Floating Bodies - Multiple Choice Questions

The weight of a floating body is equal to the weight of the displaced liquid. Questions based on this law of floatation can often be found in Medical and Engineering entrance test papers. Consider the following M.C.Q.:
A toy boat containing a piece of ice is floating in kerosene contained in a beaker. If the piece of ice is gently transferred to kerosene in the beaker, the level of kerosene in the beaker is
(a) lowered (b) raised (c) unchanged (d) first lowered and then raised (e) first raised and then lowered.
You should note that ice is denser than kerosene. So, on transferring the piece of ice in the toy boat to kerosene, it sinks, displacing a volume of kerosene equal to its own volume. When the piece of ice was in the toy boat, it could displace a greater volume of kerosene since the displaced kerosene should have the weight of the piece of ice. The level of kerosene is therefore lowered on transferring the ice to kerosene. The correct option is (a). If the ice melts without change of temperature, the correct option will still be (a), since the water so formed is certainly denser than kerosene and will sink, displacing a smaller volume of kerosene..
If you had a piece of iron instead of the piece of ice in the boat, the correct option would still be (a). But if you had a piece of cork instead, the level of kerosene will remain unchanged on transferring it to kerosene, since the density of cork is less than that of kerosene and it will float when overboard. (Inside the boat also it is a floating body and it will displace the same quantity of kerosene).
Now, consider the following question:
A toy boat containing an ice cube is floating in water contained in a beaker. The level of water in the beaker is noted. When the ice cube in the boat is gently transferred to water in the beaker, it melts without change of temperature. Then the level of water in the beaker is
(a) lowered (b) raised (c) unchanged (d) first raised and then lowered (e) first lowered and then raised.
The correct option here is (c) because the ice cube, while floating along with the boat, can displace a volume of water that has its own weight. On melting, the water produced will have the same volume and hence there is no change in the water level.
The boat in this problem is just a minor distraction. The answer is unchanged even if there is no boat and the ice cube is floating in water.
Now, suppose there is an iron nail on an ice cube floating in water contained in a beaker. If the ice melts without change of temperature, the level of water in the beaker will be lowered since the iron nail while floating along with ice can displace a greater volume of water.
A wooden log (density 700 kgm^-3) of mass 2100 kg floats in water. How much weight should be placed on it to make it just sink?
(a) 4800 kg (b) 2100 kg (c) 900 kg (d) 700 kg (e) 600 kg
The volume of the wooden log = 2100/700 =3 m³. While floating, the wooden log displaces water having weight 2100kg. Since the density of water is 1000 kgm^-3, the volume of the displaced water is 2100/1000 = 2.1 m³. The additional volume of water to be displaced by the wooden log on making it just sink is 3 – 2.1 = 0.9 m³. Weight of this additional volume of water = 0.9×1000 = 900 kg. This is the weight to be placed on the wooden log to make it just sink.

Multiple Choice Questions on Simple harmonic Motion

You will definitely find questions on simple harmonic motion in Medical and Engineering Entrance test papers as well as in GRE (Physics) question papers. You can find Multiple Choice Questions on Simple harmonic Motion with solution at physicsplus: Multiple Choice Questions on Simple Harmonic Motion . Additional questions with solution can be found here.

Multiple Choice Questions on One Dimensional Motion

Here is a question in kinematics which is a popular one and therefore requiring your attention:
The two ends of a train running with a constant acceleration passes a certain point with velocities v₁ and v₂. The velocity with which the middle point of the train passes the point is
(a) (v₁+v₂)/2 (b) √(v₁ ² +v₂²) (c) (v₁²+v₂²)/2 (d) (v₁+v₂)/√2

(e) √[(v₁²+v₂²)/2]
If ‘s’ is the length of the train, the velocity of the train changes from v₁ to v₂ when it moves through the distance ‘s’. Therefore we have,
v₂²- v₁² = 2as from which a = (v₂² - v₁² )/2s
If’ ’v’ is the velocity with which the mid point of the train passes the reference point, we have v² = v₁² +2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v = √[(v₁² +v₂ ² )/2].
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
A body dropped from a height ‘h’ with an initial velocity zero reaches the ground with a velocity 3km/hour. Another body of the same mass is dropped from the same height ‘h’ with an initial velocity 4km/hour. It will reach the ground with a velocity
(a) 3km/hour (b) 4km/hour (c) 5km/hour (d) 12km/hour (e) 8km/hour
We have, v² = u² + 2as with usual notations.
Therefore, 3² = 0 + 2gh for the first case and
v² = 4² + 2gh for the second case. These two equations yield the value v= 5km/hour. Note that we did not convert the velocities into m/s since the answer is required in km/hour. [Note that the answer is independent of the masses of the bodies].
Now consider the following simple question:
A particle starting from rest travels with uniform acceleration for t₁ seconds and then travels (continuously) with uniform retardation and comes to rest in another t₂ seconds. If the total distance traveled is ‘s’, the maximum velocity attained during the motion is
(a) s/(t₁ – t₂) (b) s/(t₁ + t₂) (c) 2s/(t₁ – t₂) (d) 2s/(t₁ + t₂) (e) s/t₁ + s/t₂
Since the acceleration and retardation are uniform, this can be easily solved using the concept of average velocity. If ‘v’ is the maximum velocity attained, the average velocity during the acceleration part as well as the deceleration part is v/2.
Therefore, s = (v/2)t₁+ (v/2)t₂ = (v/2)(t₁+t₂) from which v = 2s/(t₁+t₂).

Birla Institute of Technology & Science (BITS) -- BITSAT-2007

The BITSAT-2007 Online tests (for admission to the academic year 2007-08) will be conducted during 7th May - 10th June 2007. These tests are for admitting students to the Integrated First Degree Programmes. You will find details here

MCQ on Semiconductor Devices

The following simple question appeared in Kerala Medical Entrance 2006 Test paper:

A PN junction diode is connected to a battery of emf 5.5 V and external resistance 5.1 kΩ. The barrier potential in the diode is 0.4 V. The current in the circuit is
(a) 1.08 mA (b) 0.08 mA (c) 1 mA (d) 1 A (e) 2 mA

Out of 5.5 volts, 0.4 volt (barrier potential) is dropped across the forward biased diode and the remaining 5.1 volts appears across the 5.1 kΩ resistance. So, the current through the resistance is (5.1V/5.1kΩ), which is equal to 1mA [Option (c)].
The following MCQ on half wave rectifier also is simple:
A sinusoidal voltage of peak to peak value 310 V is connected in series with a diode and a load resistance R so that half wave rectification occurs. If the diode has negligible forward resistance and very high reverse resistance, the r.m.s. voltage across the load resistance is
(a) 310 V (b) 155 V (c) 219 V (d)109.5 V (e) 77.5 V
If V_m is the peak value (maximum value) of an alternating voltage, its r.m.s. value is Vm/√2. The r.m.s. value of full wave rectifier output also is V_m/√2. Therefore, the mean square value of the full wave rectifier output is V_m²/2. The mean square value of the half wave rectifier output is half of this, which is equal to V_m²/4. Therefore, r.m.s. value of half wave rectifier output voltage is V_m/2.
Note that the peak to peak value (310 V) is given in the question. The peak value V_m is 155 volts and hence the r.m.s. value is 155/2 = 77.5 V.
Here is a question which may confuse you and you may be tempted to pick out the wrong answer:
A common emitter low frequency amplifier has a collector supply voltage (V_CC) of 9 V. The amplifier has input resistance 1kΩ and collector load resistance 5 kΩ. The transistor used has a common emitter current gain (β) of 100. What will be the peak to peak output signal voltage if an input peak to peak signal voltage of 50 mV from a low impedence source is applied to this amplifier?
(a) 250 V (b) 25 V (c) 18 V (d) 12.5 V (e) 9 V
The voltage gain (amplification) of the amplifier is, A_v = β (R_L/R_i) where β is the common emitter current gain, R_L is the load resistance and Ri is the input resistance.Therefore, A_v = 100×5/1 = 500.
( Note that questions on calculation of voltage gain are often seen in Medical and Engineering Entrance test papers. They are simple to solve at Higher Secondary and Plus Two levels, as we hve done above).
The important thing to remember (especially in the context of the present question) is that you will get the full gain only if the input voltage is small. If the input voltage were 5 mV peak to peak for instance, you would have obtained a peak to peak output voltage of 500 times the input voltage, which is 500×5 mV = 2500 mV = 2.5 V.
But since the input is 50 mV, you cannot obtain 500×50 mV (= 25000 mV = 25 V) for the simple reason that the collector supply voltage is 9 volts only. The maximum possible peak to peak signal voltage will be 9 volts only given in option (e).
[The transistor will swing between saturation and cut off as the signal voltage swings between the positive and negative peaks and you will get a clipped output signal which has a peak to peak value of approximately 9 volts in the present case].

You will find more multiple choice questions in this section at physicsplus: Multiple Choice Questions from Electronics