datavorous/entrance-exam-dataset huggingface dataset:
3 moles of an ideal monoatomic gas performs $A B C D A$ cyclic process as shown in figure below. The gas temperatures are $T_A=400 \mathrm{~K}$, $T_B=800 \mathrm{~K}, T_C=2400 \mathrm{~K}$ and $T_D=1200 \mathrm{~K}$. The work done by the gas is (approximately) $(R=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})$

Options:
A $10 \mathrm{~kJ}$
B $20 \mathrm{~kJ}$
C $40 \mathrm{~kJ}$
D $100 \mathrm{~kJ}$

The correct answer is:
$20 \mathrm{~kJ}$

Solution:
Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of form $y=m x$.
Also,
$p=\frac{\mu R}{V} T(\mu=3)$
$p \propto T$
So, volume remains constant for the graphs $A B$ and $C D$.
So, no work is done during processes for $A$ to $B$ and $C$ to $D$.
$W_{A B}=W_{C D}=0$
and $\quad W_{B C}=p_2\left(V_C-V_B\right)$
$=\mu R\left(T_C-T_B\right)$
$=3 R(2400-800)$
$=3 R \times 1600$
$=4800 R$
$W_{D A}=p_1\left(V_A-V_D\right)$
$=\mu R\left(T_A-T_D\right)$
$=3 R(400-1200)$
$=-2400 R$
Work done in complete cycle
$W=W_{A B}+W_{B C}+W_{C D}+W_{D A}$
$=0+4800 R+0+(-2400) R$
$=2400 R$
$=19.944 \mathrm{~J}=20 \mathrm{~kJ}$

Topic: Physics, Thermodynamics, JEE Main
3 moles of an ideal monoatomic gas performs $A B C D A$ cyclic process as shown in figure below. The gas temperatures are $T_A=400 \mathrm{~K}$, $T_B=800 \mathrm{~K}, T_C=2400 \mathrm{~K}$ and $T_D=1200 \mathrm{~K}$. The work done by the gas is (approximately) $(R=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})$

Options:
A $10 \mathrm{~kJ}$
B $20 \mathrm{~kJ}$
C $40 \mathrm{~kJ}$
D $100 \mathrm{~kJ}$

The correct answer is:
$20 \mathrm{~kJ}$

Solution:
Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of form $y=m x$.
Also,
$p=\frac{\mu R}{V} T(\mu=3)$
$p \propto T$
So, volume remains constant for the graphs $A B$ and $C D$.
So, no work is done during processes for $A$ to $B$ and $C$ to $D$.
$W_{A B}=W_{C D}=0$
and $\quad W_{B C}=p_2\left(V_C-V_B\right)$
$=\mu R\left(T_C-T_B\right)$
$=3 R(2400-800)$
$=3 R \times 1600$
$=4800 R$
$W_{D A}=p_1\left(V_A-V_D\right)$
$=\mu R\left(T_A-T_D\right)$
$=3 R(400-1200)$
$=-2400 R$
Work done in complete cycle
$W=W_{A B}+W_{B C}+W_{C D}+W_{D A}$
$=0+4800 R+0+(-2400) R$
$=2400 R$
$=19.944 \mathrm{~J}=20 \mathrm{~kJ}$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2010
3 moles of metal complex with formula Co(en)2Cl3 gives 3 moles of silver chloride on treatment with excess of silver nitrate. The secondary valency of Co in the complex is (Round off to the nearest integer)

Options:

The correct answer is:
6

Solution:
3Co(en)2Cl2Cl+AgNO3 (excess)→3AgCl (white ppt. )

Secondary valency is equal to coordination number of Co=6

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (27 Jul Shift 2)
3 moles of mono-atomic gas $(\gamma=5 / 3)$ is mixed with 1 mole of a diatomic gas $(\gamma=7 / 3) .$ The value of $\gamma$ for the mixture will be

Options:
A $\frac{9}{11}$
B $\frac{11}{7}$
C $\frac{12}{7}$
D $\frac{15}{7}$

The correct answer is:
$\frac{11}{7}$

Solution:
The number of degrees of ireedom for the mixture is
$\therefore$$\begin{aligned} f_{\text {mix }} &=\frac{n_{1} f_{1}+n_{2} f_{2}}{n_{1}+n_{2}} \\ &=\frac{3 \times 3+1 \times 5}{4}=\frac{7}{2} \\ \gamma &=1+\frac{2}{f} \end{aligned}$
$\Rightarrow \quad \gamma=1+\frac{4}{7}=\frac{11}{7}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
3 particles each of mass $m$ are kept at vertices of an equilateral triangle of side $L$. The gravitational field at centre due to these particles is

Options:
A Zero
B $\frac{3 G M}{L^2}$
C $\frac{9 G M}{L^2}$
D $\frac{12}{\sqrt{3}} \frac{G M}{L^2}$

The correct answer is:
Zero

Solution:
Due to three particles net intensity at the centre
$I=I_A+I_B+I_C=0$
because out of these three intensities one equal in magnitude and the angle between each other is $120^{\circ}$.

Topic: Physics, Gravitation, JEE Main
3. The equation of the circle which intersects circles $x^2+y^2+x+2 y+3=0, x^2+y^2+2 x+4 y+5=0$ and $x^2+y^2-7 x-8 y-9=0$ at right angle, will be

Options:
A $x^2+y^2-4 x-4 y-3=0$
B $3\left(x^2+y^2\right)+4 x-4 y-3=0$
C $x^2+y^2+4 x+4 y-3=0$
D $3\left(x^2+y^2\right)+4(x+y)-3=0$

The correct answer is:
$3\left(x^2+y^2\right)+4(x+y)-3=0$

Solution:
Let circle be $x^2+y^2+2 g x+2 f y+c=0$. Then according to the conditions given,
$g+2 f=c+3...(i)$
$\begin{aligned}& 2 g+4 f=c+5 ...(ii)\\& -7 g-8 f=c-9 ...(iii)\\& \Rightarrow g=\frac{2}{3}, f=\frac{2}{3}, c=-1\end{aligned}$
Therefore, the required equation is
$3\left(x^2+y^2\right)+4(x+y)-3=0$

Topic: Mathematics, Circle, JEE Main
$\begin{array}{llll}\text { 3. The sum } & \text { of } & \text { the } & \text { series }\end{array}$ $1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$ is equal to

Options:
A $\frac{2^{n+1}-1}{n+1}$
B $\frac{3\left(2^{n}-1\right)}{2 n}$
C $\frac{2^{n}+1}{n+1}$
D $\frac{2^{n}+1}{2 n}$

The correct answer is:
$\frac{2^{n+1}-1}{n+1}$

Solution:
$1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$
$=\frac{1}{n+1}\left[(n+1)+\frac{(n+1) n}{21}\right.$
$+\frac{(n+1) n(n-1)}{3 !}+(1]$
$=\frac{1}{n+1}\left[^{n+1} C_{1}+{ }^{n+1} C_{2}+\quad+{ }^{n+1} C_{n+1}\right]$
\(=\frac{2^{n+1}-1}{n+1}\)

Topic: Mathematics, Binomial Theorem, JEE Main
$\int \frac{\log \sqrt{\mathrm{x}}}{3 \mathrm{x}} \mathrm{dx}$ is equal to

Options:
A $\frac{1}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$
B $\frac{2}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$
C $\frac{2}{3}(\log x)^2+C$
D $\frac{1}{3}(\log x)^2+C$

The correct answer is:
$\frac{1}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$

Solution:
Hints : $\mathrm{x}=\mathrm{t}^2 \Rightarrow \int \frac{\ell \mathrm{nt}}{3 \mathrm{t}^2}(2 \mathrm{tdt})=\frac{2}{3} \int \frac{\ell \mathrm{nt}}{\mathrm{t}} \mathrm{dt}=\frac{2}{3} \frac{(\ell \mathrm{nt})^2}{2}+\mathrm{c}=\frac{(\ln \sqrt{\mathrm{x}})^2}{3}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, JEE Main
3-hydroxybutanal is formed when $(X)$ reacts with $(Y)$ in dilute $(Z)$ solution. What are $X, Y$ and $Z$ ?
$\begin{array}{lll}X & Y & Y\end{array}$

Options:
A $\mathrm{CH}_3 \mathrm{CHO},\left(\mathrm{CH}_3\right)_2 \mathrm{CO}, \mathrm{NaOH}$
B $\mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{NaCl}$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{CO},\left(\mathrm{CH}_3\right)_2 \mathrm{CO}, \mathrm{HCl}$
D $\mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{NaOH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{CH}_3 \mathrm{CHO}, \quad \mathrm{NaOH}$

Solution:
Aldehydes having $\alpha$-H-atoms undergoes aldol condensation in the presence of dil. $\mathrm{NaOH}$ and yield $\beta$-hydroxy aldehydes.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
3-Methylpent-2-ene →H2O2HBrZ.The number of stereoisomers possible for the product 'Z' is ______.

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 L. The amount of heat absorbed in this expansion is L atm.

Options:

The correct answer is:
0

Solution:
In case of expansion, work is done by system.

Now, as we know that,

W=-Pext. ∆V

As the gas is expanding into vacuums which has no pressure, i.e.,

Pext.=0

∴ W=0

Hence, no work is done.

From first law of thermodynamics,

ΔU=q+W

As the system is working at constant temperature, i.e., isothermally.

∴ΔU=0

Hence q=-W=0

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (28 Jun Shift 1)
$\int_{\pi / 4}^{\pi / 2} \operatorname{cosec}^2 x d x =$

Options:
A -1
B 1
C 0
D $\frac{1}{2}$

The correct answer is:
1

Solution:
$\int_{\pi / 4}^{\pi / 2} \operatorname{cosec}^2 x d x=[-\cot x]_{\pi / 4}^{\pi / 2}=-\left[\cot \frac{\pi}{2}-\cot \frac{\pi}{4}\right]=1$

Topic: Mathematics, Definite Integration, JEE Main
$\int_{-4}^4|x+2| d x=$

Options:
A 50
B 24
C 20
D None of these

The correct answer is:
20

Solution:
$\begin{aligned} \int_{-4}^4|x+2| d x=\int_{-4}^2-(x+2) d x & +\int_{-2}^4(x+2) d x \\ & =\left|\frac{-x^2}{2}-2 x\right|_{-4}^{-2}+\left|\frac{x^2}{2}+2 x\right|_{-2}^4=20\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel.At equilibrium, 3 moles of HI were found. The equilibrium constant for H2g+I2g⇌2HIg is ........

Options:

The correct answer is:
1

Solution:
The equilibrium moles of each reactant and product can be calculated as follows,

H2g+I2g ⇌2HIginitial 4.5 4.5 0At equilibrium 4.5-1.5 4.5-1.5 3

Now, using equilibrium moles, equilibrium constant calculated as shown below.
Keq=[HI]2H2I2=(3)23×3=1

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (11 Apr Shift 2)
[4]
Aqueous solution of an organic compound, ' $A$ ' on electrolysis liberates acetylene and $\mathrm{CO}_2$ at anode. ' $A$ ' is

Options:
A potassium acetate
B potassium succinate
C potassium citrate
D potassium maleate

The correct answer is:
potassium maleate

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
4 different balls of green colour and 4 different balls of red colour are to be distributed equally among 4 people. The probability that all 4 people have balls of a different colour is Î», then the value of 7Î» is equal to

Options:

The correct answer is:
1.6

Solution:
Total cases = number of ways in which 4G and 4R balls are distributed equally among 4 people
=8!2!4Ã14!Ã4!
Favourable cases = all 4 people have balls of different colour
=4!4!
Required probability =4!Ã4!8!Ã2!4
=835

Topic: Mathematics, Probability, JEE Main
4 g equimolar mixture of NaOH and Na2CO3 contains xg of NaOH and yg of Na2CO3. The value of x is g. (Nearest integer)

Options:

The correct answer is:
1

Solution:
Total mass=4g

Now

NaOH: a mol WNaOH+WNa2CO3=4

Na2CO3: 'a' mol ⇒40a+106a=4

⇒a=4146 mol

⇒ therefore mass of NaOH is: 4146×40 g

=1.095≈1

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
4 g of copper was dissolved in concentrated nitric aci(d) The copper nitrate solution on strong heating gave $5 g$ of its oxide. The equivalent weight of copper is

Options:
A 12
B 20
C 23
D 32

The correct answer is:
32

Solution:
Copper on reacting with conc. HNO ${ }_{3}$ gives copper nitrate.
On strongly heating with copper nitrate, copper oxide is obtained.
$\therefore$ Equivalent weight of $\mathrm{Cu}$
$=$ Weight of copper $\times$ Equivalent weight of oxygen $=4 \times 8=32$

Topic: Chemistry, Redox Reactions, JEE Main
$0.4 \mathrm{~g}$ of dihydrogen is made to react with $7.4 \mathrm{~g}$ of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at $273 \mathrm{~K}$ and 1 bar pressure is

Options:
A $9.08 \mathrm{~L}$
B $4.67 \mathrm{~L}$
C $90.8 \mathrm{~L}$
D $45.4 \mathrm{~L}$

The correct answer is:
$4.67 \mathrm{~L}$

Solution:
$71 \mathrm{~g} \mathrm{Cl}_{2}$ requires $2 \mathrm{~g} \mathrm{H}_{2}$, than $7.4 \mathrm{~g} \mathrm{Cl}_{2}$ will require, how much gram $\mathrm{H}_{2}$.
$\therefore \mathrm{H}_{2} \text { required }=\frac{2 \times 7.4}{71}=0.208 \mathrm{gm}$
$\therefore \mathrm{Cl}_{2}$ is limiting reagent $\left[\because \mathrm{H}_{2}\right.$ is in excess $]$
$\therefore$ Amount of $\mathrm{HCl}$ formed will depend on $\mathrm{Cl}_{2}$ gas
$\begin{array}{rlr}
& 71 \mathrm{gm} \mathrm{Cl}_{2} & 44.8 \mathrm{~L} \mathrm{HCl} \\
& 7.4 \mathrm{gm} \mathrm{Cl}_{2} & x \mathrm{~L} \mathrm{HCl} \\
\therefore \quad & x=\frac{7.4 \times 44.8}{71} \\
& x=4.67 \mathrm{~L} \text { of } \mathrm{HCl} \text { gas is formed. }
\end{array}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. What would be the final pH of solution? (log 2 = 0.3)

Options:
A 10.3
B 11.3
C 11
D 4.3

The correct answer is:
11.3

Solution:
pH=2, HCl=10-2M & pOH=2
or NaOH=10-2 M
4 ml of 10-2 M HClâ¡4Ã10-5 moles HCl.
6 ml of 10-2 MNaOHâ¡6Ã10-5 moles NaOH
After mixing total moles of OH â = moles of NaOHâ moles of HCl
=6Ã 10 â5 â4Ã 10 â5
After mixing excess moles of OH-=2Ã10-5
Total volume =6+4=10âmlâ=â10âÃ 10 â3 âlitre
OH-=2Ã10-510Ã103=2Ã10-3
or pOH=3-logâ¡2=3-0.3=2.7
or pH=11.3

Topic: Chemistry, Ionic Equilibrium, JEE Main
4 mol of a mixture of Mohr's salt and Fe2SO43 requires 500 ml of 1 M, K2Cr2O7 for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is

Options:
A 25
B 50
C 60
D 75

The correct answer is:
75

Solution:
Cr2O72-+6Fe2+(n=1) (Mohr's salt) +14H+→2Cr3++6Fe3++7H2O
Equivalent of Fe2+= moles of Mohr's salt
= equivalent of K2Cr2O7
=500×10-3×6×1=3
Hence mole percent of Mohr's salt
=34×100=75

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
4 moles of $A$ are mixed with 4 moles of $B$. At equilibrium for the reaction $A+B \rightleftharpoons C+D, 2$ moles of $C$ and $D$ are formed. The equilibrium constant for the reaction will be

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C 1
D 4

The correct answer is:
1

Solution:
$A+B \Rightarrow C+D$
Initial conc. 4,4 0 0
After $T$ time conc. (4-2) (4-2) 2 2
Equilibrium constant $=\frac{[C][D]}{[A][B]}=\frac{2 \times 2}{2 \times 2}=1$

Topic: Chemistry, Chemical Equilibrium, JEE Main
$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

Options:
A $x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c$
B $2 x+\log (2 \sin x+3 \cos x)+c$
C $x+2 \log (2 \sin x+3 \cos x)+c$
D $\frac{1}{2} \log (4 \sin x+6 \cos x)+c$

The correct answer is:
$x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c$

Solution:
Let $I=\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$
We can write
$\sin x+8 \cos x=A(4 \sin x+6 \cos x)$
$+B \frac{d}{d x}(4 \sin x+6 \cos x)$
$\sin x+8 \cos x$
$=A(4 \sin x+6 \cos x)+B(4 \cos x-6 \sin x)$
On equating the coefficient of $\sin x$ and $\cos x$, we get
$\begin{array}{ll}1=4 A-6 B, & 8=6 A+4 B \\ \Rightarrow & A=1, B=\frac{1}{2}\end{array}$
$\begin{aligned} \therefore I & =\int \frac{(4 \sin x+6 \cos x)+\frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x \\ & =\int\left(1+\frac{1}{2} \cdot \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x}\right) d x \\ & =x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}=$

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{4}$
D None of these

The correct answer is:
$\frac{\pi}{4}$

Solution:
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}$
$\begin{aligned} & =2 \tan ^{-1}\left[\frac{\frac{2}{5}}{1-\frac{1}{25}}\right]-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =2 \tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{25}{144}}\right]-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left(\frac{120}{119}\right)+\tan ^{-1}\left[\frac{\frac{1}{99}-\frac{1}{70}}{1+\frac{1}{99} \cdot \frac{1}{70}}\right] \\ & =\tan ^{-1}\left(\frac{120}{119}\right)+\tan ^{-1}\left(\frac{-29}{6931}\right) \\ & =\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{29}{6931}=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left[\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right]=\tan ^{-1}(1)=\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
4-digit numbers are formed using the digits $4,5,6,7,8,9$ allowing repetition of the given digits. If a number is chosen at random from those numbers thus formed, then the probability that it is exactly divisible by 3 is

Options:
A $\frac{7}{36}$
B $\frac{5}{18}$
C $\frac{5}{6}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:
If we form a 4-digit number by using the digits 4, 5, 6, 7, 8, 9 allowing repetition, then we can categories the remainders on dividing by 3 as 0,1 or 2 , so the probability to form a number which is exactly divisible by 3 is $\frac{1}{3}$.

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327oC . 30% of the solid NH4SH is decomposed to NH3 and H2S as gases. The KP of the reaction at 327oC is
R=0.082 L atm mol-1K-1, Molar mass of S=32 g mol-1, Molar mass of N=14 g mol-1

Options:
A 0.242 atm2
B 0.242×10-4 atm2
C 1×10-4 atm2
D 4.9×10-3 atm2

The correct answer is:
0.242 atm2

Solution:
NH4SH s⇌NH3 g+H2S g
Number of moles=5.151=0.1 mol

NH4SH	NH3	H2S

Initial

concentration

0.1 mol	0	0

Equilibrium

concentration

0.11-α	0.1α	0.1α

α=30%=0.3= Degree of dissociation
So, number of moles at equilibrium,

NH3=H2S=0.1×0.3=0.03

NH4SH is not considered as it is a solid.

Kc=NH3H2S

Kc=0.0330.033=10-4

Kp=10-4 0.082×6002

Kp=0.242 atm2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2019 (10 Jan Shift 2)
$\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & -3 \\ 9 & 6 & -2\end{array}\right|=0$ , then $x$ is equal to

Options:
A 3
B 5
C 7
D 9

The correct answer is:
9

Solution:
We have $\Delta=\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & -3 \\ 9 & 6 & -2\end{array}\right|=0$
$\therefore \quad-10 x+90-42-81+42+9 x=0$ or $x=9$

Topic: Mathematics, Determinants, JEE Main
5 amperes of current is passed through a metallic conductor. The charge flowing in one minute in coulombs will be

Options:
A 5
B 12
C 1/12
D 300

The correct answer is:
300

Solution:
Charge $=$ Current $\times$ Time $=5 \times 60=300 \mathrm{C}$

Topic: Physics, Current Electricity, JEE Main
5 beats/ second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95 \mathrm{~m}$ or $1 \mathrm{~m}$. The frequency of the fork will be:

Options:
A
$195 \mathrm{~Hz}$

B
$251 \mathrm{~Hz}$

C
$150 \mathrm{~Hz}$

D
$300 \mathrm{~Hz}$

The correct answer is:
$195 \mathrm{~Hz}$

Solution:
Probable frequencies of tuning fork be $n \pm$ 5
Frequency of sonometer wire, $n \propto \frac{1}{l}$
$$
\begin{aligned}
&\therefore \frac{n+5}{n-5}=\frac{100}{95} \Rightarrow 95(n+5)=100(n-5) \\
&\text { or, } 95 n+475=100 n-500 \\
&\text { or, } 5 n=975 \\
&\text { or, } n=\frac{975}{5}=195 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
5 books are to be chosen from a lot of 10 books. If $m$ is the number of ways of choice when one specified book is always included and $n$ is the number of ways of choice when a specified book is always excluded, then which one of the following is correct?

Options:
A $m>n$
B $m=n$
C $m=n-1$
D $m=n-2$

The correct answer is:
$m=n$

Solution:
Number of ways when one specified book is always included. So, 4 books can be chosen from remaining 9 ways:
$=m={ }^{9} C_{4}$
$\Rightarrow m=126$
and number of ways when one specified book is always excluded $=n={ }^{9} C_{5}$
$\Rightarrow n=126$
$\Rightarrow m=n$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is

Options:
A 14
B 1564
C 2164
D 17632

The correct answer is:
1564

Solution:
Total cases =45=1024
Favourable cases =5!1!32!Ã13!Ã4! (By grouping)
=240
Hence, required probability =2401024=1564

Topic: Mathematics, Probability, JEE Main
5 - digit numbers are to be formed using $2,3,5,7$, 9 without repeating the digits. If $p$ be the number of such numbers that exceed 20000 and $q$ be the number of those that lie between 30000 and 90000 , then $p: q$ is:

Options:
A
$6: 5$

B
$3: 2$

C
$4: 3$

D
$5: 3$

The correct answer is:
$5: 3$

Solution:
$\mathrm{p}: \begin{array}{llllll}0 & 0 & 0 & 0 & 0 & \text { place } \\ 5 & 4 & 3 & 2 & 1 & \text { ways }\end{array}$
Total no, of ways $=5 !=120$
Since all numbers are $>20,000$
$\therefore$ all numbers $2,3,5,7,9$ can come at first place.
$\mathrm{q}: \begin{array}{llllll}0 & 0 & 0 & 0 & 0 & \text { place } \\ 3 & 4 & 3 & 2 & 1 & \text { ways }\end{array}$
Total no. of ways $=3 \times 4 !=72$
( $\because 2$ and 9 can not be put at first place)
So, $p: q=120: 72=5: 3$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2013 (25 Apr Online)
$0.5 \mathrm{~g}$ of an acid is neutralised by $40 \mathrm{cc}$ of $0.125 \mathrm{~N} \mathrm{NaOH}$. The equivalent mass of the acid is

Options:
A 50
B 100
C 40
D 80

The correct answer is:
100

Solution:
Equivalent moles of $\mathrm{NaOH}=$ Equivalent moles of acid
$N_{1} V_{1}=\frac{\text { Weight }}{\text { Equivalent mass }}$ $40 \times 10^{-3} \times 0.125=\frac{0.5}{\text { Equivalent mass }}$ $\therefore$ Equivalent mass of acid $=100$

Topic: Chemistry, Redox Reactions, JEE Main
$5 \mathrm{~g}$ of benzene on nitration gave $6.6 \mathrm{~g}$ of nitrobenzene. The theoretical yield of the nitrobenzene will be

Options:
A
$4.5 \mathrm{~g}$

B
$5.6 \mathrm{~g}$

C
$8.09 \mathrm{~g}$

D
$6.6 \mathrm{~g}$

The correct answer is:
$8.09 \mathrm{~g}$

Solution:
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{HNO}_3 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$ $78 \mathrm{~g} \quad 123 \mathrm{~g}$ Now since $78 \mathrm{~g}$ of benzene on nitration give $=123 \mathrm{~g}$ nitrobenzene hence $5 \mathrm{~g}$ of benzene on nitration give $=\frac{123}{78} \times 5=7.88 \mathrm{~g}$
nearest answer is (c) i.e. theoritical yield $=7.88 \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (12 May Online)
5g of Na2SO4 was dissolved in x g of H2O . The change in freezing point was found to be 3.82oC . If Na2SO4 is 81.5% ionised, the value of x
( Kf for water =1.86oC kg mol-1 ) is approximately:
(molar mass of S=32 g mol-1 and that of Na=23 g mol-1 )

Options:
A 25 g
B 65 g
C 15 g
D 45 g

The correct answer is:
45 g

Solution:
Na2SO4→2Na++SO42-
i=1+3-1 0.815=2.63
3.82=1.86×2.63×5×1000142×x
∴ x=1.86×2.63×5000142×3.82
=45 gm

Topic: Chemistry, Solutions, JEE Main, JEE Main 2017 (08 Apr Online)
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution ? Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol–1 respectively

Options:

The correct answer is:
180

Solution:
Molarity od stock solution =moles of soluteVolume of solution in L=5/400.45

M1V1=M2V2

5/400.45×V1=0.1×5001000

V1=1801000=0.18 L

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (24 Jan Shift 1)
5 g of water at 30 Â°C and 5 g of ice at -20 Â°C are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, specific heats of ice and water are 0.5 and 1 cal g-1 Â°C-1, and latent heat of ice is 80 cal g-1.

Options:
A 0 °C

B 10 °C

C -30 °C
D > 10 °C

The correct answer is:
0 °C

Solution:
Here ice will absorb heat while hot water will release it.
So if T is the final temperature of the mixture, heat given by water
Q1=mcΔT=5×1×30-T
And heat absorbed by the ice
Q2=5×120--20+5×80+5×1T-0
So, by the principle of calorimetry Q1= Q2, i.e.,
150 - 5T = 450 + 5T
T=- 30 °C
Which is impossible as a body cannot be cooled to a temperature below the temperature of the cooling body. The physical reason for this discrepancy is the heat remaining after changing the temperature of ice from -20 to 0oC with some ice left unmelted and we are taking it for granted that heat is transferred from water at 0oC to ice at 0oC so that temperature below 0oC.
However, as heat cannot flow from one body (water) to the other (ice) at the same temperature (0oC), the temperature of the system will not fall below 0oC.

Topic: Physics, Thermal Properties of Matter, JEE Main
5g of zinc is treated separately with an excess of
(a) dilute hydrochloric acid and
(b) aqueous sodium hydroxide.
The ratio of the volumes of H2 evolved in these two reactions is:

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

The correct answer is:
1:1

Solution:
Zn+2NaOH⟶Na2ZnO2+H2
Zn+2HCl→ZnCI2+H2
According to stoichiometry in both the reactions, equal number of moles of H2 are evolved

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (09 Jan Shift 2)