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**Physics Multiple Choice Questions for Medical and Engineering Entrance and AP Physics Examinations

If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

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**

(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_{m}^{2}/2. The mean square value of the half wave rectifier output is half of this, which is equal to V_{m}^{2}/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].

The following MCQ on half wave rectifier also is simple:

(a) 310 V (b) 155 V (c) 219 V (d)109.5 V (e) 77.5 V

Note that the peak to peak value (310 V) is given in the question. The peak value V

Here is a question which may confuse you and you may be tempted to pick out the wrong answer:

(a) 250 V (b) 25 V (c) 18 V (d) 12.5 V (e) 9 V

( 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**** **

Spring constant (force constant of a spring) is the force required for unit extension (or contraction) in a spring. Suppose a spring is cut into two pieces of equal length. Will the spring constant change? Don’t be doubtful. The spring constant will be doubled. If you cut a spring of spring constant ‘k’ into ‘n’ pieces of equal length, the spring constant of each piece will be nk.

Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have**springs **of constants k_{1}, k_{2}, k_{3}….etc. **in series**, the net spring constant ‘k’ of the combination is given by the reciprocal relation, **1/k = 1/k**_{1} + 1/k_{2} + 1/k_{3} + . . . . . . .etc.

If you have **springs in parallel**, the net spring constant ‘k’of the combination is given by **k = k**_{1}+ k_{2}+ k_{3}+ . . . . .etc.

Often you will encounter questions involving the potential energy of a spring. You should remember that the **potential energy of a spring** of constant ‘k’, stretched (or contracted) through a distance ‘x’ is **(½) kx**^{2}. Consider the following MCQ:

**When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be**

(a) 3U (b) 6U (c) 9U (d) U (e) 36U

Since the potential energy is ½ kx^{2}, it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).

Suppose the above question is asked in the following modified form:

**When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be**

(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U

Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.

Now, consider the following question:

**The tension in a spring of spring constant k is T. The potential energy of the spring is**

(a) T^{2}/k^{2} (b) T^{2}/k (c) 2T^{2}/k (d) T^{2}/2k (e) 2T^{2}/k^{2}

Potential energy of a spring, as you know, is (½)kx^{2}. But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.

Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:

**An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is**

(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k

When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx^{2}, from which x = 2Mg/k. [Option (b)]

Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have

(a) 3U (b) 6U (c) 9U (d) U (e) 36U

Suppose the above question is asked in the following modified form:

(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U

Now, consider the following question:

(a) T

Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:

(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k

Application Form and the Information Bulletin of the All India Engineering/Architecture Entrance Examination 2007 to be conducted by Central Board of Secondary Education (CBSE) on 29th April 2007 will be distributed from 1-12-2006 to 5-1-2007. The Information Bulletin and the Application Form can be obtained personally from designated branches of Syndicate Bank/Institutions and Regional Offices of CBSE. Designated branches of Syndicate Bank in Kerala are:

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