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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.
| 8201. |
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10 in quadratic equations |
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Answer» Let two parts be x and y.thereforex+y=20 ⇒ y=20-x Then we have 3x2-(20-x)=10 3x2+x-30=0 3x2+10x-9x-30=0 x(3x+10)-3(3x+10)=0 (x-3)(3x+10)=0 x=3 is the solution. The two parts are 3 and 17 respectively. |
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| 8202. |
In a cricket match Harbhajan took three wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Represent the above situation in the form of quadratic equation. |
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Answer» Solution: |
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| 8203. |
A school library has 280 science journals and 300 maths journals. Students were told to stack these journals in such a way that each stack contains equal number of journals. Determine the number of stacks of science and maths journals. What is the benefit of library in student life? |
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Answer» Solution: => There are 14 stacks of science journals and 15 stacks of maths journals. |
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| 8204. |
For a morning walk, three persons steps off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk to show that they can cover the distance in complete steps? Which value is preferred in this situation? |
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Answer» Solution: |
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| 8205. |
Estimate the mass of copper metal produced during the passage of 5 A current through `CuSO_(4)` solution for 100 minutes. The molar mass of Cu is 63.5 g `mol^(-1)` |
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Answer» Correct Answer - Mass of Cu metal formed `=9 .87 g` Given : Electric current = I = A Time = t = 100 min = 100 `xx 60 s = 6000 s ` Molar mass of Cu = Atomic mass of Cu = 63.5 g `mol^(-1)` Mass of cu deposited = ? Reduction half reaction at cathode . `Cu_((ag))^(++) + 2e^(-) to Cu_((s))` Quantity of electricity `= Q= I xx t = 5 xx 6000 = 3000 C` Number of moles of electrons passed `= (Q)/(F ) ` `= (3000)/(96500) "mol"` ` C "mol"^(-1)` `= 0.3109 "mol"` From the half reduction reaction 1 mol of Cu is formed by the passage of 2 moles of electrons. Molar ratio of Cu `= ("Molar of Cu formed " )/("Number of moles of electron " ) ` `= (1 " mol " ) /(2 " mol " ) =0.5 ` Number of moles of Cu formed = Number of moles of electrons passed `xx ` Mole ratio `= 0.3109 xx 0.5 = 0.1554` mol Mass of Cu formed = Moles of Cu formed `xx` Molar Mass of Cu `= 0.1554 xx 63.5 = 9.87 g` |
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| 8206. |
what is the molar conductivity of Agl at zero concentration if the `^^_(0)` values of NaI, `AgNO_(3)` and `NaNO_(3)` are respectively `126.9 Omega^(-1) cm^(2) mol^(-1), 133.4 Omega^(-1) cm^(2) mol^(-1) ` and 121.5 `Omega^(-1) cm^(2) mol^(-1)` |
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Answer» Correct Answer - Cell reaction `: Ni_((s)) + C1_(2(g)) to Ni_((ag))^(2+) + 2C1_((ag))^(-)` `E_(cell)^(0) = 1.61 V` Given : `E_(C1_(2))^(0) = 1.36 V , E_(Ni_(2+)//Ni)^(0)= - 0.25 N,` `E_("Cell ")^(0)=?` `Ni_((s))|Ni^(2+) (1M) "||" C1^(-) (1M)| C1_(2(g)) (1 atm) |Pt` `((" At LHE " Ni_((s)) to Ni_((ag)^(2+) + 2e^(-)))/(" At RHE (.(1)/(2) C1_(2(g))+ e^(-) to C1^(-)_(ag))xx2))/(Ni_((s) + C1_(2(g)) ) to Ni_((ag))^(2+) + 2C1_((ag))^(-))` The standard emf is given by `E_(" Cell ) ^(0) = E_(cathode)^(0) - E_(anode)^(0)` `=E_(C1_(2)//C1^(-))^(0) -E_(Ni^(2+)//Ni)^(0)` `= 1.36 - (- 0.25)` `=1.36 + 0. 25 = 1.61 V` |
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| 8207. |
Construct at cell consisting of `Ni_((ag))^(2+)| Ni_((s))` half cell and `H^(+) ||H_(2(g)) | Pt` half cell (a) Write the cell reaction ( b) Calculate emf of the cell if cell `[Ni^(2+)] = 0.1 ` `[ H^(+)] =0.05 M` and `E_(Ni)^(0) =- 0.257 V` |
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Answer» Correct Answer - (a) cell reaction `: Ni_((s))^(2++) + 2H_((ag))^(+) to NI_((aq))^(+) to Ni_((aq))^(+) H_(2(g))` ( b) `E_("cell ") = 0.2096 V` Given : `E_(Ni^(+ + )//Ni) =- 0.257 V , [ H^(+) ] = 0.1 M` `[Ni ^(++) ] =0.1 M` (a) Half cells : (i) `Ni_((ag))^(2+) | Ni_((s)) " and " (ii) H_((aq))^(+) | H_(2(s)) | Pt ` `E_(Ni^(2+)//Ni) = - 0.257 V , E_(H^(+) //H_(2) )^(0) E_(SHE)^(0) = 0.0V` The formulation of the cell will be `Ni_((s)) | NI_((ag))^(2+) " || " H_((ag))^(+) |H_(2) (g , 1 atm ) |Pt` `(RHE (H_((ag)^(+))) +e^(-) to .(1)/(2) H_(2(g))xx2)/(Ni_((s) + 2H_((ag))^(+) to Ni_((aq))^(+) + H_(2(g))))" " underset"(Overall cell reaction )"underset"(Reduction at cathode )"("(Oxidation at anode)")` `( b) [Ni^(2+) ] =0.1 M : [ H^(+)] = 0.05 M , n=2` `E_(" cell " )^(0) = E_(H^(+)// H_(2))^(0) = E_(Ni^(2+)// Ni)^(0) =0.0 -(0.257)` `=0.256 V` `E_("cell ") = E_(" cell ")^(0) - (0.0592)/(n) lgo _(10) .([Ni^(2+)] xx [ H_(2)])/([H^(+) ]^(2))` `=0.257 -(0.0592)/(2) log_(10) .(0.1 xx1)/((0.05)^(2))` ` =0.257 - 0.0296 log _(10) 40` `=0.257 -0.0296 xx 1.6020` `=0.257 -0.04742` `=0.2096 V` `~= 0.2096V` |
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| 8208. |
The equilibrium constant for the following reaction at `25^(@)C " is " 2.9 xx 10^(9)`. Calculate standard voltage of the cell. `C1_(2(g)) + 2Br_((aq))^(-) hArr Br_(2(1)) + 2C1_((aq))^(-)` |
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Answer» Correct Answer - Standard voltage of the cell `=E_("cell ")^(0) = 0.28 V` Given : Cell reaction `:C1_(2(g)) + 2Br_((aq))^(-) hArr Br_(2(1)) + 2C1_((aq))^(-)` Equilibrium constant = k = 2.9 `xx10^(-9) atm ^(-1)` Standard voltage of the cell = `E_("cell ")^(0) = ? ` The formulation of the cell : `Pt|Br_(2(1)) | (1M)" ||"C1_((aq))^(-) (1M)|C1_(2) (g,P_(C1_(2))) | Pt` `((LHE 2Br_((aq))^(-) to Br_(2(1)) + 2e^(-))/(RHE . C1_(2(g)) + 2e^(-) to 2C1_((aq))^(-)))/(2Br_((aq))^(-) + C1_(2(g)) to Br_(2(1))+ 2C1_((aq))^(-))" "underset"(Overall cell reaction )"underset"(Reduction at cathode )"("(Oxidation at anode)")` `:. n=2` `E_("cell ")^(0) = (0.0592)/(n) log_(10) K` `= (0.0592)/(2) log_(10) 2.9 xx 10^(9)` `=0. 0296 xx (9.4624)` `=0.28 V` |
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| 8209. |
Calculate `E_("cell ")^(0) ,Delta G^(@)` and equilibrium constant for the reaction `2Cu^(+) to Cu^(2+) +Cu` `E_(Cu)^(0) | Cu = 0.52 V " and " E_(Cu) ^(0) ,Cu^(+) = 0.16 V .` |
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Answer» Correct Answer - `E_("cell ")^(0) =0.36 V ; Delta G^(@) =- 34.74 kJ ;` Equilibrium constant = `K = 1.2 xx 10^(6) mol^(-1) dm^(-3)` Given : Cell reaction `: 2Cu_((aq))^(+) to Cu_((aq))^(2+) + Cu_((s))` `E_(Cu^(+))^(0) | Cu = 0.52 V , E_(Ca^(2+)) , Cu^(+)= 0.16 V` ` 1 F = 96500 C` `E_("cell")^(0) = ? Delta G^(@) = ? ` (i) The formulation of the cell : `Pt| Cu_((aq))^(+) Cu_((aq))^(2+) "||" Cu_((aq))^(+) | Cu_((s))` `((LHE .Cu_((aq))^(+) to Cu_((aq))^(+) + e^(-))/(RHE. Cu_((aq))^(+) to Cu_((s))))/(2Cu_((aq))^(+) to Cu_((aq))^(2+) + Cu_((s)))" "underset"(Overall cell reaction )"underset"(Reduction at cathode)"("(Oxidation at anode)")` `:. n=1` `E_("cell ")^(0) = E_(cu^(+) |Cu)- E_(Cu^(+2))^(0) .Cu^(+)` `=0.52- 0.16 =0.36 V` (ii) `Delta G^(@) =- nFE_("Cell")^(0) =- 1xx 96500 xx 0.36` `=- 34740 J` `=- 34.74 kJ` (ii) If K is the equilibrium constant for the electrochemical redox reaction then `K= ([Cu^(2+)])/([Cu^(+)]^(2)) ` `E_("cell ")^(0) = (0.0592)/(n) log_(10) k` `:. log_(10) K = (nxx E_("cell ")^(0))/(0.0592)` `=(1xx 0.36)/(0.0592)` `=6.081 ` `:. K = A16 .081 = 1x2 xx 10^(6) ("mol" dm^(-3))^(-1) (or "mol"^(-1) dm^(3))` |
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| 8210. |
Consider the reaction `2A + B to 2C`. Suppose that at a particular moment during the reaction , rate of disappearance of A is 0.076 `m//s` (a) What is the rate of formation of C ? ( b ) What is the rate of consumption of B ? (c) What is the rate of the reaction ? |
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Answer» Correct Answer - (a) Rate of formation of C= 0.076 `Ms^(-1)` ( b) Rate of consumption of B= 0.038 `Ms^(-1)` ( c) Rate of reaction = 0 .038 `Ms^(-1)` Given : `2A + B to 2C` Rate of disappearance of A= 0.076 `Ms^(-1)` (a) Rate of formation of C= ? ( b) Rate of consumption of B = ? (c ) Rate of reaction = ? Rate of disappearance of `A = (d[A])/(dt) = 0.076 Ms^(-1)` (a) Rate of formation of `C= (d[C ])/(dt) = (d[A])/(dt)` `=0.076 Ms^(-1)` ( b) Rate disappearance of `B= (d[B])/(dt) = - (1)/(2) (d[A])/(dt)` `= (1)/(2) xx 0.076 =0. 038 Ms^(-1)` ( c) Rate of reaction `=- (1)/(2) .(d[A])/(dt) = (-d[B])/(dt) = (1)/(2) (d[C])/(dt)` `= (1)/(2) (0.076) = 0.038 = (1)/(2) xx 0.076` `= 0.038 Ms^(-1)` |
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| 8211. |
The gas - phase reaction between NO and `Br_(2)` is respresented by the equation `2NO_((g)) + Br_(2(g)) to 2NOBr_((g)` (a ) Write the expressions for the rate of consumption of reactants and formation of products. ( b) Write the expression for the rate of overall reaction in terms of rates of consumption of reactions and formation of products. |
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Answer» Given : `2NO_((g)) + Br_(2(g)) to 2NOBr_((g))` (a ) Rate of consumption of NO at time `t= - (d[NO])/(dt)` Rate of consumption of `Br_(2)` at time `t = (- d[Br_(2)])/(dt)` Rate of formation of NOBr at time `t= (d[NOBr])/(dt)` ( b) Rate of reaction `= - (1)/(2) (d[NO])/(dt) = (-d[Br_(2)])/(dt) = (1)/(2) (d[NOBr])/(dt)` |
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| 8212. |
The curve passing through the point (0,1) and satisfying the equation sin `((dy)/(dx))`=a, is-A. `cos((y+1)/x)=a`B. `cos(x/(y+1))=a`C. `sin((y-1)/x)=a`D. `sin(x/(y-1))=a` |
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Answer» Correct Answer - 3 `sin((dy)/(dx))=a` `(dy)/(dx)=sin^(-1)a` `intdy=intsin^(-1)acdotdx` `y=sin^(-1)acdotx+c` passes (0,1) `1=0+c,c=1` `y=sin^(-1)acdotx+1` `y-1=sin^(-1)acdotx` `(y-1)/x=sin^(-1)a` `sin((y-1)/x)=a` |
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| 8213. |
A cadiadate has to reach the examination centre in time, Probality of him going by bus or scooter or by other means of transport are `(3)/(10),(1)/(10) ,(3)/(5)` respectively. The probability of getting late , if the travels by bus is 1/4 ,1/3 if he travells by scooter and 0 for any other medium. But he reaches in time if the uses any mode of transport . He reached late at the centre. The probability that he travelled by bus is -A. `(1)/(9)`B. `(2)/(13)`C. `(9)/(13)`D. None of these |
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Answer» Correct Answer - C Let A, B, C be the events of candidate going by bus, scooter and other means of transport, Let E be the event of getting late. `P(A)=(3)/(10),P(B)=(1)/(10),P(C)=(3)/(5)` `p((E)/(A)) =(1)/(4),P((E)/(B))=(1)/(3),P((E)/(C))=0` `underset(P("he travelled bus"))(P((A)/(E)))` `=(P(A)P((E)/(A)))/(P(A).P((E)/(A))+P(B).P((E)/(A))+P(C).P((E)/(C)))=(9)/(13)` |
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| 8214. |
Assertion :- `CH_(3)-O-C_(3)H^(7)` and `C_(2)H_(5)-O-C_(2)H_(5)` are metamer. Reason:- They are differ in alkyl group around polyvalent functional group.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - A | |
| 8215. |
A person goes to office either by car, scooter, bus or train probability of which being`1/7, 3/7, 2/7 and 1/7` respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is `2/9, 1/9, 4/9 and 1/9` respectively. Given that he reached office in time, then what is the probability that he travelled by a car?A. `1/5`B. `1/7`C. `6/7`D. None of these |
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Answer» Correct Answer - 2 Let C, S, B, T be the events that the person reaching the office in time - `P(C )=1/7,P(S)=3/7,P(B)=2/7,(T)=1/7` Let `barL` be the event that the person reaching the office in time - `P(barL//C)=7/9,P(barL//S)=8/9,` `(x-1)/(9)=(y-2)/(-1)=(z+3)/(-3)` `thereforeP(C//barL)=(P(C)P(barL//C))/(P(barL))` `=(1/7cdot7/9)/(1/7cdot7/9+3/7cdot8/9+2/7cdot5/9+1/7cdot8/9)` `=7/49=1/7` |
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| 8216. |
The species: `H_(2)O, HCO_(3)^(Θ), HSO_(4)^(Θ)` and `NH_(3)` can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base. |
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Answer» Conjugate acids: `H_(3)O^(o+), H_(2)CO_(3), H_(2)SO_(4)` Conjugate bases: `overset(Θ)(O)H, CO_(3)^(2-), SO_(4)^(2-)` |
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| 8217. |
A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an `N//20` HCl solution when titrated with phenolphthalein as an indicator.But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid.The amount of KOH present in the solution isA. 0.014 gB. 0.14 gC. 0.028 gD. 1.4 g |
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Answer» Correct Answer - A `underset(a Mili eq.)(KOH)+underset(b Mili eq.)(Na_2CO_3)` `a+b/2=15xx1/20` 2a+b=1.5 ..(i) (in presence of phenolphthalein) `a+b=25xx1/20=1.25`…(ii)(in presence of Methyl orange) by solving (i) & (ii) a = 0.25 Mili eq. mass of KOH=`0.25/1000xx56=0.014` gm |
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| 8218. |
A husband and wife appear in an interview for two vacancies for the same post. The probability of husband getting selected is 1/5 while the probability of wife getting selected is 1/7. Then the probability that any one of them getting selected is1. 11/352. 12/353. 1/354. 34/35 |
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Answer» Correct Answer - Option 1 : 11/35 The probability of husband getting selected is 1/5 and the probability of wife getting selected is 1/7. Now, the converse of the statement that “any one of them getting selected” is “none of them getting selected” \(\left( {1 - \frac{1}{5}} \right).\left( {1 - \frac{1}{7}} \right)\frac{4}{5}.\frac{6}{7} = \;\frac{{24}}{{35}}\) The probability of none them getting selected = (Probability of husband not getting selected). (Probability of wife not getting selected) = = Hence, the probability that any one of them getting selected = 1 - probability of none them getting selected = \(- \frac{{24}}{{35}} = \frac{{11}}{{35}}\) |
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| 8219. |
A parallel- plate capacitor is charged from a cell and then isolated from it. A dielectric slab of dielectric constant K is now introduced in the region between the plated, filling half of it. The electric intensity in the diaelectric is `E_(1)` and that in air is `E_(2)`.A. `E_(1)=E_(2)`B. `E_(1)=(E_(2))/(K)`C. `E_(1)=(1-(1)/(K))E_(2)`D. `E_(1)=(E_(2))/(K-1)` |
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Answer» Correct Answer - A |
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| 8220. |
Statement I: If the accelerating potential in an X-ray tube is increased, the minimum or cut-off wavelength of the X-rays produced remains unchanged. Of these statements :Statement II: lf the potential difference between electrodes in X-ray tube is increased, the intensity of the emitted radiations remains unchanged.(a) Both the statements are true and Statement II is the correct explanation of Statement I. (b) Both the statements are true, but Statement II is not the correct explanation of Statement I. (c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true. |
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Answer» (a) Both the statements are true and Statement II is the correct explanation of Statement I. |
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| 8221. |
Statement-1 If the accelerating potential in an X-ray tube is increased, the wavelength of the characterstic X-rays do not change. because Statement-2 When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - B |
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| 8222. |
Two vibrating strings of the same material but lengths `L` and `2L` have radii `2r` and `r` respectively. They are stretched under the same tension. Both the string vibrate in their fundamental nodes, the one of length `L` with freuqency `v_(1)` and the other with frequency `v_(2)`. the ratio `v_(1)//v_(2)` is given byA. 2B. 4C. 8D. 1 |
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Answer» Correct Answer - D `n_(1)=(1)/(2l) sqrt([(T)/(4 pi r^(2) rho)])` and `n_(2) =(1)/(4 l) sqrt([(T)/(pi r^(2) rho)])` `:. (n_(1))/(n_(2)) = 2xx (1)/(2) =1`. |
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| 8223. |
Two pulse in a stretched string whose centers are initially `8cm` apart are moving towards each other as shown in the figure. The speed of each pulse is `2cm//s`. After `2seconds`, the total energy of the pulse will be A. zeroB. purely kineticC. purely potentialD. partly kinetic and partly potential |
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Answer» Correct Answer - B |
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| 8224. |
When a block of iron in mercury at `0^@C,` fraction `K_1`of its volume is submerged, while at the temperature `60^@C,` a fraction `K_2` is seen to be submerged. If the coefficient of volume expansion of iron is `gamma_(Fe)` and that of mercury is `gamma_(Hg),` then the ratio `(K_1)//(K_2)` can be expressed asA. `(1+(60^(@)C)gamma_("Fe"))/(1+(60^(@)C)gamma_("HG"))`B. `(1-(60^(@)C)gamma_("Fe"))/(1+(60^(@)C)gamma_("HG"))`C. `(1+(60^(@)C)gamma_("Fe"))/(1-(60^(@)C)gamma_("HG"))`D. `(1+(60^(@)C)gamma_("HG"))/(1+(60^(@)C)gamma_("Fe"))` |
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Answer» Correct Answer - A |
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| 8225. |
Assertion: In the phenomenon of mutual induction, self- induction of each of the coil persists. Reason: Self-induction arises when strength of current in one coil changes. In mutual induction current is changing in both the individuals coils.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true statement but resaon is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - a The phenomenon of production of emf in a coil when the current in neighbouring coil changes is called mutual induction. The coil in which the current changes is called the primary coil while the neighbouring coil in which emf is induced is called secondry coil. In addition to induced emf which is produced by mutual induction, also induced emf is produced in each coil because of self-induction Hence, option (a) is correct. |
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| 8226. |
Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at `0^@C` and `90^@C`, respectively. The temperature of junction of the three rods will be (a) `45^@C` (b) `60^@C` (c) `30^@C` (d) ` 20^@C`. A. `45^(@)` CB. `60^(@)` CC. `30^(@)` CD. `20^(@)` C |
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Answer» Correct Answer - B |
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| 8227. |
What should come in place of the question mark (?) in the following number series? 3 3 12 108 ? 43200 (a) 2700 (b) 1728 (c) 972 (d) 432 (e) None of these |
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Answer» (b) The pattern of the series is as follows × 12, × 22, × 32, × 42, × 52 Hence, required number = 1728. |
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| 8228. |
A, B and C started a business by investing Rs. 13,750, Rs. 16,250 and Rs. 18,750, respectively. If B's share in the profit earned by them is Rs. 5,200, what is the total profit (in Rs.) earned by them together?1. 15,6002. 17,5003. 16,6004. 18,200 |
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Answer» Correct Answer - Option 1 : 15,600 Given: Investment of A = Rs. 13,750 Investment of B = Rs. 16,250 Investment of C = Rs. 18,750 B's share in profit = Rs. 5200 Calculation: Ratio of investment A ∶ B ∶ C = 13,750 ∶ 16,250 ∶ 18,750 ⇒ A ∶ B ∶ C = 11 ∶ 13 ∶ 15 Ratio of their profit will be same as ratio of their investment Let profit of A be 11x, B be 13x and C be 15x As, B's share = 5200 ⇒ 13x = 5200 ⇒ x = 5200/13 = 400 Total profit = 11x + 13x + 15x = 39x ⇒ Total profit = Rs. (39 × 400) ⇒ Rs. 15,600 ∴ Total profit is Rs. 15,600 |
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| 8229. |
What is the HCF of 54, 162 and 189?1. 272. 93. 34. 1 |
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Answer» Correct Answer - Option 1 : 27 Given: We have to find the HCF of 54, 162 and 189 Concept Used: Concept of HCF HCF (Highest Common Factor): HCF of x and y is the highest common factor that can divide x and y exactly For Example, 12 → Factors of 12 are 1, 2, 3, 4, 6 and 12 16 → Factors of 16 are 1, 2, 4, 8 and 16 Here common factors are 1, 2, and 4. But of these 4 is the largest/highest ∴ HCF of 12 and 16 is 4 Calculation: 54 → 2 × 3 × 3 × 3 162 → 2 × 3 × 3 × 3 × 3 189 → 3 × 3 × 3 × 7 Here, common factors amongst the above are 3 × 3 × 3 HCF of 54, 162 and 189 is 3 × 3 × 3 = 27 ∴ The required HCF is 27. |
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| 8230. |
In the following number series only one number is wrong. Find out the wrong number.48 72 108 162 243 366 (a) 72 (b) 108 (c) 162 (d) 243 (e) None of these |
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Answer» (e) The series is × 1.5 |
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| 8231. |
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 `In eAV_(d)` through both the rods is the same, i.e., `2(n) eAv_(L) = n e(2A)v_(R)` `v_(L)/v_(R) = 1 ` . |
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| 8232. |
Mohan covers a distance of 2.5 km by scooter at the rate of 30 km/h. The time taken by Mohan to cover he given distance in minute is:1. 82. 103. 54. 6 |
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Answer» Correct Answer - Option 3 : 5 Given: Speed of scooter = 30 km/h Distance covered = 2.5 km Formula used: Speed = Distance/Time Calculation: Time = Distance/Speed ⇒ Time = 2.5/30 ⇒ 1/12 hr Convert hour in minutes ⇒ 1/12 × 60 ⇒ 5 minutes ∴ The time is taken by Mohan to cover the given distance in a minute is 5 minutes. |
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| 8233. |
A sperical shell , made of material of electrical conductivity `10^9//pi(Omegam)^(-1)`, has thickness t = 2 mm and radius R = 10 cm. In an arrangement, its inside surface is kept at a lower potential than its outside surface. The resistance offered by the shell is equal to A. `5pi xx 10^(-12) Omega`B. `2.5 xx 10^(-11)Omega`C. `5 xx 10^(-12)Omega`D. `5 xx 10^(-11)Omega` |
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Answer» Correct Answer - D d. `R = 1/sigma xx t/(4piR^2)` Using values ` R = 5 xx 10^(-11) Omega` . |
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| 8234. |
The length of conductor is doubled and its radius is halved, its specific resistance is(a) Unchanged(b) Halved(c) Doubled(d) Quadrupled |
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Answer» Correct option(a) Explanation: Specific resistance doesn’t depend upon dimensions of the conductor. |
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| 8235. |
A fruit seller purchased 20 kgs of apples at Rs. 8 per kg. He found 5% of the apples to be rotten. If he sells the remaining apples at Rs. 11 per kg, what his gain percentage?1. 30.625%2. 40.625%3. 50.625%4. 60.625% |
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Answer» Correct Answer - Option 1 : 30.625% Given: A fruit seller purchased 20 kg apples. The cost price of apples = Rs. 8 per kg Wasted apple = 5% of 20 kg The selling price of remaining apples = Rs. 11 per kg Formula used: Profit% = (profit/C.P) × 100 where C.P = cost price Calculation: C.P of 20 kg of apples = 8 × 20 ⇒ 160 As 5% of apples were rotten, so good apples = [20 - 5% of 20] kg ⇒ [20 - (5/100) × 20] kg ⇒ (20 - 1) kg ⇒ 19 kg Total selling price of apples = Rs. 19 × 11 ⇒ Rs. 209 Profit = Rs. (209 - 160) ⇒ Rs. 49 Percentage profit = (49/160) × 100 ⇒ 30.625% ∴ The gain percentage of the seller is 30.625%. |
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| 8236. |
Prabhat cycled a distance of 240 kilometers at a certain speed. if he cycled 3km faster, he would have taken 4 hours fewer to reach the distance. what was the speed in km/h at which Prabhat actually cycled?1. 12 km/hr2. 10 km/hr3. 8 km/hr4. 14 km/hr |
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Answer» Correct Answer - Option 1 : 12 km/hr Given : Distance = 240 km Formula Used: Time = Distance ÷ Speed Calculations: Let the speed be x km/h If he cycled 3km faster every hour, the speed will be (x + 3) km/h Time needed when the speed is x km/h = 240/x hour Time needed when the speed is (x + 3) km/h = 240/(x + 3) hour According to Question, The difference in time taken is 4 hours 240/x - 240/(x + 3) = 4 ⇒ 240(x + 3) - 240x = 4x(x + 3) ⇒ 240x + 720 - 240x = 4x² + 12x ⇒ 4x² + 12x - 720 = 0 ⇒ x² + 3x - 180 = 0 ⇒ (x - 12)(x + 15) = 0 ⇒ x = 12 or x = -15 (rejected, since speed cannot be negative) ⇒ Speed = x = 12 km/h ∴ Prabhat is cycling at 12 km/h |
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| 8237. |
A person covers the first 176 kilometers at a speed of 16 kilometers/hour and the next 64 kilometers at a speed of 32 kilometers/hour. What would be the approximate average speed for the first 240 kilometers of the journey?A. 13 Km/hB. 27 Km/hC. 18.5 Km/hD. 21 Km/h1. B2. D3. A4. C |
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Answer» Correct Answer - Option 4 : C Given: Total distance = 240 km Speed for 176 km = 16 km/hour Speed for 64 km = 32 km/hour Formula used: Speed = Distance/Time Average speed = Total distance/total time Calculation: According to question Time required to cover 176 km, Time = Distance/speed ⇒ 176/16 = 11 hour Time required to cover 64 km, Time = Distance/speed ⇒ 64/32 = 2 hour Total time required to cover 240 km = 11 + 2 = 13 hour Average speed = Total distance/total time ⇒ 240/13 ⇒ 18.46 km/hour ∴ The approximate average speed for the first 240 kilometers is 18.5 km/hour |
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| 8238. |
Kishore covers 2/5 of the total distance by car, 8/15 by cycle and the remaining distance by walking, at 6 kilometers per hour, is covered in 40 minutes. The total distance traveled by him is …….1. 60 kilometers2. 67.5 kilometers3. 65 kilometers4. 69.5 kilometers |
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Answer» Correct Answer - Option 1 : 60 kilometers Given: Kishore covers 2/5 of the total distance by car He covers 8/15 of the total distance by cycle He covers the remaining distance by walking at 6 km/h in 40 minutes Formula Used: Distance = Speed × Time Calculation: Let the total distance covered by Kishore be D km Remaining distance covered by walking = D – 2D/5 – 8D/15 ⇒ (15D – 6D – 8D)/15 ⇒ D/15 Kishore covers the remaining distance by walking at 6 km/h in 40 minutes 40 minutes = 40/60 hours ⇒ D/15 = 6 × (40/60) ⇒ D = 15 × 4 ⇒ D = 60 Total distance covered by Kishore = D km ⇒ 60 km ∴ The total distance covered by Kishore is 60 kilometers |
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| 8239. |
Concerning detection of micro bubbles and their applications: a. The main use for micro bubbles is to enhance Doppler ultrasound imaging b. As micro bubbles are quickly destroyed by insonation, they can only be imaged intermittently c. Stimulated acoustic emission (SAE) utilizes high-energy Doppler pulses d. Characterization of liver lesions with micro bubbles is usually based on their enhancement pattern e. Micro bubbles can be used for targeted drug delivery |
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Answer» a. False. Due to their high degree of echogenicity, micro bubbles indeed enhance the Doppler signal (‘Doppler rescue’), which was the original application of these agents, and sometimes continues to be used in transcranial imaging. However, the main application is currently in contrast-specific imaging techniques utilizing non-linear resonance and production of harmonic frequencies. b. False. Real-time imaging with a low mechanical index (MI) is possible. c. True. SAE is the appearance of a random colour pattern when micro bubbles burst after insonation with high-energy Doppler pulses. According to the guidelines of the European Federation of Societies for Ultrasound in Medicine and Biology, high MI techniques are not routinely recommended due to a difficult examination technique and the superiority of real-time, low MI imaging. d. True. Scanning with a low MI allows characterization of the lesion enhancement and washout in the arterial and portal phases in real time, and has almost completely superseded high MI techniques (SAE). e. True. They can be laden with drugs and caused to burst by insonation at a specific target site. |
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| 8240. |
The speed of a boat is 10 km/hr and the speed of current is 2 km/hr. What is the distance covered by the boat in downstream in 2.5 hours?1. 30 km2. 20 km3. 24 km4. 36 km |
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Answer» Correct Answer - Option 1 : 30 km Given, Speed of a boat, b = 10 km/hr Speed of current, s = 2 km/hr Time = 2.5 hours (in downstream motion) Formula Used: Downstream speed = (b + s) Distance covered = Downstream speed × time Calculations: Let D be Downstream speed Downstream speed, D = 10 + 2 = 12 km/hr Distance covered = 12 × 2.5 Distance covered = 30 Km ∴ The distance covered by the boat in downstream in 2.5 hours is 30 Km |
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| 8241. |
Train A, with a length of 300 meters, is going in the south direction at a speed of 45 kilometers per hour. Train B, with a length of 240 meters, is going in the same direction at a speed of 60 kilometers per hour, 120 meters behind the rear end of train A. How long will Train B take to cross a person sitting exactly in the middle of Train A?1. 110 second2. 122.4 second3. 120.3 second4. 100.5 second |
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Answer» Correct Answer - Option 2 : 122.4 second Given: For Train A, Length = 300 meters Speed = 45 kilometers per hour. For train B, Length = 240 meters Speed = 60 kilometers per hour. Train B is 120 meters behind Train A. Formula Used: meter/second = (5/18) kilometer/hour Relative speed: When running in the same direction, Speed = (x – y) where x is the speed of the second train. y is the speed of the first train. Speed = Distance/Time Time = Distance/Speed Calculation: Man is sitting exactly in the middle of train A Distance that needs to be covered by B is, ⇒ Length of Train B + Distance between 2 trains + Distance of a sitting man ⇒ 240 + 120 + 150 ⇒ 510 meters Relative speed: ⇒ (60 – 45) ⇒ 15 km/hr. As the distance is in meters and Time in seconds We need to convert km/hr into m/sec ⇒ 5/18 × 15 ⇒ 25/6 meter/second Time = Distance/Speed ⇒ Time = 510/(25/6) ⇒ Time = (510 × 6)/25 ⇒ Time = 122.4 second ∴ Time taken by Train B to cross a man is 122.4 seconds. |
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| 8242. |
Regarding relaxation times and the signal of different tissues: a. Cortical bone has long T1 and T2 relaxation times b. Water is hyperintense on both T1- and T2-weighted sequences c. Grey matter has a lower signal than white matter on T1 sequences d. Melanoma metastases are usually of low T2 signal e. Hyperacute intracerebral haemorrhage is bright on both T1 and T2 images |
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Answer» a. False. Cortical bone is essentially a solid, despite a water content of 10–15%. It has an extremely short T2 of <1 ms; therefore, it appears black on conventional T1- and T2-weighted sequences. b. False. Water has a long T1 and T2; therefore, it appears hypointense on T1 and hyperintense on T2 sequences. This makes T2 sequences sensitive for pathology, as most pathological processes (e.g. oedema, tumour, infarction) are associated with increased tissue water content. c. True. Grey matter has slightly longer T1 and T2 relaxation times than white matter, and so is darker on T1 and brighter on T2 sequences. The way to remember this is that grey matter is ‘grey’ on the sequence that better shows anatomy (T1-weighted). d. True. There are only a few causes of a low T2 signal; these include melanin, calcification, fibrous tissue, high protein content and flow void. e. False. The appearance of haemorrhage on T1- and T2-weighted imaging changes with time due to the temporal degradation of blood products according to the table on p. 135. Hyperacute blood is isointense or dark on T1 and bright on T2 sequence. |
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| 8243. |
Concerning indirect conversion digital radiography: a. It typically uses a fixed-size detector b. It has a digital quantum efficiency (DQE) greater than that with computed radiography image plates c. It typically has a radiographic speed of less than 400 d. It differs from direct conversion detectors due to the use of an X-ray scintillator e. It has a resolution of approximately 3–4 line pairs mm-1 |
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Answer» a. True. b. True. The DQE is 0.6 versus 0.4. c. False. The high speed relates to the amplification of the signal by the scintillator and the conversion gain by the photo diode. d. True. e. True. |
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| 8244. |
Two trains running in opposite directions cross a pole in 30 seconds and 20 seconds respectively. They cross each other in 25 seconds. The ratio of their speed is …….1. 1 ∶ 12. 1 ∶ 73. 1 ∶ 54. 1 ∶ 2 |
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Answer» Correct Answer - Option 1 : 1 ∶ 1 Formula Used: Distance = Speed × Time Let the speed of the two trains be 'x' m/sec and 'y' m/ sec Then, Length of the first train = 30x meters And length of the second train = 20y meters Calculation: According to question, ⇒ {(30x + 20y)/x + Y} = 25 seconds ⇒ 30x + 20y = 25x + 25y ⇒ 5x = 5y ⇒ x/y = 1/1 ∴ The ratio of speed = 1 : 1 |
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| 8245. |
The price of 1 kg apple in Srinagar is reduced by Rs. 0.6 per Kg which allows a man to purchase 1.5 Kg more apple for Rs. 55. Find the original price of apple per Kg? 1. Rs. 112. Rs. 53. Rs. 64. Rs. 4 |
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Answer» Correct Answer - Option 2 : Rs. 5 Given: Decrease in price = Rs. 0.6/Kg Final purchase - Initial purchase = 1.5 Kg Total amount of apples bought = Rs. 55 Concept: Number of Kg apple bought = (Total money)/(price of 1 kg apple) Calculation: Let, the initial price of apple = Rs. a/Kg Price after reduction = Rs.(a - 0.6)/Kg [55/(a - 0.6)] - (55/a) = 1.5 ⇒ (55a - 55a + 33)/(a2 - 0.6a) = 1.5 ⇒ 33 = 1.5a2 - 0.9a ⇒ 22 = a2 - 0.6a By Values putting from the option; We get a = 5 ∴ The original price of apple per Kg is Rs. 5.
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| 8246. |
Concerning image contrast in spin–echo imaging: a. The best contrast between tissues with different T1s is obtained with short repetition times (TRs) and short echo times (TEs) b. T2 weighting is obtained with short TEs and long TRs c. An image obtained with a TE of 15 ms and a TR of 2000 ms would be proton density (PD) weighted d. Fat appears with a high signal on T1-, T2- and PD-weighted sequences e. The fat signal is partially suppressed on fast (turbo) spin–echo imaging |
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Answer» a. True. T1 weighting is obtained with a short TR, usually between 300 and 800 ms (close to the T1s of the tissues being imaged). With increasing TR, the longitudinal magnetization in all tissues would recover more, diminishing the T1 contrast. Short TEs are used to reduce the effect of T2 relaxation on contrast (see below). b. False. Relatively long TEs (90–140 ms, close to the T2s of the tissues being imaged) are used so measurable differences in the degree of transverse magnetization can develop. With short TEs, these differences are small; therefore, short TEs reduce the effect of T2 weighing. Long TRs are used to allow near-full recovery of longitudinal magnetization and eliminate differences in contrast due to T1. c. True. As described above, a short TE and long TR reduce the effect of T2 and T1, respectively. The remaining contrast is only dependent on the density of hydrogen nuclei (PD weighted). d. True. Due to a short T1 and a higher PD in comparison with most tissues, fat is bright on these sequences. In T2-weighted images, fat is less bright than water, but still brighter than most other tissues. e. False. On fast spin–echo images, fat appears hyperintense. This results from the multiple refocusing pulses, which suppress a phenomenon known as J-coupling (an interaction between different nuclei within the fat molecules), resulting in fat having a longer T2 and therefore appearing brighter. |
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| 8247. |
Concerning intensifying screens: a. They typically measure 1 cm in thickness b. They are usually made of calcium tung state phosphorc. They convert the pattern of X-ray intensities into one of light, which is detected by the film d. Films and screens should be matched e. A phosphor with a K-edge value similar to the mean photon energy should be selected |
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Answer» a. False. The thickness is much less, usually 1–2 mm. b. False. They are usually made of the rare earth metals such as gadolinium oxysulphide or lanthanum oxy bromide. c. True. To produce an image on film without a screen would require high levels of radiation due to its relative insensitivity. Screens are used to reduce the dose required to produce an adequate image. d. True. The colour of the emitted light will vary depending on the phosphor used within the screen. The film should therefore be sensitive to the emitted light to ensure that the image is obtained. For example, gadolinium oxysulphide emits green light, whereas lanthanum oxy bromide emits blue light. e. True. This ensures that there is increased attenuation of the radiation spectrum, whose intensity may then be converted into one of light and thus represented as an image on the radio graphic film. |
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| 8248. |
A shopkeeper allows an discount on cashless payments. The marked price of an article is Rs.5000. He sold two article one with discount and other with without discount. The difference between the selling price of an article is Rs.500. Find the cost price of an article if total profit percentage he had is 18.75%.1. Rs.55002. Rs.45003. Rs.48004. Rs.40005. Rs.3500 |
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Answer» Correct Answer - Option 4 : Rs.4000 Given: Let discount offered be a%. ⇒ selling price of an article after discount = 5000 × (100 - a)/100 = Rs.(5000 - 50a) ⇒ 5000 - 5000 + 50a = 500 ⇒ a = 10% Total selling price of an article = 5000 + 5000 × 90/100 = Rs.9500 Let cost price of an article be Rs.m. ⇒ 18.75 = (9500 - 2m)/2m × 100 ⇒ m = 4000 ∴ The cost price of an article is Rs.4000. |
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| 8249. |
Concerning quality assurance in film-screen radiography: a. The testing of filtration within the system is not routinely performed b. Total filtration should be equivalent to at least 1.5 mm of aluminium c. Assessment of optical density is the main measure of correct automatic exposure control (AEC) functioning d. Focal spot size may be estimated using the pinhole method e. A lower-than-expected output dose measurement is unrelated to exposure time |
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Answer» a. False. The amount of filtration is most important when the diagnostic information depends on fine contrast being demonstrated between the tissues being imaged (e.g. in mammography). This contrast may be lost if there is beam hardening if filtration is too great. b. False. It should be equivalent to 2.5 mm of aluminium. c. True. The AEC functions by terminating the exposure once a film has a received dose sufficient to produce an adequate film. Therefore, the optical density of test films can be monitored as the level at which the AEC is set and then varied while monitoring the dose. Test subject thickness (usually Perspex) may also be altered at the same time. d. True. This technique produces a magnified image of the focal spot. The focal spot size can then be calculated by determining the magnification by knowing the focus-to-film and pinhole-to-film distances. e. False. Along with changing the tube potential and tube current, a shorter exposure time may also reduce the output. |
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| 8250. |
Concerning quality assurance in film-screen radiography: a. It is insufficient to test only the film speed and contrast index b. A gradual change in the speed of the film is more likely to be related to problems during the processing of film than to problems with the film batch c. Screens should be changed routinely at yearly intervalsd. The tube potential may be tested by assessing the kV of the X-ray beam e. X-ray output should increase with increasing kV |
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Answer» a. True. The base plus fog level and maximum density (Dmax) also need to be tested. These may be affected by long-term storage of the film at incorrect temperatures. b. True. c. False. There is no set time as to when they need to be changed. When the sensitivity of a screen has reduced by 20%, then generally it should be changed. This may be tested by producing test films at set kV and mAs values and comparing the optical density produced. d. True. Although the tube potential is not usually measured directly, an electronic meter may be used to demonstrate the kV and its variation with the exposure time using an oscilloscope. e. True. |
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