Q: The relation between [E] and [B] is

In terms of dimension, Fe=Fm ⇒qE=qvB or E=vBE]=[B][LT−1]

Q: The relation between [ε0] and [μ0] is

cμ0=μ0ε01⇒c2=μ0ε01=ε0−1⋅c−2⇒[μ0]=[ε0]−1[L−2 T2]

Q: Two vectors A and B are defined as A=ai^ and B=a(cosωi^+sinωj^), where a is a constant and ω=π/6rads−1. If ∣A+B∣=3∣A−B∣ at time t=τ for the first time, the value of τ, in seconds, is

A=ai^ and B=acosωi^+asinωtj^A+B=(a+acosωt)i^+asinωj^A−B=(a−acosωt)i^+asinωtj^∣A+B∣=3∣A−B∣(a+acosωt)2+(asinωt)2=3(a−acosωt)2+(asinωt)2⇒2cos2ωt=±3×2sin2ωt⇒tan2ωt=±31⇒t=(12n±2)s=2 s,10 s,14 s and so on.

Q: Consider the ratio r=(1+a)(1−a) to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is Δa(Δa/a≪1), then what is the error Δr in determining r ?

rlnr=1+a1−a=ln(1−a)−ln(1+a)Differentiating, we getrdr=−1−ada−1+adaor, we can writerΔrrΔr or Δr=−[1−aΔa+1+aΔa]=1−a2−2Δa=−(1−a22Δa)(r)=(1+a)2−2Δa

Q: In an experiment, the initial number of radioactive nuclei is 3000. It is found that 1000±40 nuclei decayed in the first 1.0 s. For ∣x∣≪1,ln(1+x)=x up to first power in x. The error Δλ, in the determination of the decay constant λ ins −1, is

N=N0e−λtlnN=lnN0−λtDifferentiating w.r.t λ, we getN1⋅dλdN∣dλ∣or,∣Δλ∣=0−t=NtdN=NtΔNHere, N=3000−1000=2000∴∣Δλ∣=2000×140=0.02

Q: To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction,MnCl2+K2 S2O8+H2O⟶KMnO4+H2SO4+HCl(equation not balanced).Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ...........(Atomic weights in gmol−1 : Mn=55,Cl=35.5 )

The balanced equations are(1) 2MnCl2+5 K2 S2O8+8H2O→2KMnO4+4 K2SO4+6H2SO4+4HCl(2) 2KMnO4+5H2C2O4+3H2SO4→K2SO4+2MnSO4+8H2O+10CO2Given, mass of oxalic acid added =225mgSo, millimoles of oxalic acid added =90225=2.5Now from Equation 2Millimoles of KMnO4 used to react with oxalic acid=I andMillimoles of MnCl2 required initially=1∴ Mass of MnCl2 required initially =1×(55+71)=126mg

Question 1

The relation between [E] and [B] is

Accepted Answer

In terms of dimension, Fe​=Fm​ ⇒qE=qvB or E=vBE]=[B][LT−1]

Question 2

The relation between [ε0​] and [μ0​] is

Accepted Answer

cμ0​​=μ0​ε0​​1​⇒c2=μ0​ε0​1​=ε0−1​⋅c−2⇒[μ0​]=[ε0​]−1[L−2 T2]​

Question 3

Two vectors A and B are defined as A=ai^ and B=a(cosωi^+sinωj^​), where a is a constant and ω=π/6rads−1. If ∣A+B∣=3​∣A−B∣ at time t=τ for the first time, the value of τ, in seconds, is

Accepted Answer

A=​ai^ and B=acosωi^+asinωtj^​A+B=(a+acosωt)i^+asinωj^​A−B=(a−acosωt)i^+asinωtj^​∣A+B∣=3​∣A−B∣​(a+acosωt)2+(asinωt)2​=3​(a−acosωt)2+(asinωt)2​⇒2cos2ωt​=±3​×2sin2ωt​⇒tan2ωt​=±3​1​⇒t=(12n±2)s=2 s,10 s,14 s and so on. ​

Question 4

Consider the ratio r=(1+a)(1−a)​ to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is Δa(Δa/a≪1), then what is the error Δr in determining r ?

Accepted Answer

rlnr​=1+a1−a​=ln(1−a)−ln(1+a)​Differentiating, we getrdr​=−1−ada​−1+ada​or, we can writerΔr​rΔr​ or Δr​=−[1−aΔa​+1+aΔa​]=1−a2−2Δa​=−(1−a22Δa​)(r)=(1+a)2−2Δa​​

Question 5

In an experiment, the initial number of radioactive nuclei is 3000. It is found that 1000±40 nuclei decayed in the first 1.0 s. For ∣x∣≪1,ln(1+x)=x up to first power in x. The error Δλ, in the determination of the decay constant λ ins −1, is

Accepted Answer

N=N0​e−λtlnN=lnN0​−λt​Differentiating w.r.t λ, we getN1​⋅dλdN​∣dλ∣or,∣Δλ∣​=0−t=NtdN​=NtΔN​​Here, N=3000−1000=2000∴∣Δλ∣=2000×140​=0.02

Question 6

Galena (an ore) is partially oxidised by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the content undergo self-reduction. The weight (in kg ) of Pb produced per kg of O2​ consumed is .........(Atomic weights in gmol−1:O=16, S=32, Pb=207 )

Accepted Answer

The equations of chemical reactions occurring during the process areIn the presence of oxygen2PbS+3O2​⟶2PbO+2SO2​By self-reduction2PbO+PbS⟶3 Pb+SO2​Thus 3 moles of O2​ produces 3 moles of Pb i.e. 32×3=96 g of O2​ produces 3×207=621 g of Pb So 1000 g(1 kg) of oxygen will produce96621​×1000​=6468.75 g=64687 kg≈6.47 kg​

Question 7

To measure the quantity of MnCl2​ dissolved in an aqueous solution, it was completely converted to KMnO4​ using the reaction,MnCl2​+K2​ S2​O8​+H2​O⟶KMnO4​+H2​SO4​+HCl(equation not balanced).Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2​ (in mg) present in the initial solution is ...........(Atomic weights in gmol−1 : Mn=55,Cl=35.5 )

Accepted Answer

The balanced equations are(1) 2MnCl2​+5 K2​ S2​O8​+8H2​O→2KMnO4​+4 K2​SO4​+6H2​SO4​+4HCl(2) 2KMnO4​+5H2​C2​O4​+3H2​SO4​→K2​SO4​+2MnSO4​+8H2​O+10CO2​Given, mass of oxalic acid added =225mgSo, millimoles of oxalic acid added =90225​=2.5Now from Equation 2Millimoles of KMnO4​ used to react with oxalic acid=I andMillimoles of MnCl2​ required initially=1∴ Mass of MnCl2​ required initially =1×(55+71)=126mg

Question 8

In the following reaction sequence, the amount of D (in grams) formed from 10 moles of acetophenone is ....... (Atomic weights in gmol−1:H=1,C=12, N=14, O=16,Br=80. The yield (%) corresponding to the product in each step is given in the parenthesis)

Accepted Answer

Given,The products formed are10 moles of acetophenone.Molar mass of  =240+14+4+72=330Hence, amount of  produced is 330×1.5=495 g

Question 9

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2​⋅6H2​O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl2​⋅6H2​O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is            ​.(Atomic weights in gmol−1 : H=1, N=14,O=16, S=32, Cl=35.5,Ca=40,Ni=59 )

Accepted Answer

Balanced equations of reactions used in the problem are as follows (i) (NH4​)2​SO4​+Ca(OH)2​⟶1 mol132 g (ii) NiCl2​⋅6H2​O+1 mol238 g​CaSO4​⋅2H2​O+1 mol172 g6NH3​⟶6 mol102 g​2NH3​−2 mol(2×17)=34 g[Ni(NH3​)6​]Cl2​+6H2​O1 mol232 g​Now, in Eq. (i)if, 1584 g of ammonium sulphate is used.i.e., 1584 g(NH4​)2​SO4​=1321584​=12 molSo, according to the Eq. (i) given above 12 moles of (NH4​)2​SO4​ produces(a) 12 moles of gypsum(b) 24 moles of ammoniaHere, 12 moles of gypsum =12×172=2064 gand 24 moles of NH3​=24×17=408 gFurther, as given in question,24 moles of NH3​ produced in reaction (i) is completely utilised by 952 g or 4 moles of NiCl2​⋅6H2​O to produce 4 moles of [Ni(NH3​)]Cl2​.So, 4 moles of [Ni(NH3​)1​]Cl2​=4×232=928gmsHence, total mass of gypsum and nickel ammonia coordination compound [Ni(NH3​)6​]Cl2​=2064+928=2992

Question 10

Let s,t,r be non-zero complex numbers and L be the set of solutions z=x+iy(x,y∈R,i=−1​) of the equation sz+tzˉ+r=0, where zˉ=x−iy. Then, which of the following statement(s) is (are) TRUE?

Accepted Answer

We have,sz+tzˉ+r=0..(i).On taking conjugatesˉzˉ+tˉz+rˉ=0..(ii).On solving Eqs. (i) and (ii), we getz=∣s∣2−∣t∣2rˉt−rsˉ​(a) For unique solutions of z∣∣​s2∣−∣∣​t∣∣​2=0⇒∣s∣=∣t∣It is true(b) If ∣s∣=∣t∣, then rˉt−rsˉ may or may not be zero.So, z may have no solutions.∴L may be an empty set.It is false.(c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true.(d) In this case locus of z is a line, so L has infinite elements. Hence it is true.

Question 11

For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE?

Accepted Answer

(a) Letz=−1−i and arg(z)=θNow, tanθ=∣∣​Re(z)im(z)​∣∣​=∣∣​−1−1​∣∣​=1⇒θ=4π​Since, x<0,y<0∴arg(z)=−(π−4π​)=−43π​(b) We have, f(t)=arg(−1+it)arg(−1+it)={π−tan−1t,t≥0−(π+tan−1t),t<0​This function is discontinuous at t=0.(c) We have,arg(z2​z1​​)−arg(z1​)+arg(z2​) Nowarg (z2​z1​​)=arg(z1​)−arg(z2​)+2nπ∴arg(z2​z1​​)−arg(z1​)+arg(z2​)=arg(z1​)−arg(z2​)+2nπ−arg(z1​)+arg(z2​)=2nπ​So, given expression is multiple of 2π.(d) We have, arg ((z−z3​)(z2​−z1​)(z−z1​)(z2​−z3​)​)=π⇒(z−z3​z−z1​​)(z2​−z1​z2​−z3​​) is purely realThus, the points A(z1​),B(z2​),C(z3​) and D(z) taken in order would be concyclic if purely real.Hence, it is a circle.∴ (a), (b), (d) are false statement.

Question 12

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Fig. 1). If the old partition is replaced by a new partition that can slide and conduct heat but does not allow the gas to leak across (Fig. 2), the volume (in m3 ) of the compartment A after the system attains equilibrium is              ​.

Accepted Answer

Given p1​=5 bar, V1​=1 m3,T1​=400 KSo, n1​=400R5​( from pV=nRT)Similarly, p2​=1 bar, V2​=3 m3,T2​=300 K,n2​=300R3​Let at equilibrium the new volume of A will be (1+x)So, the new volume of B will be (3−x)Now, from the ideal gas equation.n1​RT1​p1​V1​​=n2​RT2​p2​V2​​and at equilibrium (due to the conduction of heat)T1​p1​​=T2​p2​​n1​V1​​=n2​V2​​ or V1​n2​=V2​n1​​So, n1​V1​​=n2​V2​​ or V1​n2​=V2​n1​After putting the values(1+x)×300R3​=(3−x)×400R5​ or (1+x)=4(3−x)5​or 4(1+x)=15−5x or 4+4x=15−5x or x=911​Hence, new volume of A i.e., (1+x) will comes as 1+911​=920​ or2.22

Question 13

Consider a hydrogen-like ionised atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n=2 to n=1 transition has energy 74.8 eV higher than the photon emitted in the n=3 to n=2 transition. The ionisation energy of the hydrogen atom is 13.6 eV. The value of Z is ......

Accepted Answer

∴∴​ΔE2−1​=13.6×Z2[1−41​]=13.6×Z2[43​]ΔE3−2​=13.6×Z2[41​−91​]=13.6×Z2[365​]ΔE2​=ΔE3−2​+74.813.6×Z2[43​]=13.6×Z2[365​]+74.813.6×Z2[43​−365​]=74.8Z2=9Z=3​

Question 14

In a photoelectric experiment, a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F=n×10−4 N due to the impact of the electrons. The value of n is ........... (Take mass of the electron, me​=9×10−31 kg and eV=1.6×10−19 J )

Accepted Answer

∴ Power =nhf (where, n= number of photons incident per second)  Since, KE=0,hf= work-function W​KE=0,hf= work-function W200=nW=n[6.25×1.6×10−19]n=1.6×10−19×6.25200​​As photon is just above threshold frequency KEmax​ is zero and they are accelerated by potential difference of 500 V.KEf​2mP2​⇒P​=qΔV=qΔV=2mqΔV​​Since, efficiency is 100%, number of electrons emitted per second = number of photons incident per second.=n(mV)=nP=n2mqΔV​=6.25×1.6×10−19200​×2(9×10−31)×1.6×10−19×500​=24​

Question 15

Consider a hydrogen-like ionised atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n=2 to n=1 transition has energy 74.8eV higher than the photon emitted in the n=3 to n=2 transition. The ionisation energy of the hydrogen atom is 13.6eV. The value of Z is

Accepted Answer

ΔE2−1​ΔE3−2​ΔΔE2​13.6×Z2[43​]13.6×Z2[43​−365​]Z2Z​=13.6×Z2[1−41​]=13.6×Z2[43​]=13.6×Z2[41​−91​]=13.6×Z2[365​]=ΔE3−2​+74.8=13.6×Z2[365​]+74.8=74.8=9Z​

Question 16

In a photoelectric experiment, a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25eV. The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F=n×10−4 N due to the impact of the electrons. The value of n is ........... (Take mass of the electron, me​=9×10−31 kg and eV=1.6×10−19 J )

Accepted Answer

∴ Power =nhf (where, n= number of photons incident per second)  Since, KE=0,hf= work-function W​
KE=0,hf= work-function W200=nW=n[6.25×1.6×10−19]n=1.6×10−19×6.25200​​
As photon is just above threshold frequency KEmax​ is zero and they are accelerated by potential difference of 500 V.
KEf​2mP2​​=qΔV=qΔV⇒P=2mqΔV​​
Since, efficiency is 100%, number of electrons emitted per second = number of photons incident per second.
=n(mV)=nP=n2mqΔV​=6.25×1.6×10−19200​×2(9×10−31)×1.6×10−19×500​=24​

Question 17

For a reaction, A⇌P, the plots of [A] and [P] with time at temperatures T1​ and T2​ are given below.If T2​>T1​, the correct statement(s) is are (Assume ΔH⊖ and ΔS⊖ are independent of temperature and ratio of lnK at T1​ to lnK at T2​ is greater than T2​/T1​. Here H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Accepted Answer

For the reaction, A⇌PGiven, T1T1T2It shows, On increasing the temperature, K decreases so reaction is exothermic i.e., ΔH∘<0Besides, the graph shows K>1So ΔG∘<0Now from equation (i)T1lnK1>T2lnK2−ΔG1∘>−ΔG∘2Likewise (−ΔH0+T1ΔS0)>(−ΔH∘+T2ΔS∘)or simply T1ΔS∘>T2ΔS∘So, (T2−T1)ΔS∘<0∴ΔS0<0In other words, increase of ΔG with increase in temperature is possible only when ΔS∘<0. Hence, options (a) and (c) are correct.

Question 18

The solubility of a salt of weak acid (AB) at pH 3 is Y×10−3 mol L−1. The value of Y is (Given that the value of solubility product of AB(Ksp​)=2×10−10 and the value of ionisation constant of HB(Ka​)=1×10−8

Accepted Answer

Given, pH=3, so [H+]=10−3Ka​=1×10−8⇒Ksp​=2×10−10after putting the values in above formula Solubility =2×10−10(10−810−3​+1)​=2×10−5​=4.47×10−3MHence, the value of y=4.47

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Solutions For Exams JEE Advanced, 2018

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