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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.
| 101. |
On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is __________. (Given data: Molar mass & molar freezing point depression is 78 g `mol^(-1) & 5.12 K kg mol^(-1)`] |
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Answer» Correct Answer - 1.02 `(P^(0)-P_(S))/(P_(S))=i[(n_("Solute"))/(n_("solvent"))]` `(650-640)/640=1xx(0.5xx78)/(Mxx39)rArr M_("Solute")=64 gm` `DeltaT_(f)=K_(f)xxm=5.12xx(0.5xx1000)/(64xx39)` `rArr DeltaT_(f)=1.02` |
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| 102. |
One mole of a monatomic ideal gas is taken along two cyclic processes `E to F to G to E and E to F to H to E` as shown in the PV diagram. The processes involved are purely isochoric , isobaric , isothermal or adiabatic . Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. A. `{:(P,Q,R,S),(4,2,3,1):}`B. `{:(P,Q,R,S),(4,3,1,2):}`C. `{:(P,Q,R,S),(3,1,2,4):}`D. `{:(P,Q,R,S),(1,3,2,4):}` |
| Answer» Correct Answer - A | |
| 103. |
A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is A. `0%`B. `20%`C. `75%`D. `80%` |
| Answer» Correct Answer - D | |
| 104. |
Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+` energy. In the core of fusion reactor, a gas of heavy hydrogen of `_(1)^(2) H` is fully ionized into deuteron nuclei and electrons. This collection of `_1^2H` nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time `t_(0)` before the particles fly away from the core. If `n` is the density (number volume ) of deuterons , the product` nt_(0) `is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following boltzmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` Result of calculations for four different design of a fusion reactor using `D-D` reaction are given below. which of these is most promising based on Lawson criterion ?A. deuteron density `=2.0xx10^(12)cm^(-3)`, confinement time `=5.0xx10^(-3)s`B. deuteron density `=8.0xx10^(14)cm^(-3)`, confinement time `=9.0xx10^(-1)s`C. deuteron density `=4.0xx10^(23)cm^(-3)`, confinement time `=1.0xx10^(-11)s`D. deuteron density `=1.0xx10^(24)cm^(-3)`, confinement time `=4.0xx10^(-12)s` |
| Answer» Correct Answer - B | |
| 105. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. The allowed energy for the particle for a particular value of n is proportional toA. `a^(-2)`B. `a^(-3//2)`C. `a^(-1)`D. `a^(2)` |
| Answer» Correct Answer - A | |
| 106. |
A piece of wire is bent in the shape of a parabola `y=kx^(2)` (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is:A. `(a)/(gk)`B. `(a)/(2gk)`C. `(2a)/(gk)`D. `(a)/(4gk)` |
| Answer» Correct Answer - B | |
| 107. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. If the mass of the particle is `m=1.0xx10^(-30)`kg and `alpha=6.6nm`, the energy of the particle in its ground state is closest toA. 0.8 meVB. 8meVC. 80meVD. 800 meV |
| Answer» Correct Answer - B | |
| 108. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)`C Q. The speed of the particle that can take discrete values is proportional toA. `n^(-3//2)`B. `n^(-1)`C. `n^(1//2)`D. `n` |
| Answer» Correct Answer - D | |
| 109. |
A thermodynamic system is taken from an initial state I with internal energy `U_i=-100J` to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the pat af, ib and bf are `W_(af)=200J, W_(ib)=50J and W_(bf)=100J` respectively. The heat supplied to the system along the path iaf, ib and bf are `Q_(iaf), Q_(ib),Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b=200J and Q_(iaf)=500J`, The ratio `(Q_(bf))/(Q_(ib))` is |
| Answer» Correct Answer - 2 | |
| 110. |
STATEMENT-1: For practical purposes, the earth is used as a reference at zero potencial in electrical circuits. and STATEMENT-2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by `(Q)/(4piepsilon_0R)`.A. Statement-1 is true, statement-2 is true,statement-2 is a correct explanation for statement-1.B. statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-3C. statement-1 is true, statement -2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 111. |
STATEMENT-1: For practical purposes, the earth is used as a reference at zero potencial in electrical circuits. and STATEMENT-2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by `(Q)/(4piepsilon_0R)`.A. Statement-1 is true, statement-2 is true,statement-2 is a correct explanation for statement-1.B. statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-4C. statement-1 is true, statement -2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - C | |
| 112. |
The carbon-based reduction method is NOT used for the extraction ofA. tin from `SnO_(2)`B. iron from `Fe_(2)O_(3)`C. aluminium from `Al_(2)O_(3)`D. magnesium from `MgCO_(3) *CaCO_(3)` |
| Answer» Correct Answer - C::D | |
| 113. |
Column I shows four systems, each of the same length L, for producing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as `lambda_(f)`. Match each system with statements given in Column II describing the nature and wavelength of the standing waves. |
| Answer» Correct Answer - A : p and t B : p and s C : q and s D : q and r | |
| 114. |
let `P_(1)=[(1,0,0),(0,1,0),(0,0,1)],P_(2)=[(1,0,0),(0,0,1),(0,1,0)],P_(3)=[(0,1,0),(1,0,0),(0,0,1)],P_(4)=[(0,1,0),(0,0,1),(1,0,0)],P_(5)=[(0,0,1),(1,0,0),(0,1,0)],P_(6)=[(0,0,1),(0,1,0),(1,0,0)]` and `X=sum_(k=1)^(6) P_(k)[(2,1,3),(1,0,2),(3,2,1)]P_(k)^(T)` Where `P_(k)^(T)` is transpose of matrix `P_(k)`. Then which of the following options is/are correct?A. X is a symmetric matrixB. if `X=[(1),(1),(1)]=alpha[(1),(1),(1)]`, then `alpha=30`C. X-30I is an invertible matrixD. The sum of diagonal entries of X is 18. |
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Answer» Correct Answer - A::B::D Clearly `P_(1)=P_(1)^(T)=P_(1)^(-1)` `P_(2)=P_(2)^(T)=P_(2)^(-1)` . . . `P_(6)=P_(6)^(T)=P_(6)^(-1)` And `A^(T)=A` where `A=[(2,1,3),(1,0,2),(3,2,1)]` Using formula `(A+B)^(T)=A^(T)+B^(T)` `X^(T)=(P_(1)AP_(1)^(T)+ . . . .+P_(6)AP_(6)T)^(T)=P_(1)A^(T)P_(1)^(T)+ . . . . . +P_(6)A^(T)P_(6)^(T)=XimpliesX` is symmetric Let `B=[(1),(1),(1)]` `XB=P_(1)AP_(1)^(T)B+P_(2)AP_(2)^(T)B+. . . .+P_(6)AP_(6)^(T)B=P_(1)AB+PAB+. . . . +P_(6)AB` `XB=(P_(1)+P_(2)+. . . .+P_(6))[(6),(3),(6)]` `=[(6xx2+3xx2+6xx2),(6xx2+3xx2+6xx2),(6xx2+3xx2+6xx2)]=[(30),(30),(30)]=30Bimpliesalpha=30` since X`[(1),(1),(1)=30[(1),(1),(1)]` `implies(X-30I)B=0` has a non trivial solution `B=[(1),(1),(1)]` `implies|X-30I|=0` `X=P_(1)P_(1)AP_(1)^(T)+. . . .+P_(6)AP_(6)^(T)` trace `(x)=t_(r)(P_(1)AP_(1)^(T))+. . . . .+t_(r)(P_(6)AP_(6)^(T))=(2+0+1)+. . . . +(2+0+1)=3xx6=18` |
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| 115. |
A set S is given by {1,2,3,4,5,6}. |X| is number of elements in set X. If A and B are independent eventsassociated with S are chosen such that each elements is equally likely and `1 le |B| le |A|` then the number of ordered pairs of (A,B) are |
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Answer» Correct Answer - 1523 Number of ordered pair (A,B) `.^(6)C_(1)(.^(6)C_(2)+.^(6)C_(3)+. . . .+.^(6)C_(6))+.^(6)C_(2)(.^(6)C_(3)+.^(6)C_(4)+.^(6)C_(5)+.^(6)C_(6))+.^(6)C_(3)(.^(6)C_(4)+.^(6)C_(5)+.^(6)C_(6))(.^(6)C_(5)+.^(6)C_(6))+.^(6)C_(5).^(6)C_(6)` `=(.^(6)C_(1).^(6)C_(2)+.^(6)C_(3)+. . . .+.^(6)C_(5).^(6)C_(6))+(.^(6)C_(1).^(6)C_(3)+.^(6)C_(2).^(6)C_(4)+.^(6)C_(3).^(6)C_(5)+.^(6)C_(4).^(6)C_(6)` `+(.^(6)C_(1).^(6)C_(4).^(6)C_(2).^(6)C_(5)+.^(6)C_(5).^(6)C_(6)+(.^(6)C_(1).^(6)C_(5)+.^(6)C_(2).^(6)C_(6))+(.^(6)C_(1).^(6)C_(6))` `=(.^(12)C_(5)-.^(6)C_(1))+(.^(12)C_(4)-.^(6)C_(2))+(.^(12)C_(3)-^(6)C_(3))+(.^(12)C_(2)-.^(6)C_(4))+(.^(12)C_(1)-.^(6)C_(1))` `=(.^(12)C_(1)+. . . .+.^(12)C_(5))-(.^(6)C_(1)+. . . .+.^(6)C_(4)+.^(6)C_(5))=1585-62=1523` |
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| 116. |
Suppose det ` [{:(sum_(k=0)^(n)k,,sum_(k=0)^(n).^nC_(k)k^2),(sum_(k=0)^(n).^nC_(k)k^k,,sum_(k=0)^(n).^nC_(k)3^(2)):}]=0` holds for some positive integer n. then `sum_(k=0)^(n)(.^nC_(k))/(k+1)` equals ............ |
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Answer» Correct Answer - `6.20` `|{:((n(n+1))/(2),n2^(n-1)+n(n-1)2^(n-2)),(n2^(n-1),):}|=0` `n=0` or `4(n+1)-2n-n(n-1)=0` `4n+4-2n-n^(2)+n=0` `4n+4-2n-n^(2)+n=0` `3n-n^(2)+4=0impliesn^(2)-3n-4=0` `(n-4)(n+1)=0` `n=4` `underset(r=0)overset(4)sum(.^(4)C_(r))/(r+1)=underset(r=0)overset(4)sum(.^(5)C_(r+1))/(5)=(2^(5)-1)/(5)=(31)/(5)=6.20` |
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| 117. |
For non-negative inger n, let `f(n)=sum_(k=)^(n) sin((k+1)/(n+1)pisin((k+2)/(n+1)pi))/(sum_(k=0)^(n)sin^(2)((k+1)/(n+1)pi))` Assuming `cos^(-1)x` takes values in `[0,pi]` which of the following options is/are correct?A. if `alpha=tan(cos^(-1)f(6))`, then `alpha^(2)+2alpha-1=0`B. `underset(ntoinfty)(lim)f(x)=(1)/(2)`C. `f(4)=(sqrt(3))/(2)`D. `sin(7cos^(-1)f(5))=0` |
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Answer» Correct Answer - A::C::D `f(n)=(underset(k=0)overset(n)sum sin((k+1)/(n+1)pi)sin((k+2)/(n+2)pi))/(underset(k=0)overset(n)sum2sin^(2)((k+1)/(n+2))pi)=(underset(k=0)overset(n)sum(cos((pi)/(n+2))-cos((2k+3)/(n+2))pi))/(underset(k=0)overset(n)sum2sin^(2)((k+1)/(n+2))pi)` `=((n+1)cos((pi)/(n+2))-(cos((n+3)/(n+2))pisin((n+1)/(n+2))pi)/(sin((pi)/(n+2))))/((n+1)-(cospisin((n+1)/(n+2))pi)/(sin((pi)/(n+2))))` `=((n+1)cos((pi)/((n+2)))+cos((n+3)/(n+2))pi)/((n+1)+1)=((n+1)cos((pi)/((n+2)))+cos((pi)/(n+2)))/(n+2)=cos((pi)/(n+2))` (A) `alpha=tan(cos^(-1)f(6))=tan|cos^(-1)(cos((pi)/(8)))|=tan((pi)/(8))` `tan((pi)/(4))=(2tan((pi)/(8)))/(1-tan^(2)((pi)/(8)))implies1=(2alpha)/(1-alpha^(2))impliesalpha^(2)+2alpha-1=0` (A) correct (C). `f(4)=cos((pi)/(6))=(sqrt(3))/(2)` correct (D). `sin(7cos^(-1)f(5))=sin(7cos^(-1)(cos((pi)/(7))))=sinpi=0` correct (A), (C), (D) correct |
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| 118. |
`sec^(-1)|(1)/(4) sum_(k=0)^(10) (sec((7pi)/(12)+(kpi)/(2))sec((7pi)/(12)+(k+1)((pi)/(2)))]` will be |
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Answer» Evaluate `sec^(-1)[(1)/(4)underset(k=0)overset(10)sumsec((7pi)/(12)+(kpi)/(12))sec((7pi)/(12)+(k+1)((pi)/(2)))]` `=sec^(-1)[-(1)/(4)underset(k=0)overset(10)sum(2)/(sin((7pi)/(6)+kpi))]=sec^(-1)[-(1)/(2)underset(k=0)overset(10)sum(1)/((1-)^(k+1)sin((pi)/(6)))]=sec^(-1)(-underset(k=0)overset(10)sum(1)/((-1)^(k+1)))=sec^(-1)(1)=0` |
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| 119. |
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image isA. virtual and at a distance of 16 cm from the mirrorB. real and at a distance of 16 cm from the mirrorC. virtual and at a distance of 20 cm from the mirrorD. real and at a distance of 20 cm from the mirror |
| Answer» Correct Answer - B | |
| 120. |
For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)` `Fe_((aq))^(2+)+S_((aq))^(2) hArr FeS(s)` When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is |
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Answer» Correct Answer - 8.92 & 8.93 `{:(,Fe_((aq))^(2+),+,S_((aq))^(2-),hArr FeS(s),K_(C)=1.6xx10^(17)),(,0.06 M,,0.2 M,,),("After mixing",0.03 M,,0.1 M,,),(,?,,0.07M,,):}` `1.6xx10^(17) =1/([Fe^(2+)]xx0.07)` or `[Fe^(2+)]=(10^(-17))/(1.6xx0.07)=(10^(-15))/11.2=100/11.2xx10^(-17)=8.928xx10^(-17) =Yxx10^(-17)` Answer after rounding of is =8.93 Answer after truncation of is 8.92 |
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| 121. |
There are three bags `B_(1),B_(2),B_(3),B_(1)` contians 5 red and 5 green balls. `B_(2)` contains 3 red and 5 green balls and `B_3` contains 5 red and 3 green balls, bags `B_(1),B_(2)` and `B_(3)` have probabilities 3/10, 3/10, and 4/10 respectively of bieng chosen. A bag is selected at randon and a ball is randomly chosen from the bag. then which of the following options is/are correct?A. Probability that the chosen ball is green equals `(39)/(80)`B. Probability that the chosen all is green, gen that selected bag is `B_(3)` equals `(3)/(8)`C. Probability that the selected bag is `B_(3)`, given that the chosen ball is green equals `(4)/(13)`D. Probability that the selected bag is `B_(3)` given that the chosen ball is green equals `(3)/(10)` |
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Answer» Correct Answer - A::B::C `{:(,"Bag"_(1),"Bag"_2,"Bag"_(3)),("Red balls",5,3,5),("Green balls",5,5,3),("Total",10,8,8):}` (A) P(Ball is green) `=P(B_(1))P(G//B_(1))+P_(B_(2))P(G//B_(2))+P_(B_(3))P_(G//B_(3))` `=(3)/(10)xx(5)/(10)+(3)/(10)xx(5)/(8)+(4)/(10)xx(3)/(8)=(39)/(80)` B(). P (ball chosen is green/ball is fro `3^(rd)` bag) `=(3)/(8)` (C,D) P (Ball is from `3^(rd)` bag/ball chosen is green) `=(P_(B_(3))P(G//B_(3)))/(P(B_(1))P_(G//B_(1))+P(B_(2))P(G//B_(2))+P(B_(3))P(G//B_(3)))` `P(B_(1))=(3)/(10)` `P(B_(2))=(3)/(10)` `P(B_3)=(4)/(10)` `=((4)/(10)xx(3)/(8))/((3)/(10)xx(5)/(10)+(3)/(10)xx(5)/(8)+(4)/(10)xx(3)/(8))=(4)/(13)` |
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| 122. |
if a matrix M is given by `[{:(0,1,2),(1,2,3),(3,1,1):}]` and if `M[{:(alpha),(beta),(gamma):}]=[{:(1),(2),(3):}]` thenA. `adj(M^(-1))+(adjM)^(-1)=-M`B. `|adj(M^(2))|=81`C. `alpha+2beta+3gamma=2`D. `beta+2gamma=3` |
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Answer» Correct Answer - A::C `[{:(0,1,2),(1,2,3),(3,1,1):}][{:(alpha),(beta),(gamma):}]=[{:(1),(2),(3):}]` `impliesbeta+2gamma=1` `alpha+2beta+3gamma=2` `3alpha+beta+gamma=3` `impliesalpha=1,beta=-1,gamma=1` `|M|=-2` `|adjM^(2)|=|M^(2)|^(2)=|M|^(4)=16` `adj(M^(-1))=|M^(-1)|M=(-M)/(2)` `(adjM)^(-1)=adj(M^-1)=-(M)/(2)` |
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| 123. |
if `a_(n=(alpha^(n)-beta^(n))/(alpha-beta)` where `alpha` and `beta` are roots of equation `x^(2)-x-1=0` and `b_(n)=a_(n+1)+a_(n-1)` thenA. `b_(n)=alpha^(n)+beta^(n)`B. `underset(n=1)overset(infty)sum(b_(n))/(10^(n))=(8)/(89)`C. `underset(n=1)overset(infty)sum(a_(n))/(10^(n))=(10)/(89)`D. `a_(1)+a_(2)+…a_(n)=a_(n+2)-1` |
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Answer» Correct Answer - A::C::D (A). `b_(n)=a_(n+1)+a_(n-1)-(alpha^(n+1)-beta^(n+_1))/(alpha-beta)+(alpha^(n-1)-beta^(n-1))/(alpha-beta)=(alpha^(n-1)(alpha^(2)+1)-beta^(n-1)(beta^(2)+1))/(alpha-beta)` `=(alpha^(n-1)(alpha+2)-beta^(n-1)(beta+2))/(alpha-beta)=(alpha^(n-1)((5+sqrt(5))/(2))-beta^(n-1)((5-sqrt(5))/(2)))/(alpha-beta)` `=(sqrt(5)alpha^n-1)((sqrt(5)+1)/(2))-sqrt(5)beta^(n-1)((sqrt(5)-1)/(2)))/(alpha-beta)=(sqrt(5)(alpha^(n)+beta^(n)))/(alpha-beta)=alpha^(n)+beta^(n)" "becausealpha-beta` (B). `underset(n=1)overset(infty)sum(b_(0))/(10^(n))=sum((alpha)/(10))^(n)+sum((beta)/(10))^(n)+sum((beta)/(10))^(n)=((alpha)/(10))/(1-(alpha)/(10))+((beta)/(10))/(1-(beta)/(10))=(alpha)/(10-alpha)+(beta)/(10-beta)` `=(10(alpha+beta)-2alphabeta)/(100-10(alpha+beta)+alphabeta)=(10+2)/(89)=(12)/(89)` (C). `underset(n=1)overset(infty)(a_(n))/(10^(n))=sum(alpha^(n)-beta^(n))/((alpha-beta)10^(n))=(1)/(alpha-beta)(((alpha)/(10))/(1-(alpha)/(10))-((beta)/(10))/(1-(beta)/(10)))(1)/(alpha-beta)((alpha)/(10-alpha)-(beta)/(10-beta))` `=(1)/(alpha-beta.((10(alpha-beta)-alphabeta+alphabeta))/(100-10(alpha+beta)+alphabeta))=(10)/(89)` (D). `a_(1)+a_(2)+...a_(n)=suma_(i)=(sumalpha^(n)-sumbeta^(i))/(alpha-beta)=((alpha(1-alpha^(n)))/((1-alpha))-(beta(1-beta^(n)))/((1-beta)))/(alpha-beta)` `=((alpha+1)(1+alpha^(n))-(beta+1)(1-beta^(n)))/((1-alpha)(1-beta)(alpha-beta))=(alpha^(2)-alpha^(n+2)-beta^(2)+beta^(n+2))/((1-alpha)(1-beta)(alpha-beta))=(sqrt(5)+beta^(n+2)-alpha^(n+2))/(beta-alpha)=-1+a_(n+2)` |
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| 124. |
`M=[{:(sin^(4)theta,-1-sin^(2)theta),(1+cos^(2)theta,cos^(4)theta):}]=alphaI+betaM^(-1)` Where `alpha=alpha(theta)` and `beta=beta(theta)` ar real numbers and I is an identity matric of `2xx2` if `alpha^(**)=` min of set `{alpha(theta):thetain[0.2pi)}` and `beta^(**)=` min of set `{beta(theta):thetain[0.2pi)}` Then value of `alpha^(**)+beta^(**)` isA. `(-37)/(16)`B. `(-17)/(16)`C. `(-31)/(16)`D. `(-29)/(16)` |
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Answer» Correct Answer - D `msin^(4)theta.cos^(4)theta+(1+sin^(2)theta)(1+cos^(2)theta)` `2+sin^(4)cos^(4)theta+sin^(2)thetacos^(2)theta` `[{:(sin^(4)theta,-(1+sin^(2)theta)),(1+cos^(2)theta,cos^(4)theta):}]=[{:(alpha,0),(0,alpha):}]+beta=(1)/(|m|)[{:(cos^(4)theta,1+sin^(2)theta),(-1-cos^(2)theta,sin^(4)theta):}]` `sin^(4)theta=(alpha+beta)/(|m|)cos^(4)theta,-1-sin^(2)theta=(beta)/(|m|)(1+sin^(2)theta)` `beta=-|m|` `beta=-[sin^(4)thetacos^(4)theta+sin^(2)thetacos^(2)theta+2]=-[t^(2)+t+2]impliesbeta_(min)=-(37)/(16)` `alpha=sin^(2)theta+cos^(4)theta=1-2sin^(2)thetacos^(2)theta=1-(1)/(2)(sin^(2)2theta)impliesminalpha=(1)/(2)` `alpha+beta=-(37)/(16)+(1)/(2)=-(37)/(16)+(8)/(16)=-(29)/(16)` |
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| 125. |
The loss of kinetic energy during the above proces isA. `(Iomega^(2))/2`B. `(Iomega^(2))/3`C. `(Iomega^(2))/4`D. `(Iomega^(2))/6` |
| Answer» Correct Answer - B | |
| 126. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then beA. `P_(0)`B. `P_(0)/2`C. `P_(0)/2+ (Mg)/(piR^(2))`D. `P_(0)/2-(Mg)/(piR^(2))` |
| Answer» Correct Answer - A | |
| 127. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top isA. `(2P_(0)piR^(2))/(piR^(2)P_(0)+Mg)(2L)`B. `(P_(0)piR^(2)-Mg)/(piR^(2)P_(0))(2L)`C. `(P_(0)piR^(2)+Mg)/(piR^(2)P_(0))`D. `(P_(0)piR^(2))/(piR^(2)P_(0)-Mg)(2L)` |
| Answer» Correct Answer - D | |
| 128. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is `rho`. In equilibrium, the height H of the water coulmn in the cylinder satisfies A. `rhog(L_(0)-H)^(2)+P_(0)(L_(0)-H)+L_(0)P_(0)=0`B. `rhog(L_(0)-H)^(2)-P_(0)(L_(0)-H)-L_(0)P_(0)=0`C. `rho g(L_(0)-H)^(2)+P_(0)(L_(0)-H)-L_(0)P_(0)=0`D. `rho g(L_(0)-H)^(2)-P_(0)(L_(0)-H)+L_(0)P_(0)=0` |
| Answer» Correct Answer - C | |
| 129. |
Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in column II and indicate your answer by darkening appropriate bubbles in the `4xx4` matrix given in the ORS. `{:(,"Coloumn-I",,"Column-II"),((A),GM_(e)M_(s),(P),("volt")("coulomb")("meter")),(,"G-universal gravitational ",,),(,"Contant",,),(,M_(e)-"mass of the earth",,),(,M_(s)-"mass of the Sun",,),((B),(3RT)/(3),(Q),("kilogram")("metre")^(3)),(,R-"universal gas constant",,),(,T-"absolute temperature",,),(,M-"molar mass",,),((C),(F^(2))/(q^(2)B^(2)),(R),("metre")^(2)("second")^(-2)),(,F-"force",,),(,q-"charge",,),(,B-"magnetic field",,),((D),(GM_(e))/(R_(e)),(S),("farad")("volt")^(2)("kg")^(-1)),(,G-"universal gravitational constant",,),(,M_(e)-"mass of the earth",,),(,R_(e)-"radius of the earth",,):}` |
| Answer» Correct Answer - A::B | |
| 130. |
In the following reaction, the product(s) formed is(are) A. P(major)B. Q(minor)C. R(minor)D. S(major) |
| Answer» Correct Answer - B::D | |
| 131. |
Two spherical planets P and Q have the same uniform density `rho,` masses `M_p and M_Q` and surface areas A and 4A respectively. A spherical planet R also has uniform density `rho` and its mass is `(M_P + M_Q).` The escape velocities from the plantes P,Q and R are `V_P V_Q and V_R` respectively. ThenA. `V_(Q) gt V_(R ) gt V_(P)`B. `V_(R ) gt V_(Q) gt V_(P)`C. `V_(R )//V_(P)=3`D. `V_(P)//V_(Q)=(1)/(2)` |
| Answer» Correct Answer - B::D | |
| 132. |
A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod change with the (t) as `T(t)=T_(0)(1+betat^(1//4))` where `beta` is a constant with appropriate dimension of temperature. the heat capacity of metal is :A. `(4P(T(t)-T_(0))^(3))/(beta^(4)T_(0)^(4))`B. `(4P(T(t)-T_(0))^(2))/(beta^(4)T_(0)^(3))`C. `(4P(T(t)-T_(0))^(4))/(beta^(4)T_(0)^(5))`D. `(4P(T(t)-T_(0)))/(beta^(4)T_(0)^(2))` |
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Answer» Correct Answer - A dQ=HdT `(dQ)/(dt)=H.(dT)/(dt)` `P=H.T_(0).beta.1/4.t^(-3//4)` `(4P)/(T_(0)beta)=t^(-3//4).H` Now `T-T_(0)=T_(0)betat^(1//4)` So, `t^(3//4) =((T-T_(0))/(T_(0)beta))^(3)` `:. H=(4P(T-T_(0))^(3))/(T_(0)^(4)beta^(4))` |
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| 133. |
STATEMENT-l : In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. STATEMENT-2 : In an elastic collision, the linear momentum of the system is conserved.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1.B. Statement-1 is true, statement-2 is true, Statement-2 is Not a correct explanation for Statement -2C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - B::D | |
| 134. |
Asseration : A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a ceratin distance. The surface is now tilted to an angle of `30^@` with the horizontal and same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is small than the first situation. Reason : The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1.B. Statement-1 is true, statement-2 is true, Statement-2 is Not a correct explanation for Statement -1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - C | |
| 135. |
Plots showing the variation of the rate constant `(k)` with temperature `(T)` are given below. The plot that follows the Arrhenius equation isA. B. C. D. |
| Answer» Correct Answer - A | |
| 136. |
The sides of a right angled triangle are inarithmetic progression. If the triangle has area 24, then what is the lengthof its smallest side? |
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Answer» a-d,a,a+d are in AP `(a+d)^2=a^2+(a-d)^2` `a^2+d^2+2ad=a^2+a^2-2ad+d^2` `a^2-2ad-2ad=0` `a^2-4ad=0` `a(a-4d)=0` `a-4d=0` `a=4d` `a!=0` So, base=4d,perpendicular=3d,hypotenuse=5d `1/2*b*h=1/2*4d*3d` `6d^2=24` `d^2=4` `d=pm2` d can not be negative So, b=8,p=6,h=10. |
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| 137. |
Word of length 10 are formed usingthe letters A,B,C,D,E,F,G,H,I,J. Let `x`be the number of such words where no letter is repeated; and let `y`be the number of such words where exactly one letter is repeated twiceand no other letter is repeated. The, `y/(9x)=` |
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Answer» x=number of such word where number letter is expected `x=10!` y= the number of such words where exactly one letter is repeated . `y=.^10C_1*.^9C_8*(10!)/(2!)` `y/(g(x))=(.^10C_1*.^9C_8*(10!)/(2!))/(9*10!)` `y/(g(x))=5`. |
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| 138. |
Let S be the set of matrices of order `3xx3` such that all elemtns of the matrix belong to `{0,1}` let `E_(1)={A in S:|A|=0}` where |A| denotes determinant of matrix A `E_(2)={A in S:` sum of elements of `A=7`} find `P(E_(1)//E_(2))` |
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Answer» Correct Answer - 0.5 `E_(2)`: sum of elements of `A=7implies `These are 7 ones and 2 zeros Number of such matrices `= .^(9)C_(2)=36` Out of all such matrices `E_(1)` eill be those when both zeros lie in te same row or in te same colume eg. `[{:(1,1,1),(0,1,1),(0,1,1):}]` `n(E_(1)capE_(2))=2xx underset(uarr)(.^(3)C_(2))xx underset(uarr)(.^(3)C_(2)=18)` so `n(E1//E2)=(n(E_(1)capE_(2)))/(n(E_(2)))=(18)/(36)=(1)/(2)` |
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| 139. |
A radioactive sample `S_1` having an activity of `5muCi` has twice the number of nuclei as another sample `S_2` which has an activity of `10muCi`. The half-lives of `S_1` and `S_2` can beA. 20 years and 5 years, respectivelyB. 20 years and 10 years, respectivelyC. 10 years eachD. 5 years each |
| Answer» Correct Answer - A | |
| 140. |
Which one of the following statement is `WRONG` in the context of X- rays generated from X- rays tube ?A. Wavelength of characteristic X rays decreases when the atomic number of the target increasesB. Cut off wavelength of the continuous X ray depenss on the atomic number of the targetC. Intensity of the characteristic X ray depends on the electrical power given to the x ray tubeD. cut off wavelength of the continous x rays depends on the energy of the electrons in the x ray tube |
| Answer» Correct Answer - B | |
| 141. |
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is ` 60degree`). In the position of minimum deviation, the angle of refraction will beA. `30^(@)` for both the colourB. greater for the violet colourC. greater for the red colourD. equal but not `30^(@)` for both the colours |
| Answer» Correct Answer - A | |
| 142. |
The compound(s) that exhibits(s) geometrical isomerism is/areA. `[Pt(en)Cl_(2)]`B. `[Pt(en)_(2)]Cl_(2)`C. `[Pt(en)_(2)Cl_(2)]Cl_(2)`D. `[Pt(NH_(3))_(2)Cl_(2)]` |
| Answer» Correct Answer - C::D | |
| 143. |
The compounds(s) formed upon combustion of sodium metal in excess air is/areA. `Na_(2)O_(2)`B. `Na_(2)O`C. `NaO_(2)`D. `NaOH` |
| Answer» Correct Answer - A::B | |
| 144. |
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle `theta` with the horizontal floor. The coefficient of friction between the wall and the ladder is `mu_1` and that between the floor and the ladder is `mu_2.` the normal reaction of the wall on the ladder is `N_1` and that of the floor is `N_2.` if the ladder is about to slip. than A. `mu_(1) = 0 mu_(2) ne 0 and N_(2) tantheta = (mg)/(2)`B. `mu_(1)ne 0 mu_(2) = 0 and N_(1) tan theta = (mg)/(2)`C. `mu_(1) ne 0" " mu_(2) ne 0 and N_(2) = (mg)/(1+mu_(1)mu_(2))`D. `mu_(1) ne 0" " mu_(2) ne 0 and N_(1) = (mg)/(1+mu_(1)mu_(2))` |
| Answer» Correct Answer - A::C::D | |
| 145. |
In the reaction the products areA. B. C. D. |
| Answer» Correct Answer - D | |
| 146. |
A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination `theta`. Various values of `theta` are given in List I. The coefficient of friction between the block `m_(1)` and the plane is always zero. The coefficient of static and dynamic friction between the block `m_(2)` and the plane are equal to `mu = 0.3` . In List II expressions for the friction on block `m_(2)` are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g [Useful information `: tan (5.5^(@)) ~~ 0.1, tan (11.5^(@)) ~~ 0.2 , tan (16.5^(@))~~ 0.3`]. A. P-1, Q-1, R-1, S-B. P-2, Q-2, R-2, S-3C. P-2, Q-2, R-2, S-4D. P-2, Q-2, R-3, S-3 |
| Answer» Correct Answer - D | |
| 147. |
Consider an expanding sphere of instantaneous radius ? whose total mass remains constant. The expansion is such that the instantaneous density `rho` remains uniform throughout the volume. The rate of fractional change in density `((dp)/(rhodp))` is constant. The velocity v of any point on the surface of the expanding sphere is proportional toA. RB. `R^(3)`C. `1/R`D. `R^(2//3)` |
| Answer» Correct Answer - A | |
| 148. |
Column I gives a listof possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs even in Column II. Indicate your answer by darkening the appropriate bubbles of the `4 xx 4` matrix given in the ORS. |
| Answer» Correct Answer - A::B::D | |
| 149. |
An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the components can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Indicate your answer by darkening the appropriate bubbles of the `4 xx 4` matrix given in the ORS. |
| Answer» Correct Answer - A::B::C::D | |
| 150. |
Let `[x]`be the greatest integer less than or equal to `xdot`Then, at which of the following point (s) function `f(x)=xcos(pi(x+[x]))`is discontinuous?(a)`x=1`(b) `x=-1`(c) `x=0`(d) `x=2` |
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Answer» `f(x)=xcosx(pi(x+[x]))` `f(x)=xcos(pix+pi[x])` `f(x)=(-1)^[[x]] xcospix` `lim_(x->o^-)f(x)=lim_(x->0^-)f(x-4)=(x-h)cos(pi(x-h+[x-h]))` `=-hcos(pi(-h-1))` `=-hcos(pih+pi)` `=0` `lim_(x->0^+)=lim_(x->o^+)f(x+h)=(x+h)cos(pi(x+h+[x+h]))` `=hcospih=0` Correct option is A,B and D. |
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