Mention two strategies evolved to prevent self-pollination in flowers.
Question: Mention two strategies evolved to prevent self-pollination in flowers. Solution: Self-pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies that have evolved to prevent self-pollination in flowers are as follows: (1)In certain plants, the stigma of the flower hasthecapability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube.It is a genetic mechanism to prevent self-pollination calledself-...
Read More →For a CE-transistor amplifier,
Question: For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. Solution: Collector resistance,RC= 2 kΩ = 2000 Ω Audio signal voltage across the collector resistance,V= 2 V Current amplification factor of the transistor,= 100 Base resistance,RB= 1 kΩ = 1000 Ω Input signal voltage =Vi Base current =IB W...
Read More →The slope of a line is double of the slope of another line.
Question: The slope of a line is double of the slope of another line.If tangent of the angle between them is $\frac{1}{3}$, find the slopes of he lines. Solution: Let $m_{1}$ and $m$ be the slopes of the two given lines such that $m_{1}=2 m$. We know that if $\theta$ isthe angle between the lines $I_{1}$ and $I_{2}$ with slopes $m_{1}$ and $m_{2}$, then $\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$. It is given that the tangent of the angle between the two lines is $\frac{1}{3}$. ...
Read More →What are chasmogamous flowers?
Question: What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer. Solution: There are two types of flowers present in plants namelyOxalisandViola chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species. Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to ...
Read More →Is the function defined by
Question: Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi ?$ Solution: The given function is $f(x)=x^{2}-\sin x+5$ It is evident that $f$ is defined at $x=\pi$. At $x=\pi, f(x)=f(\pi)=\pi^{2}-\sin \pi+5=\pi^{2}-0+5=\pi^{2}+5$ Consider $\lim f(x)=\lim \left(x^{2}-\sin x+5\right)$ Put $x=\pi+h$ If $x \rightarrow \pi$, then it is evident that $h \rightarrow 0$ $\begin{aligned} \therefore \lim _{x \rightarrow x} f(x) =\lim _{x \rightarrow x}\left(x^{2}-\sin x+5\right) \\ =\lim ...
Read More →In half-wave rectification, what is the output frequency if the input frequency is 50 Hz.
Question: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. Solution: Input frequency = 50 Hz For a half-wave rectifier, the output frequency is equal to the input frequency. Output frequency = 50 Hz For a full-wave rectifier, the output frequency is twice the input frequency. Output frequency = 2 50 = 100 Hz...
Read More →For a transistor amplifier, the voltage gain
Question: For a transistor amplifier, the voltage gain (a)remains constant for all frequencies. (b)is high at high and low frequencies and constant in the middle frequency range. (c)is low at high and low frequencies and constant at mid frequencies. (d)None of the above. Solution: The correct statement is(c). The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies....
Read More →With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Question: With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte. Solution: The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs. The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends an...
Read More →For transistor action, which of the following statements are correct:
Question: For transistor action, which of the following statements are correct: (a)Base, emitter and collector regions should have similar size and doping concentrations. (b)The base region must be very thin and lightly doped. (c)The emitter junction is forward biased and collector junction is reverse biased. (d)Both the emitter junction as well as the collector junction are forward biased. Solution: The correct statement is(b),(c). For a transistor action, the junction must be lightly doped so ...
Read More →What is meant by monosporic development of female gametophyte?
Question: What is meant by monosporic development of female gametophyte? Solution: The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspore develops into the f...
Read More →When a forward bias is applied to a p-n junction,
Question: When a forward bias is applied to a p-n junction, it (a)raises the potential barrier. (b)reduces the majority carrier current to zero. (c)lowers the potential barrier. (d)None of the above. Solution: The correct statement is(c). When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced....
Read More →With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.
Question: With a neat, labelled diagram, describe the parts of a typical angiosperm ovule. Solution: An ovule is a female megasporangium where the formation of megaspores takes place. The various parts of an ovule are (1)Funiculus It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary. (2)Hilum It is the point where the body of the ovule is attached to the funiculus. (3)Integuments They are the outer layers surrounding the ovule that provi...
Read More →In an unbiased p-n junction, holes diffuse from the p-region to n-region because
Question: In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a)free electrons in the n-region attract them. (b)they move across the junction by the potential difference. (c)hole concentration in p-region is more as compared to n-region. (d)All the above. Solution: The correct statement is(c). The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has gre...
Read More →Find the angle between the x-axis and the line joining the points
Question: Find the angle between thex-axis and the line joining the points (3, 1) and (4, 2). Solution: The slope of the line joining the points $(3,-1)$ and $(4,-2)$ is $m=\frac{-2-(-1)}{4-3}=-2+1=-1$ Now, the inclination $(\theta)$ of the line joining the points $(3,-1)$ and $(4,-2)$ is given by $\tan \theta=-1$ $\Rightarrow \theta=\left(90^{\circ}+45^{\circ}\right)=135^{\circ}$ Thus, the angle between thex-axis and the line joining the points (3, 1) and (4, 2) is$135^{\circ}$...
Read More →Show that the function defined by
Question: Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral point. Here $[x]$ denotes the greatest integer less than or equal to $X$. Solution: The given function is $g(x)=x-[x]$ It isevident thatgis defined at all integral points. Let $n$ be an integer. Then, $g(n)=n-[n]=n-n=0$ The left hand limit offatx=nis, $\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{-}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]=n-(n-1)=1$ The right hand l...
Read More →Carbon, silicon and germanium have four valence electrons each.
Question: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Siand (Eg)Ge. Which of the following statements is true? (a)(Eg)Si (Eg)Ge (Eg)C (b)(Eg)C (Eg)Ge (Eg)Si (c)(Eg)C (Eg)Si (Eg)Ge (d)(Eg)C= (Eg)Si= (Eg)Ge Solution: The correct statement is(c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The ener...
Read More →Carbon, silicon and germanium have four valence electrons each.
Question: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Siand (Eg)Ge. Which of the following statements is true? (a)(Eg)Si (Eg)Ge (Eg)C (b)(Eg)C (Eg)Ge (Eg)Si (c)(Eg)C (Eg)Si (Eg)Ge (d)(Eg)C= (Eg)Si= (Eg)Ge Solution: The correct statement is(c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The ener...
Read More →Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Question: Which of the statements given in Exercise 14.1 is true for p-type semiconductors. Solution: The correct statement is(d). In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms....
Read More →Without using distance formula,
Question: Without using distance formula, show that points (2, 1), (4, 0), (3, 3) and (3, 2) are vertices of a parallelogram. Solution: Let points (2, 1), (4, 0), (3, 3), and (3, 2) be respectively denoted by A, B, C, and D. Slope of $A B=\frac{0+1}{4+2}=\frac{1}{6}$ Slope of $C D=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$ $\Rightarrow$ Slope of $A B=$ Slope of $C D$ $\Rightarrow \mathrm{AB}$ and $\mathrm{CD}$ are parallel to each other. Now, slope of $B C=\frac{3-0}{3-4}=\frac{3}{-1}=-3$ Slope...
Read More →In an n-type silicon, which of the following statement is true:
Question: In an n-type silicon, which of the following statement is true: (a)Electrons are majority carriers and trivalent atoms are the dopants. (b)Electrons are minority carriers and pentavalent atoms are the dopants. (c)Holes are minority carriers and pentavalent atoms are the dopants. (d)Holes are majority carriers and trivalent atoms are the dopants. Solution: The correct statement is(c). In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carrier...
Read More →Arrange the following terms in the correct developmental sequence:
Question: Arrange the following terms in the correct developmental sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes Solution: The correct development sequence is as follows: Sporogenous tissue pollen mother cell microspore tetrad Pollen grain male gamete During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process ...
Read More →Find the value of x for which the points (x, –1),
Question: Find the value ofxfor which the points (x, 1), (2, 1) and (4, 5) are collinear. Solution: If points A (x, 1), B (2, 1), and C (4, 5) are collinear, then Slope of AB = Slope of BC $\Rightarrow \frac{1-(-1)}{2-x}=\frac{5-1}{4-2}$ $\Rightarrow \frac{1+1}{2-x}=\frac{4}{2}$ $\Rightarrow \frac{2}{2-x}=2$ $\Rightarrow 2=4-2 x$ $\Rightarrow 2 x=2$ $\Rightarrow x=1$ Thus, the required value ofxis 1....
Read More →Differentiate between microsporogenesis and megasporogenesis.
Question: Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events. Solution: (a) (b)Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells. (c)Microsporogenesis results in the formation of haploid microspores from a diploid microspo...
Read More →Find the slope of the line,
Question: Find the slope of the line, which makes an angle of $30^{\circ}$ with the positive direction of $y$-axis measured anticlockwise. Solution: If a line makes an angle of $30^{\circ}$ with the positive direction of the $y$-axis measured anticlockwise, then the angle made by the line with the positive direction of the $x$-axis measured anticlockwise is $90^{\circ}+30^{\circ}=120^{\circ}$. Thus, the slope of the given line is $\tan 120^{\circ}=\tan \left(180^{\circ}-60^{\circ}\right)=-\tan 6...
Read More →Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants.
Question: Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of235U to be about 200MeV. Solution: Amount of electric power to be ge...
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