IIT-JEE 2008: Linked Comprehension Type Multiple Choice Questions on Bohr Model of Hydrogen-like Atoms

The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

In a mixture of H – He⁺ gas (He⁺ is singly ionized He atom) H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

(1) The quantum number n of the state finally populated in He⁺ ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (E) of the electron in the n^th orbit is given by

E = – 13.6 Z²/n² electron volt where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of n equal to 1, 2, 3 and 4 respectively

For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of n equal to 1, 2, 3 and 4 respectively

The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3^rd excited state (n = 4)

of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

(2) The wave length of light emitted in the visible region by He⁺ ions after collision with H atoms is

(A) 6.5×10^–7 m

(B) 5.6×10^–7 m

(C) 4.8×10^–7 m

(D) 4.0×10^–7 m

The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to remember the range of energies in eV. You will definitely remember the wave length range of visible photons which you can convert into energy range using the equation, λ E = 12400  where λ is the wave length in Angstrom and E is the energy in electron volt].

The energy difference between the He⁺ ion levels for n = 3 and n = 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) =  2.64 eV

The wave length of the light emitted in the visible region is therefore 124000/2.64 = 4800 Ǻ, nearly. The answer is thus 4.8×10^–7 m [Option (C)].

(3) The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He⁺ ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n^th orbit is 13.6 Z²/n².

[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z_H²/Z_He² = ¼.

All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

You can find a few posts on hydrogen atom at apphysicsresources.blogspot.com also. One such post is here.

Kerala Engineering Entrance 2008 Questions on Heat & Thermodynamics

The following four questions on heat and thermodynamics appeared in KEAM (Engineering) 2008 question paper:

(1) Three rods made of the same material and having the same cross section have been joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

(a) 45º C

(b) 90º C

(c) 30º C

(d) 20º C

(e) 60º C

The quantity of heat flowing through the two hotter rods has to flow through the colder rod so that we have

2×KA(90º – T)/L = KA(T– 0)/L where K, A and L are respectively the thermal conductivity, area of cross section and the length of each rod and T is the temperature of the junction in degree Celsius.

This gives 180º – 2T = T from which T = 60º

(2) The PV diagram of a gas undergoing a cyclic process (ABCDA) is shown in the graph where P is in units of Nm^–2 and V is in cm^–3. Identify the incorrect statement

(a) 0.4 J of work is done by the gas from A to B

(b) 0.2 J of work is done by the gas from C to D

(c) No work is done by the gas from B to C

(d) Net work is done by the gas in one cycle is 0.2 J

(e) Work is done by the gas in going from B to C and on the gas fromD to A

BC and DA represent isochoric processes (in which volume does not change) and hence no work is done. So, the incorrect statement is (e).

[Since Kerala entrance questions are single answer type there is no need of checking the other options when you are actually facing the examination. But it will be usful to know the correctness of the other options at the present moment.

Work (PdV) which is equal to 2×10⁵×2×10^–6 = 0.4 J is done by the gas from A to B since the gas expands.

Work (PdV) which is equal to 1×10⁵×2×10^–6 = 0.2 J is done on the gas from C to D since the gas gets compressed.

Since the cyclic process is clockwise, a net amount of work is done by the gas and its value is equal to the area of the closed curve which is 1×10⁵×2×10^–6 = 0.2 J].

(3) The plots of intensity of radiation versus wave length of three black bodies at temperatures T₁, T₂ and T₃ are shown. Then

(a) T₃> T₂> T₁

(b) T₁> T₂> T₃

(c) T₂> T₃> T₁

(d) T₁> T₃> T₂

(e) T₃> T₁> T₂

According to Wien’s law the product λ_mT is a constant (equal to 0.29 cmK) in the case of any black body (where λ_m and T are respectively the wave length of emitted radiation of maximum intensity and T is the temperature of the black body). Obviously, option (d) is correct.

(4) A bubble of 8 moles of helium is submeged at a certain depth in water. The temperature of water increases by 30º C. How much heat is added approximately to helium during expansion?

(a) 4000 J

(b) 3000 J

(c) 3500 J

(d) 4500 J

(e) 5000 J

The heat δQ added to helium is given by

δQ = mC_PδT wher m is the mass of helium, C_P is its specific heat at constant pressure and δT is its rise in temperature.

[You have to use C_P (and not the specific heat at constant volume C_V) since the bubble absorbs the heat at constant pressure. Further, since we are given the amount of the gas in moles, the molar specific heat is to be used in the above relation].

Therefore, δQ = 8×(5/2)R×30, noting that helium is mono atomic so that its C_P = (5/2)R where R is universal gas constant which is approximately equal to 8.3 J mol^–1K^–1

Thus δQ = 8×(5/2) ×8.3×30 = 5000 J, nearly.

You will find a useful post on the equations to be remembered in thermodynamics here

Kerala Engineering Entrance 2008 Questions on Rotational Motion

Here are the two questions on rotational motion which appeared in KEAM (Engineering) 2008 question paper:

(1) A thin circular ring of mass M and radius R rotates about an axis through the centre and perpendicular to its plane, with a constant angular velocity ω. Four small spheres each of mass m (negligible radius) are kept gently at the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

(a) 4ω

(b) Mω/4m

(c) [(M+4m)/M]ω

(d) [M/(M–4m)]ω

(e) [M/(M+4m)]ω

The angular miomentum of the system is conserved since there is no external torque. Therefore we have I₁ω₁ = I₂ω₂ where I₁ and I₂ are the initial and final moments of inertia and ω₁ and ω₂ are the initial and final angular velocities of the system.

We have I₁= MR², ω₁ = ω and I₂ = MR²+4mR².

Therefore, MR²ω = (M+4m)R²ω₂ from which ω₂ = [M/(M+4m)]ω

[Note that the four masses can be placed anywhere on the ring and you will get the same answer as above].

(2) The angular velocity of a wheel increases from 100 rps to 300 rps in 10 seconds. The number of revolutions made during that time is

(a) 600

(b) 1500

(c) 1000

(d) 3000

(e) 2000

The angular acceleration (α) of the wheel is given by

α = Change of angular velocity/Time = (300–100)/10 = 20 revolutions per second².

[We have retained the angular velocity in rps itself since the number of revolutions is to be found out]

Substituting the known values in the equation, ω² = ω₀² + 2αθ (which is similar to v² = v₀² +2as in linear motion), we have

300² = 100² + 2×20×θ

[The angular displacement θ will be in revolutions for obvious reasons]

From the above equation, θ = 2000 revolutions.

You can find all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.

You will find many useful multiple choice questions (with soluton) involving rotational motion at apphysicsresources here as well as here

Kerala Medical Entrance 2008 Questions on Rotational Motion

Here are the two questions on angular motion which appeared in the Kerala Medical Entrance 2007 question paper:

(1) When a ceiling fan is switched off, its angular velocity reduces to half its initial value after it completes 36 rotations. The number of rotations it will make further before coming to rest is (Assume angular retardation to be uniform)

(a) 10

(b) 20

(c) 18

(d) 12

(e) 16

You have to use the equation, ω² = ω₀² + 2αθ for finding the angular acceleration α and hence the number of further rotations. Note that this equation is the rotational analogue of the equation v² = v₀² + 2as (or, v² = u² + 2as) in linear motion.

Since the angular velocity has reduce to half of the initial value ω₀ after 36 rotations, we have

(ω₀ /2) ² = ω₀² + 2α×36 from which α = – ω₀²/96

[We have expressed the angular displacement θ in rotations itself for convenience]

If the additional number of rotations is x, we have

0 = (ω₀ /2) ² + 2αx = (ω₀ /2) ² + 2×(– ω₀²/96)x

This gives

x = 12

(2) Two particles starting from a point on a circle of radius 4 m in horizontal plane move along the circle with constant speeds of 4 ms^–1 and 6 ms^–1 respectively in opposite directions. The particles will collide with each other after a time of

(a) 3 s

(b) 2.5 s

(c) 2.0 s

(d) 1.5 s

(e) 3.5 s

This is a very simple question. The sum of the distances to be traveled by the two particles is 2πR = 2π×4 = 8π.

The relative speed is 10 ms^–1 so that the time required for the particles to collide is 8π/10 = 2.5 s.

You can find all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.

IIT-JEE 2008 Questions on Direct Current Circuits

Two questions from direct current circuits appeared in the IIT-JEE 2008 question paper. Here are those questions:

Figure shows three resistor configurations R₁, R₂ and R₃ connected to 3 V battery. If the power dissipated by the configurations R₁, R₂ and R₃ are P₁, P₂ and P₃ respectively, then

(a) P₁ > P₂ > P₃

(b) P₁ > P₃ > P₂

(c) P₂ > P₁ > P₃

(d) P₃ > P₂ > P₁

For the configuration R₁, the central 1 Ω resistor can be ignored since it is connected between equipotential points (similar to the galvanometer in a balanced Wheatstone bridge). The equivalent resistance connected across the battery is therefore 1 Ω. [2 Ω in parallel with 2 Ω].

The equivalent resistance in the configuration R₂ is 0.5 Ω. [2 Ω, 2 Ω and 1 Ω in parallel]

The equivalent resistance in the configuration R₃ is 2 Ω.

Since power P = V²/R we have P₂ > P₁ > P₃ [Option (c)].

The second question which follows is Reasoning Type with 4 choices out of which only one is correct

STATEMENT-1

In a metre bridge experiment, null point null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature.. The null point can be obtained at the same point as before by decreasing the value of the standard resistance

and

STATEMENT-2

Resistance of a metal increases with increase in temperature.

(A) STATEMENT-1 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-1

(C) STATEMENT-1 is True, STATEMENT-2 is False

(D) STATEMENT-1 is False, STATEMENT-2 is True

We have R/X = l₁/l₂

If the unknown resistance X is decreased on increasing its temperature, the null point can be obtained at the sme point as before by decreasing the value of the standard resistance. You should note that nothing is mentioned about the nature of the unknown resistance. It can be a negative temperature coefficient (of resistance) device such as a thermistor. So, statement 1 is true.

Metals have positive temperature coefficient of resistance. So, statement 2 is true. But it is not a correct explanation for statement-1.

The correct option is (B).

[If the unknown resistance is a metallic wire such as copper wire or nichrome wire, the resistance will increase on heating and the correct option will then be (D).

Multiple Choice Questions (MCQ) on Elastic Collision

The figure shows three identical blocks A, B and C (each of mass m) on a smooth horizontal surface. B and C which are initially at rest, are connectd by a spring of negligible mass and force constant K. The block A is initially moving with velocity ‘v’ along the line joining B and C and hence it collides with B elastically.

Now, answer the following questions related to the system:

(1) After the collision the velocity of the block A is

(a) v/3

(b) – v

(c) – v/3

(d) v/2

(e) zero

At the moment the block A collides with the block B, you need consider these two blocks only. In one dimensional elastic collision between two bodies of the same mass, if one body is initially at rest, the moving body comes to rest and the body initially at rest moves with the velocity of the moving body. After the collision, the velocity of A is therefore zero.

[Generally, if the moving block 1 has mass m₁ and it moves with initial velocity v_1i towards the stationary block 2 of mass m₂, the final velocities of blocks 1 and 2 are given by

v_1f = v_1i(m₁ – m₂) /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(m₁ + m₂)

The above relations can be easily obtained from momentum and kinetic energy conservation equations.

(2) When the compression of the spring is maximum, the velocity of block B is

(a) v

(b) – v

(c) – v/3

(d) v/2

(e) zero

On receiving the entire momentum (mv) of the block A, the block B moves towards the block C, compressing the spring. When the compression of the spring is maximum, both B and C move with the same velocity (v’). Since B and C have the same mass (m) and momentum has to be conserved, we have

mv = (m + m) v’

Therefore v’ = v/2.

(3) When the compression of the spring is maximum, the kinetic energy of the system is

(a) (½) mv²

(b) (1/4) mv²

(c) Zero

(d) mv²

(e) (1/8) mv²

The block A has come to rest after colliding with block B. When the compression of the spring is maximum, both B and C are moving with the same velocity v/2 as shown above. Therefore, the kinetic energy of the system under this condition is 2×(1/2)m(v/2)² = (1/4) mv². [Note that the spring has negligible kinetic energy since its mass is negligible].

(4) The maximum compression of the spring is

(a) √(mv²/2K)

(b) √(mv/2K)

(c) mv/2K

(d) mv²/2K

(e) √(2mv²/K)

The total energy of the system is equal to the initial kinetic energy (½ )mv² of the colliding block A. Therfore we have

(½)mv² = KE of A + KE of B + KE of C + PE of spring. The KE of A is zero after the collision. As shown above, the total KE of B and C is (1/4) mv² when the compression of the spring is maximum. Therfore, the potential energy (PE) of the spring at maximum compression is given by the equation,

(½)mv² = (1/4) mv² + (1/2)Kx² where x is the maximum compression of the spring.

This gives (1/2)Kx² = (1/4) mv² from which x = √(mv²/2K).

You will find similar questions (with solution) from different branches of physics at apphysicsresources.blogspot.com

Two Questions from Electrostatics-- Capacitor Combinations

Here are two multiple choice questions on effective capacitance of capacitor networks. [The first question was found to be some what difficult for the students (from their response)]:

(1) An unusual type of capacitor is made using four identical plates P₁, P₂, P₃ and P_4, each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P₂ and P₄. Wires soldered to the plates P₁ and P₃ serve as terminals T₁ and T₂ of the system. What is the effective capacitance between the terminals T₁ and T₂?

(a) (2ε₀A)/3d

(b) (3ε₀A)/2d

(c) (3ε₀A)/d

(d) (ε₀A)/d

(e) (ε₀A)/2d

The arrangement contains 3 identlcal capacitors C₁, C₂ and C₃ each of capacitance (ε₀A)/d arranged as shown in the adjoining figure. Plate P₂ is common for C₁ and C₂. Likewise, plates P₃ is common for C₂ and C₃. The capacitors C₂ and C₃ are connected in parallel (to produce an effective value 2ε₀A/d) and this parallel combination is connected in series with the capacitor C₁ of value ε₀A/d . The effective capacitance C between the terminals T₁ and T₂ is therfore given by

C = [(2ε₀A/d) ×(ε₀A/d)] / [(2ε₀A/d) +(ε₀A/d)].

Therefore, C = (2ε₀A)/3d

(2) Four 2 μF capacitors and a 3 μF capacitor are connected as shown in the figure. The effective capacitance between the points A and B is
(a) 3 μF

(b) 2 μF

(c) 1.5 μF

(d) 1 μF

(e) 0.5 μF

This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance between the points A and B is 2 μF.

Two Karnataka CET Multiple Choice Questions on Direct Current Circuits

Questions on direct current circuits have been discussed on this site on various occasions. You can find all those posts by clicking on the label ‘direct current circuit’ below this post. The search box at the top of this page can also be used to access topics of your choice.

The following multiple choice questions appeared in Karnataka CET 2004 question paper:

(1) In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 Ω resistor is

(a) 0.6 A

(b) 1.2 A

(c) 0.9 A

(d) 1.5 A

In steady state, the capacitor is fully charged and no current flows into its branch. So, you can ignore the branch containing the capacitor and the 4 Ω resistor. The circuit thus reduces to the 6 V cell in series with the 2.8 Ω resistor and the parallel combined value of 2 Ω and 3 Ω [ which is 2×3/(2+3) = (6/5) Ω = 1.2 Ω].

The current sent by the cell is 6 V/ (2.8 + 1.2) Ω = (6/4) A = 1.5 A.

This current gets divided between the 2Ω and 3Ω resistor paths. The current (I) through the 2Ω resistor is given by

I = (Main current × Resistance of the other branch)/ Total resistance

[Remember the above relation for the branch current when current gets divided between two branches].

Therefore, we have

I = (1.5×3)/(2+3) = 0.9 A.

(2) An unknown resistance R₁ is connected in series with a resistance of 10 Ω. This combination is connected to one gap of a metre bridge while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. Now, when the 10 Ω resistor is removed and R₁ alone is connected, the balance point shifts to 40 cm. The value of R₁ is (in ohms)

(a) 20

(b) 10

(c) 60

(d) 40

The condition for the balance of a metre bridge (which is basically a Wheatstone bridge) is

P/Q = L/(100 – L) where P and Q are the resistances in the two gaps and the balancing length L cm is measured from the side of P.

In the above problem we have

(R₁ + 10) /R₂ = 1 since the balance point is at the middle of the wire when one gap contains the series combination of R₁ and 10 Ω and the other gap contains R₂.

With R₁ alone in the gap (on removing the 10 Ω resistance), the balance condition is

R₁/R₂ = 40/60

From the above two equations (on dividing one by the other), we have

(R₁ + 10) /R₁ = 60/40, from which R₁ = 20 Ω.