Q: The density of a solid ball is to be determined in experiment. The diameter of the ball is measured with screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions.If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

Least count of screw gauge =500.5=0.01 mm=ΔrDiameter, r=2.5 mm+20×500.5=2.70 mmrΔr=2.700.01 or rΔr×100=2.71Now, density, d=Vm=34π(2r)3mHere, r is the diameter.∴dΔd×100={mΔm+3(rΔr)}×100=mΔm×100+3×(rΔr)×100=2%+3×2.71=3.11%

Q: Taking the electronic charge as e and the permittivity as ε0, use dimensional analysis to determine the correct expression for ωp.

N= Number of electrons per unit volume∴[N][ε0]=[L−3],[e]=[q]=[It]=[AT]=[M−1 L−3 T4 A2]Substituting the dimensions, we can see thatωp=[mε0Ne2]=[T−1]Angular frequency has also the dimension [T−1].

Q: Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N≈4×1027 m−3. Take ε0≈10−11 and m≈10−30, where these quantities are in proper SI units.

ω=2πf=λ2πc⇒λ=ω2πc=mε0Ne22πcSubstituting the values, we get λ≃600 nm.

Q: The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then,

E=21mω2A2 or E∝A2,E1E2=(A1A2)2=(2aa)2 or E1=4E2

Q: Dissolving 120 g of urea (mol. wt. 60 ) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

Molarity = Volume of solution (L) Moles of solute Moles of urea =60120=2 Weight of solution = Weight of solvent + Weight of solute =1000+120=1120 g⇒ Volume =1.15 g/mL1120 g×1000 mL/L1=0.973 L⇒ Molarity =0.9732.000=2.05M

Q: The difference in the oxidation numbers of the two types of sulphur atoms in Na2 S4O6 is

Na2 S4O6 is a salt of H2 S4O6 which has the following structure⇒ Difference in oxidation number of two types of sulphur =5

Q: The work function (ϕ) of some metals is listed below. The number of metals that will show a photoelectric effect when the light of 300 nm wavelength falls on the metal isMetalΦ(cV)Li2.4Na2.3K2.2Mg3.7Cu4.8Ag4.3Fe4.7Pt6.3W4.75

Energy of photon=λhcJ=eλhceV=300×10−9×1.602×10−196.625×10−34×3×108=4.14eVFor the photoelectric effect to occur, the energy of incident photons must be greater than the work function of the metal. Hence, only Li,Na, K and Mg have work functions less than 4.14 V.

Question 1

A ball of mass (m)0.5 kg is attached to the end of a string having length (l) 0.5 m. The ball is rotated on a horizontal circular path like a physical pendulum about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

Accepted Answer

r=lsinθTcosθ component will cancel mg.Tsinθ component will provide the necessary centripetal force to the ball towards center C.∴Tsinθ=mrω2=m(lsinθ)ω2 or T=m/ω2∴ω=mlT​​ or ωmax​=mlTmax​​​=0.5×0.5324​​=36 rad/s

Question 2

The density of a solid ball is to be determined in experiment. The diameter of the ball is measured with screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions.If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

Accepted Answer

Least count of screw gauge =500.5​=0.01 mm=ΔrDiameter, r=2.5 mm+20×500.5​=2.70 mmrΔr​=2.700.01​ or rΔr​×100=2.71​Now, density, d=Vm​=34​π(2r​)3m​Here, r is the diameter.∴dΔd​×​100={mΔm​+3(rΔr​)}×100=mΔm​×100+3×(rΔr​)×100=2%+3×2.71​=3.11%​

Question 3

Taking the electronic charge as e and the permittivity as ε0​, use dimensional analysis to determine the correct expression for ωp​.

Accepted Answer

N= Number of electrons per unit volume∴[N][ε0​]​=[L−3],[e]=[q]=[It]=[AT]=[M−1 L−3 T4 A2]​Substituting the dimensions, we can see thatωp​=[mε0​Ne2​​]=[T−1]Angular frequency has also the dimension [T−1].

Question 4

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N≈4×1027 m−3. Take ε0​≈10−11 and m≈10−30, where these quantities are in proper SI units.

Accepted Answer

ω=2πf=λ2πc​⇒λ=ω2πc​=mε0​Ne2​​2πc​Substituting the values, we get λ≃600 nm.

Question 5

The phase space diagram for a ball thrown vertically up from ground is

Accepted Answer

Momentum is first positive but decreasing. Displacement (or say position) is initially zero. It will first increase. At highest point, momentum is zero and displacement is maximum. After that momentum is downwards (negative) and increasing but displacement is decreasing. Only (d) option satisfies these conditions.

Question 6

The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1​ and E2​ are the total mechanical energies respectively. Then,

Accepted Answer

E=21​mω2A2 or E∝A2,E1​E2​​=(A1​A2​​)2=(2aa​)2 or E1​=4E2​​

Question 7

Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

Accepted Answer

In all the given four figures, at mean position, the position coordinate is zero. At the same time, mass is starting from the extreme position in all four cases. In Fig. (c) and (d), the final extreme position is more than the initial extreme position. But due to viscosity position should be the less.Hence, the answer should be either option (a) or option (b). Correct answer is (b), because mass starts from positive extreme position (from uppermost position). Then, it will move downwards or momentum should be negative.

Question 8

Dissolving 120 g of urea (mol. wt. 60 ) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

Accepted Answer

Molarity = Volume of solution (L) Moles of solute ​ Moles of urea =60120​=2 Weight of solution = Weight  of solvent + Weight of solute =1000+120=1120 g⇒ Volume =1.15 g/mL1120 g​×1000 mL/L1​=0.973 L⇒ Molarity =0.9732.000​=2.05M​

Question 9

The difference in the oxidation numbers of the two types of sulphur atoms in Na2​ S4​O6​ is

Accepted Answer

Na2​ S4​O6​ is a salt of H2​ S4​O6​ which has the following structure⇒ Difference in oxidation number of two types of sulphur =5

Question 10

The work function (ϕ) of some metals is listed below. The number of metals that will show a photoelectric effect when the light of 300 nm wavelength falls on the metal isMetalΦ(cV)​Li2.4​Na2.3​K2.2​Mg3.7​Cu4.8​Ag4.3​Fe4.7​Pt6.3​W4.75​​

Accepted Answer

Energy of photon=λhc​J=eλhc​eV=300×10−9×1.602×10−196.625×10−34×3×108​=4.14eVFor the photoelectric effect to occur, the energy of incident photons must be greater than the work function of the metal. Hence, only Li,Na, K and Mg have work functions less than 4.14 V.

Question 11

The maximum number of electrons that can have a principal quantum number, n=3 and spin quantum number, ms​=−1/2, is

Accepted Answer

When n=3,l=0,1,2 i.e. there are 3s,3p and 3d-orbitals. If all these orbitals are completely occupied asTotal 18 electrons, 9 electrons with s=+21​ and 9 withs=−21​. Alternatively, In any nth orbit, there can be a maximum of 2n2 electrons. Hence, when n=3, number of maximum electrons = 18. Out of these 18 electrons, 9 can have spin −21​ and remaining nine with spin+21​

Question 12

If z is any complex number satisfying ∣z−3−2i∣≤2, then the maximum value of ∣2z−6+5i∣ is .

Accepted Answer

Given, ∣z−3−2i∣≤2To find minimum of ∣2z−6+5i∣ or 2∣∣​z−3+25​i∣∣​, using triangle inequality i.e. ∣∣z1​∣−∣z2​∣∣≤∣z1​+z2​∣∴∣∣​z−3+25​i∣∣​=∣∣​z−3−2i+2i+25​i∣∣​=∣∣​(z−3−2i)+29​i∣∣​≥∣∣z−3−2i∣∣​−29​∣∣​≥∣∣​2−29​∣∣​≥25​⇒∣∣​z−3+25​i∣∣​≥25​ or ∣2z−6+5i∣≥5​

Question 13

Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…9.

Accepted Answer

(P)  eθ⋅eiα=e(θ+iα)Given zk​=ei102kπ​⇒zk​⋅zj​=ei(102π​)(k+j)zk​ is 10 th root of unity.⇒zˉk​ will also be 10th root of unity.Taking, zj​ as zˉk​, we have zk​⋅zj​=1 (True)(Q) eiαeθ​z​=e∠θ−α)=zk​/z1​=e(10zππ​−102π​)=ej5π​(k−1)​For k=2;z=ei5π​ which is in the given set (False)(R) (i) 1−cos2θ=2sin2θ(ii) sin2θ=2sinθcosθ and(i) cos36∘=45​−1​(ii) cos108∘=45​+1∣1−z1​∣∣1−z2​∣…∣1−z9​∣​∣1−zk​∣​=∣∣​1−cos102πk​−isin102πk​∣∣​=∣∣​2sin10πk​∣∣​∣∣​sin10πk​−icos10πk​∣∣​=2∣∣​sin10πk​∣∣​​Now, required product is1029sin10π​⋅sin102π​⋅sin103π​…sin108π​⋅sin109π​​=1029(sin10π​sin102π​sin103π​sin104π​)2sin105π​​=1029(sin10π​cos10π​⋅sin102π​cos102π​)2⋅1​=1029(21​sin5π​⋅21​sin52π​)2​=1025(sin36∘⋅sin72∘)2​=22×1025​(2sin36∘sin72∘)2=522​(cos36∘−cos108∘)2=522​[(45​−1​)+(45​+1​)]2=522​⋅45​=1​(S) Sum of nth roots of unity =01+α+α2+α3+…+α91+k=1∑9​αh1+k=1∑9​(cos102kπ​+isin102kπ​)1+k=1∑9​cos102kπ​1−k=1∑9​cos102kπ​​=0=0=0=0=2​(P) → (i), (Q) → (ii), (R) → (iii), (S) → (iv)

Question 14

Let ω=ejπ/3 and a,b,c,x,y,z be non-zero complex numbers such that a+b+c=x,a+bω+cω2=y, a+bω2+cω=zThen, the value of ∣a∣2+∣b∣2+∣c∣2∣x∣2+∣y∣2+∣z∣2​ is ......

Accepted Answer

Now, consider, ∣x∣2+∣y∣2+∣z∣2=x⋅xˉ+y⋅yˉ​+z⋅zˉSubstitute the given values of x,y and z.​=(a+b+c)(a+b+c)​+(a+bω+cω2)(a+bω+cω2)​+(a+bω2+cω)(a+bω2+cω)​=(a+b+c)(aˉ+bˉ+cˉ)+(a+bω+cω2)(aˉ+bˉω+cˉ2)+(a+bω2+cω)(aˉ+bˉω2+​c)​Using 1+ω+ω2=0 and ω3=1 and simplifying, we get∣x∣2+∣y∣2+∣z∣2=3∣a∣2+3∣b∣2+3∣c∣2=3(∣a∣2+∣b∣2+∣c∣2)Then, (a)2+(b)2+∣c∣2∣x∣2∣+∣y∣2+∣z∣2​=3

Question 15

Let α and β be the roots of x2−6x−2=0, with α>β. If an​=αn−βn for n≥1, then the value of 2a9​a10​−2a8​​ is

Accepted Answer

2a9​a10​−2a8​​=2(α9−β9)(α10−β10)−2(α8−β8)​=2(α9−β9)α8(α2−2)−β8(β2−2)​∵α is root of x2−6x−2=0⇒α2−2=6α [and β is root of x2−6x−2=0⇒β2−2=6β ] =2(α9−β9)α8(6α)−β8(6β)​=2(α9−β9)6(α9−β9)​=3​

Question 16

A value of b for which the equations x2+bx−1=0, x2+x+b=0 have one root in common is

Accepted Answer

If a1​x2+b1​x+c1​=0and a2​x2+b2​x+c2​=0have a common real root, then(a1​c2​−a2​c1​)2=(b1​c2​−b2​c1​)(a1​b2​−a2​b1​)∴x2+bx−1=0x2+x+b=0​} have a common root.(1+b)2b2+2b+1b3+3bb(b2+3)b​=(b2+1)(1−b)=b2−b3+1−b=0=0=0,±3​i​

Question 17

According to kinetic theory of gases:

Accepted Answer

(a) According to a postulate of kinetic theory of gases, collision between the molecules as well as with the wall of container is perfectly elastic in nature.(b) If a gas molecule of mass m moving with speed u collide to the wall of container, the change in momentum is Δp=−2mu. Therefore, heavier molecule will transfer more momentum to the wall as there will be greater change in momentum of the colliding gas molecule. However, this is not postulated in kinetic theory.(c) According to Maxwell-Boltzmann distribution of molecular speed, very few molecules have either very high or very low speeds. Most of the molecules moves in a specific, intermediate speed range.(d) According to kinetic theory of gases, a gas molecule moves in straight line unless it collide with another molecule or to the wall of container and change in momentum is observed only after collision.

Question 18

To an evacuated vessel with a movable piston under the external pressure of 1 atm,0.1 mole of He, and 1.0 mole of an unknown compound (vapor pressure 0.68 atm at 0∘C ) are introduced. Considering the ideal gas behavior, the total volume (in liter) of the gases at 0∘C is close to

Accepted Answer

Since, the external pressure is 1.0 atm, the gas pressure is also 1.0 atm as piston is movable. Out of this 1.0 atm partial pressure due to unknown compound is 0.68 atm.Therefore, partial pressure of He=1.00−0.68=0.32 atm.⇒ Volume =p(He)n(He)RT​=0.320.1×0.082×273​=7 L⇒ Volume of container = Volume of He.​

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Solutions For Exams JEE Advanced, 2011

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