In an NPN transistor circuit, the collector current is 10mA.
Question: In an NPN transistor circuit, the collector current is 10mA. If 95 per cent of the electrons emitted reaches the collector, which of the following statements are true? (a) The emitter current will be 8 mA (b) The emitter current will be 10.53 mA (c) The base current will be 0.53 mA (d) The base current will be 2 mA Solution: (b) The emitter current will be 10.53 mA (c) The base current will be 0.53 mA...
Read More →In the figure shows the transfer characteristics
Question: In the figure shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true? (a) At Vi = 0.4V, transistor is in active state (b) At Vi = 1V, it can be used as an amplifier (c) At Vi = 0. .5V, it can be used as a switch turned off (d) At Vi = 2.5V, it can be used as a switch turned on Solution: (b) At Vi = 1V, it can be used as an amplifier (c) At Vi = 0. .5V, it can be used as a switch turned off (d) At Vi = 2.5V, it can be used as a swit...
Read More →Consider an NPN transistor with its base-emitter
Question: Consider an NPN transistor with its base-emitter junction forward biased and collector-base junction reverse biased. Which of the following statements are true? (a) Electrons crossover from emitter to collector (b) Holes move from base to collector (c) Electrons move from emitter to base (d) Electrons from emitter move out of base without going to the collector Solution: (a) Electrons crossover from emitter to collector (c) Electrons move from emitter to base...
Read More →Two dice are thrown. Find
Question: Two dice are thrown. Find (i) the odds in favor of getting the sum 6 (ii) the odds against getting the sum 7 Solution: Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), $(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$, $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$, $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$, $(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$, $(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$ Total cases where sum will be 6 is $(1,5),(2,4),(3,3),(4,2),(5,1)$ i.e. 5 Probability of getting sum 6 $=\frac...
Read More →When an electric field is applied across a semiconductor
Question: When an electric field is applied across a semiconductor (a) electrons move from lower energy level to higher energy level in the conduction band (b) electrons move from higher energy level to lower energy level in the conduction band (c) holes in the valence band move from higher energy level to lower energy level (d) holes in the valence band move from lower energy level to higher energy level Solution: (a) electrons move from lower energy level to higher energy level in the conducti...
Read More →Truth table for the given circuit in the figure is
Question: Truth table for the given circuit in the figure is (a) Solution:...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{3 x+1}{\sqrt{5-2 x-x^{2}}} d x$ Solution: Given $I=\int \frac{3 x+1}{\sqrt{-x^{2}-2 x+5}} d x$ Integral is of form $\int \frac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow 3 x+1=\lambda(-2 x-2)+\mu...
Read More →In the circuit shown in the figure,
Question: In the circuit shown in the figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is (a) 1.3 V (b) 2.3 V (c) 0 (d) 0.5 V Solution: (b) 2.3 V...
Read More →The output of the given circuit in the figure
Question: The output of the given circuit in the figure (a) would be zero at all times (b) would be like a half-wave rectifier with positive cycles in output (c) would be like a half-wave rectifier with negative cycles in output (d) would be like that of a full-wave rectifier Solution: (c) would be like a half-wave rectifier with negative cycles in output...
Read More →If 5/14 Is the probability of occurrence of an event, find
Question: If 5/14 Is the probability of occurrence of an event, find (i) the odds in favor of its occurrence (ii) the odds against its occurrence Solution: (i) We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$ Given, probability $=\frac{5}{14}$ We know, probability $=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}} .$ So, $\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}=\frac{5}{14}$ $a=5$ and $a+b=14$ i.e. $b=9$ odds in favor of its...
Read More →Prove the following
Question: Hole is (a) an anti-particle of the electron (b) a vacancy created when an electron leaves a covalent bond (c) absence of free electrons (d) an artificially created particle Solution: (b) a vacancy created when an electron leaves a covalent bond...
Read More →A 220 V A.C. supply is connected between points A and B
Question: A 220 V A.C. supply is connected between points A and B in the figure. What will be the potential difference V across the capacitor? (a) 220V (b) 110V (c) 0V (d) 220 2 V Solution: (d) 220 2 V...
Read More →In the figure, assuming the diodes
Question: In the figure, assuming the diodes to be ideal, (a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B (b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa (c) D1 and D2 are both forward biased and hence current flows from A to B (d) D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa Solution: (b) D2 is forward biased and D1 is reverse biased and hence no current flows...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{2 x+5}{\sqrt{x^{2}+2 x+5}} d x$ Solution: Given $I=\int \frac{2 x+5}{\sqrt{x^{2}+2 x+5}} d x$ Integral is of form $\int \frac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$ Writing numerator as $p x+q=\lambda\left\{\frac{d}{d x}\left(a x^{2}+b x+c\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow 2 x+5=\lambda(2 x+2)+\mu$ $\therefore \lambda=1$ and $\mu=3$ Let $2 x+5=2 x+2+3$ and ...
Read More →If the odds against the occurrence of an event be
Question: If the odds against the occurrence of an event be 4 : 7, find the probability of the occurrence of the event. Solution: We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$ similarly, if odds are not in the favor of the occurrence an event are a:b, then the probability of not occurrence of the event is $\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}$ We also know that, Probability of occurring $=1$ - the probability o...
Read More →In the figure, Vo is the potential
Question: In the figure, Vo is the potential barrier across a p-n junction, when no battery is connected across the junction (a) 1 and 3 both correspond to forward bias of junction (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction (c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction (d) 3 and 1 both correspond to reverse bias of junction Solution: (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias o...
Read More →The conductivity of a semiconductor increases
Question: The conductivity of a semiconductor increases with increase in temperature because (a) number density of free current carriers increases (b) relaxation time increases (c) both number density of carriers and relaxation time increase (d) number density of current carriers increases, relaxation time decreases but the effect of a decrease in relaxation time is much less than the increase in number density Solution: (d) number density of current carriers increases, relaxation time decreases...
Read More →The odds in favor of the occurrence of an event are
Question: The odds in favor of the occurrence of an event are 8 : 13. Find the probability that the event will occur. Solution: We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$ which indirectly came from Probability of the occurrence of an event $=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$ Where, Total no.of desired outcomes = a, and total no.of outcomes = a+b Given a = 8, b= 13 The...
Read More →Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52
Question: Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable (i) Verify this by calculating the proton separation energy Sp for Sn120 (Z = 50) and Sb121 = (Z = 51). The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by Sp = (MZ1, N + MH MZ,N) c2. Given In119= 118.9058u, Sn120= 119.902199u, Sb121=...
Read More →The activity R of an unknown radioactive
Question: The activity R of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows: (i) Plot the graph of R versus t and calculate half-life from the graph. (ii) Plot the graph of ln(R/R0) versus t and obtain the value of half-life from the graph. Solution: (i) R(MBq) t(h) From graph we can say that the activity of R has reduced by 50%. Therefore, the half-life is 40 mins. (ii) ln(R/Ro) n(h) Slope of the graph = = 1.05 h-1 Half-time = 0.693/ = ...
Read More →If 7/10 is the probability of occurrence of an event
Question: If 7/10 is the probability of occurrence of an event, what is the probability that it does not occur? Solution: We know that, Probability of occurring $=1$ - the probability of not occurring Given the probability of occurrence $=\frac{7}{10}$ Therefore, the probability of not occurrence $=1-\frac{7}{10}$ $=\frac{3}{10}$ Conclusion: Probability of not occurrence is $\frac{3}{10}$...
Read More →Before the neutrino hypothesis,
Question: Before the neutrino hypothesis, the beta decay process was thought to be the transition,nP+enP+eIf this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range. Solution: Let us consider cases before and after -decay Before -decay, En = mnc2and pn = 0 After -decay, pn = pp + pe pc = mnc2 mpc2= 938 MeV 936 MeV = 2 MeV Ep = 936 MeV Ee = 2.06 MeV...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x+1}{\sqrt{x^{2}+1}} d x$ Solution: Given $I=\int \frac{x+1}{\sqrt{x^{2}+1}} d x$ Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow x+1=\lambda(2 x)+\mu$ $\therefore \lambda=1 / 2$ and $\mu=1$ Let $x+1=1 / 2(2 x)+1$ and ...
Read More →The deuteron is bound by nuclear forces
Question: The deuteron is bound by nuclear forces just as $\mathrm{H}$-atom is made up of $\mathrm{p}$ and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e': $F=\frac{1}{4 \Pi \epsilon_{0}} \frac{e^{\prime 2}}{r}$ estimate the value of (e'/e) given that the binding energy of a deuteron is $2.2 \mathrm{MeV}$. Solution: The binding energy of H atom = E = 13.6 eV The reduced m ...
Read More →Deuteron is a bound state of a neutron and
Question: Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A -ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident -ray. If E = B, show that this cannot happen. Hence calculate how much bigger than B must E be for such a process to happen. Solution: The binding energy of a deuteron = B = 2.2 MeV Kn, Kp are the kinetic energies of neutron and proton pn and pp ar...
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