Add assignment three and four

elements_of_mech/assignment-work-and-heat/assign2.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- First Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+First law of thermodynamics suggests that $\sum Q = \sum W$.
+\begin{align*}
+	Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
+	75 - 40 + Q_3 &= -15 + 44 \\
+	Q_3 &= -6 kJ
+\end{align*}
+i.e.\ from the system to the surroundings.
+\section*{Question 2}
+We know that,
+\begin{align*}
+	Q &= \Delta E + W \\
+	Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
+	Q &= -8600 kJ \\
+	Q &= -8.6 MJ
+\end{align*}
+\section*{Question 3}
+There is no heat transfer, therefore
+\begin{align*}
+	Q &=  \Delta E + W \\
+	W &= - \Delta E - \Delta V \\
+	W &= \int_{1}^{0} C_v \cdot dT \\
+	W &= -0.718 (T_2 - T_1) \\
+	W &= -50.26 \frac{kJ}{kg} \\
+	\text{Total Work} &= 2 \times (-50.26) = -100 kJ
+\end{align*}
+\section*{Question 4}
+\begin{align*}
+	1000 &= a + b \times 0.2 \\
+	200 &= a + b \times 1.2 \\
+	b &= -800 \\
+	a &= 1000 + 2 \times 800 = 1160 \\
+	\therefore P &= 1160 - 800 V \\
+	W &= \int_{V_1}^{V_2} P \cdot dV \\
+	  &= \int_{0.2}^{1.2} (1160 - 800V)dV \\
+	  &= 6000 kJ \\
+	V_1 &= 1.5 \times  1000 \times \frac{0.2}{1.5} - 85 \\
+		&= 215 \frac{kJ}{kg} \\
+	V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
+		&= 155 \frac{kJ}{kg} \\
+	\Delta V &= V_2 - V_1 \\
+			 &= 40 \frac{kJ}{kg} \\
+	\Delta U &= m \Delta V  = 60 kJ \\
+	Q &= \Delta U + W \\
+	  &= 60 + 600 \\
+	  &= 660 kJ \\
+	U &= 1.5 PV - 85 \frac{kJ}{kg} \\
+	U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
+	  &= 800 V^2 - 85 \frac{kJ}{kg} \\
+	\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
+	\text{For maximum V,} \\
+	V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
+	V &= 0.725 \\
+	u_{\max} = 335.5 \frac{kJ}{kg} \\
+\end{align*}
+\begin{align*}
+	U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
+\end{align*}
+\section*{Question 5}
+\subsection*{Part a}
+\begin{align*}
+	Q &= \int_{273}^{373} C_p \cdot dT \\
+	t &= T - 273 K \\
+	\therefore t + 100 &= T - 173 \\
+	Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
+	Q &= 238.32 J
+\end{align*}
+\subsection*{Part b}
+\begin{align*}
+	Q &= \Delta E + \int p\cdot dV \\
+	\Delta E &= Q - p(V_2 - V_1) \\
+	\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
+	\Delta E &= 197.79J
+\end{align*}
+\end{document}

elements_of_mech/assignment-work-and-heat/assign3.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- Second Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+We know that the net power output is the difference between the heat input and the heat rejected (cyclic device).
+\begin{align*}
+	W_{net,out} &= Q_H + Q_L \\
+	W_{out} &= 90 - 55 = 35 MW
+\end{align*}
+The thermal efficiency is the ratio of net work output and the heat output.
+\[
+	\eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} =  \frac{35}{90} = 0.3889
+\]
+\section*{Question 2}
+\begin{align*}
+	\text{efficiency} = 1 - \frac{T_1}{T_2}  &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\
+	1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\
+	0.7 &> 0.6 \\
+	\text{output} &> \text{claimed $\implies$ good}
+\end{align*}
+$\therefore$ we can agree to this claim.
+\section*{Question 3}
+\[
+	\eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9%
+\]
+now,
+\begin{align*}
+	\eta &= 1 - \frac{T_2}{T_1} \\
+	T_1 &= \frac{T_2}{1 - \eta} \\
+	T_1 &= 787.5 K
+\end{align*}
+$\therefore$, the temperature at which energy is absorbed is 787.5 K
+\section*{Question 4}
+\begin{align*}
+	\eta &= 1 - \frac{T_1}{T_2} \\
+\end{align*}
+where,
+\begin{align*}
+	\eta &= 0.6 \\
+	T_2 &= 800 K \\
+	T_1 &= T_{sink} \\
+	T_{sink} &= (1 - \eta)T_2 \\
+			  &= 324K
+\end{align*}
+Also,
+\begin{align*}
+	\eta &=  \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\
+	Q_{\text{rejected}} &= Q_1 (1 - \eta) \\
+							 &= 400(1 - 0.6) \\
+							 &= 160 kJ
+\end{align*}
+\section*{Question 5}
+Max COP will be achieved only in a Carnot refrigerator.
+\[
+	{(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438
+\]
+Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$.
+\[
+	P = \frac{3 \times 57}{253} = 0.657 W
+\]
+\section*{Question 6}
+Max COP will be achieved only in a Carnot refrigerator.
+\begin{align*}
+	{(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\
+	{(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\
+	Q_L &= 83.3 \\
+\end{align*}
+\section*{Question 7}
+For process $1-2$
+\begin{align*}
+	Q_{1-2} &= U_2 - U_1 + W_{1-2} \\
+	-732 &= U_2 - U_1 - 2.8 \times 3600 \\
+	U_2 - U_1 &= 9348kJ
+\end{align*}
+For process $2-1$
+\begin{align*}
+	Q_{2-1} &= U_1 - U_2 + W_{2-1} \\
+	-732 &= -9348 - 2.8 \times 3600 \\
+	Q_{2-1} &= -708 kJ
+\end{align*}
+Now,
+\[
+	Q_{2-1} = U_1 - U_2 + W_{2-1}
+\]
+For maximum work, $Q_{2-1} = 0$.
+\[
+	\therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ
+\]
+\section*{Question 8}
+\begin{align*}
+	{(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\
+	{(COP)}_R &= \frac{5}{W} \\
+	W &= 0.56KW
+\end{align*}
+\section*{Question 9}
+\begin{align*}
+	\eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\
+	\eta &= 0.3236 \\
+	\implies 1 - 0.3236 &= 0.6764 \\
+	\text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\
+	\text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\
+	if \;Q_3 &= 1kJ \\
+	\therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\
+	W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\
+	Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\
+\end{align*}
+\section*{Question 10}
+\begin{align*}
+	\frac{10}{W} &= \frac{293}{293-273} \\
+	or\;W &= 683 W
+\end{align*}
+\end{document}

engg_mech/assignment/t_four.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Calculate Reactions:
+\begin{align*}
+	\sum M_J &= 0 \\
+	(12 kN)(4.8) + (12 kN)(2.4) - B(9.6) &= 0 \\
+	B &= 9 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN - 12 kN - 12 kN + J &= 0 \\
+	J &= 15kN
+\end{align*}
+Member CD:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN + F_{CD} &= 0 \\
+	F_{CD} &= 9 kN \to \text{compression}
+\end{align*}
+Member DF:\@
+\begin{align*}
+	\sum M_c &= 0 \\
+	F_{DF}1.8 m - 9kN \times 2.4m &= 0 \\
+	F_{DF} &= 12kN  \to \text{Tension}
+\end{align*}
+\section*{Question 2}
+Reactions:\@
+\[
+	A = N = 0
+\]
+DF member:\@
+\begin{align*}
+	\sum M_E &= 0 \\
+	(16 kN)(6m) - \frac{3}{5} F_{DF} (4m) &= 0 \\
+	F_{DF} &= 40 kN \to \text{Tension}
+\end{align*}
+
+EF member:\@
+\begin{align*}
+	\sum F&= 0 \\
+	16 kN \sin{\beta} - F_{EF} \cos{\beta} &= 0 \\
+	F_{EF} &= 16 \tan{\beta} \\
+		   &= 12 kN \to \text{Tension}
+\end{align*}
+
+EG member:\@
+\begin{align*}
+	\sum M_F &= 0 \\
+	16kN \times 9m + \frac{4}{5}F_{EG} \times 3m &= 0 \\
+	F_{EG} &= -60 kN \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Reactions
+\begin{align*}
+	\sum M_k &= 0 \\
+	36\times 2.4 - B \times 13.5 + 20 \times 9 + 20 \times 4.5 &= 0 \\
+	B &= 26.4kN \\
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	K_x &= 36 \\
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	26.4 - 20 -20 + K_y &= 0 \\
+	K_y &=  13.6 kN \uparrow \\
+\end{align*}
+\begin{align*}
+	\sum M_C &= 0 \\
+	36 \times 1.2 - 26.4 \times 2.25 - F_{AD} \times 1.2 &= 0 \\
+	F_{AD} &= 13.5 kN \to \text{compression} \\
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	\left( \frac{8}{17}F_{CD}\right)(4.5) &= 0 \\
+	F_{CD} &= 0 \\
+\end{align*}
+\begin{align*}
+	\sum M_D &= 9 \\
+	\frac{15}{17} \times F_{CE} \times 2.4 - 26.4 \times 4.5 &= 0 \\
+	F_{CE} &= 56.1 kN \to \text{Tension}
+\end{align*}
+\section*{Question 4}
+
+Support reactions
+\begin{align*}
+	\sum M_I &= 0 \\
+	2\times 12 + 5\times 8 3\times 6 + 2\times 4 - A_y \times 16 &= 0 \\
+	A_y &= 5.625 kN
+\end{align*}
+\begin{align*}
+	\sum A_x &= 0 \\
+	A_x &= 0
+\end{align*}
+Method of joints: By inspection, members BN, NC, DO, OC, HJ, LE \& JG are zero force members \\
+Method of sections:
+\begin{align*}
+	\sum M_M &= 0 \\
+	4F_{CD} - 5.625 \times 4 &= 0 \\
+	F_{CD} &= 5.625 kN \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	4F_{CM} - 2\times 4 &= 0 \\
+	F_{CM} &= 2 kN \to \text{Tension}
+\end{align*}
+\end{document}

engg_mech/assignment/t_one.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\graphicspath{{./images/}}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{14 May, 2020}
+\maketitle
+\section*{Question 1}
+
+Triangle law of vector addition states that when two vectors are represented by
+two sides of a triangle in magnitude and direction taken in same order then
+third side of that triangle represents in magnitude and direction the resultant
+of the vectors.
+
+\begin{align*}
+	P &= 48 N \\
+	Q &= 60 N \\
+	R^2 &= P^2 - Q^2 - 2PQ\cos{\ang{150}} \\
+	R^2 &= 48^2 + 60^2 - 2\times 48 \times 60 \times (-0.866) \\
+	R^2 &= 10892 \\
+	R &= 104.36 \\
+	\cos\theta &= \frac{{P^2 + R^2 - Q^2}}{2PR} \\
+	\theta &= \ang{16.70} \\
+	\text{R makes an angle of }(\ang{85} - \ang{16.70}) &= \ang{68.3}
+\end{align*}
+
+\section*{Question 2}
+
+\begin{align*}
+	\text{For the $A = 80N$ force} \\
+	A_x &= 80 \times \cos{\ang{40}} \\
+		&= 61.28N \\
+	A_y &= 80 \times \sin{\ang{40}} \\
+		&= 51.42N \\
+\end{align*}
+
+\begin{align*}
+	\text{For the $B = 120N$ force} \\
+	B_x &= 120 \times \cos{\ang{70}} \\
+		&= 41.04N \\
+	B_y &= 120 \times \sin{\ang{70}} \\
+		&= 112.76N \\
+\end{align*}
+
+\begin{align*}
+	\text{For the $C = 150N$ force} \\
+	C_x &= 150 \times \cos{\ang{165}} \\
+		&= -122.87N \\
+	C_y &= 150 \times \sin{\ang{165}} \\
+		&= 86.036N \\
+\end{align*}
+
+\section*{Question 3}
+
+\begin{figure*}[h]
+	\centering
+	\includegraphics[width=0.5\textwidth]{t_one_1}
+\end{figure*}
+
+\begin{align*}
+	\cos{\alpha} &= \frac{600}{650} \\
+				 &= 0.923 \\
+	\sin{\alpha} &= \frac{250}{650} \\
+				 &= 0.384
+\end{align*}
+\begin{align*}
+	T_1 + T_2 \sin \alpha + 360 \sin\ang{37} &= 480 N \\
+	T_2 \cos \alpha &= 360 \cos \ang{37}
+\end{align*}
+
+From here,
+\begin{align*}
+	T_1 = \text{The tension in cable } BC  &= 143.724 N \\
+	T_2 = \text{The tension in cable } AC  &= 311.5 N
+\end{align*}
+
+\section*{Question 4}
+
+Draw Free body diagram of the lower pulley.
+\begin{figure*}[h]
+	\centering
+	\includegraphics[width=0.5\textwidth]{t_one_2}
+\end{figure*}
+
+\begin{align*}
+	\cos{\theta} &= \frac{0.75}{2.514} = 0.3 \\
+	\sin{\theta} &= \frac{2.4}{2.514} = 0.954 \\
+	2P \cos{\theta} &= P \cos{\alpha} \\
+	2P \sin{\theta} + P\sin{\alpha} &= mg \\
+	\text{From this we get} \\
+	P &= 738.825 N \\
+	\alpha &= 53.13N
+\end{align*}
+
+\section*{Question 5}
+
+Moments about A
+\begin{align*}
+	F_1 &= 250 \cos{\ang{30}} \times 2 = 433 Nm \\
+	F_2 &= 300 \sin{\ang{60}} \times 5 = 1299.04 Nm \\
+	F_3 &= 500 \times 1 Nm
+\end{align*}
+
+Moments about B
+\begin{align*}
+	F_1 &= 0 Nm \\
+	F_2 &= 300 \cos{\ang{60}} \times 4 = 600 Nm \\
+	F_3 &= 0 Nm
+\end{align*}
+
+\section*{Question 6}
+
+\begin{center}
+Tension in $AD$ is $481N$ \\
+Tension in $AB$ is $T_1N$ \\
+Tension in $AC$ is $T_2N$ \\
+Length of $AD$ is $6.5m$ \\
+Length of $AB$ is $7m$ \\
+Length of $AC$ is $7.4m$ \\
+\end{center}
+
+On vertical components
+
+\begin{align*}
+	P &= 481 \cos{\ang{30.51}} + T_1 \cos{\ang{35.87}} + T_2 \frac{5.6}{7.4} \\
+	P &= 414.4 + 0.8 T_1 + 0.75 T_2
+\end{align*}
+
+On horizontal components
+
+\begin{align*}
+	T_1 \sin{\ang{36.87}} N &= T_2 \frac{4.837}{7.4} \sin{\ang{29.74}}N \\
+	0.6 \times T_1 &= 0.324 \times T_2 \\
+	481 \sin{\ang{30.51}} N &= T_2 \frac{4.837}{7.4} \cos{\ang{29.74}} N \\
+	242.2 &= 0.567 \times T_2 \\
+	T_2 &= 430.6 \\
+	T_1 &= 232.57 \\
+	P &= 414.4 + 0.8 \times 242.57 + 0.75 \times 430.6 \\
+	P &= 923.4 N
+\end{align*}
+
+\section*{Question 7}
+\begin{align*}
+	\sum{F_A} &= 0 \\
+	T_{AB} + T_{AC} + T_{AD} + P &= 0 \\
+								 & \text{where P has only one direction that is $\hat{\imath}$} \\
+	\overrightarrow{AB} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} + (380mm)\hat{k} \\
+	AB &= 1060 mm \\
+	\overrightarrow{AC} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} - (320mm)\hat{k}\\
+	AC &= 1040 mm \\
+	\overrightarrow{AD} &= - (960mm)\hat{\imath} + (720mm)\hat{\jmath} - (220mm)\hat{k}\\
+	AD &= 1220 mm \\
+	\overrightarrow{T_{AB}} &= T_{AB} \cdot \hat{AB} = T_{AB} \left(-\frac{48}{53} \hat{\imath} - \frac{12}{53} \hat{\jmath} + \frac{19}{53} \hat{k} \right) \\
+	\overrightarrow{T_{AC}} &= T_{AC} \cdot \hat{AC} = T_{AC} \left(-\frac{12}{13} \hat{\imath} - \frac{3}{13} \hat{\jmath} - \frac{4}{13} \hat{k}\right) \\
+	\overrightarrow{T_{AD}} &= T_{AD} \cdot \hat{AD} = \frac{305}{1220} \times \overrightarrow{AD} = (-240 \hat{\imath} + 180 \hat{\jmath} - 55 \hat{k}) N
+\end{align*}
+
+We know
+\begin{align*}
+	\sum F_A &= 0 \\
+	-\frac{48}{53}T_{AB} - \frac{12}{13} T_{AC} - 240 + P &= 0 \;\text{(X-axis)} \\
+	-\frac{12}{53}T_{AB} -\frac{3}{13} T_{AC} + 180 &= 0  \;\text{(Y-axis)} \\
+	-\frac{19}{53}T_{AB} + \frac{4}{13} T_{AC} + 55 &= 0\;\text{(Z-axis)} \\
+\end{align*}
+
+By solving these equations, we get
+
+\begin{align*}
+	T_{AB} &= 446.71 N \\
+	T_{AC} &= 341.71 N \\
+	P &= 960 N
+\end{align*}
+
+\section*{Question 8}
+
+Calculate the unit vector along each rope just like previous question. \\
+Coordinates of point $A, B, C, D$
+\begin{align*}
+	A &= 4 \hat{k} m \\
+	B &= -1.5 \hat{\imath} m - 2 \hat{\jmath} m \\
+	C &= 2 \hat{\imath} m + 3 \hat{\jmath} m \\
+	D &= 2.5 \hat{\jmath} m
+\end{align*}
+
+Position vectors of each rope is $reference = 0 \hat{\imath} + 0 \hat{\jmath} + 6 \hat{k}$
+\begin{align*}
+	\overrightarrow{r_{AB}} &= -1.5 \hat{\imath} - 6 \hat{k} \\
+	|r_{AB}| &= 6.5 \\
+	\overrightarrow{r_{AC}} &= 2 \hat{\imath} - 3 \hat{\jmath} - 6 \hat{k} \\
+	|r_{AC}| &= 7 \\
+	\overrightarrow{r_{AD}} &= 2.5 \hat{\jmath} - 6 \hat{k} \\
+	|r_{AD}| &= 6.5
+\end{align*}
+
+Now direction of forces are along unit vectors:
+
+\begin{align*}
+	\hat{r_{AB}} &= -\frac{1.5}{6.5} \hat{\imath} - \frac{6}{6.5} \hat{k} \\
+	\hat{r_{AC}} &= \frac{2}{7} \hat{\imath} - \frac{3}{7} \hat{\jmath} - \frac{6}{7} \hat{k} \\
+	\hat{r_{AD}} &= \frac{2.5}{6.5} \hat{\jmath} - \frac{6}{7} \hat{k}
+\end{align*}
+
+\[
+	\sum F_A = 0
+\]
+So,
+
+\begin{align*}
+	-0.231 F_{AB} + 0.286 F_{AC} &= 0 \\
+	-0.308 F_{AB} - 0.429 F_{AC} + 0.385 F_{AD} &= 0 \\
+	-0.923 F_{AB} - 0.857 F_{AC} - 0.923 F_{AD} &= -800
+\end{align*}
+
+On solving these three equations, we get
+
+\begin{align*}
+	F_{AB} &= 251.2N \\
+	F_{AC} &= 202.9N \\
+	F_{AD} &= 427.1 N
+\end{align*}
+
+\end{document}

engg_mech/assignment/t_three.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Applying the equations of the equilibrium to the FBD of the entire truss, we have
+\begin{align*}
+	\sum M_A &= 0 \\
+	N_C (2 + 2) - 4(2) - 3(1.5) &= 0 \\
+	N_C &= 3.125 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	3 - A_x &= 0 \\
+	A_x &= 3 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	A_y + 3.125 - 4 &= 0 \\
+	A_y &= 0.875 kN
+\end{align*}
+Method of joints \\
+Joint C:\@ Just assume it to be in equilibrium
+\begin{align*}
+	\sum F_y &= 0 \\
+	3.125 - F_{CD} \frac{3}{5} &= 0 \\
+	F_{CD} &= 5.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	5.208 \times \frac{4}{5} - F_{CB} &= 0 \\
+	F_{CB} &= 4.17kN  \to \text{Tension}
+\end{align*}
+Joint A:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	0.875 - F_{AD} \times \frac{3}{5} &= 0 \\
+	F_{AD} &= 1.46 kN  \to \text{Compression}
+\end{align*}
+
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} - 3 - 1.458 \times \frac{4}{5} &= 0 \\
+	F_{AB} &= 4.167 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BD} &= 4 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	4.167 - 4.167 &= 0
+\end{align*}
+\section*{Question 2}
+Analyze equilibrium of joint D, C \& E.
+Joint D
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{DE} \times \frac{3}{5} - 600 &= 0 \\
+	F_{DE} &= 1 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	1000 \times \frac{4}{5} - F_{DC} &= 0 \\
+	F_{DC} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 900 N  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CB} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EB} &= 750 N  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BA} &= 1.75 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AL} &= 28.28 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} &= 20 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BC} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BL} &= 0
+\end{align*}
+Joint L
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{LC} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{LK} &= 28.28 kN \to \text{Compression}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CD} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CK} &= 10 kN  \to \text{Tension}
+\end{align*}
+Joint K
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{KD} &= 7.454 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{KJ} &= 23.57 kN  \to \text{Compression}
+\end{align*}
+Joint J
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{IJ} &= 23.57 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{JD} &= 33.3 kN  \to \text{Tension}
+\end{align*}
+Now we know that there exists symmetry,
+\begin{gather*}
+	F_{AL} = F_{GH} = F{LK} = F{HI} = 28.3 kN \\
+	F_{AB} = F_{GF} = F_{BC} = F_{FE} = F_{CD} = F_{ED} = 20 kN \\
+	F_{BL} = F_{FH} = F_{LC} = F_{HE} = 0 \\
+	F_{CK} = F_{EI} = 10 kN \\
+	F_{KJ} = F_{IJ} = 23.6 kN \\
+	F_{KD} = F_{ID} = 7.45 kN
+\end{gather*}
+\section*{Question 4}
+To evaluate support reactions
+\begin{align*}
+	\sum M_E &= 0 \\
+	A_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	E_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	E_x &= P
+\end{align*}
+Methods of joints: By inspecting joint C, members CB \& CD are zero force members. Hence
+\[
+	F_{CB} = F_{CD} = 0
+\]
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AB} &= 2.40 P  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AF} &= 2P  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	2.404P \times \frac{1.5}{\sqrt{3.25}} - P - F_{BF} \times \frac{0.5}{\sqrt{1.25}} - F_{BD} \times \frac{0.5}{\sqrt{1.25}} &= 0 \\
+	P - 0.447 F_{BF} - 0.447 F_{BD} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	2.404P \times \frac{1}{\sqrt{3.25}} - F_{BF} \times \frac{1}{\sqrt{1.25}} + F_{BD} \times \frac{1}{\sqrt{1.25}} &= 0 \\
+	1.33P - 0.8944 F_{BF} + 0.8944 F_{BD} &= 0
+\end{align*}
+Solving the above equations, we get
+\begin{align*}
+	F_{BD} &= 0.3727 P  \to \text{Compression}\\
+	F_{BF} &= 1.863 P  \to \text{Tension}
+\end{align*}
+Joint D
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{DE} &= 0.3727 P  \to \text{Compression}
+\end{align*}
+\section*{Question 5}
+FBD of Joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension} \\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint F
+\begin{gather*}
+	\frac{F_{DF}}{2.29} = \frac{F_{EF}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{DF} = 2.29 kN  \to \text{Tension} \\
+	F_{EF} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint D
+\begin{gather*}
+	\frac{F_{BD}}{2.21} = \frac{F_{DE}}{0.6} = \frac{2.29}{2.29} kN \\
+	F_{DE} = 0.6 kN  \to \text{Compression}\\
+	F_{EF} = 2.21 kN \to \text{Tension}
+\end{gather*}
+FBD of Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+FBD of Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{EJ} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 6}
+Zero Force Members
+Analyzing joint F:\@ Note that $DF$ and $EF$ are zero force members.
+\[
+	F_{DF} = F_{EF} = 0
+\]
+Analyzing joint D:\@ Note that $BD$ and $DE$ are zero force members.
+\[
+	F_{BD} = F_{DE} = 0
+\]
+FBD of joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension}\\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BE} &= 2.7625 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BC} &= 2.25 kN  \to \text{Compression}
+\end{align*}
+FBD of joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 2.86 kN  \to \text{Compression}
+\end{align*}
+FBD of joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	f_{EJ} &= 1.657 kN  \to \text{Tension}
+\end{align*}
+
+\end{document}

engg_mech/assignment/t_two.tex

@@ -0,0 +1,135 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\graphicspath{{./images/}}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{17 May, 2020}
+\maketitle
+\section*{Question 1}
+\subsection*{Question 1.a}
+We shall balance the torque about $C$. $\angle ABC = \theta$, Tension in cable $=T$.
+\begin{align*}
+	240 (0.4) + 240 (0.8) &= T\sin{\theta} \times 0.18 \\
+	T\sin{\theta} &= 1600 \\
+	T \frac{0.24}{0.3} &= 1600 \\
+	T &= 2000
+\end{align*}
+\subsection*{Question 1.b}
+On making FBD of bracket BCD.\@
+
+\begin{align*}
+	x-component &= N_x = T\sin{\theta} = 1600 \\
+	y-component &= N_y = T\cos{\theta} + 240 +240 = 1680
+\end{align*}
+
+\section*{Question 2}
+\subsection*{Question 2.a}
+We shall balance the torque about C
+\begin{align*}
+	P \times 7.5 &= T \times 5 \\
+	T &= 150 lb
+\end{align*}
+\subsection*{Question 2.b}
+\begin{align*}
+	N_x \text{(reaction at C along x-axis)} &= - (P + T\sin{\ang{37}}) = -190 lb \\
+	N_y \text{(reaction at C along y-axis)} &= - T \cos{\ang{37}} = -120 lb
+\end{align*}
+\section*{Question 3}
+\subsection*{Question 3.a}
+\[
+	\alpha = 0
+\]
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb
+\end{align*}
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+\subsection*{Question 3.b}
+\[
+	\alpha = \ang{90}
+\]
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 12 &= 75 \times 10 \\
+	N_a &= 62.5 lb
+\end{align*}
+\subsection*{Question 3.c}
+\[
+	\alpha = \ang{30}
+\]
+Balance torque about the mid point of horizontal rod
+\begin{align*}
+	N_a \times 10 &= {(N_b)}_y \times 10 \\
+	N_a &= {(N_b)}_y
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb \\\\
+	N_a &= N_b \cos{\ang{30}} \\
+	N_b &= 43.30
+\end{align*}
+
+\section*{Question 4}
+\subsection*{Question 4.a}
+Balance torque about C
+\begin{align*}
+	120 \times 0.28 &= T \times  \frac{150}{250} \times 0.2 + T \times \frac{150}{390} \times 0.36 \\
+	33.6 &= T \times 0.26 \\
+	T &= 129.2 N \approx 130 N
+\end{align*}
+\subsection*{Question 4.b}
+\begin{align*}
+	{(N_c)}_x &= T \times \frac{200}{250} + T \times \frac{360}{390} \\
+	{(N_c)}_x &= 223 N \\
+	{(N_c)}_y &= 120 - \left(T \times  \frac{150}{250} + T \times  \frac{150}{390}\right) \\
+	{(N_c)}_y &= -7.21 N \\
+	N_c &= \sqrt{223^2 + 7.21^2} N \\
+	N_c &= 224 N
+\end{align*}
+
+\section*{Question 5}
+Balance torque about B
+\begin{align*}
+	{(N_a)}_y \times 8 &= 4000 \times 2 \\
+	{(N_a)}_y &= 1000 N
+\end{align*}
+FBD of Rod
+\begin{align*}
+	4000 &= {(N_A)}_y + {(N_B)}_y \\
+	4000 &= 1000 + {(N_B)}_y \\
+	{(N_B)}_y &= 3000 \\\\
+	{(N_B)}_y &= N_b \sin{\ang{60}} \\
+	N_B &= 3465 \\
+	{(N_A)}_x &= {(N_B)}_x \\
+	{(N_A)}_x &= N_B \sin{\ang{30}} \\
+	{(N_A)}_x &= 1732
+\end{align*}
+\section*{Question 6}
+First we have to calculate reaction at A
+\begin{align*}
+	\sum F_x &= 0 \\
+	4000 \cos{\ang{30}} &= A_x \\
+	A_x &= 3464 N
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	6000 + 4000 \cos{\ang{30}} &= A_y \\
+	A_y &= 8000 N
+\end{align*}
+\end{document}

engg_physics/lab_assignment_1/.gitignore

@@ -0,0 +1 @@
+page_00.jpg

engg_physics/lab_assignment_1/generate.sh

@@ -0,0 +1,19 @@
+#!/usr/bin/env bash
+
+text='Name: Ahmad Saalim Lone
+Enrollment No: 2019BCSE017
+Subject: Engineering Physics Lab
+Branch: Computer Science
+Unit: Semiconductor Physics
+No. of pages: 6
+Submission Date: 15 May, 2020
+Name of concerned teacher:
+Dr Mohd Zubair Ansari'
+convert -size 638x877 xc:white \
+	-font 'Fira-Code-Bold' \
+	-pointsize 30 \
+	-fill black \
+	-gravity center \
+	-draw "text 0,0 '$text'" \
+	page_00.jpg
+convert page_{00..05}.jpg assignment.pdf

evs/assign1.md

@@ -0,0 +1,119 @@
+\begin{center}
+ \section{Environmental Studies Assignment 1}
+ \subsection{Ahmad Saalim Lone, 2019BCSE017}
+\end{center}
+
+## Question 1
+
+### a) Explain the forest resources in nature and also discuss the over-exploitation methods of forest resources. How the different forests activities help to the human needs?
+
+### Solution
+
+Forest are an important renewable resource. There are many types of forests such as:
+
+- Equatorial Moist Evergreen or Rainforest
+- Mediterranean Forests
+- Tropical deciduous forest
+- Temperate board-leaved deciduous and mixed forest
+- Warm temperature broad-leaved deciduous forest
+- Coniferous Forest
+
+It is estimated that about 10% of world area is covered by forests. Forests provide food, medicine and other products to the tribal people and play a vital role in the life and economy. Forests contribute substantially to the national economy with increasing population and thus increasing demand of fuel, wood, rubber, paper and expansion of area under urban development and industries has head to over-exploitation of forest covers.
+
+### b) Classify the energy resources in nature. How renewable energy sources play an important role to save the environment?
+
+### Solution
+
+On the basis of availability and renewability of natural resources. They are classified in two categories:
+
+- Renewable resources
+- Non-renewable resources
+
+Some notes on renewable and non renewable resources are:
+
+- Renewable resources can replenish themselves naturally. E.g. Soil, Water, etc.
+- Non-renewable resources cannot be replenished once they get exhausted in a definite amount of time.
+- Renewable resources play an important role to save environment. This renewable energy resources help in decreasing the negative impact on the environment. They are clean sources of energy.
+
+## Question 2
+
+### a) Discuss the different water resources on the living planet. Explain the sustainable water management approach.
+
+### Solution
+
+Water comprises 71% of our total area. Somewhere around 97% of that water is saline which leaves only 3% fresh water. Some fresh water sources are:
+
+- Surface Water
+    - Standing Surface Water
+    - Flowing Surface Water
+- Underground Water
+
+Sustainable water management means the ability to meet the water needs of the present without compromising the ability of future generations to do the same. Water substantially also means effective and holistic management of water resources.
+
+
+### b) What are the problems associated with floods, drought and dams? How they affect the environment/humans? Highlight the national and international water conflicts.
+
+### Solution
+
+Effects of some of the natural disasters such as
+
+- **Floods**: Water submerges the low lying areas, cultivation of land gets affected, colossal damage, monetary loss.
+- **Drought**: Almost no rain for a very long time causes draughts. No irrigation, plant and animal life suffers and thus humans get affected.
+
+Unequal amount of rain distribution causes disputes among various lands such as
+
+- National Disputes: sharing of Kaveri water between Karnataka and Tamil Nadu, sharing of Krishna water between Karnataka and Andhra Pradesh.
+- International Disputes: Indus River between India and Pakistan. Colorado River between Mexico and United States of America.
+
+## Question 3
+
+### a) What are the food resources? How you can minimize world food problems? Discuss the types of nutrition.
+
+### Solution
+
+Food is what we eat and need to live. Food can be of two types to humans:
+
+1. **Primary**: Primary comprises of vegetables and plants. E.g. Vegetables, herbs, fruits, etc.
+2. **Secondary**: Secondary comprises of other animals such as chicken, goat, sheep, etc.
+
+In many developing countries, population is increasing rapidly and the government is unable to meet the daily food requirements, which results in diseases and deficiencies such as malnutrition. We can minimize world food issues by supporting domestic food production, stabilize and guarantee fair prices to farmers, maintain supply chains and entertain new and useful public policies.
+
+### Discuss the types of agriculture in India. Which factors affect the modern agriculture in India with reference to fertilizers-pesticides and water logging?
+
+### Solution
+
+Different farming practices in India include:
+
+- **Subsistence Farming**: In this type of farming nearly all the crops or livestock raised are used to maintain the farmers' own use. It is also referred to as small scale production.
+
+- **Shifting Cultivation**: Shifting Cultivation means migratory shifting agriculture. Under this system, a plot is cultivated for few years and when the crop yield declines because of soil exhaustion and is deserted, we shift to another plot for the same and back and forth over the years.
+
+- **Plantation Agriculture**: In this type of agriculture cash crops are cultivated. A single crop like rubber, sugar cane, coffee, tea is grown. These are used majorly for export.
+
+Modern agriculture has its pros and cons. Using fertilizers produce a lot of yield but at the same time cause soil pollution, irrigation is important for plants but the excess of it can cause soil erosion. So, balance must be prevailed in order to maintain soil quality and prevent health hazards. Long term thinking and approach is needed for better overall development.
+
+## Question 4
+
+### a) What are the important mineral resources in nature? What types of mining activities damages the environment?
+
+### Solution
+
+Mineral resources are defined as an occurrence of natural, solid, inorganic or fossilized organic material in/on the earth's crust in such quantity and quality that it is feasible and reasonable to extract it out.
+
+Mineral resources are divided into metallic and non-metallic resources. Mineral resources are the most important natural resources and determine a country's industrial and economic growth by supplying raw material to economy's primary, secondary and tertiary sectors.
+
+Some reasons that affect modern agriculture system affect the mineral resources such as over exploitation, waste, not recycling of materials. This not only affects us but will surely affect the coming generations.
+
+### b) Write short notes on the following:
+
+#### i. Chipko Movement
+
+#### Solution
+
+Chipko Movement (or Chipko Andolan) was a forest conservation movement in India. It began in 1970s in Uttarakhand, then a part of Uttar Pradesh and went on to become a rallying point for many future environmental movements. Chipko Andolan is a movement that practised methods of Satyagraha where both male and female activists from Uttarakhand played vital roles including Gaura Devi, Suraksha Devi, Sudesha Devi, Bachni Devi and Chandi Prasad Bhatt and others. These people after hearing that trees are being cut down in forests reached there and hugged the trees and proved their point to the whole world.
+
+#### ii. Sardar Sarovar Dam
+
+#### Solution
+
+It is a Gravity Dam on the Narmada River near Navagam, Gujarat in India. Four states -- Gujarat, Madhya Pradesh, Maharashtra and Rajasthan receive water and electricity from the dam. Foundation stone was laid out by Prime Minister Jawaharlal Nehru on 5^{th} April 1961. Narmada Bacho Andolan was initiated by environmentalists, tribal people and human right activists against the construction of the dam on the Narmada river. It had a controversial issue due to problems in the form of displacement of local people, loss of livelihood, flood deforestation, etc. Thus, by passing through many controversies, the dam was built.

maths/assignment-matrices/assign4.tex

@@ -0,0 +1,601 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Mathematics Assignment 4 --- Matrices}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{16 May, 2020}
+\maketitle
+
+\section*{Question 1}
+\begin{align*}
+	A &=
+	\begin{bmatrix}
+		1 & -1 & -1 & 2 \\
+		4 & 2 & 2 & -1 \\
+		2 & 2 & 0 & -2
+	\end{bmatrix}
+	\\
+	PAQ &= \text{Normal form} \\
+	I_3AI_4 &= A_{3 \times 4} \\
+	\begin{bmatrix}
+		1 & 0 & 0 \\
+		0 & 1 & 0 \\
+		0 & 0 & 1
+	\end{bmatrix}
+	A
+	\begin{bmatrix}
+		1 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 1
+	\end{bmatrix}
+			&=
+			\begin{bmatrix}
+				1 & -1 & -1 & 2 \\
+				4 & 2 & 2 & -1 \\
+				2 & 2 & 0 & -2
+			\end{bmatrix}
+\end{align*}
+
+Applying the following row tranformations on $I_3$ and column tranformations on $I_4$.
+
+\begin{align*}
+	C_4 &\to C_4 + C_2 \\
+	R_3 &\to \frac{R_3}{2} \\
+	R_2 &\to \frac{R_2 + 2R_1}{3} \\
+	R_1 &\to R_1 + R_3 \\
+	C_1 &\to C_1 - 2 C_4 \\
+	C_4 &\to  C_4 + C_3 \\
+	C_2 &\to C_2 - C_1 \\
+	C_3 &\rightleftharpoons C_1 \\
+	C_2 &\rightleftharpoons C_4 \\
+	R_1 &\rightleftharpoons -R_1
+\end{align*}
+
+we get
+\begin{align*}
+	P &=
+	\begin{bmatrix}
+		-1 & 0 & -\frac{1}{2} \\[6pt]
+		\frac{2}{3} & \frac{1}{3} & 0 \\[6pt]
+		0 & 0 & \frac{1}{2}
+	\end{bmatrix}
+	\\
+	Q &=
+	\begin{bmatrix}
+		0 & 0 & 1 & -1 \\
+		0 & 0 & -2 & 3 \\
+		1 & 1 & 0 & 0 \\
+		0 & 1 & -2 & 2
+	\end{bmatrix}
+	\\
+	\text{Normal Form of A} &=
+	\begin{bmatrix}
+		1 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0
+	\end{bmatrix}
+\end{align*}
+\section*{Question 2}
+\[
+	x^2yz = xy^2z^3 = x^3y^2z = e
+\]
+
+Taking $\ln$ on both sides
+\begin{align*}
+	2\ln{x} + \ln{y} \ln{z} &= 1 \\
+	\ln{x} + 2 \ln{y} + 3 \ln{z} &= 1 \\
+	3\ln{x} + 2\ln{y} + \ln{z} &= 1
+\end{align*}
+
+\[
+	\text{Augemented Matrix} = [A:B] =
+\begin{bmatrix}
+2 & 1 & 1 & : & 1 \\
+1 & 2 & 3 & : & 1 \\
+3 & 2 & 1 & : & 1
+\end{bmatrix}
+\]
+After ppling the following tranformations
+\begin{align*}
+	R_3 &\to R_3 - R_1 \\
+	R_2 &\to R_2 - R_1 \\
+	R_2 &\to R_2 - R_1 \\
+	R_1 &\to R_1 -R_3
+\end{align*}
+
+We get
+\[
+\begin{bmatrix}
+1 & 0 & 1 & : & 1 \\
+-3 & 0 & 1 & : & -1 \\
+1 & 1 & 0 & : & 0
+\end{bmatrix}
+\]
+$Rank(A) = Rank(A:B) = 3 = n$\\
+$\therefore$ unique solution.
+\section*{Question 3}
+\begin{align*}
+	x - cy - bz &= 0 \\
+	cx - y + az &= 0 \\
+	bx + ay - z &= 0 \\
+	\text{Augemented matrix} &=
+\begin{bmatrix}
+1 & -c & -b & : & 0 \\
+c & -1 & a & : & 0 \\
+b & a & -1 & : & 0
+\end{bmatrix}
+\end{align*}
+
+Applying the following tranformations
+\begin{align*}
+	R_2 &\to R_2 - cR_1 \\
+	R_3 &\to R_3 - bR_1 \\
+	R_2 &\to \frac{R_2}{c^2 - 1} \\
+	R_3 &\to R_3 - (a + bc)R_2
+\end{align*}
+we get
+\[
+	C =
+\begin{bmatrix}
+	1 & -c & -b & : & 0 \\[6pt]
+	0 & 1 & \frac{bc+c}{c^2 - 1} & : & 0 \\[6pt]
+0 & 0 & b^2 - 1 - \frac{{(a+bc)}^2}{c^2 - 1} & : & 0
+\end{bmatrix}
+\]
+
+For non trivial solutions $Rank(A) = Rank(c) \ne number\;of\;unknows$
+\begin{align*}
+	b^2 - 1 \frac{{(a+bc)}^2}{c^2 -1} &= 0 \\
+	\implies a^2 + b^2 + c^2 + 2abc &= 1 \\
+\end{align*}
+Now
+\begin{align*}
+	x - cy -bz &= 0 \\
+	y + \left(\frac{bc + a}{c^2 - 1}z\right) &= 0 \\
+\end{align*}
+We get
+\begin{align*}
+	x &= \frac{ac + b}{1-c^2}z \\
+	y &= \frac{bc + a}{1 - c^2}z \\
+	z &= z \\
+	x : y : z &= \sqrt{|1 - a^2|} : \sqrt{|1 - b^2|} : \sqrt{|1 -c^2|}
+\end{align*}
+
+\section{Question 4}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+3 & -4 & 4 \\
+1 & -2 & 44 \\
+1 & -1 & 3
+\end{bmatrix} \\
+	|A - \lambda I| &= 0\\
+\begin{vmatrix}
+3 - \lambda & -4 & 4 \\
+1 & -2 - \lambda & 44 \\
+1 & -1 & 3 - \lambda
+\end{vmatrix} &= 0
+\end{align*}
+
+\( \lambda = -1, 2, 3 \) are Eigen Values
+
+For $\lambda = -1$
+\begin{align*}
+	\begin{bmatrix}
+		4 & -4 & 4 \\
+		1 & -1 & 4 \\
+		1 & -1 & 4
+	\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		1 \\
+		1 \\
+		0
+	\end{bmatrix}
+\end{align*}
+For $\lambda = 2$
+\begin{align*}
+\begin{bmatrix}
+1 & -4 & 4 \\
+1 & -4 & 4 \\
+1 & -1 & 1
+\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		0 \\
+		1 \\
+		1
+	\end{bmatrix}
+\end{align*}
+For $\lambda = 3$
+\begin{align*}
+\begin{bmatrix}
+0 & -4 & 4 \\
+1 & -5 & 4 \\
+1 & -1 & 0
+\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		1 \\
+		1 \\
+		1
+	\end{bmatrix}
+\end{align*}
+\section*{Question 5}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+2 & 2 & 0 \\
+2 & 1 & 1 \\
+-7 & 2 & -3
+\end{bmatrix}
+\\
+	|A - \lambda I| &=  0 \\
+\begin{vmatrix}
+2 - \lambda & 2 & 0 \\
+2 & 1 - \lambda & 1 \\
+-7 & 2 & -3 - \lambda
+\end{vmatrix} &= 0
+\end{align*}
+
+Eigen values $\lambda = 1, 3, -4$ \\
+
+1\textsuperscript{st} eigen value of $A^2 -2 A  + I = 1^2 - 2(1) + 1 = 0$
+
+2\textsuperscript{nd} eigen value of $A^2 -2 A  + I = 3^2 - 2(3) + 1 = 4$
+
+3\textsuperscript{rd} eigen value of $A^2 -2 A  + I = {(-4)}^2 - 2(-4) + 1 = 25$
+
+\section*{Question 6}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+7 & 3 \\
+2 & 6
+\end{bmatrix}
+\\
+	| A - \lambda I| &= 0 \\
+\begin{vmatrix}
+	7 - \lambda & 3 \\
+	2 & 6 - \lambda
+\end{vmatrix} &= 0 \\
+	(7 - \lambda)(6 - \lambda) - 5 &= 0 \\
+	(\lambda - 4)(\lambda - 9) &= 0 \\
+	A^2 - 13 A + 36 &= 0 \\
+	A^2 &= 13 A - 36 \\
+	A^2\cdot A &= (13 A - 36)A \\
+	A^3 &= 13 A^2 - 36A \\
+	A^3 &=
+\begin{bmatrix}
+715 & 507 \\
+338 & 546
+\end{bmatrix}
+-
+\begin{bmatrix}
+252 & 108 \\
+72 & 216
+\end{bmatrix} \\
+	A^3 &=
+\begin{bmatrix}
+463 & 399 \\
+266 & 330
+\end{bmatrix}
+\end{align*}
+
+\section*{Question 7}
+Suppose that $\lambda$ is a (possibly complex) eigen value of the real symmetric matrix $A$. Thus, there is a non-zero vector $V$, also with complex entries such that $AV = \lambda V$. By taking the complex conjugate of both sides and noting that $\overline{A} = A$ since $A$ has real entries, we get $\overline{AV} = \overline{\lambda V} \implies A\overline{V} = \overline{\lambda}\;\overline{V}$. Then using that $A^T = A$,
+\begin{gather*}
+	\overline{V}^T AV = \overline{V}^t(AV) = \overline{V}(\lambda V) = \lambda( \overline{V}V ) \\
+	\overline{V}^T AV = {(A \overline{V})}^T V = {( \overline{\lambda}\;\overline{V} )}^T V = \overline{\lambda} ( \overline{V} V )
+\end{gather*}
+
+Since $V \ne 0$, we have $ \overline{V}V \ne 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in R$.
+
+\section*{Question 8}
+Quadratic Form $ax_1^2 + cx_2^2 - 2bx_1x_2$
+\[
+\begin{bmatrix}
+a & -b \\
+-b & c
+\end{bmatrix}
+\]
+Convert it to diagonal matrix by applying
+\begin{align*}
+	R_2 &\to R_2 + \frac{b}{a}R_1 \\
+	C_2 &\to C_2 + \frac{b}{a}C_1
+\end{align*}
+Now, we get
+\[
+\begin{bmatrix}
+a & 0 \\
+0 & c - \frac{b^2}{a}
+\end{bmatrix}
+\]
+
+Nature $\to$ positive definite $\to$ when $rank(r) = index(s)$ or when all eigen values are positive i.e. $a > 0$ \& $c - \frac{b^2}{a} > 0 \implies ac -b^2 > 0$. Hence proved.
+
+\section*{Question 9}
+\(
+A =
+\begin{bmatrix}
+	\lambda & 1 & 1 \\
+1 & \lambda & -1 \\
+1 & -1 & \lambda
+\end{bmatrix}
+\) is a symmetric matrix obtained when compared to Quadratic form $\lambda(x^2+ y^2 + z^2) + 2xy + 2zx -2yz$.
+
+Now, convert $A$ into diagonal matrix by:
+\begin{align*}
+	C_1 &\to C_1 + C_3 \\
+	C_1 &\to \frac{C_1}{\lambda + 1} \\
+	R_3 &\to R_3 - R_1 \\
+	C_3 &\to C_3 - C_1 \\
+	C_2 &\to C_2 -C_1 \\
+	C_3 &\to C_3 + \frac{C_2}{\lambda} \\
+	R_3 &\to R_3 + \frac{2R_2}{\lambda}
+\end{align*}
+
+we get,
+
+\[
+\begin{bmatrix}
+1 & 0 & 0 \\
+0 & \lambda & 0 \\
+0 & 0 & \lambda - \frac{2}{\lambda} - 1
+\end{bmatrix}
+\]
+For definite positive nature, all Eigen values must be positive i.e. $\lambda > 0$, $\lambda - \frac{2}{\lambda} - 1 > 0$. Taking intersection of these two, we get $\lambda \in (2, \infty)$.
+
+\section*{Question 10}
+
+Multiplication of all the eigen values = determinant of the matrix. For singular matrix, determinant value = 0.
+\begin{align*}
+	\text{Eigen Values} &= 2, 3, a \\
+	6a &= 0 \\
+	a &= 0
+\end{align*}
+
+\section*{Question 11}
+
+Quadratic Form $x_2^2 + 2x_2^2 - 5x_3^2$
+\[
+\begin{bmatrix}
+1 & 0 & 0 \\
+0 & 2 & 0 \\
+0 & 0 & -5
+\end{bmatrix}
+\]
+\begin{align*}
+	index(s) &= 2\;\text{(No of positive terms)} \\
+	rank(r) &= 3 \\
+	signature &= 2s -r = 4 - 3 = 1
+\end{align*}
+
+\section*{Question 12}
+
+Quadratic form $ax^2 + 2bcy + cy^2$.
+\[
+\begin{bmatrix}
+a & b \\
+b & c
+\end{bmatrix}
+\]
+Convert it into diagonal matrix by doing:
+\begin{align*}
+	R_2 &\to R_2 - \left(\frac{b}{a}\right)R_1 \\
+	C_2 &\to C_2 - \left(\frac{b}{a}\right)C_1
+\end{align*}
+
+Finally, we get
+\[
+\begin{bmatrix}
+a & 0 \\
+0 & c - \frac{b^2}{a}
+\end{bmatrix}
+\]
+For positive definite, $a>0$ and $ac -b^2 > 0$. \\
+For negative definite, $a<0$ and $ac -b^2 > 0$. \\
+Roots of the quadratic equation ($ax^2 + 2bx + c = 0$) are imaginary when $D < 0$. \\
+${(2b)}^2 - 4ac = 4( b^2 - ac )$ is always negative
+
+\section*{Question 13}
+\begin{align*}
+	X_1 &=
+\begin{bmatrix}
+1 \\
+2 \\
+-3 \\
+4
+\end{bmatrix}
+\\
+	X_2 &=
+\begin{bmatrix}
+1 \\
+-5 \\
+8 \\
+-7
+\end{bmatrix}
+\\
+	X_3 &=
+\begin{bmatrix}
+1 \\
+-5 \\
+8 \\
+-7
+\end{bmatrix}
+\\
+	\lambda_1 X_1 + \lambda_2 X_2 + \lambda_3 X_3 &= 0 \\
+\begin{bmatrix}
+1 & 1 & 1 \\
+2 & -5 & -5 \\
+-3 & 8 & 8 \\
+4 & -7 & -7
+\end{bmatrix}
+\begin{bmatrix}
+\lambda_1 \\
+\lambda_2 \\
+\lambda_3
+\end{bmatrix} &=
+\begin{bmatrix}
+0 \\
+0 \\
+0 \\
+0
+\end{bmatrix}
+\end{align*}
+Applying the following tranformations
+\begin{align*}
+	R_2 &\to R_2 - 2R_1 \\
+	R_2 &\to -\frac{R_2}{7} \\
+	R_3 &\to R_3 + R_4 \\
+	R_3 &\to R_3 - R_1 \\
+	R_1 &\to R_1 - R_2 \\
+	R_3 &\to R_3 - 4 R_1 \\
+	R_4 &\to -\frac{R_4}{7} \\
+	R_4 &\to R_4 - R_2
+\end{align*}
+we get
+\begin{align*}
+\begin{bmatrix}
+1 & 2 & 0 \\
+0 & 1 & 1 \\
+0 & 0 & 0 \\
+0 & 0 & 0
+\end{bmatrix}
+\begin{bmatrix}
+\lambda_1 \\
+\lambda_2 \\
+\lambda_3
+\end{bmatrix}
+&=
+\begin{bmatrix}
+0 \\
+0 \\
+0 \\
+0
+\end{bmatrix} \\
+	\lambda_1 + 2\lambda_2 &= 0 \\
+	\lambda_2 + \lambda_3 &= 0
+	\therefore \\
+	\lambda_1 &= t \\
+	\lambda_2 &= -\frac{t}{2} \\
+	\lambda_3 &= \frac{t}{2}
+\end{align*}
+Putting the values, we get
+
+\[
+	2 X_1 - X_2 + X_3 = 0
+\]
+
+\section*{Question 14}
+
+\begin{align*}
+	(2 - \lambda)x_1 + (-2)x_2 + x_3 &= 0 \\
+	2x_1 - (\lambda + 3) x_2 + 2x_3 &= 0 \\
+	-x_1 + 2x_2 - \lambda x_3 &= 0
+\end{align*}
+
+\(
+	Rank(A) = Rank(\text{augemented matrix}) < 3
+\) for non trivial solutions.
+
+Check the determinant first, $\Delta = 0$
+
+$\Delta = 0$ gets us $\lambda = 1, 3$
+
+Now, for augemented matrix $[A:B]$, put $\lambda = 1, -3$ in \(
+\begin{bmatrix}
+2-\lambda & -2 & 1 & : & 0 \\
+2 & -(\lambda + 3) & 2 & : & 0 \\
+-1 & 2 & -\lambda & : & 0
+\end{bmatrix}
+\)
+
+For $\lambda = 1$
+\[
+\begin{bmatrix}
+1 & -2 & 1 & : & 0 \\
+2 & -4 & 2 & : & 0 \\
+-1 & 2 & -1 & : & 0
+\end{bmatrix}
+\]
+
+Doing transformations to form row echelon form, we get
+\[
+\begin{bmatrix}
+1 & -2 & 1 & : & 0 \\
+0 & 0 & 0 & : & 0 \\
+0 & 0 & 0 & : & 0
+\end{bmatrix}
+\]
+\begin{align*}
+	x_1 -2x_2 + x_3 &= 0 \\
+	if\;x_2 = k,\;x_3 = t, then\;x_1 &= 2k - t
+\end{align*}
+For $\lambda = -3$
+\[
+\begin{bmatrix}
+5 & -2 & 1 & : & 0 \\
+2 & 0 & 2 & : & 0 \\
+-1 & 2 & 3 & : & 0
+\end{bmatrix}
+\]
+
+Doing transformations to form row echelon form, we get
+\[
+\begin{bmatrix}
+5 & -2 & 1 & : & 0 \\
+0 & \frac{4}{5} & \frac{8}{5} & : & 0 \\
+0 & 0 & 0 & 0 & 0
+\end{bmatrix}
+\]
+\begin{align*}
+	5x_1 - 2x_2 + x_3 &= 0 \\
+	\frac{4x_2}{5} + \frac{8x_3}{5} &= 0 \\
+	if\;x_3 = t, then \\
+	x_1 &= - t \\
+	x_2 &= -2t \\
+	x_3 &= t
+\end{align*}
+
+\section*{Question 15}
+\subsection*{Part i}
+Since $A + A^T = 0$, $A$ must either be skew symmetric. If A is skew symmetric, we know that the rank of an odd order skew symmetric matrix must be even. $\therefore Rank \leq 2020$
+\subsection*{Part ii}
+Inverse does not exist as $A$ is singular matrix.
+\end{document}