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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.
| 251. |
A particle of mass `m` is moving along the side of a square of side `a`, with a uniform speed v in `XY` plane as shown in Fig. Which of the followig statements is false for the angular momentum `overset rarr(L)` about the origin? A. `vecL=mv[(R)/(sqrt(2))-a]hatk` when the particle is moving from C to D.B. `vecL=mv[(R)/(sqrt(2))+a]hatk` when the particle is moving from B to C.C. `vecL=(mv)/(sqrt(2)) R vec k` when the particle is moving from D to A.D. `vecL=-(mv)/(sqrt(2)) R vec k` when the particle is moving from A to B. |
| Answer» Correct Answer - A::C | |
| 252. |
A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance `(2A)/3` from equilibrium position. The new amplitude of the motion is:A. 3AB. `A sqrt(3)`C. `(7A)/(3)`D. `(A)/(3) sqrt(41)` |
| Answer» Correct Answer - C | |
| 253. |
When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is `-V_(0)//2`. When the surface is illuminated by monochromatic light of frequency `v//2`, the stopping potential is `-V_(0)`. The threshold frequency gor photoelectric emission is :A. `(5v)/(3)`B. `(4)/(3)v`C. `2v`D. `(3v)/(2)` |
| Answer» Correct Answer - A | |
| 254. |
Formation of real image using a biconves lens is shown below: If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the secreen ?A. Image disappearsB. magnified imageC. Erect real imageD. no change |
| Answer» Correct Answer - A | |
| 255. |
Formation of real image using a biconves lens is shown below: If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the secreen ?A. Image disappearsB. magnified imageC. Erect real imageD. no change |
| Answer» Correct Answer - A | |
| 256. |
The value of `cos^-1(cos12)-sin^-1(sin14)` is.(1) `-2` (2) `8pi-26` (3) `4pi+2` (4) None of these |
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Answer» `cot^(-1)(cos12)-sin^(-1)(sin14)` `cos^(-1)(cos(4pi-12))-sin^(-1)(sin(4pi))` `(4pi-12)-(14-4pi)` `8pi-26` |
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| 257. |
Two deuterons udnergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)A. 23.6 MeVB. 25.8 MeVC. 30.2 MeVD. 32.4 MeV |
| Answer» Correct Answer - A | |
| 258. |
Which of the following pairs represents linkage isomers?A. `[Cu(NH_(3))_(4)] [Pt Cl_(4)] and [Pt (NH_(3))_(4)] [CuCl_(4)]`B. `[Pb(P " "Ph_(3))_(2) (NCS)_(2)] and [Pb (P Ph_(3))_(2) (SCN)_(2)]`C. `[Co(NH_(3))_(5) NO_(3)] SO_(4) and [CO (NH_(3))_(5) SO_(4)] NO_(3)`D. `[Pt Cl_(2) (NH_(3))_(4] Br_(2) and [Pt Br_(2) (NH_(3))_(4)] Cl_(2)` |
| Answer» Correct Answer - B | |
| 259. |
Which of the following does not have S-S linkage?A. `H_(2)S_(2)O_(5)`B. `H_(2)S_(2)O_(7)`C. `H_(2)S_(2)O_(3)`D. `H_(2)S_(2)O_(6)` |
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Answer» Correct Answer - B |
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| 260. |
Solution of `(2x - 10y^3)(dy)/(dx) + y = 0` is: |
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Answer» `(2x-10y^3)dy/dx + y = 0` `=>(2x-10y^3)dy/dx =- y` `=>(2x)/y - 10y^2 = -dx/dy` `=>dx/dy +(2x)/y = 10y^2` Comparing the given equation with first order differential equation, `dx/dy+Px = Q(y)`, we get,`P = 2/y and Q(y) = 10y^2` So, Integrating factor `(I.F) = e^(int 2/y) dy` `I.F.= e^(2logy) = e^(logy^2) = y^2` We know, solution of differential equation, `y(I.F.) = intQ(I.F.)dy` `:.`Our solution will be, `x(y^2) = int 10y^2(y^2)dy` `=>xy^2 = 10y^5/5+c` `=>xy^2 = 2y^5+c` |
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| 261. |
The value of the infinite product `6^(1/2)xx6^(2/4)xx6^(3/8)xx6^(4/16)xx...` |
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Answer» `6^(1/2)xx6^(2/4)xx6^(3/8)xx6^(4/16)xx...` `=6^(1/2+2/4+3/8+4/16)` Now, we will solve `(1/2+2/4+3/8+4/16+...)`. Let `S = 1/2+2/4+3/8+4/16+...` `S = 1/2+2/2^2+3/2^3+...->(1)` `S/2 = 1/2^2+2/2^3+3/2^4+...->(2)` Subtracting `(1) - (2)`, `S/2 = 1/2+1/2^2+1/2^3+...` `=>S/2 = (1/2)/(1-1/2) = 1` `=> S = 2` `:. 6^(1/2)xx6^(2/4)xx6^(3/8)xx6^(4/16)xx... = 6^S = 6^2 = 36` |
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| 262. |
If one root of the quadratic equation `x^(2)+px+q=0` is `2-sqrt(3)` : where `P,QsubQ`.Then which of the following is true ?A. `p^(2)-4q+12=0`B. p^(2)-4q-12=0`C. `q^(2)-4q+12=0`D. `q^(2)-4q-12=0` |
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Answer» Correct Answer - B |
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| 263. |
Number of ways in which 9 different toys be distributed among 4 children belonging to different age group in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is: |
| Answer» Number of ways=` 2C_2*(4C_2)*(6C_2)*(9C_2)`=`1*((4!)/(2!*2!))*((6!)/(4!*4!))*((9!)/(7!*2!))`=` ((9!)/((2!)^3*(3!)))` | |
| 264. |
In the expansion at `((2)/(x)+x^(log_(e)x))^(6)` if `T_(4)=20xx8^(7)` then value of x isA. `8^((1)/(2))`B. `8^(2)`C. `8^(3)`D. `8^(4)` |
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Answer» Correct Answer - B |
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| 265. |
The fix point through which the line `x(a+2b)+y(a+3b)=a+b` always passes for all values of a and b is: |
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Answer» `x(a+2b)+y(a+3b) = a+b` Here, both positive values of `x` and `y` can not be the answer as it can not make both sides equal. So, we can reject option `(1) and (2)`. We check for third option that is `x = 2, y = -1` Then, `L.H.S. = x(a+2b)+y(a+3b) = 2(a+2b)+(-1)(a+3b)` `=2a+4b - a-3b = a+b = R.H.S.` So, option `(3)` is the correct option. Option`(4)` also can not be the answer as `y = -2`, will make left side negative. So, option `(3)` is the only correct option. |
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| 266. |
if `intx^(5)e^(-x^(2))+C` then the value of `g(-1)` isA. `(3)/(2)`B. `(1)/(2)`C. `-(5)/(2)`D. `(e)/(2)` |
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Answer» Correct Answer - C |
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| 267. |
What is the greatest value of the positive integer n satisfying the condition `1+1/2+1/4+1/8+.....+1/2^[n-1] |
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Answer» `=1+1/2+1/4+1/8...+1/2^(n-1)` there are total n terms where a=1,r=1/2. `=(a(1-r^n))/(1-r)` Putting the values of a and r `=(1(1-(1/2)^n))/(1-(1/2))` `=2-1/2^(n-1)` `2-1/2^(n-1)<2-1/1000` `1/2^(n-1)>1/1000` `2^(n-1)<1000` `2^(n-1)=2^9` `n-1=9` `n=10`. |
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| 268. |
The solution of differential equation `dy/dx=(4x+6y+5)/(3y+2x+4)` is |
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Answer» `dy/dx = (2(2x+3y)+5)/(3y+2x+4)` let `2x+3y= t` `2 + 3 dy/dx = dt/dx` `dy/dx= 1/3(dt/dx- 2)` `1/3 dy/dx -2/3 = (2t+5)/(t+4)` `1/3dt/dx = (2t +5)/(t+4) +2/3 => (6t + 15 + 2t + 8)/(3(t+4))` `1/3 dt/dx = (8t +23)/(3(t+4))` `int ((t+4)/(8t +23)) dt = int dx` `1/8 int((8t +32)/(8t+32)) dt = x+ c` `= 1/8 int((8t + 23 +9)/(8t +23))dt= x+ c` `= 1/8 int dt + int 9/(8t+23)dt= x+ c ` `1/8[ t] + 9/8 int (du)/u ` `1/8[2x+3y + 9/8ln(8t+23)]= x+c` `` |
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| 269. |
Find area bounded by the curves `x^(2)leylex+2`A. `(11)/(2)`B. `(7)/(2)`C. `(9)/(2)`D. `(5)/(2)` |
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Answer» Correct Answer - C |
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| 270. |
Let `f(x)=15-|x-10|` and `g(x)=f(f(x))` then `g(x)` is non differentiable atA. `{5,10,15}`B. `{5,10,15,20}`C. `{10}`D. `{5,15}` |
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Answer» Correct Answer - A |
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| 271. |
let `y(x)` satisfying the differential equation `x(dy)/(dx)+2y=x^(2)` given `y(1)=1` then `y(x)=`A. `(x^(2))/(4)-(3)(4x^(2))`B. `(x^(3))/(4)+(3)/(4x^(2))`C. `(x^(2))/(4)+(3)/(4x)`D. `(x^(2))/(4)+(3)/(4x^(2))` |
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Answer» Correct Answer - D |
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| 272. |
The solution differential equation `(dy)/(dx)+ytanx=2x+x^(2)tanx` isA. `y=x^(2)+c cosx`B. `y=2x^(2)-c cosx`C. `y+x^(2)=c cosx`D. `y+2x^(2)=c cosx` |
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Answer» Correct Answer - A |
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| 273. |
The solution of the differential equation `dy/dx=sin(x+y)+cos(x+y)` is: |
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Answer» `dy/dx=sin(x+y)+cos(x+y)` Let x+y=t `dt/dx=1+sint+cost` `intdt/(1+sint+cost)=intdx` `intdt/(1+(2tant/2)/(1+tan^2t/2)+(1-tan^2t/2)/(1+tan^2t/2))=x` `int(sec^2t/2dt)/(2(tant/2+1))=x` `1+tant/2=h` `1/2sec^2t/2dt=dh` `int(dh)/h=x` `ln|h|+c=x` `ln(1+tanh/2)+c=0` `ln(1+tan((x+y)/2)+c=x` `ln|tan((x+y)/2)+1|=x+c`. |
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| 274. |
What is the order of the differential equation `dx/dy+ int y dx=x^3`? |
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Answer» `dx/dy + int ydx = x^3` As we do not know about `y` ,so we can not integrate it. So, we can not determine the order of the given differential equation with `dx/dy`. Now, we will try to reverse the given equation in the form of `dy/dx`. `int ydx = x^3-dx/dy` For this, we have to find `d/dx(dx/dy)` which is again not possible. Hence, we can not determine the order of the given differential equation. |
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| 275. |
A triangle is formed by the lines whose combined equation is given by `(x-y-4)(xy-2x-y+2)=0` The equation of its circumcircle is: |
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Answer» `(x-y-4)(xy-2x-y+2)=0` `x-y=4` `xy-2x-y+2=0` `x(y-2)-1(y-2)=0` `(x-1)(y-2)=0` AB is perpendicular to BC AC is diameter of the circle. `(x-1)(x-2)+(y-3)(y-2)=0` `x^2-3x+2+y^2-5y+6=0` `x^2+y^2-3x-5y+8=0`. |
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| 276. |
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity `A=(P^(3)Q^(2))/(sqrtRS)`. The maximum percentage error in the value of A will be :A. 0.085B. 0.06C. 0.065D. 0.075 |
| Answer» Correct Answer - C | |
| 277. |
An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1 g of chlorohydrocarbon are : (Atomic wt. of Cl = 35.5 u, Avogardo constant `=6.023xx10^(23) "mol"^(-1)`)A. `6.023xx10^(23)`B. `6.023xx10^(9)`C. `6.023xx10^(20)`D. `6.023xx10^(21)` |
| Answer» Correct Answer - C | |
| 278. |
Among the oxides of nitrogen `N_(2)IO_(3), N_(2)O_(4) and N_(2)O_(5)` , the molecule(s) having nitrogen-nitrogen bond is/are :A. `N_(2)O_(3) andN_(2)O_(4)`B. `N_(2)O_(4) andN_(2)O_(5)`C. Only`N_(2)O_(5)`D. `N_(2)O_(3) andN_(2)O_(5)` |
| Answer» Correct Answer - A | |
| 279. |
The molecule having smallest bond angle isA. `NCI_(3)`B. `AsCI_(3)`C. `SbCI_(3)`D. PCI_(3)` |
| Answer» Correct Answer - C | |
| 280. |
Prove `tan^(-1)(1/(p+q))+tan^(-1)(q/(p^2+pq+1))=cot^(-1)p` |
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Answer» LHS `tan^(-1)(1/(p+q))+tan^(-1)(q/(p^2+pq+1))` `tan^(-1)(((1/(p+q))+(q/(p^2+pq+1)))/(1-(1/(p+q))*(q/(p^2+pq+1))))` `tan^(-1)(((p+q)^2+1)/(p((p+q)^2+1)))` `tan^(-1)(1/p)` `cot^(-1)(p)` RHS. |
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| 281. |
Prove: `sin^(-1)(1/sqrt5)+cot^(-1)3=pi/4` |
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Answer» Let `cot^-1 3 = theta`. Then, `cot theta = 3` `=>costheta/sintheta = 3 =>cos^2theta/sin^2theta = 9` `=>(1-sin^2theta)/sin^2theta = 9 =>sintheta = 1/sqrt10` `=> theta = sin^-1(1/sqrt10)` `:. L.H.S. = sin^-1(1/sqrt5) +sin^-1(1/sqrt10)` We know, `sin^-1x+sin^-1y = sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))` So, our expression becomes, `= sin^-1(1/sqrt5sqrt(1-1/10)+sqrt(1-1/5)1/sqrt10)` `=sin^-1(3/sqrt50+2/sqrt50)` `=sin^-1(5/(5sqrt2))` `=sin^-1(1/sqrt2)` `=pi/4 = R.H.S.` |
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| 282. |
The solution of the differential equation `dy/dx=y tanx-2 sinx` is |
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Answer» `dy/dx = ytanx-2sinx` `=>dy/dx - ytanx = -2sinx` Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = -tanx and Q(x) = -2sinx` So, Integrating factor `(I.F) = e^(int -tandx)` `I.F.= e^(-ln|secx|) = 1/secx = cosx` we know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ycosx = int cosx(-2sinx)dx` `=>ycosx = int -sin2xdx` `=>ycosx = cos(2x)/2+c` `=>y = cos(2x)/2cosx + csecx` `=>y = (2cos^2x-1)/2cosx + csecx` `=>y = cosx - secx/2+csecx` `=>y = cosx + secx(c-1/2)` `=>y = cosx + Csecx`, where `C = (c-1/2)` is a constant. |
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| 283. |
A point P moves inside a triangle formed by `A(0,0), B(1,sqrt3), C(2,0)` such that `min{PA,PB,PC}=1`, then the area bounded by the curves traced by P is: |
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Answer» AB=2,BC=2,AC=2 Area`=sqrt3/4(2)^2-3pi*60/360` `=sqrt3-pi/2` Option 2 is correct. |
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| 284. |
`sqrt(-sqrt3+sqrt(3+8sqrt(7+4sqrt3)))=` |
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Answer» `sqrt(-sqrt3+sqrt(3+8sqrt(7+4sqrt3` `sqrt(-sqrt3+sqrt(3+8sqrt(4+4sqrt3+3` `sqrt(-sqrt3+sqrt(3+8sqrt((2+sqrt3)^2` `sqrt(-sqrt3+sqrt(3+8(2+sqrt3)` `sqrt(-sqrt3+sqrt(3+16+8sqrt3` `sqrt(-sqrt3+sqrt((4+sqrt3)^2` `sqrt(-sqrt3+4+sqrt3` `sqrt4` `2` |
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| 285. |
The sum of the first ten terms of the geometric progression is `S_1` and the sum of the next ten terms (11th term to 20th term) is `S_2` then the common ratio will be: |
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Answer» `S_1=(a(r^10-11))/(r-1)-(1)` r-common ratio a= first number `S_2=(a(r^n-1))/(r-1)-(a(r^10-1))/(r-1)` `S_2=(a(r^20-1))/(r-1)-S_1` `S_1+S_2=(a(r^20-1))/(r-1)-(2)` `(r-1)=(a(r^10-1))/S_1` `S_1+S_2=(a(r^10-1)(r^10+1))/(a(r^10-1))*S_1` `r^10+1=1+S_2/S_1` `r=10sqrt(S_2/S_1`. |
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| 286. |
The rational number, which equals the number `2.bar 357` with recurring decimal is: |
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Answer» `x=2.overline357` `1000x=2357.overline357` `1000x-x=2355` `999x=2355` `x=2355/999`. |
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| 287. |
If A.M. between positive numbers p and `q(p>=2)` is two times the GM.then `p:q` is |
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Answer» `AM=(p+q)/2` `GM=sqrt(pq)` `(p+q)/2=2sqrt(pq)` `p+q=4sqrt(pq)` squaring both side `p^2+q^2+2pq=16pq` `p^2+q^2-14pq=0` `p=(14qpmsqrt(1969^2-4q^2))/2` `p=(142pmsqrt(192q^2))/2` `p/2=7pm4sqrt3` `p/q=(2+sqrt3)^2` `((2+sqrt3)(2+sqrt3))/((2-sqrt3)(2+sqrt3))` `p/q=(2+sqrt3)/(2-sqrt3)`. |
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| 288. |
In the exteraction of copper from its sulphide ore, the metal is fanally obtained by the reduction of caprous oxide withA. `SO_(2)`B. COC. `Fe_(2)O_(3)`D. `Cu_(2)O` |
| Answer» Correct Answer - D | |
| 289. |
The following polar compound among the following is :A. B. C. D. |
| Answer» Correct Answer - C | |
| 290. |
A capacitor of capacitance `5muF` is charged with `5muC` charge its capacitance is changed to `2muF` by some external agent. The work done by external agent isA. `40.5xx10^(-7)J`B. `42.5xx10^(-7)J`C. `37.5xx10^(-7)J`D. `30.5xx10^(-7)J` |
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Answer» Correct Answer - C |
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| 291. |
A working transitor with its three legs marked `P, Q and R` is tested using a multimeter No conduction is found between `P, Q `by connecting the common (negative) terminal of the multimeter to `R`and the other (positive) terminal to or `Q` some resistance is seen on the multimeter . Which of the following is true for the transistor ?A. It is an npn transistor with R as baseB. It is a pnp transistor with R as collectorC. It is a pnp transistor with R as emitterD. It is an npn transistor with R as collector |
| Answer» Correct Answer - B | |
| 292. |
This question contains Statement - 1 and Statement -2 Of the four choice given after the Statements , choose the one that best decribes the two Statements Statement- 1: Energy is released when heavy undergo fission or light nuclei undergo fusion and Statement- 2: for nuclei , Binding energy nucleon increases with increasing `Z` while for light nuclei it decreases with increasing `Z`A. Statement-1 is false, Statement-2 is trueB. Statement-1 is true, Statement-2 is true, Statement -2 is correct explanation for Statement-1C. Statement-1 is true, Statement-2 is true, Statement -2 is not a correct explanation for Statement-1D. Statement-1 is true, Statement-2 is false |
| Answer» Correct Answer - D | |
| 293. |
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between in limits 589.0V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of `60^(@)` with the direction of the field ?A. 589.4VB. 589.5VC. 289.2VD. 289.6V |
| Answer» Correct Answer - A | |
| 294. |
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between in limits 589.0V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of `60^(@)` with the direction of the field ?A. 589.4VB. 589.5 VC. 589.2 VD. 589.6 V |
| Answer» Correct Answer - A | |
| 295. |
5 g of `Na_(2)SO_(4)` was dissolved in x g of `H_(2)O` . The change in freezing point was found to be `3.82^(@)C` . If `Na_(2)SO_(4)` is `81.5%` ionised , the value of x (`k_(f)` for water =`1.86^(@)C` kg `"mol"^(-1)`) is apporximately : (molar mass of S=32 g `"mol"^(-1)` and that of Na=23 g `"mol"^(-1)`)A. 25 gB. 65 gC. 15 gD. 45 g |
| Answer» Correct Answer - D | |
| 296. |
Consider the following standard electrode potentials (`E^(@)` in volts) in aqueous solution: `{:(underline("Element"), M^(3+)//M, M^(+)//M), ("Al", -1.66, + 0.55), (TI, +1.26, -0.34):}` Based on these data, which of the following statements is correct ?A. `Tl^(3+)` is more stable than `Al^(3+)`B. `Al^(+)` is more stable then `Al^(3+)`C. `Tl^(3+)` is more stable than `Al^(3+)`D. `Tl^(3+)` is more stable than `Al^(3+)` |
| Answer» Correct Answer - D | |
| 297. |
The major product expected from the following reaction is : A. B. C. D. |
| Answer» Correct Answer - C | |
| 298. |
The major product expected from the following reaction is : A. B. C. D. |
| Answer» Correct Answer - C | |
| 299. |
If `A+B=90` then prove that `sqrt((tanAtanB+tanAcotB)/(sinAsecB))=secA` |
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Answer» `tan(90-A)=cotA` `tan B= cotA` `sinB=cosA` `cosecB=secA` `tanA=cotB` `sinA=cosB` `coseccA=secB` LHS `=sqrt((tanAcotA+tanAtanA)/(sinAcosecA))` `=sqrt(1+tan^2A)` `=sqrt(sec^2A)` `=secA` |
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| 300. |
What approximate value will come in place of question mark (?) in the given question? (you are not expected to calculate exact value):`105.27% of 1200.11+11.80% of 2360.85=21.99%` of `?` `+1420.99` |
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Answer» `105% of 1200+12% of 2360` `(105*1200)/100+(12*2360)/100` `1260+283.2` `1543.2` `(x*22)/100+1421=1543.2` `x*11/50=122.2` x=555.45. |
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