datavorous/entrance-exam-dataset huggingface dataset:
$(0,0,0),(\mathrm{a}, 0,0),(0, \mathrm{~b}, 0)$ and $(0,0, \mathrm{c})$ are four distinct points. What are the coordinates of the point which is equidistant from the four points?

Options:
A $\left(\frac{a+b+c}{3}, \frac{a+b+c}{3}, \frac{a+b+c}{3}\right)$
B $(\mathrm{a}, \mathrm{b}, \mathrm{c})$
C $\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)$
D $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$

The correct answer is:
$\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)$

Solution:
$\mathrm{A}(0,0,0), \mathrm{B}(\mathrm{a}, 0,0), \mathrm{C}(0, \mathrm{~b}, 0), \mathrm{D}(0,0, \mathrm{c})$
Let the equidistant point be $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$
i.e, $\mathrm{AP}=\mathrm{BP}, \mathrm{AP}=\mathrm{CP}, \mathrm{AP}=\mathrm{DP}$
$\Rightarrow \sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$
$=\sqrt{(x-a)^{2}+y^{2}+z^{2}}$
$\Rightarrow x^{2}+y^{\prime}+z^{\prime}=(x-a)^{2}+y^{\prime}+z^{\prime}$
$\Rightarrow x^{2}=x^{2}-2 a x+a^{2}$
$\Rightarrow a^{2}-2 a x=0$
$\Rightarrow a(a-2 x)=0$
Since $a \neq 0, a=2 x \Rightarrow x=\frac{a}{2}$
Similarly, we will get $\mathrm{y}=\frac{\mathrm{b}}{2}, \mathrm{z}=\frac{\mathrm{c}}{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 1)
$\int_{0}^{\pi / 2} e^{\sin } x \cos x d x$ is equal to

Options:
A $e+1$
B $e-1$
C $e+2$
D $e$

The correct answer is:
$e-1$

Solution:
$\int_{0}^{\frac{\pi}{2}} e^{\sin x} \cdot \cos x d x$
Let $\sin x=t \Rightarrow \cos x \cdot d x=d t$
$\therefore \int_{0}^{\frac{\pi}{2}} e^{\sin x} \cdot \cos x d x=\int_{0}^{1} e^{t} . d t$
$=\left(e^{t}\right)_{0}^{1}=e^{1}-e^{0}=e-1$

Topic: Mathematics, Definite Integration, NDA, NDA 2019 (Phase 1)
$\int_{0}^{2 \pi} \sin ^{5}\left(\frac{\mathrm{x}}{4}\right) \mathrm{d} \mathrm{x}$ is equal to

Options:
A $\frac{8}{15}$
B $\frac{16}{15}$
C $\frac{32}{15}$
D 0

The correct answer is:
$\frac{32}{15}$

Solution:
$\int_{0}^{2 \pi} \sin ^{5}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi}\left(1-\cos ^{2} \frac{x}{4}\right)\left(1-\cos ^{2} \frac{x}{4}\right) \sin \frac{x}{4} d x$
Put $\cos \left(\frac{x}{4}\right)=t$
$\Rightarrow-\sin \left(\frac{x}{4}\right) \cdot \frac{d x}{4}=d t$
$\Rightarrow \sin \left(\frac{x}{4}\right) d x=-4 d t$
$\Rightarrow \int_{0}^{2 \pi} \sin ^{5}\left(\frac{x}{4}\right) d x=-4 \int\left(1-t^{2}\right)\left(1-t^{2}\right) \mathrm{dt}$
$=-4 \int\left(1+t^{4}-2 t^{2}\right) \mathrm{dt}$
$=-4\left[t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]$
$=-4\left[\cos \left(\frac{x}{4}\right)+\frac{\cos ^{5}\left(\frac{x}{4}\right)}{5}-\frac{ \left.2 \cos ^{3}\left(\frac{x}{4}\right)\right]_{0}^{2 \pi}}{3}\right]$
$=-4\left[(0+0-0)-\left(1+\frac{1}{5}-\frac{2}{3}\right)\right]=\frac{32}{15}$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 2)
$\int_{0}^{\pi / 2}|\sin x-\cos x| d x$ is equal to.

Options:
A 0
B $2(\sqrt{2}-1)$
C $2 \sqrt{2}$
D $2(\sqrt{2}+1)$

The correct answer is:
$2(\sqrt{2}-1)$

Solution:
$\begin{aligned} & \int_{0}^{\pi / 2}|\sin x-\cos x| d x \\ &=\int_{0}^{\pi / 4}(\sin x-\cos x) \cdot d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ &=\int_{0}^{\pi / 4}(\cos x-\sin x) \cdot d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) \cdot d x \\ &=(\sin x+\cos x)_{0}^{\pi / 4}+(-\cos x-\sin x)_{\pi / 4}^{\pi / 2} \\ &=2 \sqrt{2}-2=2(\sqrt{2}-1) \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2019 (Phase 1)
$\int_{-1}^{1} x|x| d x$ is equal to

Options:
A 0
B $\frac{2}{3}$
C 2
D $-2$

The correct answer is:
0

Solution:
$\begin{aligned} & \int_{-1}^{1} x|x| d x \\ &=\int_{-1}^{0} x|x| d x+\int_{0}^{1} x|x| d x \\ &=\int_{-1}^{0} x(-x) d x+\int_{0}^{1} x \cdot x d x \\ &=-\int_{-1}^{0} x^{2} d x+\int_{0}^{1} x^{2} d x \\ &=-\left[\frac{x^{3}}{3}\right]_{-1}^{0}+\left[\frac{x^{3}}{3}\right]_{0}^{1} \\ &=-\left[0-\frac{(-1)^{3}}{3}\right]+\frac{1}{3}\left[(1)^{3}-(0)^{3}\right] \\ &=-\frac{1}{3}+\frac{1}{3}=0 \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 2)
14-54·8+5·94·8·12-…=

Options:
A 33-2593
B 23-3293
C 214-1214
D 23-3593

The correct answer is:
214-1214

Solution:
We have,

14-54·8+1·5·94·8·12-……

=14-1·1+42!142+1·1+41+2·43!143-……

=114-1·1+42!142+1·1+41+2·43!143-…… ...i

Now, we know that

1+x-pq=1-pxq+pp+q2!xq2-…………

Therefore, from i, we have

=114-1·1+42!142+1·1+41+2·43!143-……

=1-1-114+1·1+42!142-1·1+41+2·43!143+……

Comparing with the above standard series, we get

p=1, xq=14, q=4⇒x=1

Therefore,

1+x-pq=1+1-14=2-14

Therefore, required sum is

=1-2-14

=214-1214

Topic: Mathematics, Sequences and Series, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
$\int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}}$ is equal to

Options:
A $1+\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
B $\ln \left(1+\mathrm{e}^{-x}\right)+\mathrm{c}$
C $\ln \left(1+\mathrm{e}^{x}\right)+\mathrm{c}$
D $2 \ln \left(1+\mathrm{e}^{-x}\right)+\mathrm{c}$
where $c$ is the constant of integration

The correct answer is:
$\ln \left(1+\mathrm{e}^{x}\right)+\mathrm{c}$

Solution:
$\int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}}$
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx}$
Let $\mathrm{e}^{\mathrm{x}}+1=\mathrm{t}$
$\mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$=\int \frac{\mathrm{dt}}{\mathrm{t}}$
$\Rightarrow \log \mathrm{t}+\mathrm{c} \Rightarrow \log \left(\mathrm{e}^{\mathrm{x}}+1\right)+\mathrm{c}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
$\sqrt{1+\sin \mathrm{A}}=-\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$ is true if

Options:
A $\frac{3 \pi}{2} < \mathrm{A} < \frac{5 \pi}{2}$ only
B $\frac{\pi}{2} < \mathrm{A} < \frac{3 \pi}{2}$ only
C $\frac{3 \pi}{2} < \mathrm{A} < \frac{7 \pi}{2}$
D $0 < \mathrm{A} < \frac{3 \pi}{2}$

The correct answer is:
$\frac{3 \pi}{2} < \mathrm{A} < \frac{7 \pi}{2}$

Solution:
$\sqrt{1+\sin A}=-\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)$
We know, $1+\sin A=\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)^{2}$
$\therefore \sqrt{1+\sin A}=\left|\cos \frac{A}{2}+\sin \frac{A}{2}\right|$
We know, $|x|=\left\{\begin{array}{l}x \text { if } x \geq 0 \\ -x \text { if } x < 0\end{array}\right.$
$\sin \frac{A}{2}+\cos \frac{A}{2} \mid=\left\{\begin{array}{l}\sin \frac{A}{2}+\cos \frac{A}{2}, \quad \text { if } 2 n \pi-\frac{\pi}{4} \leq \frac{A}{2} \leq 2 n \pi+\frac{3 \pi}{4} \\ -\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right),\end{array}\right.$
$\quad \therefore \sqrt{1+\sin A}=-\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)$ when
$\frac{3 \pi}{4} < \frac{A}{2} < \frac{7 \pi}{4}$
$\Rightarrow \frac{3 \pi}{2} < A < \frac{7 \pi}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 2)
1-Propanol can be distinguished from 2-propanol by which test?

Options:
A 2, 4-DNP test
B Tollen's test
C Lucas test
D Fehling's test

The correct answer is:
Lucas test

Solution:
Primary and Secondary alcohols can be distinguished by the Lucas test. It involves heating the alcohol with anhydrous $\mathrm{ZnCl}_2$ in concentrated $\mathrm{HCl}$.
Turbidity is observed for $2^{\circ}$ alcohols after some time while no turbidity is observed for a $1^{\circ}$ alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
(1-\sin A+\cos A)^{2}$ is equal to

Options:
A $2(1-\cos A)(1+\sin A)$
B $2(1-\sin A)(1+\cos A)$
C $2(1-\cos A)(1-\sin A)$
D None of the above

The correct answer is:
$2(1-\sin A)(1+\cos A)$

Solution:
$(1-\sin A+\cos A)^{2}$
$=1+\sin ^{2} A+\cos ^{2} A-2 \sin A$
$\quad-2 \sin A \cdot \cos A+2 \cos A$
$=2-2 \sin A-2 \sin A \cos A+2 \cos A$
$=2(1-\sin A)+2 \cos A(1-\sin A)$
$=2(1+\cos A)(1-\sin A)$
$\therefore \quad$ Option (b) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
2, 3-dimethyl hexane contains tertiary...... secondary and primary carbon atoms, respectively:

Options:
A $2, 2, 4$
B $2,4,3$
C $4,3,2$
D $3,2,4$

The correct answer is:
$2, 2, 4$

Solution:
The structure of 2, 3-dimethyl hexane is
So, the number of tertiary carbon atoms $=2$
The number of secondary carbon atoms $=2$
The number of primary carbon atoms $=4$

Topic: Chemistry, General Organic Chemistry, AP EAMCET, AP EAMCET 2003
2, 3-dimethyl hexane contains tertiary...... secondary and primary carbon atoms, respectively:

Options:
A $2, 2, 4$
B $2,4,3$
C $4,3,2$
D $3,2,4$

The correct answer is:
$2, 2, 4$

Solution:
The structure of 2, 3-dimethyl hexane is
So, the number of tertiary carbon atoms $=2$
The number of secondary carbon atoms $=2$
The number of primary carbon atoms $=4$

Topic: Chemistry, General Organic Chemistry, TS EAMCET, TS EAMCET 2003
2Ag+(aq)+Cu(s)⇌Cu2+(aq)+2Ag(s)
The standard potential E° for this reaction is 0.46 V. Which change will increase the potential the most?

Options:
A Doubling the [Ag+]
B Halving the [Cu2+]
C Doubling the size of the Cu(s) electrode
D Decreasing the size of the Ag electrode by one-half

The correct answer is:
Doubling the [Ag+]

Solution:
E=E°-0.0592log[Cu2+][Ag+]2
The cell potential will increase the most by doubling the concentration of Ag+ ion.

Topic: Chemistry, Electrochemistry, NEET
$2 x^{2}+3 x-\alpha-0$ has roots $-2$ and $\beta$ while the equation $x^{2}-3 m x+$ $2 \mathrm{~m}^{2}=0$ has both roots positive, where $\alpha>0$ and $\beta>0$.
If $\beta, 2,2 \mathrm{~m}$ are in GP, then what is the value of $\beta \sqrt{\mathrm{m}}$ ?

Options:
A 1
B 2
C 4
D 6

The correct answer is:
1

Solution:
$\beta=\frac{1}{2}$
$\Rightarrow \frac{2}{\beta}=\frac{2 m}{2}$
$\Rightarrow \mathrm{m}=\frac{2}{\beta}=2 \times \frac{2}{1}$
$\Rightarrow \mathrm{m}=4$
$\Rightarrow \beta \sqrt{m}=\frac{1}{2} \times \sqrt{4}=1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)
$2 x^{2}+3 x-\alpha-0$ has roots $-2$ and $\beta$ while the equation $x^{2}-3 m x+$ $2 \mathrm{~m}^{2}=0$ has both roots positive, where $\alpha>0$ and $\beta>0$.
What is the value of $\alpha ?$

Options:
A $1 / 2$
B 1
C 2
D 4

The correct answer is:
2

Solution:
$2 x^{2}+3 x-\alpha=0$
Its roots are: $-2 \& \beta .$
i.e., $\frac{-3}{2}=\beta-2 \Rightarrow \beta=2-\frac{3}{2}=\frac{1}{2} \Rightarrow \beta=\frac{1}{2}$
$\frac{\alpha}{2}=2 \beta \Rightarrow \alpha=4 \times \frac{1}{2} \Rightarrow \alpha=2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)
3-Pentanol on reaction with aluminium tertiary butoxide in the presence of acetone gives

Options:
A 3-pentanal
B 2-pentanal
C 3-pentanone
D 2-pentanone

The correct answer is:
3-pentanone

Solution:
Aluminium tertiary butoxide is an oxidising agent used for the oxidation of secondary alcohols into ketones.

Topic: Chemistry, Aldehydes and Ketones, NEET
8 coins are tossed simultaneously. The probability of getting at least 6 heads is

Options:
A $\frac{7}{64}$
B $\frac{57}{64}$
C $\frac{37}{256}$
D $\frac{229}{256}$

The correct answer is:
$\frac{37}{256}$

Solution:
In tossing of coin getting $r$ head out of $n$ tossing
$={ }^{n} C_{r} \cdot\left(\frac{1}{2}\right)^{n}$
$\therefore \quad$ Required probability
$=\left({ }^{8} \mathrm{C}_{6}+{ }^{8} \mathrm{C}_{7}+{ }^{8} \mathrm{C}_{8}\right)\left(\frac{1}{2}\right)^{8}$
$=(28+8+1) \times \frac{1}{256}=\frac{37}{256}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
$100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid is completely neutralized using a standard solution of $\mathrm{NaOH}$. The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is

Options:
A $56 \mathrm{~mL}$.
B $224 \mathrm{~mL}$
C $560 \mathrm{~mL}$
D $112 \mathrm{~mL}$.

The correct answer is:
$112 \mathrm{~mL}$.

Solution:
Acetic acid on electrolysis gives ethane as
Given moles of acetic acid,
$$
\mathrm{CH}_{3} \mathrm{COOH}=\frac{0.1 \times 100}{1000}=0.01
$$
$\because 2 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{COOH}$ gives $\mathrm{C}_{2} \mathrm{H}_{6}=22.4 \mathrm{~L}$
$\therefore 0.01 \mathrm{~mol} \mathrm{CH} \mathrm{CHOH}_{3} \mathrm{COOH}$ give
$$
\begin{aligned}
\mathrm{C}_{2} \mathrm{H}_{6} &=\frac{22.4 \times 0.01}{2} \\
&=0.112 \mathrm{~L} \\
&=112 \mathrm{~mL}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, KCET, KCET 2012
$100 \mathrm{~mL}$ of $0.2 \mathrm{M}$ acetic acid is completely neutralised using a standard solution of $\mathrm{NaOH}$. The volume of ethane obtained at STP after complete electrolysis of the resulting solution is

Options:
A $11.2 \mathrm{~L}$
B $2.24 \mathrm{~L}$
C $0.224 \mathrm{~L}$
D $22.4 \mathrm{~L}$

The correct answer is:
$0.224 \mathrm{~L}$

Solution:
Given moles of acetic acid
$$
\begin{array}{r}
=\frac{0.2 \times 100}{1000}=0.02 \text { moles } \\
2 \mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{C}_2 \mathrm{H}_6+2 \mathrm{CO}_2+\mathrm{H}_2
\end{array}
$$
2 moles of $\mathrm{CH}_3 \mathrm{COOH}$ gives 1 mole of ethane $=22.4 \mathrm{~L}$
0.02 moles of $\mathrm{CH}_3 \mathrm{COOH}$ will give ethane
$$
=\frac{22.4 \times 0.02}{2}=0.224 \mathrm{~L}
$$
$\therefore 0.224 \mathrm{~L}$ of ethane is obtained.

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
$18.0 \mathrm{~g}$ of water completely vaporizes at $100^{\circ} \mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is $40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What will be the enthalpy change for vaporizing two moles of water under the same conditions? What is the standard enthalpy of vaporization for water?

Options:

None

Solution:
$+81.58 \mathrm{~kJ}, \Delta_{\text {vap }} \mathrm{H}^{\circ}=+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Thermodynamics (C)
18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Options:
A Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆vapH = +40.79 kJ mol–1
B Enthalpy change for vapourising two moles of water under the same conditions is +69.58 kJ and ∆vapH = +40.79 kJ mol–1
C Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆vapH = +45.79 kJ mol–1
D Enthalpy change for vapourising two moles of water under the same conditions is +69.58 kJ and ∆vapH = +45.79 kJ mol–1

The correct answer is:
Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆vapH = +40.79 kJ mol–1

Solution:
Correct Option is : (A)
Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆vapH = +40.79 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
$2.33 \mathrm{~g}$ of compound $\mathrm{X}$ (empirical formula $\left.\mathrm{CoH}_{12} \mathrm{~N}_{4} \mathrm{Cl}_{3}\right)$ upon treatment with excess $\mathrm{AgNO}_{3}$ solution produces $1.435 \mathrm{~g}$ of a white precipitate. The primary and secondary valences of cobalt in compound $\mathrm{X}$, respectively, are
[Given : Atomic mass : $\mathrm{Co}=59, \mathrm{Cl}=35.5, \mathrm{Ag}=108$ ]

Options:
A 3,6
B 3,4
C 2,4
D 4,3

The correct answer is:
3,6

Solution:
$$
\begin{array}{l}
\text { M.wt }=59+12+14 \times 4+35.5 \times 3=233.5 \\
\mathrm{C}_{0} \mathrm{H}_{12} \mathrm{~N}_{4} \mathrm{Cl}_{3} \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} 1.435 \mathrm{~g} \mathrm{AgCl} \\
\frac{2.33}{233.5} \mathrm{~g}=0.01 \text { mole } \quad(0.01 \mathrm{~mole})
\end{array}
$$
(i) $0.01$ mole molecule produce $0.01$ mole $\mathrm{AgCl}$ $\therefore$ one replaceable $\mathrm{Cl}^{\underline{\Theta}}$ ion so formula of complex is
$$
\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}
$$
(ii) Oxidation no. of Co is $+3$ so primary valency is 3 .
(iii) Coordination no. is 6 so sec. valency is 6 so ans. is 3,6

Topic: Chemistry, Some Basic Concepts of Chemistry, KVPY, KVPY 2016 (SB/SX)
$\mathrm{A}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right]$, then which of the following is/are correct?
$1.$ $\mathrm{AB}\left(\mathrm{A}^{-1} \mathrm{~B}^{-1}\right)$ is a unit matrix.
$2.$ $(\mathrm{AB})^{-1}=\mathrm{A}^{-1} \mathrm{~B}^{-1}$
Select the correct answer using the code given below.

Options:
A 1 only
B 2 only
C Both 1 only 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
Here, $A=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 3 \\ -1 & -2\end{array}\right]$
$|A|=3-(-2)=5$ and $|B|=-4-(-3)=-1$
$\Rightarrow A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$ and $B^{-1}=-1\left[\begin{array}{cc}-2 & -3 \\ 1 & 2\end{array}\right]$
$A B=\left[\begin{array}{ll}3 & 5 \\ 1 & 0\end{array}\right]$ and $A^{-1} B^{-1}=\frac{1}{5}\left[\begin{array}{cc}5 & 7 \\ -5 & 8\end{array}\right]$
$\Rightarrow A B\left(A^{-1} B^{-1}\right)=\frac{1}{5}\left[\begin{array}{cc}-10 & -61 \\ 5 & 7\end{array}\right] \neq 1 .$
$|A B|=0-5=-5$
$\therefore(A B)^{-1}=\frac{-1}{5}\left[\begin{array}{cc}0 & -5 \\ -1 & 3\end{array}\right] \neq A^{-1} B^{-1}$

Topic: Mathematics, Matrices, NDA, NDA 2016 (Phase 2)
$A=\left|\begin{array}{ccc}2 a & 3 r & x \\ 4 b & 6 s & 2 y \\ -2 c & -3 t & -z\end{array}\right|=\lambda\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$, then what is the value of $\lambda ?$

Options:
A 12
B $-12$
C 7
D $-7$

The correct answer is:
$-12$

Solution:
Given, $\left|\begin{array}{ccc}2 a & 3 r & x \\ 4 b & 6 s & 2 y \\ -2 c & -3 t & -z\end{array}\right|=\lambda\left|\begin{array}{ccc}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
Taking 2 common from $C_{1}$ and 3 from $C_{2}$ in LHS
$\therefore 2 \times 3\left|\begin{array}{ccc}a & r & x \\ 2 b & 2 s & 2 y \\ -c & -t & -z\end{array}\right|=\lambda\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
Taking 2 common from $R_{2}$ and $-1$ from $R_{3}$ in LHS
$\therefore-12\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|=\lambda\left|\begin{array}{ccc}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
$\Rightarrow \lambda=-12$

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 2)
(a, $2 \mathrm{~b}$ ) is the mid-point of the line segment joining the points $(10,-6)$ and $(\mathrm{k}, 4)$. If $\mathrm{a}-2 \mathrm{~b}=7$, then what is the value of $\mathrm{k}$ ?

Options:
A 2
B 3
C 4
D 5

The correct answer is:
2

Solution:
$M \equiv$ mid point of line segment $P Q$
$\frac{K+10}{2}=a \Rightarrow K=(2 a-10)$
$2 b=\frac{4-6}{2}=-1$
given $a-2 b=7$...(1)
Put the values of $a$ \& $b$ in $e q(1)$, we get
$\frac{K+10}{2}+1=7$
$\frac{K+10}{2}=6 \Rightarrow K=12-10$
$K=2$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
$A=\left[\begin{array}{ll}3 & 1 \\ 0 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]$, then which of the following is/are correct?
I. $A B$ is defined
II. $B A$ is defined
III. $A B=B A$
Select the correct answer using the codes given below.

Options:
A Only I
B Only II
C Both I and II
D I, II and III

The correct answer is:
I, II and III

Solution:
Let $\mathrm{A}=\left[\begin{array}{ll}3 & 1 \\ 0 & 4\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]$
Now, $\mathrm{AB}=\left[\begin{array}{ll}3 & 1 \\ 0 & 4\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}3 & 5 \\ 0 & 8\end{array}\right]$
and $\mathrm{BA}=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 0 & 4\end{array}\right]=\left[\begin{array}{ll}3 & 5 \\ 0 & 8\end{array}\right]$
$\Rightarrow \quad \mathrm{AB}=\mathrm{BA}$
Hence, all the three statements are correct.

Topic: Mathematics, Matrices, NDA, NDA 2011 (Phase 1)
$\mathrm{A}(3,4)$ and $\mathrm{B}(5,-2)$ are two points and $\mathrm{P}$ is a point such that $\mathrm{PA}=\mathrm{PB}$. If the area of triangle $\mathrm{PAB}$ is 10 square unit, what are the coordinates of P ?

Options:
A $(1,0)$ only
B $(7,2)$ only
C $(1,0)$ or $(7,2)$
D Neither $(1,0)$ nor $(7,2)$

The correct answer is:
$(1,0)$ or $(7,2)$

Solution:
Given $\mathrm{A}(3,4)$ and $\mathrm{B}(5,-2)$ Let, $\mathrm{P}(\mathrm{x}, \mathrm{y})$
Given that, $\mathrm{PA}=\mathrm{PB}$ $\Rightarrow \mathrm{PA}^{2}=\mathrm{PB}^{2}$
$\Rightarrow(x-3)^{2}+(y-4)^{2}=(x-5)^{2}+(y+2)^{2}$
$\Rightarrow x^{2}-6 x+9+y^{2}-8 y+16$
$=x^{2}-10 x+25+y^{2}+4 y+4$
$\Rightarrow 4 x-12 y=4$
$\Rightarrow x-3 y=1$ ...(i)
$\therefore$ Area of $\Delta \mathrm{PAB}=10$
$\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 3 & 4 & 1 \\ 5 & -2 & 1\end{array}\right|=\pm 10$
$\Rightarrow x(4+2)-y(3-5)+1(-6-20)=\pm 20$
$\Rightarrow 6 \mathrm{x}+2 \mathrm{y}-26=\pm 20$
$\Rightarrow 6 \mathrm{x}+2 \mathrm{y}-26=20$
or, $6 x+2 y-26=-20$
$\Rightarrow 6 x+2 y=46$...(ii)
or $6 x+2 y=6$...(iii)
From eqs. (i) and (ii), we get $x=7, y=2$
Similarly, from eqs. (i) and (iii), we get $\mathrm{x}=1, \mathrm{y}=0$
Hence, coordinates of $\mathrm{P}$ are $(7,2)$ or $(1,0)$

Topic: Mathematics, Straight Lines, NDA, NDA 2014 (Phase 2)
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil?

Options:
A 0.54 A
B 0.27 A
C 1.08 A
D 2.16 A

The correct answer is:
0.27 A

Solution:
Correct Option is : (B)
0.27 A

Topic: Chemistry, Alternating Current, NEET
A $60 \mathrm{~W}$ load is connected to the secondary of a transformer whose primary draws line voltage. If a current of $0.54 \mathrm{~A}$ flows in the load, what is the current in the primary coil? Comment on the type of tansformer being used.

Options:

None

Solution:
As given that,
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W} \\
&\mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A}
\end{aligned}
$$
Primary current $\mathrm{I}_{\mathrm{p}}=$ ?
Taking line voltage as $220 \mathrm{~V}\left(\mathrm{~V}_{\mathrm{P}}=220 \mathrm{~V}\right)$
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} \\
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W}, \mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A} \\
&\mathrm{~V}_{\mathrm{S}}=\frac{60}{0.54}=110 \mathrm{Volt} .
\end{aligned}
$$
Voltage in the secondary $\left(\mathrm{V}_{\mathrm{S}}\right)$ is less than voltage in the primary $\left(\mathrm{V}_{\mathrm{p}}\right)$.
So, the transformer is step down transformer.
since, the transformation ratio factor is
$$
\begin{aligned}
&r=\frac{\text { Secondary voltage }}{\text { Primary voltage }} \\
&r=\frac{V_S}{V_P}
\end{aligned}
$$
$\therefore$ For transformer $\frac{V_S}{V_{\mathrm{p}}}=\frac{I_P}{I_s}$
Substituting the values,
$$
\begin{aligned}
\frac{110 \mathrm{~V}}{220 \mathrm{~V}} &=\frac{\mathrm{I}_{\mathrm{P}}}{0.54 \mathrm{~A}} \\
\Rightarrow \quad \quad \mathrm{I}_{\mathrm{P}} &=0.27 \text { Ampere }
\end{aligned}
$$

Topic: Physics, Alternating Current
A 90 g of ethyl amine on reaction with methyl chloride produced a tertiary amine as an exclusive product. The amount of methyl chloride required is
[Given mass in amu : H=1, C=12, N=14, Cl=35.5]

Options:
A 50.5 g
B 101 g
C 202 g
D 303 g

The correct answer is:
202 g

Solution:
The Standard Reaction is:

CH3CH2NH2 + 2CH3Cl → CH33N + HCl

Molar mass of ethyl amine = 45 g

The molar mass of methyl chloride = 50.5 g

Given Moles ethyl amine= 9045=2

As per the Standard reaction,

1 mole of CH3CH2NH2 react with 2 mole of CH3Cl,

So, 2 moles of CH3CH2NH2 react with 4 moles of CH3Cl

Hence, the amount methyl chloride obtained is 4×50.5 = 202 g

Topic: Chemistry, Some Basic Concepts of Chemistry, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)