Find the mean deviation about the mean for the following data :
Question: Find the mean deviation about the mean for the following data : Solution: we make the following table from the given data:br data-mce-bogus="1"/ppimg src="https://www.esaral.com/qdb/uploads/2022/02/22/image21561.png" alt="" Therefore, $\overline{\mathrm{x}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}}=\frac{12500}{100}=125$...
Read More →Find the mean deviation about the mean for the following data :
Question: Find the mean deviation about the mean for the following data : Solution: we make the following table from the given data:br data-mce-bogus="1"/ppimg src="https://www.esaral.com/qdb/uploads/2022/02/22/image86178.png" alt="" Therefore, $\overline{\mathrm{X}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}}=\frac{1350}{50}=27$ Thus, the required mean deviation about the mean is M. $D(\bar{x})=\frac{\sum_{i=1}^...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$ Solution: $\left.\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x=\int(\sin x-\cos x) / \sqrt{(}(\sin x+\cos x)^{2}-1\right) d x$ Let $\sin x+\cos x=t$ $(\cos x-\sin x)=d t$ Therefore, $\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^{2}-1}} d x=\int-\frac{d t}{\sqrt{t^{2}-1}}$ Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}-a^{2}\right)}\right]+c$ $=\int-\frac{d t...
Read More →Find the mean deviation about the median for the following data :
Question: Find the mean deviation about the median for the following data : Solution: The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get, Now, $N=50$ which is even. Median is the mean of the $25^{\text {th }}$ observation and $26^{\text {th }}$ observation. Both of these observations lie in the cumulative frequency 29 , for which the corresponding observation is 30 . Median $(M)=\frac{25^{\text {th }} \text { observation ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \sqrt{\operatorname{cosec} x-1} d x$ Solution: $\int \sqrt{\operatorname{cosec} x-1} d x$ Since $\operatorname{cosec} x=1 / \sin x$ $\int \sqrt{\operatorname{cosec} x-1} d x=\int \sqrt{\frac{1}{\sin x}-1} d x=\int \sqrt{\frac{1-\sin x}{\sin x}} d x$ Multiply with $(1+\sin x)$ both numerator and denominator $=\int \sqrt{\frac{1-\sin x}{\sin x}} d x=\int \sqrt{\frac{1-\sin x *(1+\sin x)}{\sin x *(1+\sin x)}} d x$ Since $(a+b) \times(a-b)=a^{2}-b^{2...
Read More →Find the mean deviation about the median for the following data :
Question: Find the mean deviation about the median for the following data : Solution: The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get, Now, $\mathrm{N}=50$ which is even. Median is the mean of the $25^{\text {th }}$ observation and $26^{\text {th }}$ observation. Both of these observations lie in the cumulative frequency 30 , for which the corresponding observation is 13 . Median $(M)=\frac{25^{\text {th }} \text { obs...
Read More →Find the mean deviation about the median for the following data :
Question: Find the mean deviation about the median for the following data : Solution: The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get, Now, $N=29$ which is odd. Since, $15^{\text {th }}$ observation lie in the cumulative frequency 21 , for which the corresponding observation is 30 . $\operatorname{Median}(\mathrm{M})=\left(\frac{29+1}{2}\right)^{\text {th }}$ or $15^{\text {th }}$ observation $=30$ Now, absolute values...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x$ Solution: Let $\sin x=t$ $\cos x d x=d t$ $\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x=\int \frac{d t}{\sqrt{t^{2}-2 t-3}}$ Add and subtract $1^{2}$ in denominator $=\int \frac{d t}{\sqrt{t^{2}-2 t-3}}=\int \frac{d t}{\sqrt{t^{2}-2 t+1^{2}-1^{2}-3}}=\int \frac{d t}{\sqrt{\left((t-1)^{2}-2^{2}\right)}}$ Let $t-1=u$ $\mathrm{dt}=\mathrm{du}$ $=\int \frac{d t}{\sqrt{\left((t-1)^{2}-2^{2}\ri...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{\left(1-x^{2}\right)\left\{9+\left(\sin ^{-1} x\right)^{2}\right.}} d x$ Solution: Let $\sin ^{-1} x=t$ $\mathrm{dt}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}$ Therefore, $\int \frac{1}{\sqrt{\left(1-x^{2}\right)\left\{9+\left(\sin ^{-1} x\right)^{2}\right.}} d x=\int \frac{1}{\sqrt{3^{2}-t^{2}}} d t$ Since we have, $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left.\left(x^{2}+a^{2}\right)\right]+c}\r...
Read More →Find the mean deviation about the mean for the following data :
Question: Find the mean deviation about the mean for the following data : Solution: We have, Therefore, $\overline{\mathrm{X}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}}=\frac{528}{66}=8$ Now, Thus, the required mean deviation about the mean is M. $D(\bar{x})=\frac{\sum_{i=1}^{6} f_{i}\left|x_{i}-\bar{x}\right|}{\sum_{i=1}^{6} f_{i}}=\frac{138}{66}=2.09$...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}}-4}} d x$ Solution: Let $\mathrm{x}^{\frac{1}{3}}=\mathrm{t}$ So, $\mathrm{dt}=1 / 3 \mathrm{x}^{\frac{1}{2}-1} \mathrm{dx}$ $=\mathrm{dt}=\frac{1}{3} \mathrm{x}^{\frac{1}{3}-1} \mathrm{dx}=\frac{1}{3} \mathrm{x}^{-\frac{2}{3}}$ $O r, \frac{d x}{x^{\frac{2}{3}}}=3 d t$ $\int \frac{1}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}}-4}} d x=3 \int \frac{d t}{\sqrt{t^{2}-2^{2}}}$ Since, $\int \frac{1}{\sqrt{\left(...
Read More →Verify the Gauss’s law for magnetic field
Question: Verify the Gausss law for magnetic field of a point dipole of dipole moment m at the origin for the surface which is a sphere of radius R. Solution: P is the point at a distance r from O and OP, then the magnetic field is given as: $B=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{r}$ dS is the elementary area of the surface P, then dS = r2(r2sin d r) $\oint B . d S=\oint \frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{r}\left(r^{2} \sin \theta d \theta \hat{r}\right...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x$ Solution: Let $\sin x=t$ $d t=\cos x d x$ therefore, $\int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{2^{2}-t^{2}}}$ Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$ $=\int \frac{d t}{\sqrt{2^{2}-t^{2}}}=\sin ^{-1}\left(\frac{t}{2}\right)+c=\sin ^{-1}\left(\frac{\sin x}{2}\right)+c$...
Read More →Find the mean deviation about the mean for the following data :
Question: Find the mean deviation about the mean for the following data : Solution: We have, Therefore $\overline{\mathrm{X}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}}=\frac{300}{40}=7.5$ Now Thus, the required mean deviation about the mean is $\overline{\mathrm{x}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sin 2 x}{\sqrt{\cos ^{4} x-\sin ^{2} x+2}} d x$ Solution: Let $t=\cos ^{2} x$ $d t=2 \cos x \sin x d x=-\sin 2 x d x$ therefore, $\int \frac{\sin 2 x}{\sqrt{\cos ^{4} x-\sin ^{2} x+2}} d x=\int-\frac{d t}{\sqrt{t^{2}-\left(1-t^{2}\right)+2}}$ since, $\left[\sin ^{2} x=1-\cos ^{2} x\right]$ $\int-\frac{d t}{\sqrt{t^{2}-\left(1-t^{2}\right)+2}}=\int-\frac{d t}{\sqrt{t^{2}+t+1}}=\int-\frac{d t}{\sqrt{t^{2}+t+\frac{1}{4}+\frac{3}{4}}}$ $=\int-...
Read More →A ball of superconducting material
Question: A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. (i) In which direction will it move? (ii) What will be the direction of its magnetic moment? Solution: (i) The superconducting material will move away from the bar magnet. (ii) The direction of the magnetic moment will be from left to right....
Read More →Find the mean deviation about the mean for the following data :
Question: Find the mean deviation about the mean for the following data : Solution: We have, Therefore, $\overline{\mathrm{X}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}}=\frac{756}{36}=21$ Now, Thus, the required mean deviation about the mean is $\overline{\mathrm{x}}=\frac{\sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum_{\mathrm{i}=1}^{6} \mathrm{f...
Read More →From molecular view point,
Question: From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism, and ferromagnetism. Solution: The temperature dependence of susceptibility for a diamagnetism is not much affected by the temperature. The temperature dependence of susceptibility for a paramagnetism and ferromagnetism is affected by the temperature that is as the temperature rises, the magnetic moments get disturbed....
Read More →Explain quantitatively the order of magnitude
Question: Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N2and Cu. Solution: Density of nitrogen = 28 g/ 22400 cc Density of copper = 8 g/ 22400 cc Ratio of densities = 16 10-4 Diamagnetic susceptibility = density of nitrogen/density of copper = 1.6 10-4...
Read More →A permanent magnet in the shape of a thin
Question: A permanent magnet in the shape of a thin cylinder of length 10 cm has M = 106 A/m. Calculate the magnetisation current Im. Solution: Intensity of magnetisation = 106A/m Length, l = 0.1 m M = IM/l IM= Ml = 105A...
Read More →A proton has spin and magnetic moment
Question: A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials? Solution: The comparison between the spinning of a proton and an electron is done by comparing their magnetic dipole moment which is given as p= eh/4mp e= eh/4me p/e = me/mp = 1/1837 1 p e...
Read More →Let the magnetic field on the earth be modelled
Question: Let the magnetic field on the earth be modelled by that of a point magnetic dipole at the centre of the earth. The angle of dip at a point on the geographical equator (a) is always zero (b) can be zero at specific points (c) can be positive or negative (d) is bounded Solution: (b) can be zero at specific points (c) can be positive or negative (d) is bounded...
Read More →Find the mean deviation about the median for the following data :
Question: Find the mean deviation about the median for the following data : 70, 34, 42, 78, 65, 45, 54, 48, 67, 50, 56, 63 Solution: Here the number of observations is 12 which is odd. Arranging the data into ascending order, we have 34, 42, 45, 48, 50, 54, 56, 63, 65, 67, 70, 78 Now, Median $(M)=\left(\frac{6^{\text {th }} \text { observation }+7^{\text {th }} \text { observation }}{2}\right)=\frac{54+56}{2}=55$ The respective absolute values of the deviations from median, i.e' $\left|\mathrm{x...
Read More →Essential difference between electrostatic
Question: Essential difference between electrostatic shielding by a conducting shell and magneto static shielding is due to (a) electrostatic field lines can end on charges and conductors have free charges (b) lines of B can also end but conductors cannot end them (c) lines of B cannot end on any material and perfect shielding is not possible (d) shells of high permeability materials can be used to divert lines of B from the interior region Solution: (a) electrostatic field lines can end on char...
Read More →A long solenoid has 1000 turns per meter and carries a current of 1 A.
Question: A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of r = 1000. The core is heated beyond the Curie temperature Tc (a) the H field in the solenoid is unchanged but the B field decreases drastically (b) the H and B fields in the solenoid are nearly unchanged (c) the magnetisation in the core reverses direction (d) the magnetisation in the core diminishes by a factor of about 108 Solution: (a) the H field in the solenoid is unchanged but the B ...
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