Moment of inertia - Two Multiple Choice Questions

The value of a man resides in what he gives and not in what he is capable of receiving

– Albert Einstein

You will find many questions involving moment of inertia on this site and you can access all of them by clicking on the label ‘MOMENT OF INERTIA’ below this post. I give you two more questions here, the first being the one which appeared in the AIEEE 2008 question paper. You will have to apply the parallel axis theorem to answer these questions.

(1) Consider a uniform square plate of side ‘a’ and mass ‘m’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

(1) (5/6) ma²

(2) (1/12) ma²

(3) (7/12) ma²

(4) (2/3) ma²

The moment of inertia (I_CM) of the square plate about an axis through its centre of mass (CM) passing perpendicular to its plane is given by

I_CM = m(a² + a²)/12 = ma²/6.

[Note that generally for a bar of rectangular cross section, the moment of inertia about an axis passing through the centre, perpendicular to its length is m(l² + b²)12 where l is the length and b is the breadth].

The moment of inertia (I) about a parallel axis through one corner O is given by

I = I_CM + m(a/√2)² since the corner is at a distance a/√2 from the centre of mass (fig).

Thus I = ma²/6 + ma²/2 = (2/3) ma²

(2) A thin uniform rod of mass 6m is bent to form a regular hexagon of side a. What is the moment of inertia of the hexagon about an axis passing through its centre and perpendicular to its plane?

(a) 5 ma²

(b) 5 ma²/6

(c) 5 ma²/3

(d) 6 ma²

(e) 6 ma²/5

Consider one side (such as AB) of the hexagon. Since its mass is m and its length is a, its moment of inertia about an axis through its centre and perpendicular to its length is given by

I_CM = ma²/12.

The moment of inertia of this side about a parallel axis through the centre C of the hexagon is given by

I = I_CM + m[(^√3/₂)a] ² since the side is at a distance (^√3/₂)a from the centre of the hexagon (fig).

Thus I = ma²/12 + 3ma²/4 = 5 ma²/6.

Since there are six sides, the total moment of inertia of the hexagon is 6×5 ma²/6 = 5 ma².

Resistive Networks- AIPMT 2008 Questions

The following questions on resistive networks appeared in All India Pre-Medical/Pre-Dental Entrance Examination 2008 question paper:

(1) In the circuit shown, the current through the 4 Ω resistor is 1 amp when the points P and M are connected to a d.c. voltage source. The potential difference between the points M and N is

(1) 1.0 volt

(2) 0.5 volt

(2) 3.2 volt

(2) 1.5 volt

Since the current through the 4 Ω resistor is 1 ampere, the potential difference between the points P and M is 4 volt. The 1 Ω resistor and the two parallel resistors (each of value 0.5 Ω) make the total resistance in the branch PNM equal to 1.25 Ω. The potential drop across the 1 Ω resistor is therefore equal to 4×1/1.25 = 3.2 volt.

[The p.d. of 4 volt gets divided between 1 Ω and 0.25 Ω]

(2) A current of 3 amp flows through the 2 Ω resistor shown in the circuit. The power dissipated in the 5 Ω resistor is

(1) 2 watt

(2) 1 watt

(3) 5watt

(4) 4watt

The potential difference across the 2 Ω resistor is 2×3 = 6 volt.

The same p.d. appears across the branch containing 1 Ω and 5 Ω. Therefore, the current through 1 Ω and 5 Ω is 6 V/6 Ω = 1 A.

Therefore, the power dissipated in the 5 Ω resistor is I²R = 1²×5 = 5 watt.

AP Physics & Degree Entrance Kinematics- Questions on One Dimensional Motion

The following questions numbered 1, 2 and 3 are based on the velocity-time graph of a particle in one dimensional motion shown in the adjoining figure.

(1) The displacement of the particle during the first second of its motion is nearly

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.5 m

(e) 2 m

The area under the velocity-time curve gives the displacement. The portion of the graph for the interval from zero to 1 second is straight and the area under the curve is triangular and is equal to (½)×1×4 = 2 m.

(2) The average accelerations during the 3^rd second and 7^th second respectively are nearly

(a) 0.5 ms^–2 and 2 ms^–2

(b) 1 ms^–2 and –2 ms^–2

(c) 1 ms^–2 and 2 ms^–2

(d) 1 ms^–2 and –2 ms^–2

(e) zero and –2 ms^–2

The 3^rd second is the interval from 2 seconds to 3 seconds and during this time the velocity increases from 7 ms^–1 to 8 ms^–1. The acceleration is therefore (8 – 7)/1 = 1 ms^–2.

The 7^th second is the interval from 6 seconds to 7 seconds and during this time the velocity decreases from 8 ms^–1 to 6 ms^–1.

The acceleration is therefore (6 – 8)/1 = – 2 ms^–2. [Option (b)]

(3) Which one among the following acceleration–time graphs most closely represents the motion of the particle?

The acceleration is positive and uniform initially. Afterwards the acceleration becomes zero (since the velocity remains constant) for some time and then becomes negative (since the velocity goes on decreasing) and uniform. The curve shown in (b) therefore represents the motion of the particle.

Let us leave the velocity time graph here and consider a couple of different questions:

(4) Two boys running at uniform speeds v₁ and v₂ respectively along a straight line path in opposite directions get 9 m closer each second. While running along the same direction with their speeds reduced by 50%, they get 0.5m closer each second. The speeds v₁ and v₂ are respectively

(a) 6 ms^–1 and 3 ms^–1

(b) 5 ms^–1 and 4 ms^–1

(c) 5 ms^–1 and 4.5 ms^–1

(d) 5 ms^–1 and 3.5 ms^–1

(e) 5.5 ms^–1 and 3.5 ms^–1

This is a simple question involving relative velocity. Wile running along opposite directions, we have

v₁+ v₂ = 9

While running along the same direction, we have

v₁/2 – v₂/2 = 0.5, from which v₁– v₂ = 1

Solving the above equations, we obtain v₁ = 5 ms^–1 and v₂ = 4 ms^–1

(5) The rear end of a train running on a straight track with uniform acceleration has velocities 6 ms^–1 and 10 ms^–1 respectively when passing points A and B in its path. The velocity of the rear end midway between these points is approximately

(a) 7 ms^–1

(b) 7.5 ms^–1

(c) 8ms^–1

(d) 8.2 ms^–1

(e) 8.4 ms^–1

We have v² = u² + 2as where u and v are the initial ans final velocities respectively, a is the acceleration and s is the displacement.

If the distance between A and B is s, we have

10² = 6² + 2as from which 2as = 64

If v₁ is the velocity midway between A and B we have

v₁² = 6² + 2a(s/2) = 6² + 32 = 68

Therefore, v₁ = 8.2 ms^–1 nearly.

You will find more questions (with solution) on one dimensional motion and other sections at AP Physics Resources

Multiple Choice Questions (MCQ) Involving Rotation

Questions on rotational motion have been discussed on many occasions on this site. You can access them by clicking on the label ‘rotation’ below this post. Now, see the following questions involving rotation.

(1) A light inextensible string is wound round a wheel of moment of inertia I and radius R. A mass m is attached to its hanging end as shown. The wheel and the mass are initially at rest. Assume that there are no frictional forces opposing the motion. When the mass is released, it moves down and rotates the wheel. When the mass has fallen down through a distance h, what is the angular velocity of the wheel?
(a) [2gh/(I + mR)]^1/2

(b) [2mgh/ (I + 2m)]^1/2

(c) [2mgh/ (I + 2mR²)]^1/2

(d) [2mgh/ (I + mR²)]^1/2

(e) [2mgh/ (I + m)]^1/2

The mass loses gravitational potential energy and gains kinetic energy. The wheel also gains kinetic energy so that we have

mgh = ½ mv² + ½ Iω² where v is the velocity of the mass after falling through the distance h and ω is the angular velocity of the wheel.

Since v =ωR the above equation becomes

mgh = ½ mω²R ² + ½ Iω²

This yields ω = [2mgh/ (I + mR²)]^1/2

(2) A simple pendulum has a spherical bob of mass 200 g. The string of the pendulum has negligible mass and it can withstand a maximum tension of 22 N. By holding the string in hand a student whirls the bob in a vertical circle of radius 1 m. The maximum possible angular velocity the bob can have throughout its motion (in radian per second) will be nearly

(a) 2

(b) 4

(c) 6

(d) 8

(e) 10

The tension in the string will be maximum when the bob is at the bottom point of the vertical circle. This can be seen by writing the expression for the resultant force on the bob and equating it to the centripetal force which makes the bob move along the circle:

T – mg = mrω², where m is the mass of the bob, T is the tension in the string, r is the radius of the circular path and ω is the angular velocity of revolution of the bob.

This gives ω = [(T – mg)/mr]^1/2

= [(22 – 0.2×10)/(0.2×1)]^1/2 = 10 radian/sec

We have substituted the maximum tension the string can withstand and hence the above angular velocity is the maximum value possible.

You will find some useful posts on rotational motion at AP Physics Resources: AP Physics B and C– Multiple Choice Questions on Circular Motion and Rotation

AIEEE 2008 and AIPMT 2008 Questions on Digital Circuits

Questions from digital circuits at your level are simple and you should never omit them. The following question appeared in All India Engineering/Architecture Entrance Examination 2008 question paper:

In the circuit below, A and B represent two inputs and C represents the output.

The circuit represents

(1) AND gate

(2) NAND gate

(3) OR gate

(4) NOR gate

If the inputs A and B are low, the diodes nwill not conduct and the potential at the output point C will be low (ground potential). If either A or B is high (at high voltage level), the corresponding diode will conduct and will place the output point C at the high voltage level. If both A and B are high, again the output will be high. The circuit therefore implements OR operation [Option (3)].

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In the above question we have taken, as usual, the high voltage level as the logic 1 level and the low voltage level as the logic 0 level. This system is the positive logic system.

If the low voltage is taken as logic 1 level and the high voltage is taken as logic 0 level, the systerm is called negative logic system. Unless specified otherwise, all systems are positive logic systems.

In the negative logic system, the above circuit will become AND gate.

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The following question appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

The circuit

is equivalent to

(1) NAND gate

(2) NOR gate

(3) OR gate

(4) AND gate

The second gate is a NAND gate. Since its inputs are shorted, it functions as a NOT gate. So the circuit is a NOR gate followed by Two NOT gates. NOR followed by NOT is OR and OR followed by NOT is once again NOR. So the correct option is (2).

AIPMT 2008 Questions on Newton’s Laws of Motion

The following questions which appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 are worth noting:

(i) A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

(1) 80 ms^–1

(2) 40 ms^–1

(3) 120 ms^–1

(4) 100 ms^–1

This question is meant for testing your grasp of the laws of conservation of momentum and energy. If v and V represent the initial velocity of the shell of mass m and the recoil velocity of the gun of mass M respectively, we have (from the law of conservation of momentum)

mv + MV = 0 from which V = – (m/M)v = – (0.2/4)v = – v/20 ms^–1

[The negative sign shows that the recoil velocity of the gun is opposite to the velocity of the shell].

The energy of the explosion is used to impart kinetic energy to the shell and the gun so that we have

½ (mv² + MV²) = 1.05×10³

Or, ½ [0.2 v² + (4v²/400 )] = 1.05×10³

This gives v² = 10⁴ so that v = 100 ms^–1.

(ii) Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v ms^–1 will be

(1) 2Mv newton

(2) Mv/2 newton

(3) zero

(4) Mv newton

Since there is a time rate of change of momentum equal to Mv per second, the force required is Mv Newton.

IIT – JEE 2008 Question on Errors in Measurement

Today we will discuss a few questions on errors in measurement. Students are generally found to commit mistake in working out questions involving the calculation of error. The following question appeared in IIT – JEE 2008 question paper:

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm.

Least count for time = 0.1 s.

Student .Length of pend(cm).No.of osc (n).Total time (s)..Time period (s)

....I.............64.0............ .........8...................128.0 ...............16.0....

....II ...........64.0......................4.....................64.0................16.0....

....III..........20.0......................4.....................36.0..................9.0.....

If E_I, E_II and E_III are the percentage errors in g, i.e., (∆g/g)×100 for students I, II and III respectively,

(a) E_I = 0

(b) E_I is minimum

(c) E_I = E_II

(d) E_II is maximum

The period of oscillation (T) of a simple pendulum of length ℓ is given by

T = 2π√(ℓ/g)

Therefore, g = 4π² ℓ/T² so that the fractional error in g is given by

∆g/g = (∆ℓ/ℓ) + 2(∆T/T)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the maximum possible error].

Here ∆ℓ = 0.1 cm and ∆T = 0.1 s

The percentage error is 100 times the fractional error so that

E_I = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

E_II = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

E_III = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus E_I is minimum so that the correct option is (b).

Now, consider the following MCQ:

A firm manufacturing electronic watches claims that the maximum error in the time indicated by their watches is 5 seconds in six months. What will be maximum possible difference between the times indicated by two watches in one year?

(a) 5 s

(b) 10 s

(c) 20 s

(d) 40 s

(e) zero

This is a very simple question. But you must remember that both positive and negative errors are possible. The maximum possible difference between the times indicated by the watches will be obtained when one watch loses time (slow) and the other watch gains time (fast).

The maximum time lost in one year is 10 s. The maximum time gained in one year also is 10 s. Therefore, the maximum possible difference between the times indicated by the clocks in one year will be 20 s.

Here is another question:

A 200 Ω carbon film resistor with 1% tolerance is connected in series with a 100 Ω carbon film resistor with 5% tolerance. The effective resistance of the combination is (in Ω)

(a) 300 ± 7 Ω

(b) 300 ± 6 Ω

(c) 300 ± 3 Ω

(d) 300 + 18 Ω

(e) 300 +7 Ω

The effective resistance (R) is given by

R = R₁ + R₂ where R₁ = (200 ± 2) Ω and R₂ = (100 ± 5) Ω

Thus R =(200 ± 2) Ω + (100 ± 5) Ω = 300 ± 7 Ω

Multiple Choice Questions involving Lens Maker’s Equation

While attempting to solve problems in optics related to spherical mirrors and lenses, many among you will have some confusion about the application of the inevitable sign convention. Get your confusion cleared by going through the post on this site here. You can find all posts in optics on this site by clicking on the label ‘OPTICS’ below this post. Your doubts regarding the application of the sign convention can be cleared completely only by working out questions involving the sign convention.

We will now discuss a few questions involving lens maker’s equation:

(1) Crown glass of refractive index 1.5 is used to make a plano concave lens of focal length 40 cm in air. What should be the radius of curvature of the curved face?

(a) 20 cm

(b) 40 cm

(c) 60 cm

(d) 80 cm

(e) 120 cm

This is a situation where you have to use the lens maker’s equation,

1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are its radii of curvature, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed.

Since the lens is plano concave and hence diverging, its focal length is negative according to the Cartesian sign convention. The above equation on substituting the known values (n₁ = 1, n₂ = 1.5 and R₂ = ∞) becomes

–1/40 = (1.5 – 1)( 1/R₁ – 0)

This gives R₁ = –20 cm [Option (a)].

You obtain the answer as negative in accordance with the sign convention, indicating that the curved face is concave towards the incident ray. If you had taken the second face as curved, you would have put R₁ = ∞ and the answer would be R₂ = 20 cm. This too conforms to the Cartesian sign convention, indicating that the second face is convex towards the incident ray and hence the radius of curvature is positive.

In the present problem, there will be no need of worrying too much about the sign convention if you remember that the focal length (in air) and the radius of curvature will have the same value in the case of biconvex and biconcave lenses of equal radii of curvature if the refractive index of the lens is 1.5. In the case of plano convex and plano concave lenses made of material of refractive index 1.5, the radius of curvature of the curved face will be half the value of the focal length.

(2) A biconvex lens has the same radii of curvature for its faces. If its focal length in air is equal to the radii of curvature of its faces, its focal power when immersed in a liquid of refractive index 1.5 will be (in dioptre)

(a) 0.667
(b) 1.5
(c) 0.5
(d) zero
(e) data insufficient.
You may substitute the same values for the focal length in air and the radii of curvature of the faces in the lens maker’s equation and satisfy yourself that the refractive index of the lens is 1.5.[Since it is a biconvex lens, f is positive, R₁ is positive and R₂ is negative. Also, put the numerical values of R₁ and R₂ equal to f].Since the refractive index of the liquid is the same as that of the lens, there is no convergence or divergence for the rays of light and the focal power will be zero.
(3) The radius of curvature of the convex face of a plano convex lens is 15 cm and the refractive index of the material is 1.4. Then the power of the lens in dioptre is
(a) 1.6
(b) 1.566
(c) 2.6
(d) 2.66
(e) 1.4
The above question appeared in Kerala Engineering Entrance 2008 question paper.

You have to use the lens maker’s equation,

1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are its radii of curvature, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed.

Note that you have to substitute the focal length in metre in this equation which gives you the focal power 1/f itself. Substituting the given values,

1/f = (1.4 – 1)(1/0.15 – 0) = 2.66 nearly.

You will find more useful questions at apphysicsresources.

I have no special talents. I am only passionately curious. 

                                                                    – Albert Einstein