Questions (MCQ) Involving Gravitation

"Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn't have it in the beginning."

– Mahatma Gandhi

Today we shall discuss a few multiple choice questions involving gravitation. The first question pertains to a simple binary star system.

In a simple binary star system two stars orbit around their centre of mass under their mutual gravitational attractive force. The two stars constituting the binary star system need not be of the same mass. Note that Kepler’s laws are applicable to binary stars also. Click to see this.

Star systems with more than two stars (multiple star systems) are also present. About half of the stars in our galaxy, the milky way, are part of binary star systems or multiple star systems. Here is the first question:

(1) Consider a simple binary star system (Fig.) in which the member stars are identical, orbiting in a circle of radius r. If each star has mass m, the orbital speed of each star is

(a) √(Gm/r)

(b) √(Gm/2r)

(c) √(Gm/4r)

(d) √(2Gm/r)

(e) √(4Gm/r)

The gravitational force between the stars supplies the centripetal force required for the circular motion. Therefore we have

             Gmm/(2r)² = mv²/r where G is the gravitational constant and v is the orbital speed.

The above equation gives

            v = √(Gm/4r)

The following question is related to a simple multiple star system: 

(2) Consider a multiple star system containing four stars of the same mass m arranged symmetrically and orbiting around their centre of mass C (Fig.) in the clockwise direction in a circle of radius r. What is the orbital speed of each star?

(a) 4[(Gm/r)(1+2√2)]^1/2

(b) 2[(Gm/2r)(1+2√2)]^1/2

(c) [(Gm/r)(1+2√2)]^1/2

(d) (½)[(Gm/r)(1 + 2√2)]^1/2

(e) (¼)[(Gm/r)(1+2√2)]^1/2

The gravitational force F₁ on a star due to the diametrically opposite star is given by

            F₁ = Gmm/(2r)² = Gm²/4r²

[Note that the distance between the diametrically opposite stars is 2r]

Evidently F₁ is directed towards the centre of mass C of the star system.

The other two stars exert gravitational forces F₂ and F₃ as shown in the figure. These forces are directed at 45º with respect to F₁ and they have the same magnitude given by

                        F₂= F₃ = Gmm/(r√2)² = Gm²/2r²

[Note that the distance between adjacent stars is (√2)r]

            F₂ and F₃ have equal components (Gm²/2r²)cos 45º directed towards the centre of mass C of the star system. The net radial force acting on a star is therefore equal to F₁ + F₂ cos 45º + F₃ cos 45º = Gm²/4r² + (Gm²/2r²)cos 45º + (Gm²/2r²)cos 45º.

The radial force thus works out to (Gm²/r²)[(1/4) + (1/√2)] = (Gm²/r²)[(1 + 2√2)/4]

            Since the above radial force supplies the centripetal force required for circular motion, we have

            mv²/r =  (Gm²/r²)[(1 + 2√2)/4]  

This gives the orbital speed, v = (½)[(Gm/r)(1 + 2√2)]^1/2, as given in option (d).

(3) A particle of mass 4m is kept fixed at point P (Fig.) in the xy-plane. Another free particle of mass m at the origin is found to be unperturbed when a third fixed particle of mass 6m also is present in the xy-plane. If the x-coordinate and y-coordinate of the particle of mass 4m are –√2 and +√2 respectively, what are the x-coordinate and y-coordinate respectively of the particle of mass 6m?

(a) +√3 and –√3

(b) –√3 and +√3

(c) –√6 and +√6

(d) +√2 and –√2

(e) +3 and –3

The free particle of mass m at the origin is unperturbed since the gravitational forces on it due to the fixed particles are equal in magnitude and opposite in direction. This can happen only if the three particles are on the same straight line, with the free particle of mass m in between the fixed particles.

Equating the magnitudes of the gravitational forces, we have

            G(4m)m/2² = G(6m)m/r² where G is the constant of gravitation and r is the distance of the particle of mass 6m from the origin.

[Note that the distance of the particle of mass 6m from thr origin is {(√2)² + (√2)²}^1/2 = 2

The above equation gives

            Gm² = 6Gm²/r² from which r = √6

Since the line joining the particles evidently bisects the angle XOY’ the x-coordinate and y-coordinate of the particle of mass 6m must be of equal value d such that

            √(d² + d²) = √6

Therefore d =√3

Since the particle of mass 6m is located in the 4^th quadrant, its x-coordinate and y-coordinate are +√3 and –√3 respectively [Option (a)].

Multiple Choice Questions on Transistors including EAMCET Engineering 2004 and 2005 Questions

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives.”

– Albert Einstein

A very important semiconductor device you need to study is the transistor, the invention of which brought about a revolution in electronics. Questions involving transistors have been discussed earlier on this site. You can access them by trying a search for ‘transistor’ using the search box provided on this page or by clicking on the label ‘transistor’ below this post. Today we shall discuss a few more questions on transistors

(1) In an amplifier circuit using a transistor the collector current changes by1.224 mA when the emitter current changes by 1.228 mA. What is the common emitter small signal current gain of the transistor used in the circuit?

(a) 0.997 (very nearly)

(b) 1.003 (very nearly)

(c) 122 (very nearly)

(d) 307

(e) 306

The small signal common emitter current gain is the ratio of the change in collector current to the change in base current. Since the emitter current is the sum of the collector current and the base current, the change in base current in the present case is 1.228 mA – 1.224 mA which is equal to 0.004 mA.

Therefore common emitter small signal current gain (β_ac) = 1.224/0.004 = 306.

(2) In an n-p-n transistor in CE configuration

(i) the emitter is more heavily doped than the collector

(ii) emitter and collector can be interchanged

(iii) the base region is very thin but is heavily doped

(iv) the conventional current flows from base to emitter

(a) (i) and (ii) are correct

(b) (i) and (iii) are correct

(c) (i) and (iv) are correct

(d) (ii) and (iii) are correct

The above question appeared in EAMCET Engineering 2004 question paper.

Option (c) is the answer. Note that the configuration deoes not matter.

(3) An n-p-n transistor power amplifier in CE configuration gives

(a) voltage amplification only

(b) current amplification only

(c) both current and voltage amplifications

(d) only power gain of unity

This question appeared in EAMCET engineering 2005 question paper. In common emitter configuration a transistor (n-p-n as well as p-n-p) gives both current and voltage amplifications so that there will be appreciable power gain. Therefore the correct option is (c).

(4) The adjoining figure shows an n-p-n transistor operated in the common emitter configuration in which the potential divider resistors R₁ and R₂ provide the base voltage required for forward biasing the base emitter junction. A student wrongly selected too small a value for R₁ so that the transistor got saturated. (When a transistor is saturated, its emitter to collector voltage is nearly zero). The collector load resistor R_C is 5 KΩ and the emitter resistor R_E (used for bias stabilization) is 1 KΩ. If β_dc of the transistor is 200 what is the minimum base current required for saturating the transistor?

Even though this question may appear to be difficult at the first glance, it is really simple. You have to first calculate the current I which will produce a voltage drop of 12 volts across 6 KΩ.

[Since the emitter to collector voltage is nearly zero when the transistor is saturated, the entire supply voltage of 12 volts must be dropped across the resistors R_C and R_E].

Therefore we have

            I × 6 KΩ = 12 V so that I = 2mA

The emitter current and collector current are nearly the same (because of large current gain β_dc) so that we can take the collector current to be 2 mA. The base current required for producing a collector current (I_C) of 2 mA is the saturating base current in this circuit. Therefore the minimum base current I_B required for saturating the transistor is given by

            I_B = I_C/β_dc = 2 mA/200 = 0.01 mA = 10 μA.

Questions from Kinematics (Including Karnataka CET 2008 Question)

“God used beautiful mathematics in creating the world.”

– P.A.M. Dirac

Today we shall discuss a few questions (MCQ) in the section, ‘kinematics in one dimension’. The questions I give you are meant for testing your knowledge, comprehension and the ability for applying what you have learned in this section.

(1) The velocity-time graphs of a car and a motor bike traveling along a straight road are shown in the adjoining figure. At time t = 0 they have the same position co-ordinate.  Pick out the correct statement from the following:

(a) At time t = 0 the car and the motor bike are at rest.

(b) At time t = 0 the car is at rest but the motor bike is moving.

(c) The distances traveled by the car and the motor bike in time t₁ are equal.

(d) The distance traveled by the bike in time t₁ is twice the distance traveled by the car in the same time.

(e) The distance traveled by the bike in time t₁ is half the distance traveled by the car in the same time.

You can easily conclude that options (a) and (b) are wrong.

To check the remaining options we use the equation of linear motion, s = ut + ½ at² where s is the displacement in time t, u is the initial velocity and a is the uniform acceleration.

The distance traveled by the car in time t₁ is v_ct₁ where v_c (let us say) is the constant velocity of the car.

[The velocity of the car is constant since the velocity-time graph of the car is parallel to the time axis].

The motion of the motor bike is uniformly accelerated and the acceleration a is given by

            a =  v_c/t₁

[Note that the velocity of the motor bike changes from 0 to v_c in time t₁.

The distance s traveled by the motor bike in time t₁ is given by

            s = ut + ½ at² = 0 + (½)×(v_c/t₁) t₁²

Or, s = v_ct₁/2

Therefore the distance traveled by the bike in time t₁ is half the distance traveled by the car in the same time [Option (e)].

(2) A student standing at the edge of a cliff throws a stone of mass m vertically upwards with speed v. It strikes the ground at the foot of the cliff with speed v₁. The student then throws another stone of mass m/4 vertically downwards with speed v. It strikes the ground at the foot of the cliff with speed v₂. If air resistance is negligible, v₁ and v₂ are related as

(a) v₁ = v₂

(b) v₁ = v₂/2

(c) v₁ = 2v₂

(d) v₁ = v₂/4

(e) v₁ = 4v₂

While returning, the sphere of mass m has downward speed v when it passes the edge of the cliff. For calculating the speed with which it strikes the ground at the foot of the cliff, the situation is similar to that of the stone of mass m/4 thrown downwards. Obviously both stones will strike the ground with the same speed [Option (a)].  

(3) A body is projected vertically upwards. The times corresponding to height h while ascending and descending are t₁ and t₂ are respectively. Then the velocity of projection is (g is acceleration due to gravity)

(1) g√(t₁t₂)

(2) gt₁t₂/(t₁+t₂)

(3) g√(t₁t₂)/2

(4) g(t₁+t₂)/2

The above question appeared in Karnataka CET 2008 question paper.

We have the following two equations to give h:

            h = ut₁ –  (½) g t₁²…………….(i)

            h = ut₂ –  (½) g t₂²…………….(ii)

(We have taken the displacement h and the initial velocity u (both upwards) as positive and that’s why the acceleration due to gravity g is negative).

(ii) – (i) gives u(t₂ – t₁) = (½) g(t₂² – t₁²)

Therefore u = (½) g(t₂² – t₁²)/(t₂ – t₁) = g(t₁+t₂)/2

You can find a few more multiple choice practice questions (with solution) in this section here.

Multiple Choice Questions on Metre Bridge

“If I have seen a little further it is by standing on the shoulders of Giants.”

– Sir Isaac Newton

The metre bridge is basically a Wheatstone bridge, the condition of balance of which is conveniently used in measuring unknown resistances. Today we shall discuss a few questions (MCQ) involving metre bridge:

(1) Resistances R₁ and R₂ are connected respectively in the left gap and right gap of a metre bridge. If the balance point is located at 55 cm, the ratio R₂/R₁ is

(a) 4/5

(b) 5/4

(c) 9/11

(d) 11/9

(e) 6/7

The distance of the balance point from the left end of the wire in the metre bridge is 55 cm. Therefore the distance of the balance point from the right end of the wire is 45 cm. Thus we have

            R₁/R₂ = 55/45

Or, R₂/R₁ = 45/55 = 9/11

(2) A 15 Ω resistance is connected in the left gap and an unknown resistance less than 15 Ω is connected in the right gap of a metre bridge. When the resistances are interchanged, the balance point is found to shift by 20 cm. The unknown resistance is

(a) 5 Ω

(b) 6 Ω

(c) 8 Ω

(d) 10 Ω

(e) 12 Ω

Initially the balance point will be at J₁ as indicated in the figure. On interchanging the resistances the balance point will shift to J₂ so that the length of the bridge wire between J₁ and J₂ is 20 cm. The balance points J₁ and J₂ must be equidistant from the mid point (50 cm mark) of the wire so that J₁ is at 60 cm and J₂ is at 40 cm.

Therefore we have

            R/X =  60/40 = 6/4

Or, 15/X = 6/4

This gives X = 10 Ω

(3) Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a metre bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the shift in the balance point is

(a) 10 cm

(b) 15 cm

(c) 20 cm

(d) 25 cm

(e) 30 cm

Initially when the left gap and right gap contain 4 Ω and 6 Ω respectively, the condition of balance is

            4/6 = L/(100 – L) where L is the balancing length

This gives L = 40 cm.

When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the balancing length becomes 50 cm (since the gaps contain equal resistances).

Therefore the shift in the balance point is (50 cm – 40 cm) = 10 cm.