datavorous/entrance-exam-dataset huggingface dataset:
At a given temperature, the pressure of an ideal gas of density ${ }^\rho$ is proportional to

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

The correct answer is:
$\rho$

Solution:
$v_{r m s}=\sqrt{\frac{3 P}{\rho}}=P \propto \rho$
[ $\boldsymbol{V}_{\text {vous }}$ is constant for fixed temperature]

Topic: Physics, Kinetic Theory of Gases, JEE Main
At a metro station, a girl walks up a stationary escalator in $20 \mathrm{~s}$. If she remains stationary on the escalator, then the escalator take her up in $30 \mathrm{~s}$. The time taken by her to walk up on the moving escalator will be

Options:
A $25 \mathrm{~s}$
B $60 \mathrm{~s}$
C $12 \mathrm{~s}$
D $10 \mathrm{~s}$

The correct answer is:
$12 \mathrm{~s}$

Solution:
Let $h$ be the height of platform, then speed of walking $\left(v_{g}\right)=\frac{h}{20}$
Speed of escalator $\left(v_{e}\right)=\frac{h}{30}$
Time taken by girl when she walk up on the moving escalator is
$t=\frac{h}{v_{g}+v_{e}}=\frac{h}{\frac{h}{20}+\frac{h}{30}}=\frac{20 \times 30}{(30+20)}=12 \mathrm{~s}$

Topic: Physics, Motion In One Dimension, KCET, KCET 2020
At a metro station, a girl walks up a stationary escalator in time $t_1$. If she remains stationary on the escalator, then the escalator take her up in time $t_2$. The time taken by her to walk up on the moving escalator will be

Options:
A
$\frac{t_1+t_2}{2}$

B
$\frac{t_1 t_2}{t_2-t_1}$

C
$\frac{t_1 t_2}{t_2+t_1}$

D
$t_1-t_2$

The correct answer is:
$\frac{t_1 t_2}{t_2+t_1}$

Solution:
Let us consider, displacement is $L$, then velocity of girl with respect to ground,
$$
v_g=\frac{L}{t_1}
$$
Velocity of escalator with respect to ground,
$$
v_e=\frac{L}{t_2}
$$
Net velocity of the girl on moving escalator with respect to ground
$$
=v_g+v_e=\frac{L}{t_1}+\frac{L}{t_2} \Rightarrow v_{g e}=L\left[\frac{t_1+t_2}{t_1 t_2}\right]
$$
Now, if $t$ is total time taken by girl on moving escalator in covering distance $L$, then
$$
t=\frac{\text { distance }}{\text { speed }}=\frac{L}{L\left(\frac{t_1+t_2}{t_1 t_2}\right)}=\frac{t_1 t_2}{t_1+t_2}
$$

Topic: Physics, Motion In One Dimension
At a particular station, the TV transmission tower has a height of 100 m. To triple its coverage range, height of the tower should be increased to

Options:
A 200 m
B 300 m
C 600 m
D 900 m

The correct answer is:
900 m

Solution:
Let d be range of TV transmission.

From above figure,

d2=h+R2-R2

⇒d2=h2+R2+2RH-R2

⇒d2=h2+2Rh

As R≫≫h then, d≈2Rh ...1

Now, if coverage is to be increased 3 times

3d=2Rh' ...2

Dividing equation 2 by 1, we have

3d d=2Rh'2Rh

⇒9=h'h

9h=h'

If h=100 m, then tower of height 900 m is required.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (27 Jul Shift 1)
At a place the true value of angle of dip is 60°. If dip circle is rotated by ϕ° from magnetic meridian, the angle of dip is found to be tan-1⁡(2). Then the value of ϕ is

Options:
A 45°
B 15°
C 60°
D 30°

The correct answer is:
30°

Solution:
tan⁡δ′=tan⁡δcos⁡xo
tan⁡tan-1⁡2=tan⁡60ocos⁡xo
Or, cos⁡x°=32
∴x=30°

Topic: Physics, Magnetic Properties of Matter, JEE Main
At a place where the acceleration due to gravity is ${ }^{10 \mathrm{msec}^{-2}}$ a force of $5 \mathrm{~kg}-\mathrm{wt}$ acts on a body of mass $10 \mathrm{~kg}$ initially at rest. The velocity of the body after 4 second is

Options:
A $5 m \mathrm{mec}^{-1}$
B $10 \mathrm{msec}^{-1}$
C $20 \mathrm{msec}^{-1}$
D $50 \mathrm{msec}^{-1}$

The correct answer is:
$20 \mathrm{msec}^{-1}$

Solution:
$\text { Acceleration }=\frac{\text { Force }}{\text { Mass }}=\frac{50 \mathrm{~N}}{10 \mathrm{~kg}}=5 \mathrm{~m} / \mathrm{s}^2$
From $v=u+a t=0+5 \times 4=20 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main
At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit
(i) electrons
(ii) protons
(iii) $\mathrm{He}^{2+}$
(iv) neutrons
The emission at instant can be

Options:
A
i, ii, iii

B
i, ii, iii, iv

C
iv

D
ii, iii

The correct answer is:
i, ii, iii

Solution:
Neutrons can’t be deflected by a magnetic field

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2002
At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

Options:
A
$\frac{6}{5^{\mathrm{e}}}$

B
$\frac{5}{6}$

C
$\frac{6}{55}$

D
$\frac{6}{e^5}$

The correct answer is:
$\frac{6}{e^5}$

Solution:
$P(X=r)=\frac{e^{-m} m^r}{r !}$
$P(X \leq 1)=P(X=0)+P(X=1)$
$=e^{-5}+5 \times e^{-5}=\frac{6}{e^5}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2006
At absolute zero, Si acts as

Options:
A
non metal

B
metal

C
insulator

D
none of these

The correct answer is:
insulator

Solution:
Semiconductors are insulators at low temperature

Topic: Physics, Semiconductors, JEE Main, JEE Main 2002
At an angle of 30° to the magnetic meridian, the apparent dip is 45°. Find the true dip:

Options:
A tan-13
B tan-113
C tan-123
D tan-132

The correct answer is:
tan-132

Solution:
Atanδ=tanδ'cosθ

=tan45°cos30°

tanδ=1×32

δ=tan-132

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (20 Jul Shift 2)
At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is

Options:
A
5040

B
6210

C
385

D
1110

The correct answer is:
385

Solution:
${ }^0 C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{10} C_4$
$=10+45+120+210=385$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2006
At an election, a voter may vote for any number of candidates not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote, is

Options:
A 6210
B 385
C 1110
D 5040

The correct answer is:
385

Solution:
Total number of ways
= 10 C 1 + 10 C 2 + 10 C 3 + 10 C 4
=10+45+120+210=385

Topic: Mathematics, Permutation Combination, JEE Main
At an incident radiation frequency of v1, which is greater than the threshold frequency, the stopping potential for a certain metal is V1. At frequency 2v1 the stopping potential is 3V1. If the stopping potential at frequency 4v1 is nV1, then n is

Options:
A 2
B 3
C 6
D 7

The correct answer is:
7

Solution:
We know,

hv=ϕ+K.E

From the given data,

hv1=ϕ+eV1

h2v1=ϕ+3eV1

On dividing 1 by 2, 12=ϕ+eV1ϕ+3eV1

⇒ ϕ+3eV1=2ϕ+2ev1

∴ ϕ=eV1

From 1

hv1=eV1+eV1

⇒ hv1=2eV1

Given, h4v1=ϕ+nV1

⇒ 42eV1=eV1+nV1

7eV1=nV1

∴ n=7

Topic: Physics, Dual Nature of Matter, JEE Main
At an instant, a plane electromagnetic wave has its magnetic field in the direction of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and its electric field is in the direction of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. The wave is travelling along which direction?

Options:
A $+x$-direction
B $-x$-direction
C +z-direction
D -z-direction

The correct answer is:
-z-direction

Solution:
Direction of magnetic field,
$\hat{\mathbf{B}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$
Direction of electric field,
$\hat{\mathbf{E}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$
Direction of propagation of EM wave is given according to poynting vector as
$\hat{\mathbf{n}}=\hat{\mathbf{E}} \times \hat{\mathbf{B}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})$
$=\hat{\mathbf{i}} \times \hat{\mathbf{i}}+\hat{\mathbf{j}} \times \hat{\mathbf{i}}-\hat{\mathbf{i}} \times \hat{\mathbf{j}}-\hat{\mathbf{j}} \times \hat{\mathbf{j}}$
$=0+(-\hat{\mathbf{k}})-\hat{\mathbf{k}}-0=-2 \hat{\mathbf{k}}$
Hence, electromagnetic wave will travel along $-z$-direction.

Topic: Physics, Electromagnetic Waves, JEE Main
At any instant the velocity of a particle of mass 500 g is 2t i^ + 3t2 j^ m s-1 . If the force acting on the particle at t=1 s is (i^ + x j^) N . Then the value of x will be:

Options:
A 3
B 4
C 2
D 6

The correct answer is:
3

Solution:
The equation for velocity is

v→=2t i^ + 3t2 j^

Thus, the acceleration is

a→=dv→dt=d(2t i^+3t2 j^)dt=2 i^+6t j^

At t=1 s, a→=2 i^ + 6 j^

The force is written as F→=ma→=2 i^+6 j^2=i^ +3 j^

Comparing it with the given value of force, x=3.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (08 Apr Shift 1)
At constant pressure, addition of helium to the reaction system :

N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)

Options:
A Favours the formation of ammonia
B Reduces the formation of ammonia
C Does not affect the position of equilibrium
D Reduces the dissociation of ammonia.

The correct answer is:
Reduces the formation of ammonia

Solution:
Kp =n2NH3nN2×nH23× PΣnΔn
At constant pressure, with the introduction inert gas Σn increases. Since Δn=–2, therefore on increasing Σn PΣnΔn increases, but Kp remain constant, hence no. of
moles of N2 g & H2 g are increased and backward reaction is favoured.

Topic: Chemistry, Chemical Equilibrium, NEET
At constant T and P, 5.0 L of SO2 are reacted with 3.0 L of O2 according to the following equation, 2  SO 2 (g) +  O 2 (g)→2  SO 3 (g) . The volume of the reaction mixture at the completion of the reaction is

Options:
A 0.5 L
B 8.0 L
C 5.5 L
D 5 L

The correct answer is:
5.5 L

Solution:
Total volume of reaction mixture =0.5+5=5.5 L

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
At constant $T$ and $P$, which one of the following statements is correct for the reaction, $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$

Options:
A $\Delta H$ is independent of the physical state of the reactants of that compound
B $\Delta H\gt\Delta E$
C $\Delta H \lt \Delta E$
D $\Delta H=\Delta E$

The correct answer is:
$\Delta H \lt \Delta E$

Solution:
$\Delta n_g=1-\frac{3}{2}=\frac{-1}{2}$, As $\Delta n_g$ is negative, thus $\Delta H \lt \Delta E$

Topic: Chemistry, Thermodynamics (C), JEE Main
At constant temperature, a gas is at a pressure of 940.3 mm Hg. The pressure at which its volume decreases by 40% is ______ mm Hg. (Nearest integer)

Options:

The correct answer is:
1567

Solution:
From Boyle's Law,

P2V2=P1V1

P1 =940.3 mm of Hg

P2 = Pressure at which volume is reduced by 40%

V2 = Volume decreased by 40%

V2=V1-0.4 V1=0.6 V1

Then,P2=940.3×V10.6V1

=1567.16

≈1567 mm Hg

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2023 (10 Apr Shift 1)
At constant temperature, the equilibrium constant (Kp) for the decomposition reaction N2O4(g)â2NO2(g) is expressed by Kp=(4x2P)(1-x2), where P = pressure, x = extent of decomposition. Which one of the following statements is true?

Options:
A Kp remains constant with change in P and x
B Kp increases with decrease of x
C Kp increases with increase of x
D Kp increases with increase of P

The correct answer is:
Kp remains constant with change in P and x

Solution:
Kp can be written as:

K p = NO 2 2 N 2 O 4

With change of pressure, x will change in such a way that Kp remains a constant.

Topic: Chemistry, Chemical Equilibrium, NEET
At constant temperature, the pressure of gas is decreased by $20 \%$. The percentage change in volume

Options:
A increase by $29 \%$
B decreases by $25 \%$
C increases by $25 \%$
D decreases by $20 \%$

The correct answer is:
decreases by $25 \%$

Solution:
Ideal gas equation states that: $P V=n R T$
As temperature remains constant, so $P V=$ constant
As pressure decreases by $20 \%$, so $P_2=0.8 P$
$\begin{aligned} & \Rightarrow P V=(0.8 P) V_2 \\ & \Rightarrow V_2=\frac{P V}{0.8 P}=\frac{V}{0.8}=1.25 \mathrm{~V}\end{aligned}$
So, volume of gas increase by $25 \%$

Topic: Physics, Thermodynamics, MHT CET, MHT CET 2022 (10 Aug Shift 1)
At constant volume, 4 mol of an ideal gas when heated from 300K to 500K changes its internal energy by 5000J. The molar heat capacity at constant volume is ________

Options:

The correct answer is:
6.25

Solution:
For ideal gas :
ΔU=nCVT2−T1
⇒ 5000=4× C V [500−300]
⇒CV=5000800=6.25 J mole −1K−1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (08 Jan Shift 2)
At constant volume, \( 4 \mathrm{~mol} \) of an ideal gas when heated from \( 300 \mathrm{~K} \) to \( 500 \mathrm{~K} \) changes its internal energy by \( 5000 \mathrm{~J} \). The molar heat capacity at constant volume is

Options:

The correct answer is:
6.25

Solution:
For ideal gas :
ΔU=nCVT2−T1
⇒ 5000=4× C V [500−300]
⇒CV=5000800=6.25 J mole −1K−1

Topic: Chemistry, Thermodynamics (C), JEE Main
At constant volume the specific heat of a gas is $\frac{3 R}{2}$, then the value of ' $\gamma$ ' will be

Options:
A $\frac{3}{2}$
B $\frac{5}{2}$
C $\frac{5}{3}$
D None of the above

The correct answer is:
$\frac{5}{3}$

Solution:
$C_V=\frac{R}{(\gamma-1)} \Rightarrow$ $\gamma=1+\frac{R}{C_V}=1+\frac{R}{\frac{3}{2} R}=\frac{5}{3}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
At high temperature, the following reaction occurs
4CuOs⇌2Cu2Os+O2g
Consider the following statements regarding reaction
1. Kp=pO2
2. Kp depends upon the amounts of CuO and Cu2O
3. Kp<<Kc
4. Kp>Kc
Of the statements

Options:
A 1 and 2 are correct
B 1, 2 and 3 are correct
C 1 and 4 are correct
D 2 and 3 are correct

The correct answer is:
1 and 4 are correct

Solution:
Kp=pO2 is correct.
O2g∝pO2
Kp=Kc(RT)Δng
Δng= Number of mol gaseous (Products – Reactants)
Here Δng=1
So Kp>Kc

Topic: Chemistry, Chemical Equilibrium, JEE Main
At infinite dilution, the percentage ionization for both strong and weak electrolytes is

Options:
A 1%
B 20%
C 50%
D 100%

The correct answer is:
100%

Solution:
At infinite dilution, both weak and strong electrolyte are completely ionized.
Hence, the percentage of ionization is 100%.

Topic: Chemistry, Ionic Equilibrium, JEE Main
At low pressure, the van der Waals' equation is reduced to

Options:
A Z =  PV m RT  = 1− aP RT
B Z =  PV m RT  = 1 +  bP RT
C PV m  = RT
D Z =  PV m RT  = 1− a V m RT

The correct answer is:
Z =  PV m RT  = 1− a V m RT

Solution:
Van der Waals' equation (at low pressure),
Pâ+â a V 2 Vâb â=âRT
at low pressure V m will be high
∴ b can be neglected Pâ+â a V m 2 â V m â=âRT
PV m â=âRTâ a V m
â´ Zâ=â PV m RT â=â1â a V m RT

Topic: Chemistry, States of Matter, JEE Main
At low pressure, the Vander Waal's equation is reduced to

Options:
A $Z=\frac{p V_m}{R T}=1-\frac{a p}{R T}$
B $Z=\frac{p V_m}{R T}=1+\frac{b}{R T} p$
C $p V_m=R T$
D $Z=\frac{p V_m}{R T}=1-\frac{a}{R T}$

The correct answer is:
$Z=\frac{p V_m}{R T}=1-\frac{a p}{R T}$

Solution:
When pressure is low
$\left[P+\frac{a}{V^2}\right](V-b)=R T$
or $P V=R T+P b-\frac{a}{V}+\frac{a b}{V^2}$ or $\frac{P V}{R T}=1-\frac{a}{V R T}$
$Z=-\frac{a}{V R T}\left(\because \frac{P V}{R T}=Z\right)$

Topic: Chemistry, States of Matter, JEE Main
At menopause there is rise in urinary excretion of:

Options:
A Estrogen
B Progesterone
C MSH
D FSH

The correct answer is:
FSH

Solution:
Menopause is the time in a woman's life when monthly cycles of menstruation cease forever and the level of hormones (estrogens) produced by the ovaries decreases. Menopause usually occurs in the late 40s or early 50s. When estrogen production falls a critical value, estrogens can no longer inhibit the production of the gonadotropins. FSH and LH. Instead, the gonadotropins FSH and LH (mainly FSH) are produced after menopause in large and continuous quantities and these (mainly FSH) are excreted out in the urine.

Topic: Biology, Human Reproduction, NEET
At metaphase, chromosomes are attached to the spindle fibres by their

Options:
A satellites
B secondary constrictions
C kinetochores
D centromere

The correct answer is:
kinetochores

Solution:
Kinetochores are small disc-shaped proteinous structures present at the surface of the centromere. These structures serve as the sites of attachment of spindle fibres to the chromosomes that are moved into position at the centre of the cell.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2011 (Mains)
At NTP the density of a gas is \( 1.3 \mathrm{~kg} / \mathrm{m}^{3} \) and the velocity of sound propagation in the gas is \( 330 \mathrm{~m} / \mathrm{s} \). The degree of freedom of gas molecule is-

Options:
A \( 3 \)
B \( 5 \)
C \( 6 \)
D \( 7 \)

The correct answer is:
\( 5 \)

Solution:
V=γPρ
330=γ×1×1051.3
332×100×1.31×105=γ
1.089×103×102×1.31×105=γ
2f+1=γ=1.4=75
2f=25
f=5

Topic: Physics, Kinetic Theory of Gases, JEE Main
At ordinary temperatures, the electrical conductivity of semi conductors in mho/meter is in the range

Options:
A $10^{-3}$ to $10^{-4}$
B $10^6$ to $10^9$
C $10^{-6}$ to $10^{-10}$
D $10^{-10}$ to $10^{-16}$

The correct answer is:
$10^6$ to $10^9$

Solution:
A ordinary temperatures, the electrical conductivity of semi conductors in mho/meters is in the range
$10^6 \text { to } 10^9 \text {. }$

Topic: Physics, Semiconductors, JEE Main
At pH=2,EQuinhydrone0=1·30V,EQuinhydrone will be

Options:
A 1·36 V
B 1Â·32 V
C 1Â·42 V
D 1Â·26 V

The correct answer is:
1Â·42 V

Solution:
E=E0-0.0592logH+2=1Â·30-0.0592log10-22
=1Â·30+0.2362=1Â·418 V

Topic: Chemistry, Electrochemistry, JEE Main
At present, a firm is manufacturing 2 00 0 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dPdx=100-12x. If the firm employs 2 5 more workers, then the new level of production of items is

Options:
A 3500
B 4500
C 2500
D 3000

The correct answer is:
3500

Solution:
dPdx=100-12x

⇒∫dP=∫100-12xdx

⇒P=100x-8x3/2+C

Now, x=0

⇒P=2000⇒C=2000

hence, P=100x-8x32+2000

∵ x=25

P=2500-1000+2000

P=3500

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2013 (07 Apr)
At present, a firm is manufacturing 2 00 0 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dP dx = 1 0 0 - 1 2 x . If the firm employs 2 5 more workers, then the new level of production of items is

Options:
A 3500
B 4500
C 2500
D 3000

The correct answer is:
3500

Solution:
dP dx = 1 0 0 - 1 2 x
â â« dP = â« 1 0 0 - 1 2 x dx
â P = 1 0 0 x - 8 x 3 /2 + C
Now, x=0
âP=2000 â C=2000
hence, P=100x-8x32+2000
âµ x=25
P=2500-1000+2000
P=3500

Topic: Mathematics, Differential Equations, JEE Main
At puberty, a female has more than 400,000 immature egg cells in her ovaries. Calculate the percentage of these eggs that will ovulate, assuming that 13 ovarian cycles occur per year in her reproductive life span of 40 years.

Options:
A 52%
B 0.13%
C 5.2%
D 13%

The correct answer is:
0.13%

Solution:
At the time of birth, there will be approximately 1 million eggs in the ovary of the female child. By the time of puberty, only about 300,000 remains and other undergo atresia. Around 300 to 400 only will be ovulated during a woman's reproductive lifetime. Hence, the percentage will be around 0.1 to 0.2.

Topic: Biology, Human Reproduction, NEET
At puberty, only

primary follicles remain are left in each ovary.

Options:
A 200-300 million
B 60000-80000
C 1 million
D 6000-8000

The correct answer is:
60000-80000

Solution:
A large number of primary follicles degenerate during the phase from birth to puberty. Therefore, at puberty, only 60,000-80,000 primary follicles are left in each ovary. The primary follicles get surrounded by more layers of granulosa cells and a new theca and are called secondary follicles.

Topic: Biology, Human Reproduction, NEET
At room temperature $\left(27^{\circ} \mathrm{C}\right)$, the resistance of a heating element is $50 \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. The temperature of the element, when its resistance is $62 \Omega$, is _____ ${ }^{\circ} \mathrm{C}$.

Options:

The correct answer is:
1027

Solution:
$\begin{aligned} & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & 62=50\left[1+2.4 \times 10^{-4} \Delta \mathrm{T}\right] \\ & \Delta \mathrm{T}=1000^{\circ} \mathrm{C} \\ & \Rightarrow \mathrm{T}-27^{\circ}=1000^{\circ} \mathrm{C} \\ & \mathrm{T}=1027^{\circ} \mathrm{C}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2024 (09 Apr Shift 2)
At room temperature a diatomic gas is found to have an r.m.s. speed of $1930 \mathrm{~ms}^{-1}$. The gas is:

Options:
A
$\mathrm{H}_2$

B
$\mathrm{Cl}_2$

C
$\mathrm{O}_2$

D
$\mathrm{F}_2$

The correct answer is:
$\mathrm{H}_2$

Solution:
$$
\begin{aligned}
&\because \quad \mathrm{C}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
&\left(1930^2\right)=\frac{3 \times 8.314 \times 300}{\mathrm{M}} \\
&\mathrm{M}=\frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 \times 10^{-3} \mathrm{~kg}
\end{aligned}
$$
The gas is $\mathrm{H}_2$.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2014 (12 Apr Online)
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg , lowering of vapour pressure will be:
(molar mass of urea =60 g mol-1 )

Options:
A 0.028 mm Hg
B 0.027 mm Hg
C 0.031 mm Hg
D 0.017 mm Hg

The correct answer is:
0.017 mm Hg

Solution:
Relative lowering in vapour Pressure
⇒ Mole of urea nB=0.660=10-2 mole ;

Mole of water nA=36018=20

Po-PsPo=xB=Po-PsPo=nBnn+nB

Here =Po=V.P. of pure solvent

Ps=V.P of solution.

nA+nB≃nA

Po-PsPo=nBnA
lowering of vapour pressure(Po-Ps)=nBnA×Po

Po-Ps=10-220×35

Po-Ps=0.0175  mm of Hg

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (10 Apr Shift 1)
At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance. Molal elevation constant (Kb) for water 0.52 K kg mol-1 ?

Options:
A 53
B 65
C 84
D 79

The correct answer is:
53

Solution:
ΔTb=kb×wM×1000W
M=Kb×w×1000ΔTb×W
MB=Kb×WB×1000ΔTb×WA
According to available data
Mass of solute WB=12.5 g
Mass of solute WA=175 g
Molal elevation constant Kb=0.52 K kg mol-1
Elevation in boiling point temperature ΔTb=0.70 K
∴M=0.52 K kg mol-1×12.5 g×10000.70 K×175 g
=53 g mol-1

Topic: Chemistry, Solutions, JEE Main
At some instant, a radioactive sample S1 having an activity 5 Î¼Ci has twice the number of nuclei as another sample S2 which has an activity of 10 Î¼Ci . The ratio of half lives of S1 and S2 is

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

The correct answer is:
4:1

Solution:
Given: N 1 =2. N 2
Now, activity A1=λ1N1=5μCi .....(1)
and A2=λ2N2=10μCi .....(2)
Dividing ( 1 ) ( 2 ) we get
λ 1 λ 2 . N 1 N 2 = 5 10
⇒ t 2 t 1 .2= 1 2 ⇒ t 1 t 2 = 4 1

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2018 (16 Apr Online)
At some location the horizontal component of earth's magnetic field is 18×10-6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angles with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

Options:
A 1.8×10-5 N
B 3.6×10-5 N
C 6.5×10-5 N
D 1.3×10-5 N

The correct answer is:
6.5×10-5 N

Solution:
The horizontal and vertical components of earth's magnetic field (BH and BV) are related as

BVBH=tanθ

Here, θ=45° and BH=18×10-6 T

⇒BV=BHtan45°

⇒BV=BH=18×10-6 T ∵ tan45°=1

Now, when the external force F is applied, to keep the needle stays in horizontal position is shown below,

Taking torque at point P, we get

mBV×2l=Fl

∴ F=2×mBV

Substituting the given values, we get

F=2×1.8×18×10-6

F=6.48×10-5=6.5×10-5 N.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (10 Jan Shift 2)
At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation
2H2Sg+SO2g→2H2Ol+3Ss
Which of the following shows the volume of the gas remaining after the reaction?

Options:
A 0.2 litres of SO2g
B 0.4 litres of H2g
C 1.2 litres of H2Sg
D 1.2 litres of SO2g

The correct answer is:
0.2 litres of SO2g

Solution:
So, Sol 2 vol  H 2 S≡1 vol  SO 2
or 2.8 L  H 2 S = 1.4 L  SO 2
Hence, after reaction, volume of remaining SO2
=1.6−1.4 L
=0.2 L

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main