Monday, April 14, 2014

JEE (Main) 2014 Multiple Choice Question involving Gravitation and Circular Motion




“God used beautiful mathematics in creating the world”.
– P. A. M. Dirac



The following question involving gravitation and circular motion appeared in JEE (Main) 2014 question paper:

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

(1) √(GM/R)

(2) √(2√2 GM/R)

(3) √[(GM/R)(1+2√2)]

(4) (½)√[(GM/R)(1+2√2)]


The answer is (½)√[(GM/R)(1+2√2)] given in option (4).


This question was worked out in the post dated 6th January 2013 on this site. The statement of the question is slightly different; but the essential points are the same. Click here to see the solution (Question No. 2 in the post)

You can access all questions on gravitation on this site by clicking on the label ‘gravitation’ below this post or by trying a search for ‘gravitation’ using the search box provided on this page. 

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Thursday, February 13, 2014

Multiple Choice Questions on Semiconductor Electronics



“Man perfected by society is the best of all animals; he is the most terrible of all when he lives without law and without justice.”
– Aristotle



Questions on semiconductor electronics are generally simple and you can answer most of the questions correctly without consuming much time. But you need to understand the simple fundamental principles of operation of various semiconductor devices so that common mistakes are avoided.

Here are a few multiple choice questions involving semiconductor electronics:

(1) A square wave of peak to peak voltage 100 volt is applied between the input terminals A and B of the circuit shown in the adjoining figure. What is the voltage across the capacitor C?
(a) 100 V
(b) 0 V
(c) 50 V
(d) 100√2 V
(e) 50√2 V
The diode will pass only the negative half cycles of the square voltage wave so that the voltage across the capacitor will be 50 V.
(2) If a sine wave of peak to peak voltage 100 volt is applied between the input terminals A and B of the circuit shown in the above question, what will be the voltage across the capacitor C?
(a) 100√2 V
(b) 50√2 V
(c) 50 V
(d) 100 V
(e) 0 V
In this case also the diode will pass only the negative half cycles of the applied voltage wave. Even though the voltage varies during the half cycle, the capacitor can retain the charge on it so that the voltage across the capacitor will be 50 V itself.
(3) If a 230 volt house supply alternating voltage is applied between the terminals A and B of the circuit shown in Question No. 1, what will be the voltage across the capacitor C?
(a) 230√2 V
(b) 115√2 V
(c) 230 V
(d) 150 V
(e) 0 V
Alternating voltages and alternating currents (A.C.) supplied to houses are specified in terms of root mean square (RMS) values. Thus a specified value of 230 volt a.c. means that the root mean square value of the voltage is 230 V. Therefore, the peak value of the alternating voltage is 230√2 V.
The diode will conduct during negative half cycles only and the capacitor will be charged to the peak value of the applied voltage. Assuming that the capacitor is ideal, the charge will be retained through out the entire cycle of the applied voltage so that the voltage across the capacitor will be 230√2 V.

(4) The truth table for the digital circuit shown in the figure above is



When both A and B are zero, the NAND gate on the input side produces an input at level 1 for the NOR gate on the output side. Therefore, the final output Y = 0.
When A = 0 and B = 1, the NAND gate on the input side again produces an input at level 1 for the NOR gate on the output side. In this case also the final output Y = 0.
When A = 1 and B = 0, the NAND gate on the input side once again produces an input at level 1 for the NOR gate on the output side. Once again the final output Y = 0.
When both A and B are at level 1, the NAND gate on the input side produces an input at level 0 for the NOR gate on the output side.
The inverter connected to the other input of  the NOR gate also supplies an input at level 0 for the NOR gate on the output side. Since both inputs of the NOR gate are at zero level, its output Y = 1.
The correct truth table is (b).

(5) The transfer characteristic curve of an npn transistor used as a switch is shown in the figure. Vi is the base bias voltage and Vo is the collector voltage. Which portions of the characteristic curve are used for operating the transistor as a switch?

(a) AB and EF

(b) BC, CD and DE

(c) CD only

(d) AB only

(e) EF only

When a transistor is use as a switch, it operates in the saturation region EF for full conduction (switch on) and in the cut off region AB for no conduction (switch off). The correct option is (a).

[When the transistor operates in the saturation region, the base current is large so that the collector current also is large. Almost the entire collector supply voltage will then be dropped across the resistance connected to the collector lead so that the emitter to collector voltage will be very small. The transistor will operate in the cut off region when the emitter to base voltage is insufficient to forward bias the emitter base junction and to make the transistor conduct. Consequently there is no voltage drop across the resistance connected to the collector lead so that the entire collector supply voltage appears on the collector].

You can access all posts on electronics on this site by clicking on the label, ‘electronics’ below this post.