Consider sunlight incident on a pinhole of width 103A.
Question: Consider sunlight incident on a pinhole of width 103A. The image of the pinhole seen on a screen shall be (a) a sharp white ring (b) different from a geometrical image (c) a diffused central spot, white in colour (d) diffused coloured region around a sharp central white spot Solution: The correct answer is (b) different from a geometrical image (d) diffused coloured region around a sharp central white spot...
Read More →A die is thrown. Find the probability of
Question: A die is thrown. Find the probability of getting a 2 or a 3 Solution: We know that, Probability of occurrence of an event $=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$ Total outcomes are $1,2,3,4,5,6$, and the desired outcomes are 2,3 Therefore, total no.of outcomes are 6 , and total no.of desired outcomes are 2 Probability of getting a 2 or 3 $=\frac{2}{6}$ $=\frac{1}{3}$ Conclusion: Probability of getting 2 or 3 when a die is thrown is $\frac{1}{3}$...
Read More →A die is thrown. Find the probability of
Question: A die is thrown. Find the probability of getting a 5 Solution: We know that, Probability of occurrence of an event $=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$ Total outcomes are $1,2,3,4,5,6$, and the desired outcome is 5 Therefore, total no.of outcomes are 6 , and total no.of desired outcomes are 1 Probability of getting $5=\frac{1}{6}$ Conclusion: Probability of getting 5 when die is thrown is $\frac{1}{6}$...
Read More →Two sources S1 and S2 of intensity I 1 and I 2 are placed
Question: Two sources S1 and S2 of intensity I 1 and I 2 are placed in front of a screen in the figure (a). The pattern of intensity distribution seen in the central portion is given by the figure (b). In this case which of the following statements are true. (a) S1 and S2 have the same intensities (b) S1 and S2 have a constant phase difference (c) S1 and S2 have the same phase (d) S1 and S2 have the same wavelength Solution: The correct answer is (a) S1 and S2 have the same intensities (b) S1 an...
Read More →A coin is tossed once. Find the probability of getting a tail.
Question: A coin is tossed once. Find the probability of getting a tail. Solution: We know that Probability of occurrence of an event $=\frac{\text { Total no. of Desired outcomes }}{\text { Total no.of outcomes }}$ Total outcomes of the coin are tails and heads Hence the total no.of outcomes are 2 (i.e. heads and tails) And the desired output is tail. Hence no. of desired outcomes $=1$ Therefore, the probability of getting a tail is $=\frac{1}{2}$ Conclusion: Probability of getting a tail when ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}+x+1}{x^{2}-x} d x$ Solution: Given $I=\int \frac{x^{2}+x+1}{x^{2}-x} d x$ Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$ $\Rightarrow \int \frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}-1) \mathrm{x}} \mathrm{dx}$ $\Rightarrow \int\left(\frac{2 ...
Read More →The figure shows a standard two-slit arrangement
Question: The figure shows a standard two-slit arrangement with slits S1, S2. P1, P2 are the two minima points on either side of P. At P2 on the screen, there is a hole and behind P2 is a second 2- slit arrangement with slits S3, S4 and a second screen behind them (a) There would be no interference pattern on the second screen but it would be lighted (b) The second screen would be totally dark (c) There would be a single bright point on the second screen (d) There would be a regular two-slit pat...
Read More →In a Young’s double-slit experiment,
Question: In a Youngs double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case (a) there shall be alternate interference patterns of red and blue (b) there shall be an interference pattern for red distinct from that for blue (c) there shall be no interference fringes (d) there shall be an interference pattern for red mixing with one for blue Solution: The correct answer is (c) there shall be no interference fringes...
Read More →Consider a ray of light incident from air
Question: Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is (a) $\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\pi$ (b) $\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}$ (c) $\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\frac{\pi}{2}$ (d...
Read More →Consider sunlight incident on a slit of width 104 A.
Question: Consider sunlight incident on a slit of width 104A. The image seen through the slit shall (a) be a fine sharp slit white in colour at the centre (b) a bright slit white at the centre diffusing to zero intensities at the edges (c) a bright slit white at the centre diffusing to regions of different colour s(d) only be a diffused slit white in colour Solution: The correct answer is (a) be a fine sharp slit white in colour at the centre...
Read More →The sum and the sum of squares of length x (in cm) and weight y (in g) of 50 plant products are given below:
Question: The sum and the sum of squares of length x (in cm) and weight y (in g) of 50 plant products are given below: $\sum_{i=1}^{50} x_{i}=212, \sum_{i=1}^{50} x_{i}^{2}=902.8, \sum_{i=1}^{50} y_{i}=261$ and $\sum_{i=1}^{50} y_{i}^{2}=1457.6$ Which is more variable, the length or weight? Solution: To find the more variable, we again need to compare the coefficients of variation (CV). Here the number of products are n = 50 for length and weight both. For length Mean $=\frac{\sum \mathrm{x}_{\m...
Read More →Consider a light beam incident from air
Question: Consider a light beam incident from air to a glass slab at Brewsters angle as shown in the figure. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid (a) For a particular orientation, there shall be darkness as observed through the polaroid (b) The intensity of light as seen through the polaroid shall be independent of the rotation (c) The intensity of light as seen through ...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{x^{3}-3 x}{x^{4}+2 x^{2}-4}$ Solution: Let, $I=\int \frac{x^{3}-3 x}{x^{4}+2 x^{2}-4} d x$ $I=\int \frac{\left(x^{2}-3\right) x}{\left(x^{2}\right)^{2}+2 x^{2}-4} d x$ If we assume $x^{2}$ to be an another variable, we can simplify the integral as derivative of $x^{2}$ i.e. $x$ is present in numerator. Let, $x^{2}=u$ $\Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{du}$ $\Rightarrow \mathrm{x} \mathrm{dx}=1 / 2 \mathrm{du}$ $\therefore I=\frac{1}{2} \int...
Read More →The following results show the number of workers and the wages paid to them in two factories A and B of the same industry.
Question: The following results show the number of workers and the wages paid to them in two factories A and B of the same industry. (i) Which firm pays a larger amount as monthly wages? (ii) Which firm shows greater variability in individual wages? Solution: (i) Both the factories pay the same mean monthly wages. For factory A there are 560 workers. And for factory B there are 650 workers. So, factory A totally pays as monthly wage $=(5460 \times 560)$ Rs. $=3057600 \mathrm{Rs}$. Factory B tota...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$ Solution: Let, $I=\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$ $I=\int \frac{x^{2} x}{\left(x^{2}\right)^{2}+x^{2}+1} d x$ If we assume $x^{2}$ to be an another variable, we can simplify the integral as derivative of $x^{2}$ i.e. $x$ is present in numerator. Let, $x^{2}=u$ $\Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{du}$ $\Rightarrow \mathrm{x} \mathrm{dx}=1 / 2 \mathrm{du}$ $\therefore I=\frac{1}{2} \int \frac{u}{u^{2}+u+1} d u$...
Read More →The mean and variance of the heights and weights of the students of a class are given below:
Question: The mean and variance of the heights and weights of the students of a class are given below: Which shows more variability, heights or weights? Solution: In case of heights, Mean $=63.2$ inches and $S D=11.5$ inches So, the coefficient of variation, $\mathrm{CV}=\frac{\mathrm{SD}}{\text { Mean }} \times 100$ $\Rightarrow \mathrm{CV}=\frac{11.5}{63.2} \times 100=18.196$ In case of weights, Mean $=63.2$ inches and $S D=5.6$ inches. So, the coefficient of variation, $\mathrm{CV}=\frac{\mat...
Read More →Coefficient of variation of the two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively.
Question: Coefficient of variation of the two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively. Find their arithmetic means Solution: Given: Coefficient of variation of two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively. Need to find: Arithmetic means of the distributions. For the first distribution, Coefficient of variation (CV) is $60 \%$, and the standard deviation (SD) is 21 . We know th...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{x+7}{3 x^{2}+25 x+28} d x$ Solution: $I=\int \frac{x+7}{3 x^{2}+25 x+28} d x$ As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $3 x^{2}+13 x-10$ and I can be reduced to a fundamental integration. As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}+25 \mathrm{x}+28\right)=6 \mathrm{x}+25$ $\therefore$ Let, $x+7=A(6 x+25)+B$ $\Rightarrow x+7=6 A x...
Read More →The photoelectric cut-off voltage in a certain
Question: The photoelectric cut-off voltage in a certain experiment is 1.5 V.What is the maximum kinetic energy of photoelectrons emitted? Solution: Photoelectric cut-off voltage,Vo= 1.5 V For emitted photoelectrons, the maximum kinetic energy is: Ke= eVo Where, e = charge on an electron= 1.6 x 10-19C Therefore, Ke= 1.6 x 10-19x 1.5= 2.4 x 10-19J Therefore, 2.4 x 10-19J is the maximum kinetic energy emitted by the photoelectrons....
Read More →The work function of caesium metal is 2.14 eV.
Question: The work function of caesium metal is 2.14 eV. When light of frequency 6 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons (b) Stopping potential (c) maximum speed of the emitted photoelectrons? Solution: Work function of caesium, $\Phi_{o}=2.14 \mathrm{eV}$ Frequency of light, $\mathrm{v}=6.0 \times 10^{14} \mathrm{~Hz}$ (a) The maximum energy (kinetic) by the photoelectric effect: $\mathrm{K}=\m...
Read More →Find the: (a) Maximum frequency, and
Question: Find the: (a) Maximum frequency, and (b) The minimum wavelength of X-rays produced by 30 kV electrons. Solution: Electron potential, V = 30 kV= 3 104V Hence, electron energy,E = 3 104eV Where, e = Charge on one electron= 1.6 10-19C (a)Maximum frequency by the X-rays = The energy of the electrons: E = h Where, h = Plancks constant= 6.626 10-34Js Therefore, $v=\frac{E}{h}$ $=\frac{1.6 \times 10^{-19} \times 3 \times 10^{4}}{6.626 \times 10^{-34}}=7.24 \times 10^{18} \mathrm{~Hz}$ Hence, ...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x$ Solution: $I=\int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x=\int \frac{(3 \sin x-2) \cos x}{13-\left(1-\sin ^{2} x\right)-7 \sin x} d x$ $\Rightarrow I=\int \frac{(3 \sin x-2) \cos x}{12+\sin ^{2} x-7 \sin x} d x$ Let, $\sin x=t \Rightarrow \cos x d x=d t$ $\therefore I=\int \frac{(3 t-2)}{t^{2}-7 t+12} d t$ As we can see that there is a term of $\mathrm{t}$ in numerator and derivative o...
Read More →In deriving the single slit diffraction pattern,
Question: In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n/a. Justify this by suitably dividing the slit to bring out the cancellation. Solution: Let a be the width of a single slit. The single slit is further divided into n smaller slits of width a. a = a/n Each of the smaller slits should produce zero intensity for the single slit to produce zero intensity. For this to happen, theangle of diffraction, = /a ⇒ = /(a/n) or, = n/a Therefore, ...
Read More →The following results show the number of workers
Question: The following results show the number of workers and the wages paid to them in two factories $F_{1}$ and $F_{2}$ Which factory has more variation in wages? Solution: Mean wages of both the factories are the same, i.e., Rs. 5300. To compare variation, we need to find out the coefficient of variation (CV). We know, CV $\frac{\mathrm{SD}}{\text { Mean }} \times 100$ where SD is the standard deviation. The variance of factory $A$ is 100 and the variance of factory $B$ is 81 . Now, $S D$ of...
Read More →In deriving the single slit diffraction pattern,
Question: In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n/a. Justify this by suitably dividing the slit to bring out the cancellation. Solution: Let a be the width of a single slit. The single slit is further divided into n smaller slits of width a. a = a/n Each of the smaller slits should produce zero intensity for the single slit to produce zero intensity. For this to happen, theangle of diffraction, = /a ⇒ = /(a/n) or, = n/a Therefore, ...
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