Q: A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (Take g=10 m/s2 )

W=mg=0.1×10=1 NF1= component of weight =1⋅sinθ=sinθF2= component of applied force =1⋅cosθ=cosθNow, at θ=45∘:F1=F2, and block remains stationary without the help of friction.For θ>45∘,F1>F2, so friction will act towards Q.For θ F1 and friction will act towards P.

Q: In the determination of Young's modulus (Y=π/d24MLg) by using Searle's method, a wire of length L=2 m and diameter d=0.5 mm is used. For a load M=2.5 kg, an extension l=0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the Y measurement is

Δd=Δl=1000.5 mm=0.005 mmY=πld24MLg⇒(YΔY)max=(lΔl)+2(dΔd)(lΔl)=0.250.5/100=0.02 and dΔd=0.5(0.5/100)=0.01 or lΔl=2⋅dΔd

Q: 29.2%(w/W) HCl stock solution has density of 1.25 g mL −1. The molecular weight of HCl is 36.5 g mol−1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4MHCl is

Mass of HCl in 1.0 mL stock solution=1.25×10029.2=0.365 gMass of HCl required for 200 mL0.4MHCl=1000200×0.4×36.5=0.08×36.5 g∴0.365 g of HCl is present in 1.0 mL stock solution.0.08×36.5 gHCl will be present in 0.3650.08×36.5=8.0 mL

Q: The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ a0 is Bohr radius]

According to Bohr's model,mvr=2πnh⇒(mv)2=4π2r2n2h2⇒KE=21mv2=8π2r2mn2h2Also, Bohr's radius for H-atom is, r=n2a0Substituting ' r ' in Eq. (i) givesKE=8π2n2a02mh2 when n=2,KE=32π2a02mh2

Q: For one mole of a van der Waals' gas when b=0 and T=300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a( atm L mol−2)

The van der Waals' equation of state is(p+V2n2a)(V−nb)=nRTFor one mole and when b=0, the above equation condenses to(p+V2a)V⇒pV=RT=RT−Va.........(i)Eq. (i) is a straight equation between pV and V1 whose slope is . - a′, Equating with slope of the straight line given in the graph.−a=3−220.1−21.6=−1.5⇒a=1.5

Q: Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

IP>IQIn case of pure rolling, a=1+I/mR2gsinθaQ>aP as its moment of inertia is less. Therefore, Q reaches first with more linear speed and more translational kinetic energy.Further, ω=Rv or ω∝v∴ωQ>ωP as vP>vQ

Q: A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω, as shown in the figure. At time t=0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t=T and stops. The angular speed of the system remains ω throughout.The magnitude of the torque ∣τ∣ on the system about O, as a function of time is best represented by which plot?

∣L |or L=Iω (about axis of rI=Irod+mx2=Irod+mv2t2Here, m= mass of insect∴L=(Irod+mv2t2)ωNow ∣τ∣=dtdL=(2mv2tω) or ∣τ∣∝t i.e. the graph is a straight line passing through the origin.After time T,L= constant∴∣τ∣ or dtdL=0Maximum friction acting at the bottommost point will convert backward slip into pure rolling. The torque of this friction about the bottommost point is zero.∴ From the conservation of angular momentum about bottom most pointvmr2ω0=mvcr⊥+Icω=mvr+mr2×v/r=2ω0r

Question 1

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (Take g=10 m/s2 )

Accepted Answer

W=mg=0.1×10=1 NF1​= component of weight =1⋅sinθ=sinθF2​= component of applied force =1⋅cosθ=cosθNow, at θ=45∘:F1​=F2​, and block remains stationary without the help of friction.For θ>45∘,F1​>F2​, so friction will act towards Q.For θ<45∘,F2​>F1​ and friction will act towards P.

Question 2

In the determination of Young's modulus (Y=π/d24MLg​) by using Searle's method, a wire of length L=2 m and diameter d=0.5 mm is used. For a load M=2.5 kg, an extension l=0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the Y measurement is

Accepted Answer

Δd=Δl=1000.5​ mm=0.005 mmY=πld24MLg​⇒(YΔY​)max​=(lΔl​)+2(dΔd​)(lΔl​)=0.250.5/100​=0.02 and dΔd​=0.5(0.5/100)​=0.01 or lΔl​=2⋅dΔd​​

Question 3

29.2%(w/W) HCl stock solution has density of 1.25 g mL −1. The molecular weight of HCl is 36.5 g mol−1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4MHCl is

Accepted Answer

Mass of HCl in 1.0 mL stock solution=1.25×10029.2​=0.365 gMass of HCl required for 200 mL0.4MHCl=1000200​×0.4×36.5=0.08×36.5 g∴0.365 g of HCl is present in 1.0 mL stock solution.0.08×36.5 gHCl will be present in 0.3650.08×36.5​=8.0 mL

Question 4

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ a0​ is Bohr radius]

Accepted Answer

According to Bohr's model,mvr=2πnh​⇒(mv)2=4π2r2n2h2​⇒KE=21​mv2=8π2r2mn2h2​​Also, Bohr's radius for H-atom is, r=n2a0​Substituting ' r ' in Eq. (i) givesKE=8π2n2a02​mh2​ when n=2,KE=32π2a02​mh2​

Question 5

The number of radial nodes in 3s and 2p respectively are

Accepted Answer

The number of radial nodes is given by expression (n−l−1).For 3s, number of nodes =3−0−1=2For 2p, number of nodes =2−1−1=0

Question 6

Let z be a complex number such that the imaginary part of z is non-zero and a=z2+z+1 is real. Then, a cannot take the value

Accepted Answer

As imaginary part of z=x+iy is non-zero.⇒y=0Method I Let z=x+iy∴⇒⇒​a=(x+iy)2+(x+iy)+1(x2−y2+x+1−α)+i(2xy+y)=0(x2−y2+x+1−α)+iy(2x+1)=0..(i),​It is purely real, if y(2x+1)=0 but imaginary part of z, i.e. y is non-zero.⇒2x+1=0 or x=−1/2From Eq. (i), 41​−y2−21​+1−a=0⇒a=−y2+43​⇒a<43​Method II Here, z2+z+(1−a)=0∴z=2×1−1±1−4(1−a)​​⇒z=2−1±4a−3​​​For z do not have real roots, 4a−3<0⇒a<43​

Question 7

For one mole of a van der Waals' gas when b=0 and T=300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a( atm L mol−2)

Accepted Answer

The van der Waals' equation of state is(p+V2n2a​)(V−nb)=nRTFor one mole and when b=0, the above equation condenses to(p+V2a​)V⇒pV​=RT=RT−Va​.........(i)​Eq. (i) is a straight equation between pV and V1​ whose slope is . - a′, Equating with slope of the straight line given in the graph.−a=3−220.1−21.6​=−1.5⇒a=1.5

Question 8

Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Accepted Answer

IP​>IQ​In case of pure rolling, a=1+I/mR2gsinθ​aQ​>aP​ as its moment of inertia is less. Therefore, Q reaches first with more linear speed and more translational kinetic energy.Further, ω=Rv​ or ω∝v∴ωQ​>ωP​ as vP​>vQ​

Question 9

A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω, as shown in the figure. At time t=0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t=T and stops. The angular speed of the system remains ω throughout.The magnitude of the torque ∣τ∣ on the system about O, as a function of time is best represented by which plot?

Accepted Answer

∣L |or L=Iω (about axis of rI=Irod​+mx2=Irod​+mv2t2Here, m= mass of insect∴L=(Irod​+mv2t2)ωNow ∣τ∣=dtdL​=(2mv2tω) or ∣τ∣∝t i.e. the graph is a straight line passing through the origin.After time T,L= constant∴∣τ∣ or dtdL​=0Maximum friction acting at the bottommost point will convert backward slip into pure rolling. The torque of this friction about the bottommost point is zero.∴ From the conservation of angular momentum about bottom most pointv​mr2ω0​=mvc​r⊥​+Ic​ω=mvr+mr2×v/r=2ω0​r​​

Question 10

A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2 . as shown in the figure. The moment inertia of this lamina about ax passing through O and P is IO​ and I respectively. Both these axes are perpendicular to the p of the lamina. The ratio SIP​​ to the nearest integer is

Accepted Answer

T= Total portion R= Remaining portion C= Cavity and let σ= mass per unit area. ​Then,mT​=π(2R)2σ=4πR2σmC​=π(R)2σ=πR2σ​ For IP​IR​​=IT​−IC​=23​mT​(2R)2−[21​mC​R2+mC​r2]=23​(4πR2σ)(4R2)−[21​(πR2σ)+(πR2σ)(5R2)]=(18.5πR4σ)​ For IO​IR​=IT​−IC​=21​mT​(2R)2−23​mC​R2=21​(4πR2σ)(4R2)−23​(πR2σ)(R2)=6.5πR4σ∴IO​IP​​=6.5πR4σ18.5πR4σ​=2.846​Therefore, the nearest integer is 3 .

Question 11

Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed ω. The discs are in the same horizontal plane.

Accepted Answer

Language of question is wrong because relative speed is not the correct word. Relative speed between two is always zero. The correct word is magnitude of relative velocity.Corresponding to above values, the correct graph is (a).

Question 12

A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x−y plane with centre at O and constant angular speed ω. If the angular momentum of the system, calculated about O and P are denoted by LO​ and LP​ respectively, then

Accepted Answer

Angular momentum of a particle about a point is given byL=r×p=m(r×v)For L0​∣L∣=(mvrsinθ)=m(Rω)(R)sin90∘= constantDirection of LO​ is always upwards. Therefore, complete LO​ is constant, both in magnitude as well as direction.For LP​∣LP​∣=(mvrsinθ)=(m)(Rω)(l)sin90∘=(mRlω)Magnitude of LP​ will remain constant but direction of LP​ keeps on changing.

Question 13

If a1​,a2​,a3​,… are in a harmonic progression with a1​=5 and a20​=25. Then, the least positive integer n for which a0​<0, is

Accepted Answer

nth term of HP. tn=a+(n−1)n1Here, a1=5,a20=25 for HP∴a1=5 and a+19d1=25 ⇒51+19d =251⇒19d=251−51=−254 ∴d=19×25−4Since, an<0⇒51+(n−1)d<0⇒51−19×254(n−1)<0⇒(n−1)>495⇒n>1+495 or n>24.75∴ Least positive value of n=25

Question 14

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball, is

Accepted Answer

Objects  Distinct  Distinct ​ Groups  Distinct  Identical ​ Objects  Identical  Identical ​ Groups  Identical  Distinct ​Description of Situation Here, 5 distinct balls are distributed amongst 3 persons so that each gets at least one ball. i.e. Distinct → DistinctSo, we should make casesCase I  A 1​B1​C2​} Case II A1​B2​C2​}Number of ways to distribute 5 balls=(5C1​⋅4C1​⋅3C3​×2!3!​)+(5C1​⋅4C2​⋅2C2​×2!3!​)=60+90=150​

Question 15

Four fair dice D1​,D2​,D3​ and D4​ each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4​ shows a number appearing on one of D1​,D2​ and D3​, is

Accepted Answer

As one of the dice shows a number appearing on one of P1​,P2​ and P3​Thus, three cases arise.(i) All show same number.(ii) Number appearing on D4​ appears on any one of D1​,D2​ and D3​.(iii) Number appearing on D4​ appears on any two of D1​⋅D2​ and D3​.Sample space =6×6×6×6=64 favourable events= Case I or Case II or Case IIICase I First we should select one number for D4​ which appears on all i.e. 6C1​×1.Case II For D4​ there are 6C1​ ways. Now, it appears on any one of D1​,D2​ and D3​ i.e. 3C1​×1.For other two there are 5×5 ways.⇒6C1​×3C1​×1×5×5Case III For D4​ there are 6C1​ ways now it appears on any two of D1​,D2​ and D3​⇒3C2​×12For other one there are 5 ways.⇒6C1​×3C2​×12×5Thus, probability =646C1​+6C1​×3C1​×52+6C1​×3C2​×5​=646(1+75+15)​=21691​​

Question 16

If limx→∞​(x+1x2+x+1​−ax−b)=4, then

Accepted Answer

(∞∞​) formlim(x+1x2+x+1​−ax−b)=4⇒limx→∞​x+1x2+x+1−αx2−ax−bx−b​=4⇒limx→∞​x+1x2(1−a)+x(1−a−b)+(1−b)​=4​Here, we make degree of numerator= degree of denominator∴ and ​limx(1−a−b)+(1−b)​1−a=0⇒a=4​=1⇒⇒​​1−a−bb​=4=−4[∵(1−a)=0]​

Question 17

Let α(a) and β(α) be the roots of the equation (31+α​−1)x2−(1+a​−1)x+(61+α​−1)=0, where a>−1. Then, lima→0+​α(a) and lima→0+​β(a) are

Accepted Answer

To make the quadratic into simple form we should eliminate radical sign.Description of Situation As for given equation, when a→0 the equation reduces to identity in x. i.e. ax2+bx+c=0,∀x∈R or a=b=c→0Thus, first we should make above equation independent from coefficients as 0 .Let a+1=t6. Thus, when a→0,t→1.2x2+3x+12x2+2x+x+1(2x+1)(x+1)​=0=0=0​Thus, x=−1,−1/2 or  and ​limα→0+​α(α)=−1/2lima→0+​β(α)=−1​

Question 18

For every integer n, let an​ and bn​ be real numbers. Let function f:R→R be given byf(x)={an​+sinπx,bn​+cosπx,​ for x∈[2n,2n+1] for x∈(2n−1,2n)​,for all integers n.If f is continuous, then which of the following hold(s) for all n ?

Accepted Answer

f(2n)f(2n−)​=an​,f(2n+)=an​=bn​+1​an​−bn​f(2n+1)f{(2n+1)−}f{(2n+1)+}​=1=an​=an​=bn+1​−1​⇒ or ​an​=bn+1​−1 or an​−bn+1​=−1an−1​−bn​=−1​

Question 19

Let f(x)={x2∣∣​cosxπ​∣∣​,0,​x=0,x∈R, then f is x=0​

Accepted Answer

To check differentiability at a point we use RHD and LHD at a point and if RHD=LHD, then f(x) is differentiable at the point.Description of SituationAs,R{f′(x)}=​limh→0​hf(x+h)−f(x)​L{f′(x)}=limh→0​−hf(x−h)−f(x)​​Here, students generally gets confused in defining modulus. To check differentiable at x=0,R{f′(0)}=limh→0​hf(0+h)−f(0)​=limh→0​hh2∣coshπ​∣−0​=limh→0​h⋅∣∣​coshπ​∣∣​=0L{f′(0)}=limh→0​−hf(0−h)−f(0)​=limh→0​−hh2∣cos(−hπ​)∣−0​=0So, f(x) is differentiable at x=0.To check differentiability at x=2,R{f′(2)}=limf(2+h)−f(2)​=limh→0​h(2+h)2∣cos(2+hπ​)∣−0​=limh→0​h(2+h)2⋅cos(2+hπ​)​=limh→0​h(2+h)2⋅sin(2π​−2+hπ​)​=limh→0​π(2+h)2⋅sin(2(2+h)πh​)​⋅2(2+h)π​=π​and L{f′(2)}​=h→0lim​−hf(2−h)−f(2)​=h→0lim​−h(2−h)2⋅∣∣​cos2−hπ​∣∣​−22⋅∣∣​cos2π​∣∣​​)=h→0lim​−h(2−h)2−(−cos2−hπ​)−0​)=h→0lim​h−(2−h)2⋅sin(2π​−2−hπ​)​=h→0lim​h⋅(2−h)2⋅sin(−2(2−h)πh​)​×2(2−h)−π​=−π​Thus, f(x) is differentiable at x=0 but not at x=2.

Question 20

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 2/3​ from the point(3,1,−1) is

Accepted Answer

Equation of plane through intersection of two planes, i.e. (a1​x+b1​y+c1​z+d1​)+λ(a2​x+b2​y+c2​z+d2​)=0​Distance of a point (x1​,y1​,z1​) fromax+by+cz+d=0=a2+b2+c2​∣ax1​+by1​+cz1​+d∣​Equation of plane passing through intersection of two planes x+2y+3z=2 and x−y+z=3 is(x+2y+3z−2)+λ(x−y+z−3)=0⇒(1+λ)x+(2−λ)y+(3+λ)z−(2+3λ)=0​whose distance from (3,1,−1) is 3​2​.⇒(1+λ)2+(2−λ)2+(3+λ)2​∣3(1+λ)+1⋅(2−λ)−1(3+λ)−(2+3λ)∣​=3​2​⇒3λ2+4λ+14​∣−2λ∣​=3​2​​⇒3λ2=3λ2+4λ+14⇒λ=−27​∴(1−27​)x+(2+27​)y+(3−27​)z−(2−221​)=0−25x​+211​y−21​z+217​=0 or 5x−11y+z−17=0​

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Solutions For Exams JEE Advanced, 2012

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