Q: A particle is placed at the origin and a force F=kx is acting on it (where, k is a positive constant). If U(0)=0, the graph of U(x) versus x will be (where, U is the potential energy function)

From F=−dxdU∫0U(x)dU=−∫0xFdx=−∫0x(kx)dx∴U(x)=−2kx2 as U(0)=0

Q: A wire has a mass (0.3±0.003) g, radius (0.5±0.005) mm and length (6±0.06) cm. The maximum percentage error in the measurement of its density is

Density, ρ=πr2Lm∴ρΔρ×100=(mΔm+2rΔr+LΔL)×100or maximum percentage error in density=[0.30.003×100+0.52×0.005×100+60.06×100]=4%

Q: In the relation, p=βαe−kθαZ, p is pressure, Z is distance, k is Boltzmann's constant and θ is the temperature. The dimensional formula of β will be

[kθαZ]=[M0 L0 T0][α]=[Zkθ] Further [p]=[βα]∴[β]=[pα]=[Zpkθ]Dimensions of kθ are that to energy as E=23kT. Hence,[β]=[LML−1 T−2ML2 T−2]=[M0 L2 T0]

Q: Which hydrogen-like species will have the same radius as that of the Bohr orbit of hydrogen atom?

Expression for Bohr's orbit is, rn=Za0n2=a0 when n=2,Z=4.

Q: (a) The Schrodinger wave equation for the hydrogen atom isψ2x=4(2π)1/21(a01)3/2(2−a0r)e−r/2a0where, a0 is Bohr's radius. Let the radial node in 2s be at r0. Then, find r in terms of a0 -(b) A base ball having mass 100 g moves with velocity 100 m/s. Find out the value of the wavelength of baseball.

(a) At radial node, ψ2 must vanishes, i.e.ψ2s2=0=[42π1]2(2−a0r0)2e−a0r0⇒2−a0r0=0⇒r0=2a0 (b) λ=mvh=100×10−3×1006.625×10−34=6.625×10−35 m=6.625×10−25A˚ (negligibly small)

Q: For a monatomic gas kinetic energy =E. The relation with rms velocity is:

Kinetic energy (E)=23kT RMS speed (u)=m3kT⇒u=m2E

Q: (a) In the following equilibrium N2O4( g)⇌2NO2( g) when 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given thatΔGf∘(N2O4)=100 kJ,ΔGf∘(NO2)=50 kJ(i) Find ΔG of the reaction.(ii) The direction of the reaction in which the equilibrium shifts.(b) A graph is plotted for a real gas that follows van der Waals' equation with pVm taken on Y-axis and p on X-axis. Find the intercept of the line where Vim is molar volume.

(a)N2O4(g)⇌2NO2(g)ΔG∘=2ΔGf∘(NO2)−ΔGf∘(N2O4)=0Also ΔG∘=−RTlnK=0,K=1Let the reaction shift in the forward direction.N2O4( g)⇌2NO2( g) Total 5−x5+2x10+xpi=10+x5−x×2010+x5+2x×20⇒K=(10+x)2(5+2x)2×5−x10+x×20=1⇒81x2+405x+450=0x=−1.66 and −3.33Both values of x indicate that the reaction actually proceeds in the backward direction.(b) (p+Vm2a)(Vm−b)=RT(p+(pV)2ap2)(ppV−b)=RT⇒[(pV2)p+ap2][(pV)−b]=p(pV)2RT⇒p⋅[pV2+ap](pV−bp)=p(pV2)RT But p=0 Intercept =RT⇒(pV)3=(pV)2RT

Question 1

A particle is placed at the origin and a force F=kx is acting on it (where, k is a positive constant). If U(0)=0, the graph of U(x) versus x will be (where, U is the potential energy function)

Accepted Answer

From F=−dxdU​∫0U(x)​dU=−∫0x​Fdx=−∫0x​(kx)dx∴U(x)=−2kx2​ as U(0)=0​

Question 2

A wire has a mass (0.3±0.003) g, radius (0.5±0.005) mm and length (6±0.06) cm. The maximum percentage error in the measurement of its density is

Accepted Answer

Density, ρ=πr2Lm​∴ρΔρ​×100=(mΔm​+2rΔr​+LΔL​)×100or maximum percentage error in density=[0.30.003​×100+0.52×0.005​×100+60.06​×100]=4%​

Question 3

In the relation, p=βα​e−kθαZ​, p is pressure, Z is distance, k is Boltzmann's constant and θ is the temperature. The dimensional formula of β will be

Accepted Answer

[kθαZ​]=[M0 L0 T0][α]=[Zkθ​] Further [p]=[βα​]∴[β]=[pα​]=[Zpkθ​]​Dimensions of kθ are that to energy as E=23​kT. Hence,[β]=[LML−1 T−2ML2 T−2​]=[M0 L2 T0]

Question 4

Which of the following sets have different dimensions?

Accepted Answer

Dipole moment =( charge )×( distance )Electric flux =( electric field )×( area )

Question 5

Which hydrogen-like species will have the same radius as that of the Bohr orbit of hydrogen atom?

Accepted Answer

Expression for Bohr's orbit is, rn​=Za0​n2​=a0​ when n=2,Z=4.

Question 6

(a) The Schrodinger wave equation for the hydrogen atom isψ2x​=4(2π)1/21​(a0​1​)3/2(2−a0​r​)e−r/2a0​where, a0​ is Bohr's radius. Let the radial node in 2s be at r0​. Then, find r in terms of a0​ -(b) A base ball having mass 100 g moves with velocity 100 m/s. Find out the value of the wavelength of baseball.

Accepted Answer

(a) At radial node, ψ2 must vanishes, i.e.ψ2s2​=0=[42​π1​]2(2−a0​r0​​)2e−a0​r0​​⇒2−a0​r0​​=0⇒r0​=2a0​ (b) λ=mvh​=100×10−3×1006.625×10−34​=6.625×10−35 m=6.625×10−25A˚ (negligibly small) ​

Question 7

For a monatomic gas kinetic energy =E. The relation with rms velocity is:

Accepted Answer

Kinetic energy (E)=23​kT RMS speed (u)=m3kT​​⇒u=m2E​​

Question 8

A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is

Accepted Answer

From conservation of angular momentum (Iω= constant ), angular velocity will remain half. As, K=21​Iω2The rotational kinetic energy will become half.

Question 9

A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP​,vQ​ and vC​ be the magnitude of velocities of points P,Q and C respectively, then

Accepted Answer

In case of pure rolling bottom most point is the instantaneous centre of zero velocity.Velocity of any point on the disc, v=rω, where r is the distance of point from O.rQ​>rC​>rP​vQ​>vC​>vP​​

Question 10

(a) In the following equilibrium N2​O4​( g)⇌2NO2​( g) when 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given thatΔGf∘​(N2​O4​)=100 kJ,ΔGf∘​(NO2​)=50 kJ(i) Find ΔG of the reaction.(ii) The direction of the reaction in which the equilibrium shifts.(b) A graph is plotted for a real gas that follows van der Waals' equation with pVm​ taken on Y-axis and p on X-axis. Find the intercept of the line where Vim ​ is molar volume.

Accepted Answer

(a)​N2​O4​(g)⇌2NO2​(g)ΔG∘=2ΔGf∘​(NO2​)−ΔGf∘​(N2​O4​)=0​Also ΔG∘=−RTlnK=0,K=1Let the reaction shift in the forward direction.N2​O4​( g)⇌2NO2​( g) Total 5−x5+2x10+xpi​=10+x5−x​×2010+x5+2x​×20⇒K=(10+x)2(5+2x)2​×5−x10+x​×20=1⇒81x2+405x+450=0x=−1.66 and −3.33​Both values of x indicate that the reaction actually proceeds in the backward direction.(b) (p+Vm2a​)(Vm​−b)=RT(p+(pV)2ap2​)(ppV​−b)=RT⇒[(pV2)p+ap2][(pV)−b]=p(pV)2RT⇒p⋅[pV2+ap](pV−bp)=p(pV2)RT But p=0 Intercept =RT⇒(pV)3=(pV)2RT​

Question 11

HX is a weak acid (Ka​=10−5). It forms a salt NaX(0.1M) on reacting with caustic soda. The degree of hydrolysis of NaX is

Accepted Answer

Kh​(X−)=Ka​Kw​​=10−510−14​=10−9⇒α=CKh​​​=0.1010−9​​=10−4% hydrolysis =100α=0.01​

Question 12

The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47 th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area ( in 2 cm2 ) of the wire in appropriate number of significant figures.

Accepted Answer

Least count of screw gauge =1001 mm​=0.01 mmDiameter of wire =(1+47×0.01)mm=1.47 mmCurved surface area ( in cm2)=(2π)(2d​)(L)orS​=πdL=(π)(1.47×10−1)(5.6)cm2=2.5848 cm2​Rounding off to two significant digitsS=2.6 cm2

Question 13

In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

Accepted Answer

Young's modulus of elasticity is given byY= strain  stress ​=l/LF/A​=lAFL​=l(4πd2​)FL​Substituting the values, we getY​=(1.25×10−3)×π×(5.0×10−4)250×1.1×4​=2.24×1011 N/m2​Now,YΔY​ΔY​=LΔL​+lΔl​+2dΔd​=(1100.1​)+(0.1250.001​)+2(0.050.001​)=0.0489=(0.0489)Y=(0.0489)×(2.24×1011)N/m2=1.09×1010 N2​

Question 14

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be

Accepted Answer

Area under acceleration-time graph gives the change in velocity.vi​ is zero. Therefore Δv or vf​−vi​ is vf​ or vmax​.Hence, vmax​= area of graph =(1/2)×10×11=55 m/s

Question 15

A small block slides without friction down an inclined plane starting from rest. Let sn​ be the distance travelled from t=n−1 to t=n. Then, sn+1​sn​​ is

Accepted Answer

Distance travelled in nth  second is, sn​=u+an−21​aGiven, u=0∴sn+1​sn​​=a(n+1)−21​aan−21​a​=2n+12n−1​∴ Correct option is (c).

Question 16

If the lines 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then the value of k is

Accepted Answer

Since, the lines intersect, therefore they must have a point in common, i.e. and ⇒​2x−1​=3y+1​=4z−1​=λ1x−3​=2y−k​=1z​=μx=2λ+1,y=3λ−1z=4λ+1​and1x−3​=2y−k​=1z​=μand x=μ+3,y=2μ+k,z=μ are same.⇒2λ+1=μ+33λ−1=2μ+k⇒4λ+1=μ​On solving Ist and IIIrd terms, we get,∴⇒∴​λ=−23​ and μ=−5k=3λ−2μ−1k=3(−23​)−2(−5)−1=29​k=29​​

Question 17

A plane is parallel to two lines whose direction ratios are (1,0,−1) and (−1,1,0) and it contains the point (1,1,1). If it cuts coordinate axes at A,B,C. Then find the volume of the tetrahedron OABC.

Accepted Answer

Let the equation of plane through (1,1,1) having a,b,c as DR's of normal to plane,a(x−1)+b(y−1)+c(z−1)=0and plane is parallel to straight line having DR's.⇒ and ⇒​(1,0,−1) and (−1,1,0)a−c=0−a+b=0a=b=c​∴ Equation of plane is or ​3x​+3y​+3z​​x−1+y−1+z−1=1.​=0Its intercept on coordinate axes areA(3,0,0),B(0,3,0),C(0,0,3)Hence, the volume of tetrahedron OABC=61​[abc]=61​∣∣​300​030​003​∣∣​=627​=29​ cu units

Question 18

T is a parallelepiped in which A,B,C and D are vertices of one face and the face just above it has corresponding vertices A′,B′,C′,D′,T is now compressed to S with face ABCD remaining same and A′,B′,C′,D′ shifted to A′′,B′′,C′′,D′′ in S. The volume of parallelepiped S is reduced to 90% of T. Prove that locus of A′′ is a plane.

Accepted Answer

Let the equation of the plane ABCD be ax+by+cz+d=0, the point A′′ be (α,β,γ) and the height of the parallelepiped ABCD be h.⇒a2+b2+c2​∣aα+bβ+cγ+d∣​=90%h⇒aα+bβ+cγ+d=±0.9ha2+b2+c2​∴ Locus is αx+by+cz+d=±0.9ha2+b2+c2​​Hence, locus of A′′ is a plane parallel to the plane ABCD.

Question 19

Number of lone pair(s) in XeOF4​ is/are

Accepted Answer

At the central atom (Xe), there is one lone pair.

Question 20

On the basis of ground state electronic configuration, arrange the following molecules in increasing O−O bond length order. KO2​,O2​,O2​[AsF6​]

Accepted Answer

Bond order : O2−​=1.5,O2​=2,O2+​=2.5Bond length: O2+​<O2​<O2−​

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Solutions For Exams JEE Advanced, 2004

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