IIT-JEE 2007 Matrix Match Type Question from Units and Dimensions

In the post dated 30^th April 2007, a Matrix Match Type question based on topics in Modern Physics, which appeared in the IIT-JEE 2007 question paper, was discussed. Let us discuss now the question (which appeared in the same examination) based on units and dimensions. Here is the question:

Some physical quantities are given in column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physicsl quantities in column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4×4 matrix given in the ORS.

column I column II

(A) GM_eM_s (p) (volt) (coulomb) (metre)

G – universal gravitational constant,

M_e – mass of the earth,

M_s– mass of the sun

(B) 3RT / M (q) (kilogram)(metre)³(second)^–2

R – universal gas constant,

T – absolute temperature,

M – molar mass

(C) F²/q²B² (r) (metre)²(second)^–2

F – force,

q – charge,

B – magnetic field

(D) GM_e/R_e (s) (farad) (volt)²(kg)^–1

G – universal gravitational constant,

M_e – mass of the earth,

R_e – radius of the earth

(A) has to be matched with (p) and (q). This can be done in various ways. But it will be convenient to do it by noting that the quantity GM_eM_s is force×distance² since the gravitational force between the earth and the sun is GM_eM_s/r², with usual notations.

But, force×distance² = energy×distance. The quantity (volt) (coulomb) (metre) also is energy×distance (since ‘Vq’ is energy). The quantity (kilogram)(metre)³(second)^–2 also is energy×distance (remembering F = ma and energy = Fs).

(B) has to be matched with (r) and (s). This can be proved by noting that 3RT / M is the square of the r.m.s. velocity of molecules. Its unit is therefore (metre)²(second)^–2.

The quantity (farad) (volt)²(kg)^–1 also has the dimensions of velocity² since farad metre² has the dimensions of energy (remember E = ½ CV²) and energy per kg has dimensions of velocity² (E = p²/2m so that E/m = p²/2m² = m²v²/2m² which has dimensions of velocity²).

(C) is to be matched with (r) and (s) since F²/q²B² has dimensions of velocity². (Remember the expression F = qvB for magnetic force on a moving charge).

(D) is to be matched with (r) and (s) since GM_e/R_e is gR_e which has dimensions of [(metre/sec²) ×metre] and therefore those of velocity².

It will be difficult for you to remember the dimensions of all the physical quantities, especially those you come across in branches of physics other than mechanics. But you will remember the important expressions you derive in all the branches. The Matrix Match Type question we discussed is aimed at checking your knowledge of those expressions and your understanding of the significance of dimensions.

KEAM 2007- Engineering Entrance Examination Questions from Optics

The following questions on Optics were asked in the Kerala Government Engineering Entrance test of 2007:

(1) The position of final image formed by the given lens combination from the third lens will be at a distance of

(a) 15 cm (b) infinity (c) 45 cm (d) 30 cm (e) 35 cm

Applying the law of distances [1/f = 1/v – 1/u] to the first lens,

1/10 = 1/v₁ –1/(–30), noting that the object distance ‘u’ is negative in accordance with the Cartesian sign convention. From this, the distance of the image produced by the first lens is v₁ = 15 cm.

The image produced by the first lens is therefore on the right side of the concave lens and the object distance for it is 15 – 5 = 10 cm. This is positive according to the sign convention. The image distance for the second lens (v₂) is given by

1/(–10) = 1/v₂ – 1/10, from which 1/v₂ = 0 so that v₂ = infinity.

This means that the rays emerging from the second lens are parallel to the principal axis. Therefore, the image produced by the third lens is at its focus, at a distance of 30 cm from it.

(2) A slit of width ‘a’ is illuminated by red light of wave length 6500 Ǻ. If the first minimum falls at θ = 30º, the value of ‘a’ is

(a) 6.5×10^–4 mm (b) 1.3 micron (c) 3250 Ǻ (d) 2.6×10^–4 mm (e) 1.3×10^–4 mm

In the single slit diffraction pattern, the first minimum is obtained at an angle θ given by

sin θ = λ/a where λ is the wave length of light used and ‘a’ is the width of the slit.

[Usually, the angle θ is small so that sinθ is approximated to θ and the above relation is written as θ = λ/a].

Therefore, sin 30º = (6500×10^–10)/a, from which a = 1.3×10^–6 m = 1.3 micron

[Even if you use the approximated relation, θ = λ/a, you will arrive at option (b) since 30º = 30 × π/180 radian and hence 30 × π/180 = (6500×10^–10)/a, from which a = 1.2414 micron, which is nearest to the value given in option (b)].

(3) Two beams of light of intensity I₁ and I₂ interfere to give an interference pattern. If the ratio of maximum intensity to minimum intensity is 25/ 9, then, I₁/ I₂ is

(a) 5/3 (b) 4 (c) 81/ 625 (d) 16 (e) ½

If a₁ and a₂ are the amplitudes of the two interfering waves,

I_max/ I_min = (a₁ + a₂)² / (a₁ – a₂)², since the intensity is directly proportional to the square of the amplitude.

Therefore, (a₁ + a₂) / (a₁ – a₂) = √(25/9) = 5/3

From this, a₁/a₂ = 4 so that I₁/I₂ = a₁²/a₂² = 16.

(4) Magnification at least distance of distinct vision of a simple microscope having its focal length 5 cm is

(a) 2 (b) 4 (c) 5 (d) 6 (e) 7

The answer to this very simple question ( which is asked to boost your morale!) is 6 since m = 1 + D/f where D is the least distance of distinct vision (which is 25 cm).

You can find all posts in Optics by clicking on the label 'OPTICS' below this post

Two Kerala Engineering/Medical Entrance 2007 Questions from Electrostatics

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same, the distance between the plates is

(a) 8 mm (b) 6 mm (c) 4 mm (d) 12 mm (e) 10 mm

If a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space become respectively q/ε₀A and q/Kε₀A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε₀A)(d–t)+(q/Kε₀A)t = (q/ε₀A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε₀A/[d–t+(t/K)] = ε₀A/[d–(t – t/K)].

Since the capacitance of the capacitor with air filling the entire space between the plates is ε₀A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).

Therefore, the separation between the plates is to be increased by t–(t/K) = 4 – (4/2) = 2 mm.

The distance between the plates should be 8 + 2 = 10 mm.

The following question appeared in Kerala Medical Entrance 2007 question paper:

The capacitance of a parallel plate capacitor with air as medium is 3 μF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 μF. The permittivity of the medium is

(a) 5 (b) 15 (c) 0.44×10^–10C²N^–1m^–2

(d) 8.854×10^–10C²N^–1m^–2 (e) 5C²N^–1m^–2

The capacitance of a parallel plate capacitor with air as the medium between the plates is ε₀A/d and the capacitance with a medium of dielectric constant K instead of air is Kε₀A/d where ε₀, A and d are the relative permittivity of free space (or air very nearly), the area of each plate and the separation between the plates respectively. The capacitance therefore increases to K times the original value on replacing air with the dielectric medium.

Since the increase here is from 3 μF to 15 μF, the value of K is 5.

The permittivity of the medium is ε₀K = 8.85×10^–12×5 = 0.44×10^–10C²N^–1m^–2.

[The value of ε₀ was not given in the question paper. Some of you may not remember the value 8.854×10^–12C²N^–1m^–2, but all of you are expected to remember the value of 1/4πε₀, which is 9×10⁹, very nearly. This will be enough to get the value of ε₀].

Two Questions Involving Kinetic Theory of Gases

Questions based on specific heat and energy of gases often find place in all Medical and Engineering Entrance Examination question papers. See the following two typical questions:

(1) In an isobaric process, the increase in temperature of 0.4 mole of oxygen is 200K. The work done by the gas is (Universal gas constant, R = 1.99 calorie-mol ^–1K^–1)

(a) 88 cal (b) 124.6 cal (c) 79.6 cal (d) 159.2 cal (e) 318.4 cal

Some of you may get confused on seeing the unit of work in calories. Further, R also is given in terms of calorie. Work and energy can be expressed in calorie as well as joule.

Since the pressure is constant (isobaric process), the work done by one mole of gas on getting heated through 1K is the difference (C_p – C_v) between its molar specific heats. But C_p – C_v = R, so that the work done by 0.4 mole of the gas on getting heated through 200 K is 0.4×200×R = 0.4×200×1.99 = 159.2 calorie.

(2) At what temperature will the translational kinetic energy of an ideal gas molecule be half of the value at 100º C?

(a) 323 K (b) – 179.75º C (c) –186.5º C (d) –86.5º C (e) 50º C

This is a very simple question. The translational kinetic energy of any ideal gas molecule is (3/2) kT where ‘k’ is Boltzman constant and T is the absolute (Kelvin) temperature. So, the translational kinetic energy is directly proportional to the absolute temperature. [The total kinetic energy also is directly proportional to the absolute temperature since total kinetic energy = (n/2)kT where ‘n’ is the number of degrees of freedom of the gas molecule].

Now, 100º C = 373 K. So, the energy becomes half at 186.5 K = (186.5 – 273)º C = – 86.5º C.

IIT-JEE 2007 Matrix-Match Type Question on Modern Physics

Section IV of Part I Physics question paper of IIT-JEE 2007 contained 3 Matrix-Match Type questions. Each question contained statements given in two columns. Statements A,B,C,D in Column I had to be matched with statements p,q,r,s in Column II. The answers to the questions had to be appropriately bubbled as shown at the end of this post.

Here is the question involving topics in modern physics:

Some laws / processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 ×4 matrix in the ORS

Column I Column II

(A) Transition between (p) Characteristic X- rays

two atomic energy levels
(B) Electron emission from (q) Photoelectric effect

a material

(C) Mosley’s law (r) Hydrogen spectrum

(D) Change of photon (s) β-decay

energy

Characteristic X-rays and hydrogen spectrum are produced by electron transition between two energy levels in an atom. So, (A) is to be matched with (p) and (r).

In photo electric effect and β-decay, electrons are emitted from a material. So, (B) is to be matched with (q) and (s).

Mosley’s law [√f α Z ] relates the frequency ‘f’ of a particular characteristic X-ray (e.g., K_α ) to the atomic number Z of the target in the X-ray tube. So, (C) is to be matched with (p).

In photoelectric effect, the energy of the incident photon is used in dislodging electrons from a photo sensitive surface. So, (D) is to be matched with (q).

The appropriate bubbles darkened are shown in figure.

Five Kerala Engineering Entrance 2007 Questions (MCQ) from Electronics

Five questions from Electronics and three questions from Communication Systems were included in the Kerala Engineering Entrance 2007 (KEAM) question paper. Here are the questions from Electronics:

(1) The real time variation of input signals A and B are as shown below. If the inputs are fed in to NAND gate, then select the output signal from the following

The NAND gate will give low output when all inputs are high. The output will be high in all other cases. The inputs A and B are high during the interval from 6 seconds to 8 seconds only. So, the output must be low during this interval only. This is indicated in option (b).

(2) In a common emitter amplifier, the current gain is 62. The collector resistance and the input resistance are 5 kΩ and 500 ohms respectively. If the input voltage is 0.01 V, the output voltage is

(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V (e) 0.01 V

The collector current is given by I_c = βI_B with usual notations.

But, the base current I_B = V_i/Ri where V_i and R_i are the base voltage (input voltage) and input resistance respectively. Therefore, I_B = 0.01/ 500 = 2 ×10^–5 A.

The corresponding collector current is βI_B = 62×2 ×10^–5 A = 1.24×10^–3 A

The output voltage is the voltage drop across the collector resistance (output resistance) R_o.

So, output voltage = I_cR_o = 1.24×10^–3×5000 = 6.2 V.

[If you remember the simple expression for the voltage gain of a low frequency common emitter amplifier, the above result can be arrived at in a shorter time:

Voltage gain A_v = βR_o/R_i = 62×5000/500 = 620. Therefore output voltage = 620×0.01V = 6.2 V].

(3) The current gain of a transistor in common base mode is 0.995. The current gain of the same transistor in common emitter mode is

(a) 197 (b) 201 (c) 198 (d) 202 (e) 199

The expression for the current gain in common emitter mode is β = α /(1–α) where α is the current gain in common base mode.

Therefore, β = 0.995 /(1–0.995) = 199.

(4) When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. The forward bias resistance is

(a) 0.01 Ω (b) ) 0.1 Ω (c) ) 10 Ω (d) ) 100 Ω (e) ) 0.2 Ω

Forward resistance, R = (Change in forward biasing voltage) /(Change in forward current) = (0.1 V)/ 10 mA = 0.1/(10×10^–3) Ω = 10 Ω.

(5) The energy gap between conduction band and the valence band of a material is of the order of 0.7 eV. Then it is

(a) an insulator (b) ) a conductor (c) a semiconductor

(d) an alloy (e) a superconductor

If the energy gap between the conduction band and the valence band is less than 3 eV, the substance is a semiconductor. So, the correct option is (c).

Two Questions on Specific Heat Capacity of a Gas

Questions similar to the following one are often found in Medical and Engineering entrance examination question papers:

One mole of an ideal mono atomic gas is mixed with two moles of an ideal diatomic gas. The ratio of specific heats of the mixture is

(a) 1.5 (b) 1.4 (c) 10/6 (d) 15/11 (e) 19/13

You should remember that the values of molar specific heats at constant volume C_v) for mono atomic and diatomic gases are respectively (3/2)R and (5/2)R where R is universal gas constant. The values of molar specific heat at constant pressure C_p) are therefore (5/2)R and (7/2)R respectively, in accordance with Meyer’s relation [C_p – C_v =R].

Therefore, C_v of the mixture = [1×(3/2)R + 2×(5/2)R]/(1+2) = (13/6)R

C_p of the mixture = C_v + R = (19/6)R.

Ratio of specific heats of the mixture, γ = C_p/C_v = 19/13.

[Generally, if n₁ moles of a gas having ratio of specific heats γ₁ is mixed with n₂ moles of a gas having ratio of specific heats γ₂, the ratio of specific heats of the mixture is given by the relation, (n₁+ n₂)/(γ– 1) = n₁/( γ₁ –1) + n₂/( γ₂ –1). You can easily arrive at this result].

If one mole of an ideal mono atomic gas is mixed with one mole of an ideal diatomic gas, the ratio of specific heats of the mixture is 1.5. As an exercise, check this.

Now consider the following MCQ:

The following sets of experimental values of C_v and C_p of a given sample of gas were reported by five groups of students. The unit used is calorie mole^–1 K^–1. Which set gives the most reliable values?

(a) C_v = 3, C_p = 4.5 (b) C_v = 2, C_p = 4 (c) C_v = 3, C_p = 4.9 (d) C_v = 2.5, C_p = 4.5 (e) C_v = 3, C_p = 4.2

Since the minimum value of C_v is (3/2)R which is the value for a mono atomic gas, when you express it in calorie mole^–1 K^–1, the minimum value is approximately 3. [R = 8.3 J mole^–1 K^–1 = 2 calorie mole^–1 K^–1, approximately]. Options (b) and (d) are therefore not acceptable. Out of the remaining three options, (c) is the most reliable since C_p – C_v = R, which should be 2 calorie mole^–1 K^–1 very nearly.

Direct current Circuits- Two IIT-JEE 2007 MCQ’s

The following MCQ’s appeared in IIT-JEE 2007 question paper:

(1) A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is

(a) 0 (b) 54 μC (c) 27 μC (d) 81 μC

When S is open, the P.D. across 3 μF and 6 μF capacitors are 6V and 3V respectively. [The charges on them are equal and the P.D. across them are inversely proportional to their capacities].

When S is closed, the P.D. across them become 3V and 6V respectively since the PD across the 3Ω and 6Ω resistors are 3V and 6V respectively. Positive charges have to flow from Y to X to achieve this condition.

Since the potential at point X is to be made 6V for this, the charge flowing to the 3 μF capacitor is 3μF×3V = 9μC. The charge flowing to the 6μF capacitor is 6μF×3V = 18μC. The total charge = 9 μC +18 μC =27 μC.

[You may have certain doubts regarding this solution. Once you note that the potentials of the lower potential plate of the 3 μF capacitor and the higher potential plate of the 6 μF capacitor are raised by 3 volts, by the charges flowing from Y to X, your doubts will be cleared].

(2) A resistance of 2Ω is connected across one gap of a metre bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is

(a) 3Ω (b) 4Ω (c) 5Ω (d) 6Ω

If the balancing length is measured initially on the side of the unknown resistance X, it will shift from 60 cm to 40 cm. [Remember that the balancing length is measured from the same side before and after interchanging. You might have noted that the balance point in a meter bridge shifts symmetrically with respect to the mid point of the bridge wire].

Therefore, we have X/2 = 60/40, from which X = 3Ω.

[If the balancing length is measured on the side of the known resistance, it will change from 40 cm to 60 cm. In this case, 2/X = 40/60, from which X =3Ω].

Three Karnataka CET 2005 Questions on Electrostatics

The following three questions which appeared in Karnataka CET 2005 are aimed at testing your understanding of basic principles in electrostatics.

(1) The electric flux for Gaussian surface A that encloses the charged particles in free space is ( given that q₁ = –14 nC, q₂ = 78.85 nC, q₃ = – 56 nC)

(a) 10³ Nm²C^–1 (b) 10³ CN^–1m^–2

(c) 6.32×10³ Nm²C^–1 (d) 6.32×10³ CN^–1m^–2

The Gaussian surface B is just for distracting you. Remember, Gaussian surface is an imagined surface and it has no action on the charge configuration. When you consider the Gaussian surface A, the net charge enclosed by the surface is –14 nC + 78.85 nC + (– 56 nC) = 8.85 nC.

The electric flux over the closed surface A, according to Gauss theorem, is q/ε₀ where q is the net charge enclosed by the surface.

Therefore, electric flux = 8.85×10^–9/8.85×10^–12 = 10³ Nm²C^–1 [Option (a)].

[You should note that the electric flux is the product of electric field and area so that its unit is (N/C)×m² = Nm²C^–1].

(2) The work done in carrying a charge ‘q’ once round a circle of radius ‘r’ with a charge ‘Q’ at the centre is

(a) qQ/4πε₀r (b) qQ/4πε₀²r² (c) qQ/4πε₀r² (d) none of these

The work done is zero since there is no potential difference between the initial and final positions of the charge q. The correct option therefore is (d). Note that this is the case for any closed path of any shape.

(3) An air filled parallel plate condenser has a capacity of 2 pF. The separation of the plates is doubled and the inter space between the plates is filled with wax. If the capacity is increased to 6 pF, the dielectric constant of wax is

(a) 2 (b) 3 (c) 4 (d) 6

The capacitance of air cored capacitor is ε₀A/d where A is the area of the plates and ‘d’ is the separation between the plates. When the separation is doubled, the capacitance is halved and becomes 1 pF.

The capacitance when the inter space is filled with a dielectric of dielectric constant ‘K’, the capacitance is Kε₀A/d so that it is increased to K times the value with air as the dielectric. Since the increment is from 1 to 6, K = 6.

Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm²r]^1/2 (b) [GMmr]^1/2 (c) [GMm/r]^1/2 (d) [GMm²/r²]^1/2 (e) [GMm²r³]^1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr²ω = mr²v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv²/r = GMm/r²

From this, m²v²r² = GMm²r so that angular momentum mvr = [GMm²r]^1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m²e²r/4πε₀]^1/2 (b) [m²er/4πε₀]^1/2 (c) [me²r²/4πε₀]^1/2

(d) [me²r/4πε₀]^1/2 (e) [me²r³/4πε₀]^1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv²/r = (1/4πε₀)e²/r² from which m²v²r² = (1/4πε₀) ×me²r, so that orbital angular momentum, mvr = [me²r/4πε₀]^1/2

Two Multiple Choice Questions on Digital Circuits

Most of you will like the section on digital circuits, especially because at the class 12 level you have to study very simple circuits. Consider the following MCQ:

For the circuit shown, logic level 1 is +5 volts and logic level 0 is 0 volt. This circuit is

(a) an EXOR gate (b) an AND gate (c) an INVERTER

(d) an OR gate (e) a NOR gate

If both inputs A and B are zero, the diodes will not conduct and the output point will be at ground potential so that the output Y = 0.

If at least one input is at logic 1 level (+5 volts), the diode connected to that input will conduct. The diode connected to the output also will conduct making the output high (+5V). Thus the output Y=1. The same thing happens if both inputs are high.

So, the circuit is an OR gate.

Now, consider the following question:

The digital circuit shown in the figure implements

(a) OR operation (b) EXOR operation

(c) NAND operation (d) NOR operation (e) AND operation

The first gate is a NAND gate. The second gate also is a NAND gate whose inputs are shorted. But when the inputs of a NAND gate are shorted, it becomes an inverter (NOT gate). So, the circuit is a NAND followed by an inverter which is altogether an AND gate [Option (e)].

An IIT-JEE 2007 Question on Centre of Mass

Here is an assertion-reason type MCQ (involving centre of mass motion) which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant

because

STATEMENT-2

The linear momentum of an isolated system remains constant.

(a) Statement-1 is True, Statement-2 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

This is a very simple question meant for checking your understanding of basic principles. But if you are not careful, you are liable to pick out a wrong answer!

Generally, if there are external torques or forces or both, the velocity of the centre of mass will change. So, the condition of no external torque alone is not sufficient to ensure the constancy of the velocity of the centre of mass. Statement-1 is therefore false.

[You should also note that an external torque about the centre of mass will not change the velocity of the centre of mass].

Statement-2 is evidently true.

The correct option therefore is (d).

IIT-JEE 2007 Questions on Rotational Motion

The following straight objective type multiple choice question appeared in IIT-JEE 2007 question paper:

A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

(a) ring (b) solid sphere (c) hollow sphere (d) disc

The body has translational and rotational kinetic energies and these are completely converted in to gravitational potential energy at the maximum height so that we can write

½ Mv² + ½ Iω² = Mgh Where M is the mass, I is the moment of inertia and ω is the angular velocity of the body and h is the maximum height reached. Since ω = v/R where R is the radius of the rolling body, the above equation can be rewritten as

½ Mv² + ½ I(v²/R²) = Mg×(3v²/4g)

From this, I = MR²/2, which is the value for a disc [Option (d)].

The following three questions which appeared in IIT-JEE 2007 question paper are Linked Comprehension Type multiple choice questions:

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential enargy of a spring compressed by a distance x₁. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x₂. Both the discs rotate in the clockwise direction.

Question(1):

The ratio x₁/x₂ is

(a) 2 (b) ½ (c) √2 (d) 1/√2

Equating the potential energies of the springs to the kinetic energies of the discs, we have

½ k x₁² = ½ I×4ω² and

½ k x₂² = ½ ×2I×ω² for the two cases. Here ‘k’ is the spring constant.

From these equations, x₁/x₂ = √2

Question 2:

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

(a) 2Iω/3t (b) 9Iω/2t (c) 9Iω/4t (d) 3Iω/2t

Since the angular momentum is conserved, we have

I×2ω + 2I×ω = (I+2I)×ω’ where ω’ is the common angular velocity of the discs. [We have added the angular momenta since they are in the same direction].

From this, ω’ = (4/3)ω

Disc A will have an angular retardation of magnitude ‘α₁’ during the time ‘t’ where as disc B will have an angular acceleration of different magnitude ‘α₂’ during the time ‘t’.

Considering disc A, we have ω’ = 2ω – α₁t from which α₁ = (2ω – ω’)/t = 2ω/3t since ω’ = (4/3)ω.

The average frictional torque exerted on A by B = Iα₁ = 2Iω/3t

[An equal and opposite torque will be exerted on B by A. Check by finding α₂ and hence 2Iα₂].

Question 3:

The loss of kinetic energy during the above process is

(a) Iω²/2 (b) Iω²/3 (c) Iω²/4 (d) Iω²/6

Loss of kinetic energy = Initial kinetic energy – Final kinetic energy

= ½ I×(2ω)² + ½ ×2I×ω² – ½ ×3I×(4ω/3)² = (Iω²)/3