datavorous/entrance-exam-dataset huggingface dataset:
At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is - [integer answer]

Options:

The correct answer is:
0

Solution:
t1/21t1/22=P1P21-n

⇒340170=55.527.81-n

21=21-n

1=1-n⇒n=0

For zero order reaction half life is directly proportional to the initial concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
At 363 K, the vapour pressure of A is 21 kPa and that of B is 18kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is___________ kPa. (Round of to the Nearest Integer).

Options:

The correct answer is:
19

Solution:
Given PA0=21 kPa ⇒PB0=18 kPa

→ An Ideal solution is prepared by mixing 1 mole A and 2 mole B.

→XA=13 and XB=23

→ According to Raoult's law

PT=XAPA0+XBPB0

⇒ PT=13×21+23×18

⇒ PT=7+12=19 kPa

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (16 Mar Shift 2)
At 380Â°C, the half life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ mol-1 . Calculate the time in minutes required for 75% decomposition at 450Â°C.
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
20.00

Solution:
K1=0.693/360 min-1 at 653 K=1.93Ã10-3 min-1
Ea=200Ã103J; K2=? at 723 K, R=8.314 J
2.303log K 2 K 1 = E a R 1 T 1 â 1 T 2
â´2.302 logK21.93Ã10-3=200Ã1038.314723-653723Ã653
K2=0.068 min-1
Now, t=2.3030.068log10025=20.39 min

Topic: Chemistry, Chemical Kinetics, JEE Main
At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is

Options:

The correct answer is:
4

Solution:
vrms of X=3RTxMx;vmp of Y =2RTyMy
Given vrms=vmp
3RTxMx=2RTyMy
My=2RTyMx3RTx=2×60×403×400=4

Topic: Chemistry, States of Matter, JEE Main
At 407 K the rate constant of a chemical reaction is 9.5Ã10-5s-1 and at 420 K , the rate constant is 1.9Ã10-4s-1 . The frequency factor of the reaction is xÃ105s-1 .The value of 'x' is.
Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
5

Solution:
The Arrhenius equation is,
log 10 k 2 k 1 = E a 2.303ÃR T 2 â T 1 T 1 T 2
Given k 1 =9.5Ã 10 â5 s â1 ;â k 2 =1.9Ã 10 â3 s â1
R=8.314 Jmol â1 K â1 ;
T1=407 K and T2=420 K
Substituting the values in Arrhenius equation.
log101.9Ã10-49.5Ã10-5=Ea2.303Ã8.314420-407420Ã407
Ea=75782.3 Jmol-1
Applying now logâ¡k1=logâ¡A-Ea2.303 RT1
logâ¡9.5Ã10-5=logâ¡A-75782.32.303Ã8.314Ã407
log A 9.5× 10 −5 = 75782.3 2.303×8.314×407 =9.7246
So, A=5.0Ã105s-1

Topic: Chemistry, Chemical Kinetics, JEE Main
At $500 \mathrm{~K}$, for a reversible reaction $A_2(g)+B_2(g) \rightleftharpoons 2 A B(g)$ in a closed container, $K_C=2 \times 10^{-5}$. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant $K_C$ in the presence of catalyst at the same temperature is

Options:
A $2 \times 10^{-4}$
B $2 \times 10^{-6}$
C $2 \times 10^{-10}$
D $2 \times 10^{-5}$

The correct answer is:
$2 \times 10^{-5}$

Solution:
The value of the $K_C$ remains same as catalyst
does not affects the $K_C$. Only temperature can alter the value of the $K_C$ of a given reaction. Thus, the equilibrium constant $K_C$ in the presence of catalyst at the same temperature is $2 \times 10^{-5}$.

Topic: Chemistry, Chemical Equilibrium, KCET, KCET 2023
At 500 K, the half life period of a gaseous reaction at an initial pressure of 80 kPa is 350 s. When the pressure is 40 kPa, the half life period is 175 sec; the order of the reaction is :

Options:
A zero
B one
C two
D three

The correct answer is:
zero

Solution:
( T 1 2 T 1 2 )= ( C 1 C 2 ) 1−n ⇒( 350 175 )= ( 80 40 ) 1−n =2= ( 2 ) 1−n ⇒1−n=1                           ⇒n=0

Topic: Chemistry, Chemical Kinetics, JEE Main
At 518oC, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr was 1.00 Torr s-1 when 5% had reacted and 0.50 Torr s-1 when 33% had reacted. The order of the reaction is:

Options:
A 0
B 2
C 3
D 1

The correct answer is:
2

Solution:
Rate∝concentrationn
Let the initial concentration be 1 M.

According to the question,

Concentration Rate0.95 M10.67 M 0.5

Substituting in the general rate expression,

0.950.67n=10.5

⇒n=2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2018 (08 Apr)
At 518oC the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s-1 when 5% had reacted and 0.5 Torr s-1 when 33% had reacted. The order of the reaction is

Options:
A 0
B 2
C 3
D 1

The correct answer is:
2

Solution:
CH3CHOPi=363 torr→CO+CH4
Now we know that
r∝PRn ...(i)
where PR= reactant pressure
n= order of reaction
Now rate of reaction is 1.00 torr s-1 , when reactant pressure is 363-363×5100torr=344.85 torr, similarly rate of reaction is 0.5 torr s-1 , when reactant pressure is 363-363×33100torr=243.21 torr
Therefore, applying equation (i)
10.5=344.85243.21n
2=(1.418)n
21n=1.418
So n≈2

Topic: Chemistry, Chemical Kinetics, JEE Main
At $550 \mathrm{~K}$, the $K_c$ for the following reaction is $10^4 \mathrm{~mol}^{-1} \mathrm{~L} X(g)+Y(g) \rightleftharpoons Z(g)$
At equilibrium, it was observed that $[X]=\frac{1}{2}[Y]=\frac{1}{2}[Z]$
What is the value of [Z] (in $\mathrm{mol} \mathrm{L}^{-1}$ ) at equilibrium?

Options:
A $2 \times 10^{-4}$
B $10^{-4}$
C $2 \times 10^4$
D $10^4$

The correct answer is:
$2 \times 10^{-4}$

Solution:
$\begin{aligned} & X(g)+Y(g) \rightleftharpoons Z(g) \\ & K_c=\frac{[Z]}{[X][Y]} \text { or } 10^4=\frac{[Z]}{[X][Y]} \\ & \therefore \quad[Z]=10^4[X][Y]=10^4 \times \frac{1}{2}[Z][Z] \\ & \text { (Given, }[X]=\frac{1}{3}[Y]=\frac{1}{2}[Z] \text { ) } \\ & \therefore \quad[Z]=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \\ & \end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
At 600 K, the root mean square (rms) speed of gas X (molar mass =40) is equal to the most probable speed of gas Y at 90 K. The molar mass of the gas Y is _____ g mol–1. (Nearest integer)

Options:

The correct answer is:
4

Solution:
The root mean square velocity Vrms=3RTM

The most probable speed Vmp=2RTM

T=Temperature in Kelvin

M=Molar mass

Vrms of XVmp of Y=3TXMY2TYMX

1=3×600×MY2×90×40=MY4

MY=4 gm mol-1

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2023 (12 Apr Shift 1)
At $780 \mathrm{~K}$ and 10 atmosphere pressure the equilibrium constant for the reaction $2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ is 3.52 . At the same temperature and 7.04 atmosphere pressure, the equilibrium constant for the same reaction is

Options:
A $7.04$
B $3.52$
C $10.56$
D $5.23$

The correct answer is:
$3.52$

Solution:
The value of the equilibrium constant is dependent on temperature but is independent of the pressure. Thus, the value of the equilibrium constant will remain 3.52 .

Topic: Chemistry, Chemical Equilibrium, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
At 800Â°C, the following equilibrium is established as
F 2 g â 2 F g
The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that 800Â°C and 1 atm mixture effuses 1.6 times as fast as SO2 effuses under the similar conditions. (At. wt. of F=19). What is the value of Kp (in atm) ?

Options:
A 0.315
B 0.685
C 0.46
D 1.49

The correct answer is:
1.49

Solution:
According to Graham's law of effusion
r â 1M
r mix r SO 2 = M SO 2 M mix
â 2.56 = 6 4 M mix
â M mixture = 2 5
Let mole fraction of F2 is x. Therefore, mole fraction of F will be 1âx
2 5 = 3 8 Ã x + 1 - x Ã 1 9 1
x = 0.315 ;
â K p = P F 2 P F 2 = 0.685P 2 0.315P â 1.49 atm

Topic: Chemistry, Chemical Equilibrium, JEE Main
At $1000 \mathrm{~K}$, if the equilibrium constant $K_p$ for the reaction.
$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)$
is $4.157 \times 10^{-4}$ bar, the $K_C\left(\right.$ in $\mathrm{mol} \mathrm{L}^{-1}$ ) is $\left(R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A $4.16 \times 10^{-7}$
B $4.16 \times 10^{-4}$
C $50 \times 10^{-4}$
D $50 \times 10^{-6}$

The correct answer is:
$50 \times 10^{-6}$

Solution:
$K_p=K_c(R T)^{\Delta n}$
Where,
$\begin{aligned}
& K_C \text { and } K_p=\text { Equilibrium constants } \\
& R=\text { Gas constant }\left(0.083 \mathrm{~L} \mathrm{bar}^{-1} \times \mathrm{mol}^{-1}\right) \\
& T=\text { Temperature }(1000 \mathrm{~K})
\end{aligned}$
$\Delta n=$ Number of gaseous moles of product
-Moles of reactant
$\begin{aligned}
K_c & =4.157 \times 10^{-4} \\
\Delta n & =3-2=1 \\
K_C=\frac{K_b}{R T} & =\frac{4.157 \times 10^{-4}}{0.083 \times 1000} \\
& =5.0 \times 10^{-6}
\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
At \(1000 \mathrm{~K}\), the partial pressures of \(\mathrm{CO}_2(g)\) and \(\mathrm{CO}(g)\) for the reaction \(\mathrm{CO}_2(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) in a closed vessel at equilibrium are 0.15 and 0.60 bar respectively. The \(K_c\) for this reaction at the same temperature is approximately

Options:
A \(2.0 \times 10^{-4}\)
B \(2.89 \times 10^{-2}\)
C \(2.89 \times 10^{-3}\)
D \(5.78 \times 10^{-3}\)

The correct answer is:
\(2.89 \times 10^{-2}\)

Solution:
For the reaction,
\(\begin{aligned}
\mathrm{CO}_2(g)+\mathrm{C}(s) & \rightleftharpoons 2 \mathrm{CO}(g) \\
K_{\mathrm{sp}}= & \frac{\left[\mathrm{CO}^2(g)\right.}{\left[\mathrm{CO}_2\right](g)}=\frac{p\left[\mathrm{CO}^2(g)\right.}{p\left[\mathrm{CO}_2\right](g)} \\
& {\left[\therefore p V=n R T \text { or } \frac{n}{V}=\frac{p}{R T}, \frac{n}{V} \propto p(R, T)\right] }
\end{aligned}\)
Given,
\(\begin{aligned}
p[\mathrm{CO}] & =0.6 \text { bar } \\
p\left[\mathrm{CO}_2\right] & =0.15 \text { bar } \\
K_p & =\frac{[0.6]^2}{[0.15]}=2.4
\end{aligned}\)
Also, \(K_p=K_c \cdot(R T)^{\Delta n}\)
where, \(\Delta n=\) moles of gaseous products moles of gaseous reactants \(=2-1=1\)
\(\therefore \quad K_p=K_c \times R \times T\)
where,
\(\begin{aligned}
K_c & =\frac{K_p}{R T} \\
K_p & =2.4 \\
R & =0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}
\end{aligned}\)
or \(K_c=2.89 \times 10^{-2}\)
Hence, option (b) is the correct answer.

Topic: Chemistry, Chemical Equilibrium, JEE Main
At 3000 K, the equilibrium partial pressure of CO2,CO and O2 are 0.6, 0.4 and 0.2 atm respectively. Kp for the reaction, 2CO2â2CO+O2 is

Options:
A 0.088
B 0.0533
C 0.133
D 0.177

The correct answer is:
0.088

Solution:
2CO2 â 2CO+O2
At equili. 0.6 atm 0.4 atm 0.2 atm
â´ Kp=pCO2ÃpO2pCO22
=0.42(0.2)0.62=0.088

Topic: Chemistry, Chemical Equilibrium, JEE Main
At a certain depth d below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R above earth surface. Where R is Radius of earth (Take R = 6400 km). The depth d is equal to

Options:
A 5260 km
B 640 km
C 2560 km
D 4800 km

The correct answer is:
4800 km

Solution:
The magnitude of acceleration due to gravity g1 at a depth of d below the surface of Earth is

g1=GMR3R-d=GMR21-dR ....(1)

The magnitude of acceleration due to gravity g2 at a height d above the surface of Earth is

g2=GMR+d2 ....(2)

Given here, g1=4g2

GMR21-dR=4GMR+3R2

⇒GMR21-dR=4GM4R2

⇒1-dR=14⇒dR=34⇒d=34R

⇒d=4800 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (31 Jan Shift 1)
At a certain instant of time the mass of a rocket going up vertically is $100 \mathrm{~kg}$. If it is ejecting $5 \mathrm{~kg}$ of gas per second at a speed of $400 \mathrm{~m} / \mathrm{s}$, the acceleration of the rocket would be (taking $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $20 \mathrm{~m} / \mathrm{s}^2$
B $10 \mathrm{~m} / \mathrm{s}^2$
C $2 \mathrm{~m} / \mathrm{s}^2$
D $1 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$10 \mathrm{~m} / \mathrm{s}^2$

Solution:
The acceleration of a rocket is given by
$\begin{aligned}
& a=\frac{v}{m}\left(\frac{\Delta m}{\Delta t}\right)-g=\frac{400}{100}\left(\frac{5}{1}\right)-10 \\
& =(20-10)=10 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will be

Options:
A $4 \mathrm{~s}$
B $2 \mathrm{~s}$
C $\frac{1}{2} \mathrm{~s}$
D $\sqrt{2} \mathrm{~s}$

The correct answer is:
$\sqrt{2} \mathrm{~s}$

Solution:
$\begin{aligned} T=2 \pi & \sqrt{\frac{I}{M B_H}} \\ \therefore \quad \frac{T_1}{T_2} & =\sqrt{\frac{\left(B_H\right)_2}{\left(B_H\right)_1}} \\ T_2 & =T_1 \sqrt{\frac{\left(B_H\right)_1}{\left(B_H\right)_2}} \\ n_1 & =30 \text { oscillation } / \mathrm{min} \\ n_1 & =\frac{1}{2} \text { oscillation } / \mathrm{s} \\ T_1 & =2 \mathrm{~s} \\ T_2 & =2 \sqrt{\frac{B_H}{2 B_H}} \\ = & 2 \times \frac{1}{\sqrt{2}}=\sqrt{2} \mathrm{~s}\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count of 1 s measures the time taken by the pendulum for 40 oscillations to be 50 s. The accuracy in the measurement of g is

Options:
A 4.40%
B 3.40%
C 2.40%
D 5.40%

The correct answer is:
4.40%

Solution:
ΔTT=12Δgg+ΔLLΔgg=2ΔTT+ΔLL; =2150+0.125.0=4.4%

Topic: Physics, Oscillations, JEE Main
At a certain place, the angle of dip is $30^{\circ}$ and the horizontal component of earth's magnetic field is 0.50 Oersted. The earth's total magnetic field is (in Oersted)

Options:
A $\sqrt{3}$
B 1
C $\frac{1}{\sqrt{3}}$
D $\frac{1}{2}$

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

Solution:
Horizontal component of earth's magnetic field is given by,
$\begin{aligned}
& \mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \phi, \\
& \therefore \mathrm{B}=\frac{\mathrm{B}_{\mathrm{H}}}{\cos \phi}=\frac{0.5}{\cos 30^{\circ}}=\frac{0.5}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
At a certain place the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is

Options:
A 13
B 12
C 3
D 1

The correct answer is:
13

Solution:
Angle of dip, δ=30°

Angle of dip or magnetic dip is the angle the net earth's magnetic field and it's horizontal component.

BH=Bcosδ

B=BHcosδ

Magnetic field, B=0.532=1232=13 G.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2022 (29 Jun Shift 1)
At a certain place, the angle of dip is 30° and the horizontal component of earth's magnetic field is 50 μT. The total magnetic field (in μT) of the earth at this place, is

Options:
A 1003 μT
B 100 μT
C 1003 μT
D 200 μT

The correct answer is:
1003 μT

Solution:
B=Hcos⁡δ=50cos⁡30o=50 ×23=1003 μT
(H = horizontal component of the earth's magnetic field and B = total magnetic field of earth)

Topic: Physics, Magnetic Properties of Matter, JEE Main
At a certain place, the horizontal component of earth's magnetic field is $\sqrt{3}$ times the vertical component. The angle of dip at that place is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\tan \theta=\frac{V}{H}=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
At a certain temperature and at infinite dilution, the equivalent conductances of sodium benzoate, hydrochloric acid and sodium chloride are 240, 349 and $229 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ respectively. The equivalent conductance of benzoic acid in $\mathrm{ohm}^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ at the same conditions is

Options:
A 80
B 328
C 360
D 408

The correct answer is:
360

Solution:
Given, $\wedge_{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}}^{\infty}=240 \Omega^{-1} \mathrm{~cm}^2$ equiv $^{-1}$
$\wedge_{\mathrm{HCl}}^{\infty}=349 \Omega^{-1} \mathrm{~cm}^2$ equiv $^{-1}$
$\wedge_{\mathrm{NaCl}}^{\infty}=229 \Omega^{-1} \mathrm{~cm}^2$ equiv $^{-1}$
$\wedge_{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}}^{\infty}=\wedge_{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}}^{\infty}+\wedge_{\mathrm{HCl}}^{\infty}-\wedge_{\mathrm{NaCl}}^{\infty}$
$=240+349-229$
$=360 \Omega^{-1} \mathrm{~cm}^2$ equiv $^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
At a certain temperature in a 5L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, $\mathrm{CO}+\mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2$. At equilibrium, if one mole of $\mathrm{CO}$ is present then equilibrium constant $\left(K_{\mathrm{c}}\right)$ for the reaction is:

Options:
A
$2.5$

B
4

C
2

D
3

The correct answer is:
$2.5$

Solution:
Initially 2 moles of $\mathrm{CO}$ are present.
At equilibrium, 1 mole of $\mathrm{CO}$ is present Hence, $2-1=1$ moles of $\mathrm{CO}$ has reacted. 1 mole of CO will react with 1 mole of $\mathrm{Cl}_2$ to form 1 mole of $\mathrm{COCl}_2$. $3-1=2$ moles of $\mathrm{Cl}_2$ remains at equilibrium The equilibrium constant
$$
K_{\mathrm{c}}=\frac{\left[\mathrm{COCl}_2\right]}{[\mathrm{CO}]\left[\mathrm{Cl}_2\right]}=\frac{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}}}{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}} \times \frac{2 \mathrm{~mol}}{5 \mathrm{~L}}}=2.5
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium constant is :

Options:
A 0.5
B 4.0
C 1.0
D 0.25

The correct answer is:
0.25

Solution:
2HIg⇌H2g+I2gMoles1-αα2α2Concentration1-αVα2Vα2V
KC=H2I2HI2=α2Vα2V1-αV2
∴α=0.50α2=0.25
=α241-α2=0.50×0.504×0.50×0.50=14
K C = 0.25

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2014 (09 Apr Online)
At a certain temperature, the dissociation constants of formic and acetic acids are 1.8×10-4 and 1.8×10-5 respectively. The concentration of acetic acid solution in which the hydrogen ion has the same concentration as in 0.001 M formic acid solutions is equal to

Options:
A 0.001 M
B 0.01 M
C 0.1 M
D 0.0001 M

The correct answer is:
0.01 M

Solution:
H+=Ka⋅C

For acetic acid H+=1.8×10−5×C

For Formic acid H+=1.8×10−4×10−3

1.8×10−5×C=1.8×10−4×10−3

C= 10 −2 =0.01 M

Topic: Chemistry, Ionic Equilibrium, JEE Main
At a constant pressure, what should be the percentage increase in the temperature, in Kelvin, for a \( 10 \% \) increase in the volume of a given mass of a gas?

Options:
A \( 5 \% \)
B \( 10 \% \)
C \( 20 \% \)
D \( 50 \% \)

The correct answer is:
\( 10 \% \)

Solution:
According to Charles' law, at a constant pressure, the volume of the gas is directly proportional to the temperature in Kelvin.

V∝T

At a constant pressure, the percentage increase in the temperature, in Kelvin, should be 10% for a 10% increase in the volume.

Topic: Chemistry, States of Matter, JEE Main
At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?
(Mol.âwt.âNaCl = 58.5,â H 2 SO 4 â=â98 .0 g mol â1 )

Options:
A 1 molal NaCl (aq)
B 1 molar NaCl (aq)
C 1 molal  H 2 SO 4 aq
D 1 molar  H 2 SO 4 aq

The correct answer is:
1 molal NaCl (aq)

Solution:
One molar (1 M) aqueous solution is more concentrated than one molal aqueous solution of the same solute. In solution, H2SO4 provides three ions. While NaCl provides two ions. Hence, vapour pressure of solution of NaCl is higher (as it gives less ions). Therefore, 1 molal NaCl will have the maximum vapour pressure.

Topic: Chemistry, Solutions, JEE Main
At a definite temperature, the equilibrium constant for a reaction, A + B â 2 C , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

Options:
A 9 1 1
B 1 1 1
C 2 1 1
D 7 1 1

The correct answer is:
9 1 1

Solution:
A + B â 2 C 1 1 0 1 - x 1 - x 2 x
â´ 8 1 = 4 x 2 1 - x 2
or 2 x 1 - x = 9
â´ x = 9 1 1
Thus, A + B â 2 C 2 1 1 2 1 1 18 1 1
( Total mole = 2 1 1 + 2 1 1 + 18 1 1 = 22 1 1 )
â´ Mole fraction of C= 18 1 1 2 2 1 1 = 9 1 1

Topic: Chemistry, Chemical Equilibrium, JEE Main
At a distance of $10 \mathrm{~cm}$ from a long straight wire carrying current, the magnetic field is $0.04 \mathrm{~T}$. At the distance of 40 $\mathrm{cm}$, the magnetic field will be

Options:
A $0.01 \mathrm{~T}$
B $0.02 T$
C $0.08 T$
D $0.16 \mathrm{~T}$

The correct answer is:
$0.01 \mathrm{~T}$

Solution:
$B \propto \frac{1}{r} \Rightarrow \frac{B_1}{B_2}=\frac{r_2}{r_1} \Rightarrow \frac{0.04}{B_2}=\frac{40}{10} \Rightarrow B_2=0.01 T$

Topic: Physics, Magnetic Effects of Current, JEE Main
At a given instant of time two particles are having the position vectors $4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+57 \hat{\mathbf{k}}$ metres and $2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ respectively. If the velocity of the first particle be $0.4 \hat{\mathbf{i}} \mathrm{ms}^{-1}$, the velocity of second particle in metre per second if they collide after $10 \mathrm{sec}$ is

Options:
A $6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
B $0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
C $6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
D $0.6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\frac{1}{3} \hat{\mathbf{k}}\right)$

The correct answer is:
$0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$

Solution:
$\begin{aligned} s_1+\left(u_1\right) t & =s_2+\left(u_2\right) t \\ & =(4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+(0.4 \hat{\mathbf{i}}) 10 \\ & =(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\left(u_2\right) 10 \\ \Rightarrow \quad u_2 & =\frac{(6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{10} \\ & =0.6\left[\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right]\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
At a given instant, say t=0, two radioactive substance A and B have equal activities. The ratio RBRA of their activities after time t itself decays with time t as e-3t. If the half-life of A is ln2, the half-life of B is:

Options:
A 2ln2
B 4ln2
C ln24
D ln22

The correct answer is:
ln24

Solution:
Given,
Half life of A=ln2
Initial number of nucleus is, R0A=R0B=R0

Half life is related with decay constant as, t1/2A=ln2λ
λA=1

At t=0 RA=RB
From activity relation, we have,
R0e-λAT=R0E-λBT

at T=t, ⇒RBRA=R0e-λBtR0e-λAt

As per given question, RBRA=e-3t, On comparing with above relation, we have
⇒e-λB-λAt=e-t
⇒λB-λA=3
⇒λB=3+1=4
⇒t1/2B=λn2λB=λn24

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
At a given point of time the value of displacement of a simple harmonic oscillator is given as y=A cos30°. If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant 1.0×10x N m–1. The value of x is _____.

Options:

The correct answer is:
4

Solution:
The formula to calculate the kinetic energy K of a particle executing SHM is given by

K=12kA2-y2 ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the value of the unknown parameter.

K=12×k×A2-A2cos230°= 12×k×A2sin230°⇒200=12×1.0×10x×0.42×122= 2×10x100⇒102=10x-2⇒x-2=2⇒x=4

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (13 Apr Shift 1)
At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as p=RTV-b at T. Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs p?

Options:
A Ne
B Kr
C Ar
D Xe

The correct answer is:
Xe

Solution:
As the b increases the slope increases. We know that b represents the effective size of the molecules. Since, Xe has the biggest size among given noble gases because on moving down the group size increases due to increase in shell, hence Xe will have the highest slope.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 2)