point on the orbit of the electron |
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Answer» Correct Answer - `= 1.088 xx 10^(-3)A` `I = ev = (1.6 xx 10^(-19)) xx (6.8 xx 10^(15)) ` `= 1.088 xx 10^(-3)A` |
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| 1935. |
In Bohr model of hydrogen atom, the electron revolves around the nucleus in a circular orbit of radius `5.0xx10^(-11)` m with a speed `2.2 xx10^(6) ms^(-1)`. Find the equivalnet current. (Electronic charege = `1.6 xx 10^(-19)`coulomb) |
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Answer» Here, `r=5.0 xx 10^(-11)m`, `v=2.2xx10^(6)ms^(-1), e=1.6xx10^(-19)C` Let v be the frequency of revolution of electron in hydorgen, then `v=1/T= (1)/((2 pi r//v)) = v/2 pi r` `I= ev = (ev)/(2 pi r) =((1.6xx10^(-19))xx(2.2 xx 10^(6)))/(2 xx (22/7) xx 5.0 xx 10^(-11))` `=1.12 xx10^(-3)A` |
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| 1936. |
A wire of resistance `R` is elongated `n-fold` to make a new uniform wire. The resistance of new wire.A. nRB. `n^(2)R`C. `2nR`D. `2n^(2)R` |
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Answer» Correct Answer - B The resistance of a wire is directly proportional to the length of the wire. Thus `R prop l`, so the length increases n-fold also since the volume of the wire remains constant, so the area of the wire decreases n-fold. Hence, the resistance increases to `n^(2)R` as the area is inversely proportional to resistance. |
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| 1937. |
The equivalent resistance of series combination of four equal resistors is S. If they are joined in parallel, the total resistance is P. The relation between S and P is given by S = nP. Then the minimum possible value of n isA. 12B. 14C. 16D. 10 |
| Answer» Correct Answer - C | |
| 1938. |
A wire has resistance 12 ohms. If it is bent in the form of a equilateral triangle. The resistance between any two terminals in ohms is:A. `8//3`B. `3//4`C. 4D. 3 |
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Answer» Correct Answer - A `1/(R_(eq))=1/8+1/4=3/8 rArr R_(eq)=8/3 Omega` |
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| 1939. |
When copper voltmeter is connected with a battery of emf 12 V, 2 g of copper is deposited in 30 min. If the same voltmeter is connected across a 6 V battery, then the mass of copper deposited in 45 min would beA. 1 gB. 1.5 gC. 2 gD. 2.5 g |
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Answer» Correct Answer - B m=zit m=1.5g |
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| 1940. |
With resistances P and Q in the left gap and right gap respectively, of a meter bridge. The null point is obtained at 60 cm from left end. When Q is increased by `10Omega` the null point is obtained at 40 cm from left end. The values of P and Q areA. `12Omega, 8Omega`B. `8Omega, 12Omega`C. `8Omega, 8Omega`D. `12Omega, 12 Omega` |
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Answer» Correct Answer - A (i) `(P)/(Q)=(60)/(40) " " P=(3Q)/(2)` (ii) `(P)/(Q+10)=(40)/(60)` `(3Q)/(20+20)=2/4 " " 9Q=4Q+40 " " 5Q=40` `:.P=(3xx8)/(2)=12Omega` |
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| 1941. |
A ideal gas `(gamma=1.5)` is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules `2.0` timesA. 4 timesB. 16 timesC. 8 timesD. 2 times |
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Answer» Correct Answer - B |
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| 1942. |
A sphete of deamrter 7.0 cm and mass 266.5 g float in a bath of liquid. As the temperature is raised, the sphere begins to sink at a temperature of `35^@ C`. If the density of liqued is `1.527 g cm^(-3)` at` 0^@C`, find the coeffiecient of cubical expamsion of the liquid. Neglect the expansion of the sphere.A. `9.52xx10^(-4)per^(@)C`B. `6.23xx10^(-5)per^(@)C`C. `8.486xx10^(-6)per^(@)C`D. `4.12xx10^(-3)per^(@)C` |
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Answer» Correct Answer - A |
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| 1943. |
One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A,B and C are 400K, 800K and 600K respectively. Choose the correct statement: A. The change in internal energy in the process AB is `-350R`.B. The change in internal energy in the process BC is `-500`R.C. The change in internal energy in whole cyclic process is 250 R.D. The change in internal energy in the process CA is 700R. |
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Answer» Correct Answer - B |
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| 1944. |
A 100 W bulb `B_(1)` and two 60 W bulbs `B_(2)` and `B_(3)`, are connected to a 250V 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 thenA. `W_(1)gtW_(2)=W_(3)`B. `W_(1)gtW_(2)gtW_(3)`C. `W_(1)ltW_(2)=W_(3)`D. `W_(1)ltW_(2)ltW_(3)` |
| Answer» Correct Answer - D | |
| 1945. |
Given T-p curve for three processes. Work done in process 1, 2 and 3 (if initial and final pressure are same for all processes) is `W_(1),W_(2)` and `W_(3)` respectively. Correct order is A. `W_(1)ltW_(2)gtW_(3)`B. `W_(1)gtW_(2)gtW_(3)`C. `W_(1)ltW_(2)ltW_(3)`D. `W_(1)=W_(2)=W_(3)` |
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Answer» Correct Answer - B |
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| 1946. |
One mole of an ideal monatomic gas (intial temperature `T_(0)`) is made to go through the cycle abca as shown in the figure. If U denotes the internal energy, then choose the correct alternatives A. `U_(c )-U_(a)=10.5RT_(0)`B. `U_(b)-U_(a)=4.5RT_(0)`C. `U_(c )gt U_(b)gt U_(a)`D. `U_(C )-U_(b)=6RT_(0)` |
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Answer» Correct Answer - A::B::C::D |
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| 1947. |
One mole of an ideal monoatomic has is taken through cyclic process ABC as shown in the graph with pressure and temperature as coordinate axes. Process AB is defined as PT = constant. Take universal gas constant to be R. Then: work done on gas in the process AB isA. 675 RB. 975 RC. 300 RD. 600 R |
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Answer» Correct Answer - C |
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| 1948. |
The current through a wire depends on time as `i=(2+3t)A`. Calculate the charge crossed through a cross section of the wire in 10sec. |
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Answer» `i=(dq)/(dt)rArrdq=(2+3t)dt` `int_(0)^(q)dq=int_(0)^(10)(2+3t)dtrArrq=(2t+(3t^(2))/(2))_(0)^(10)` `q=2xx10 +(3)/(2)xx100=20+150=170C` |
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| 1949. |
Five equal resistors each of `R Omega` are connected in a network. Calculate the equivalent resistance between the points A and B. |
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Answer» The equivalent circuit will be as shown in figure, which is a balanced Wheatstoen bridge. Therefore the resistance of arm CD is ineffective. Hence we have the resistance of arm ACB = R +R = 2R, in parallel with the resistance of arm ADB = R +R = 2R. Effeective resistance between A and B is ` =(2 R xx 2R)/(2R+ 2R) = R Omega` |
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| 1950. |
In figure, AB is a potentiometer wire, length 10 m and resistance `2 Omega` with open the balancing length is 5.5 m. However, on closing key K the balancing length reduces to 5 m. The initial resistance of the cell `E_(1)` is A. `0.01 Omega`B. `0.1 Omega`C. `0.2 Omega`D. `1 Omega` |
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Answer» Correct Answer - B `l_(AB)=10m, R_(AB)=2 Omega, R_(AJ)=1 Omega` ` E_(1)=rho L` (When key is open) `E_(1)=rhoxx5.5 implies V_(1)=5rho` `(E-V)/(V) R=r(5.5rho-5rho)/(5rho)xx1=r(0.5)/(5)xx1=r` ` r=0.1 Omega` |
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