datavorous/entrance-exam-dataset huggingface dataset:
$6.02 \times 10^{20}$ molecules of urea are present in $100 \mathrm{ml}$ of its solution. The concentration of urea solution is

Options:
A
$0.001 \mathrm{M}$

B
$0.1 \mathrm{M}$

C
$0.02 \mathrm{M}$

D
$0.01 \mathrm{M}$

The correct answer is:
$0.01 \mathrm{M}$

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2004
$62.5 \times 10^{18}$ electrons per second are flowing through a wire of area of cross-section $0.1 \mathrm{~m}^2$, the value of current flowing will be

Options:
A $1 \mathrm{~A}$
B $0.1 \mathrm{~A}$
C $10 \mathrm{~A}$
D $0.11 \mathrm{~A}$

The correct answer is:
$10 \mathrm{~A}$

Solution:
$i=\frac{n e}{t}=\frac{62.5 \times 10^{18} \times 1.6 \times 10^{-19}}{1}=10$ ampere

Topic: Physics, Current Electricity, JEE Main
$75.2 \mathrm{~g}$ of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ (phenol) is dissolved in a solvent of $K_f=14$. If the depression in freezing point is $7 \mathrm{~K}$, then find the $\%$ of phenol that dimerises.

Options:

The correct answer is:
35

Solution:
Phenol is dimerised as follows
At $t=0$
$$
2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right)_2
$$
At equilibrium
$$
(1-\alpha) \quad \alpha / 2
$$
Total moles after dimerisation $=1-\alpha+\alpha / 2=1-\alpha / 2$
$$
\begin{aligned}
\text { Van't Hoff factor (i) } & =\frac{\text { Normal molecular wt. of phenol }}{\text { Abnormal molecular wt. of phenol }}=\frac{94}{m_{a b}} \\
m_{a b} & =\frac{1000 \times K_f \times w}{\Delta T_f \times W}
\end{aligned}
$$
Given that
$$
\begin{aligned}
& K_f=14 \\
& w=72.5 \mathrm{gm} \\
& W=1000 \mathrm{gm} \\
& \therefore \quad m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145 \\
&
\end{aligned}
$$
$$
\therefore \quad i=\frac{94}{145}
$$
$$
\text { (i) }=\frac{\text { No. of particles after dimerisation }}{\text { No. of particles before dimerisation }}=\frac{1-\alpha / 2}{1}
$$
From eq. (i) and (ii)
$$
\begin{aligned}
1-\alpha / 2 & =\frac{94}{145}=0.65 \\
\frac{\alpha}{2} & =1-0.65=0.35
\end{aligned}
$$
Hence, $35 \%$ phenol is present in dimeric form.

Topic: Chemistry, Solutions, JEE Main
"925 fine silver" means an alloy of

Options:
A $7.5 \% \mathrm{Ag}$ and $92.5 \% \mathrm{Cu}$
B $92.5 \% \mathrm{Ag}$ and $7.5 \% \mathrm{Cu}$
C $80 \% \mathrm{Ag}$ and $20 \% \mathrm{Cu}$
D $90 \% \mathrm{Ag}$ and $10 \% \mathrm{Cu}$

The correct answer is:
$92.5 \% \mathrm{Ag}$ and $7.5 \% \mathrm{Cu}$

Solution:
" 925 fine silver" means 925 parts of pure Ag in 1000 parts of an alloy. Therefore, in terms of percentage, it will be $92.5 \% \mathrm{Ag}$ and $7.5 \% \mathrm{Cu}$.

Topic: Chemistry, d and f Block Elements, JEE Main
$9.65 \mathrm{C}$ of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus, obtained is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained is

Options:
A $5 \times 10^{-4}$
B $1 \times 10^{-4}$
C $5 \times 10^{-5}$
D $1 \times 10^{-5}$

The correct answer is:
$5 \times 10^{-5}$

Solution:
$\mathrm{MgCl}_{2} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}$
$\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{Mg}}$ (at cathode)
$\because 2 \mathrm{~F}(2 \times 96500 \mathrm{C}) \text { deposits } \mathrm{Mg}=1 \mathrm{~mol}$
$\therefore 9.65 \mathrm{C}(2 \times 96500 \mathrm{C}$ ) deposits $\mathrm{Mg}=1 \mathrm{~mol}$
$=\frac{1 \times 9.65}{2 \times 96500}$
$=5 \times 10^{-5} \mathrm{~mol}$
In order to prepare Grignard reagent, one mole of $\mathrm{Mg}$ is used per mole of reagent obtained. Thus, by $5 \times 10^{-5} \mathrm{~mol} \mathrm{Mg}, 5 \times 10^{-5}$ mole of Grignard reagent are obtained.

Topic: Chemistry, Hydrocarbons, JEE Main
1000 kHz carrier wave is amplitude modulated by the signal frequency 200-4000 Hz. The channel width in this case is

Options:
A 8 kHz
B 4 kHz
C 7.6 kHz
D 400 kHz

The correct answer is:
8 kHz

Solution:
Bandwidth in communication System for carrier wave
Band width =2Ãfm=8 kHz

Topic: Physics, Communication System, JEE Main
$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is:

Options:
A 1.52
B 0.995
C 190
D 9

The correct answer is:
9

Solution:
Let $E$ be the equivalent weight of the metal
$$
\text { So, } \frac{E+17}{E+8}=\frac{1.52}{0.995}
$$
[17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen]
$$
\begin{array}{l}
\Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\
\Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\
\therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9
\end{array}
$$

Topic: Chemistry, Redox Reactions, JEE Main
$19.85 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{NaOH}$ reacts with $20 \mathrm{~mL}$ of $\mathrm{HCl}$ solution for complete neutralisation. The molarity of $\mathrm{HCl}$ solution is

Options:
A $9.9$
B $0.99$
C $0.099$
D $0.0099$

The correct answer is:
$0.099$

Solution:
$\begin{array}{l}
\text { Molarity of base }=\frac{\text { normality }}{\text { Acidity }}=\frac{0.1}{1}=0.1 \\
\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} \\
0.1 \times 19.85=\mathrm{M}_{2} \times 20 \\
\mathrm{M}_{2}=0.09925 \approx 0.099
\end{array}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
\( 0.2325 \mathrm{~g} \) of an organic compound was analysed for nitrogen by Dumas method. \( 31.7 \mathrm{~mL} \) of moist nitrogen was collected at \( 25^{\circ} \mathrm{C} \) arid \( 755.8 \mathrm{~mm} \) of \( \mathrm{Hg} \) pressure. Calculate the percentage of nitrogen in the sample. (Aqueous tension of water at \( 25^{\circ} \mathrm{C} \) is \( 23.8 \mathrm{~mm} \) of \( \mathrm{kg} \) )

Options:
A \( 18 \% \)
B \( 19.7 \% \)
C \( 15.1 \% \)
D \( 20.2 \% \)

The correct answer is:
\( 15.1 \% \)

Solution:
At given condition,

P1=755.8-23.8=732.0 mm

V1=31.7 mL

T1=25+273=298 K

For STP,

P2=760 mm

T2=273 K

R2=?

Applying the gas equation,

P1×V1T1=P2×V2T2

732×31.7298=760×V2273

V2=28 mL

% of N=28×28×10022400×0.2375=15.1%

Topic: Chemistry, General Organic Chemistry, JEE Main
8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

Options:
A 1 : 10

B 1 : 15

C 1 : 20

D 1 : 25

The correct answer is:
1 : 20

Solution:
As volume remains constant R 3 =8000 r 3 ∴R=20r

Surface energy of one big drop Surface energy of 8000 small drop = 4π R 2 T 8000 4π r 2 T

= R 2 8000 r 2 = ( 20r ) 2 8000 r 2 = 1 20

Topic: Physics, Mechanical Properties of Fluids, NEET
A 0.4 kg mass takes 8 s to reach ground when dropped from a certain height P above surface of earth. The loss of potential energy in the last second of fall is ______ J. [Take g=10 m s-2]

Options:

The correct answer is:
300

Solution:
Displacement in nth second is given by Sn=u+a22n-1

Here, displacement is 8th second is

S8=0+12×10×(2×8-1)

⇒S8=75 m

Loss of potential energy is

ΔU=mgS8=0.4×10×75

⇒ΔU=300 J.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A 0.5 kg block moving at a speed of 12 ms-1 compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be _____ Nm-1

Options:

The correct answer is:
600

Solution:
Given: Mass of the block m=500 g, initial velocity of the block u=12 m s-1, and final velocity of the block v=u2=6 m s-1.

As non-conservative forces are not acting on the system we can conserve energy.

Thus, loss in kinetic energy of block = Gain in potential energy of spring

12mu2-12mv2=12kx2

12×0.5×122-12×0.5×62=12k×0.32

k=600 N m-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A 0.5 percent solution of potassium chloride was found to freeze at -0.24 °C. The percentage dissociation of potassium chloride is (Nearest integer)
(Molal depression constant for water is 1.80 K kg mol-1 and molar mass of KCl is 74.6 g mol-1)

Options:

The correct answer is:
98

Solution:
KCl→K++Cl-

1 - -

1-α α α

i=1-α+2α=1+α

∆Tf=i Kf·m

0.24=1+α1.80×0.574.6×100099.5

α=0.98α%=98 %

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A 0.010 M solution of maleic acid, a monoprotic organic acid, is 14% ionized. What is K a 1 for maleic acid?

Options:
A 2.3×10-3
B 2.3×10-4
C 2.0×10-4
D 2.0×10-6

The correct answer is:
2.3×10-4

Solution:
Degree of dissociation, α=14100=0.14
Concentration of maleic acid, C=0.010 M
The equilibrium reaction is

Ka=C2α2C(1-α)=Cα2(1-α)
Ka=0.010×0.14×0.140.86=2.3×10-4.

Topic: Chemistry, Ionic Equilibrium, JEE Main
$(a, 0)$ and $(b, 0)$ are centres of two circles belonging to a coaxial system of which $y$-axis is the radical axis. If radius of one of the circles is ' $r$ ', then the radius of the other circle is

Options:
A $\left(r^2+b^2+a^2\right)^{1 / 2}$
B $\left(r^2+b^2-a^2\right)^{1 / 2}$
C $\left(r^2+b^2-a^2\right)^{1 / 3}$
D $\left(r^2+b^2+a^2\right)^{1 / 3}$

The correct answer is:
$\left(r^2+b^2-a^2\right)^{1 / 2}$

Solution:
Let the equation of circle whose centre $(a, 0)$ and radius $(r)$ is
$$
\begin{gathered}
(x-a)^2+(y-0)^2=r^2 \\
\Rightarrow \quad S_1 \equiv x^2+a^2-2 a x+y^2-r^2=0
\end{gathered}
$$
and the equation of circle whose centre $(b, 0)$ and radius $R$ is
$$
\begin{gathered}
(x-b)^2+(y-0)^2=R^2 \\
\Rightarrow \quad S_2 \equiv x^2+b^2-2 b x+y^2-R^2=0
\end{gathered}
$$
$\therefore$ Equation of radical axis is
$$
\begin{gathered}
S_1-S_2=0 \\
\Rightarrow \quad a^2-b^2+2 b x-2 a x+R^2-r^2=0 \\
\Rightarrow \quad R^2=r^2-a^2+b^2-2 b x+2 a x
\end{gathered}
$$
Since, radical axis is $y$-axis.
Therefore, putting $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& R^2=r^2-a^2+b^2-0+0 \\
\Rightarrow \quad & R=\left(r^2+b^2-a^2\right)^{1 / 2}
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main
$A=\left[\begin{array}{ccc}1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4\end{array}\right]$ is nilpotent of index

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

The correct answer is:
$2$

Solution:
Since $A^2=O$ (Zero matrix) and 2 is the least +ve integer for which $A^2=0$. Thus, $A$ is nilpotent of index 2 .

Topic: Mathematics, Matrices, JEE Main
$A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ then $A^3-4 A^2-6 A$ is equal to :

Options:
A 0
B $A$
C $-A$
D I

The correct answer is:
$-A$

Solution:
$\because \quad A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$\therefore \quad A^2=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]$
$A^3=A^2 \cdot A=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]$
$\therefore A^3-4 A^2-6 A$
$=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]-4\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-6\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6\end{array}\right]$
$=\left[\begin{array}{lll}-1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1\end{array}\right]$
$=-\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]=-A$

Topic: Mathematics, Matrices, JEE Main
$\mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{1,2,3,4,5,6\}$ are two sets, and function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ is defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}+2 \forall \mathrm{x} \in \mathrm{A}$, then the function $\mathrm{f}$ is

Options:
A bijective
B onto
C one-one
D many-one

The correct answer is:
one-one

Solution:
Hints: $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y}) \Rightarrow \mathrm{x}+2=\mathrm{y}+2 \Rightarrow \mathrm{x}=\mathrm{y} \quad \therefore$ one-one

Topic: Mathematics, Functions, JEE Main
A 1.5Ã109 years old rock contains U238 which disintegrates to form Pb206. Assume that there was no lead in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of the number of nuclei of lead to that of uranium. [Given, half-life of U238 = 4.5Ã109 years, 213â1.26]

Options:

The correct answer is:
0.26

Solution:
Let N0 be the initial number of uranium nuclei. After time t, let NU be the number of uranium nuclei and NPb be the number of lead nuclei.

The number of half-lives passed is

n=tThalf=1.5Ã1094.5Ã109=13

NU=N012n=N01213

NPb=N0-N01213
NPbNU=213-1=0.26

Topic: Physics, Nuclear Physics, JEE Main
A 1.84mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of H2 gas at STP. The number of alcoholic hydrogen present in compound 'X' is

Options:

The correct answer is:
6

Solution:
Number of moles of alcoholic compound =Given massMolar mass = 1.84×10-392.0 = 2×10-5 moles

Number of moles of hydrogen gas produced = Volume of gas at STP22400 ml = 1.34422400 = 6×10-5 moles

Monohydric alcohol will react with active metal to give 12 mole of hydrogen gas.

Since per mole of given alcohol is giving 3 moles of hydrogen gas.

So number of alcoholic hydrogen present in compound should be =6.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (29 Jul Shift 2)
$\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$ and $A_1, B_1, C_1$ denote the co-factors of $a_1, b_1, c_1$ respectively, then the value of the determinant $\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|$ is

Options:
A $\Delta$
B $\Delta^2$
C $\Delta^3$
D $0$

The correct answer is:
$\Delta^2$

Solution:
We know that $\Delta \Delta^{\prime}=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right| \cdot\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|$
$\begin{aligned} & =\left|\begin{array}{ccc}\Sigma a_1 A_1 & 0 & 0 \\ 0 & \Sigma a_2 A_2 & 0 \\ 0 & 0 & \Sigma a_3 A_3\end{array}\right|=\left|\begin{array}{ccc}\Delta & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & \Delta\end{array}\right|=\Delta^3 \\ & \Rightarrow \Delta^{\prime}=\Delta^2 .\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
$\left|\begin{array}{ccc}a-1 & a & b c \\ b-1 & b & c a \\ c-1 & c & a b\end{array}\right|=$

Options:
A 0
B $(a-b)(b-c)(c-a)$
C $a^3+b^3+c^3-3 a b c$
D None of these

The correct answer is:
None of these

Solution:
$\left|\begin{array}{lll}a-1 & a & b c \\ b-1 & b & c a \\ c-1 & c & a b\end{array}\right|=\left|\begin{array}{lll}a & a & b c \\ b & b & c a \\ c & c & a b\end{array}\right|-\left|\begin{array}{ccc}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
$=-\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right|=-\left|\begin{array}{ccc}a & a^2 & 1 \\ b-a & b^2-a^2 & 0 \\ c-a & c^2-a^2 & 0\end{array}\right|$ [By $\left.R_2 \rightarrow R_2-R_1 ; R_3 \rightarrow R_3-R_1\right]$
$=-(a-b)(b-c)(c-a)$

Topic: Mathematics, Determinants, JEE Main
A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to a 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is

Options:
A 2 Hz
B 14 Hz
C 122 Hz
D 12 Hz

The correct answer is:
12 Hz

Solution:
The frequency of 1 kg block is f=12πkm=1 Hz
4π2=km , m=1
Spring constant, k=4π2
In parallel connection of springs, the equivalent spring constant is keq=2k.
f=12π K×28
Frequency of 8 kg block is=12 Hz

Topic: Physics, Oscillations, JEE Main, JEE Main 2017 (08 Apr Online)
A 1 kg block B rests as shown on a bracket A of the same mass. Constant forces F1=20 N and F2=8 N start to act at a time t=0 when the distance of block B from the pulley is 50 cm. Time (in s) when block B reaches the pulley is t. Find the value of 10t.

Options:

The correct answer is:
5

Solution:
First, find out the acceleration of A so for this
⇒a=20-2F2=20-2×8
aA=4 m s-2
Now use pseudo concept (in which A is the non-inertial frame)
⇒8-4=4 m s-2
Now 50100=12×4×t2
t=12=0.5 s
⇒ 10t=5 s

Topic: Physics, Laws of Motion, JEE Main
A $0.1 \mathrm{~kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1 \mathrm{~m}$ and its cross-sectional area is $4.9 \times 10^{-7} \mathrm{~m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 \mathrm{rad} \mathrm{s}^{-1}$. If the Young's modulus of the material of the wire is $n \times 10^9 \mathrm{Nm}^{-2}$, the value of $n$ is

Options:

The correct answer is:
4

Solution:
$\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$
Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get,
$$
n=4
$$
$\therefore$ Answer is 4 .

Topic: Physics, Mechanical Properties of Solids, JEE Main
A $1 \mathrm{~kg}$ mass is suspended from the ceiling by a rope of length $4 \mathrm{~m}$. A horizontal force ' $F$ ' is applied at the mid point of the rope so that the rope makes an angle of $45^{\circ}$ with respect to the vertical axis as shown in figure. The magnitude of $F$ is :
(Assume that the system is in equilibrium and $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $10 \mathrm{~N}$
B $\frac{10}{\sqrt{2}} \mathrm{~N}$
C $1 \mathrm{~N}$
D $\frac{1}{10 \times \sqrt{2}} N$

The correct answer is:
$10 \mathrm{~N}$

Solution:
$\begin{aligned} & \mathrm{T}_1 \sin 45^{\circ}=\mathrm{F} \\ & \mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g} \\ & \therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}} \\ & \therefore \mathrm{F}=10 \mathrm{~N}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A $1 \mathrm{~kg}$ object strikes a wall with velocity $1 \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ with the wall and reflects at the same angle. If it remains in contact with wall for $0.1 \mathrm{~s}$, then the force exerted on the wall is

Options:
A $10 \sqrt{3} \mathrm{~N}$
B $20 \sqrt{3} \mathrm{~N}$
C $30 \sqrt{3} \mathrm{~N}$
D Zero

The correct answer is:
$10 \sqrt{3} \mathrm{~N}$

Solution:
$\begin{aligned} & F=\frac{\Delta p}{\Delta t} \\ & \Delta p=2 m v \cos 30^{\circ} \\ & =2 \times 1 \times \frac{\sqrt{3}}{2}=\sqrt{3} \\ & F=\frac{\sqrt{3}}{0.1}=10 \sqrt{3} \mathrm{~N}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2023 (Manipur)
A 1 L glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in this flask if coefficient of linear expansion of glass is 9Ã 10 â6 / C â while of volume expansion of mercury is 1.8Ã 10 â4 / C o ?

Options:
A 50 cc
B 100 cc
C 150 cc
D 200 cc

The correct answer is:
150 cc

Solution:
It is given that the volume of air in the flask remains the same. This means that the expansion in the volume of the vessel is exactly equal to the volume expansion of mercury.
i.e., Î V G = Î V L
or V G Î³ G Î Î¸ = V L Î³ L Î Î¸
â´ V L = V G Î³ G Î³ L = 1 0 0 0 Ã 3 Ã 9 Ã 1 0 - 6 1.8 Ã 1 0 - 4 = 1 5 0 cc

Topic: Physics, Thermal Properties of Matter, JEE Main
A 1 m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X andY.

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

The correct answer is:
1:2

Solution:
âââââââ

Resistance of a wire is given by R=ρLA, where, ρ is conductivity, L is length of wire and A is cross-sectional area of wire.

For a single wire, A and ρ are constant.

Thus, for parts X and Y, resistance, RXRY=lXlY.

When wire is stretched to double of its length, then resistance becomes four times, thus, RW=4RX=2RY.

Hence, the ratio is RXRY=12.

So, lXlY=12

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A 0.1 molal aqueous solution of a weak acid is $30 \%$ ionised. If $K_f$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be

Options:
A $-0.18^{\circ} \mathrm{C}$
B $-0.54^{\circ} \mathrm{C}$
C $-0.36^{\circ} \mathrm{C}$
D $-0.24^{\circ} \mathrm{C}$

The correct answer is:
$-0.24^{\circ} \mathrm{C}$

Solution:
Freezing point depression $\left(\Delta T_f\right)=i K_f m$
$$
\begin{aligned}
& i=1-0.3+0.3+0.3 \\
& i=1.3 \\
& \therefore \quad \Delta T_f=1.3 \times 1.86 \times 0.1=0.2418^{\circ} \mathrm{C} \\
& T_f=0-0.2418^{\circ} \mathrm{C} \\
& =-0.2418^{\circ} \mathrm{C} \\
&
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2011 (Mains)
A 1 molal K4Fe(CN)6 solution has a degree of dissociation of 0.4 . Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is _____ u. (Round off to the Nearest Integer). [Density of water =1.0 g cm-3

Options:

The correct answer is:
85

Solution:
K4Fe(CN)6 ⇌ 4 K+ + Fe(CN)64-

Initial conc. 1 m 0 0

Final conc. (1-0.4)m 4×0.4m 0.4 m

=0.6 m =1.6 m =0.4 m

Effective molality =0.6+1.6+0.4=2.6 m

For same boiling point, the molality of another solution should also be 2.6 m.

Now, 18.1 weight percent solution means

18.1gm solute is present in 100gm solution and hence, (100-18.1=)81.9gm water. Now, 2.6=18.1/M81.9/1000

∴ Molar mass of solute, M=85

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A2,6,2,B-4,0,λ,C2,3,-1 and D4,5,0, λ≤5 are the vertices of a quadrilateral ABCD. If its area is 18 square units, then 5-6λ is equal to _____ .

Options:

The correct answer is:
11

Solution:
Given:

A2,6,2,B-4,0,λ,C2,3,-1D4,5,0

Area of quadrilateral

=12AC→×BD→=18

⇒AC→×BD→36

Now,

AC→×BD→=i^j^k^0-3-385-λ

⇒AC→×BD→=3λ+15i^-0+24j^+0+24k^

⇒AC→×BD→=3λ+15i^-24j^+24k^

So,

AC→×BD→=36

⇒3λ+152+242+242=36

⇒9λ2+225+90λ=1296-1152

⇒9λ2+90λ-81=0

=λ2+10λ+9=0

⇒λ=-1,-9

Since,

λ≤5⇒λ=-1

Therefore,

5-6λ=5-6-1=11

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A 2 A current carrying straight metal wire of resistance $1 \Omega$, resistivity $2 \times 10^{-6} \Omega \mathrm{m}$, area of cross-section $10 \mathrm{~mm}^2$ and mass $500 \mathrm{~g}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is _____$\times 10^{-1} \mathrm{~T}$ (given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ).

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5 \\ & \mathrm{mg}=\mathrm{Bi} \ell\end{aligned}$
$\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} \ell}=\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \mathrm{Tesla}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (05 Apr Shift 1)
A2+B2→2AB.ΔHf=-200 kJ mol-1

AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, B2 and AB are in the ratio1:0.5:1, then the bond enthalpy of A2 is_________kJ mol–1 (Nearest integer)

Options:

The correct answer is:
800

Solution:
First write balanced reaction for the formation of one mole of AB with the help of diatomic molecule:

12A2+12B2→AB ΔHf=-200 kJ/mol

Now,

∆Hreaction=12∆HA-A+12∆HB-B-∆HAB let, ∆HAB=x

12x+12(0.5x)-x=-200 ⇒ x2+0.25x-x=-200

⇒ -0.25x=-200 ⇒ x=800 kJ/mol

Bond enthalpy =800 kJ/mol.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (13 Apr Shift 1)