datavorous/entrance-exam-dataset huggingface dataset:
A weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $\mathrm{NaOH}$ solution. The $\mathrm{pH}$ at the point of one-third neutralisation of the acid will be

Options:
A $5+\log 2-\log 3$
B $5-\log 2$
C $5-\log 3$
D $5-\log 6$

The correct answer is:
$5-\log 2$

Solution:
Hints: $\mathrm{K}_{\mathrm{a}}=10^{-5} \Rightarrow \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=-\log 10^{-5}=5$
$$
\begin{array}{llcc}
& \mathrm{HA}+\mathrm{NaOH} \longrightarrow \mathrm{NaA}+\mathrm{H}_2 \mathrm{O} \\
\text { Initial } & 1 \mathrm{~mole} 0 & 0 & 0 \\
\text { Final } & (1-1 / 3) \mathrm{mole} & 1 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole} \\
& =2 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole}
\end{array}
$$
(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided)
Final solution acts as an acidic buffer.
$$
\Rightarrow \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { Acid }]} \Rightarrow \mathrm{pH}=5+\log \frac{\frac{1}{3}}{\frac{2}{3}}=5+\log \frac{1}{2} \Rightarrow \mathrm{pH}=5-\log 2
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
A weak field complex of Ni2+ has magnetic moment of 2.82 BM. The number of electron in the t2g level of Ni2+ will be

Options:

The correct answer is:
6

Solution:
Ni2+=Ar3d8
and as μ=2.82 BM,
So n = 2
Thus, t2g6eg2 configuration is present here.
Hence number of electrons in t2g are 6.

Topic: Chemistry, Coordination Compounds, JEE Main
A weather balloon filled with hydrogen at $1 \mathrm{otm}$ and $27^{\circ} \mathrm{C}$ has volume equal to 12000 litres, On ascending it reaches a place where the temperature is $-23^{\circ} \mathrm{C}$ and pressure is $0.5 \mathrm{~atm}$. The volume of the balloon is

Options:
A 24000 litres
B 20000 litres
C 10000 litres
D 12000 litres

The correct answer is:
20000 litres

Solution:
$V_2=\frac{P_1}{P_2} \frac{T_2}{T_1} \cdot V_1=\frac{1}{0.5} \times \frac{250}{300} \times 12000 \mathrm{lit} .=20000 \mathrm{lit}$.

Topic: Chemistry, States of Matter, JEE Main
A wedge of mass M=4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:

Options:
A v2g
B v22g
C 2v25g
D 2v27g

The correct answer is:
2v25g

Solution:
From conservation of linear momentum,

mv=4m+mvf
vf=v5
From conservation of mechanical energy,
12mv2=124m+mvf2+mgh
v2=5vf2+2gh
v2=5×v52+2gh
h=2v25g

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A wedge of mass m, lying on a rough horizontal plane, is acted upon by a horizontal force F1 and another force F2, inclined at an angle Î¸ to the vertical. The block is in equilibrium, then the minimum coefficient of friction between it and the surface is

Options:
A F2sinθ(mg+F2cosθ)
B (F1cosθ+F2)mg-F2sinθ
C (F1+F2sinÎ¸)(mg+F2cosÎ¸)
D (F1sinÎ¸-F2)(mg-F2cosÎ¸)

The correct answer is:
(F1+F2sinθ)(mg+F2cosθ)

Solution:
We know f=Î¼N and N=mg+F2cosÎ¸

Now, since wedge is in equilibrium,

(F1+F2sinÎ¸) = Î¼(mg+F2cosÎ¸)

Î¼=F1+F2sinÎ¸mg+F2cosÎ¸

Topic: Physics, Laws of Motion, JEE Main
A weight lifter lifts $300 \mathrm{~kg}$ from the ground to a height of 2 meter in 3 second. The average power generated by him is

Options:
A 5880 watt
B 4410 watt
C 2205 watt
D 1960 watt

The correct answer is:
1960 watt

Solution:
$P=\frac{\text { Workdone }}{\text { Time }}=\frac{m g h}{t}=\frac{300 \times 9.8 \times 2}{3}=1960 \mathrm{~W}$

Topic: Physics, Work Power Energy, JEE Main
A weighted coin with the probability of showing a head is 23 is tossed. If head appears, then a number is selected from the first 20 natural numbers otherwise a number is selected from the first 9 natural numbers. The probability of getting an even number is

Options:
A 12
B 1327
C 1027
D 513

The correct answer is:
1327

Solution:
PH=23,PT=13
A= the selected number is an even number
PA=PHPAH+PT⋅PAT
=23×1020+13×49=13+427=1327

Topic: Mathematics, Probability, JEE Main
A weightless rigid rod with a small iron bob at the end is hinged at point A by a ball and socket joint to the wall so that it can rotate in all directions. The rod is kept in the horizontal position by a vertical inextensible string of length 20 cm, fixed at its midpoint. The bob is displaced slightly, perpendicular to the plane of the figure. If the period of small oscillations of the system is xÏ10 s, then what is the value of x? [Take g=10 m s-2]

Options:

The correct answer is:
4

Solution:
The small oscillation of the bob occur relative to a stationary axis passing through AB. The line CD is perpendicular to the axis AB. The bob behaves as a simple pendulum of length CD oscillating due to component of its weight mgcosθ.

If the length AD is equal to L, then the length CD=Lsinθ.

T=2πLsinθg cosθ

T=2πLtanθg

T=2π2lg =2π5=4π10

⇒x=4

Topic: Physics, Oscillations, JEE Main
A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A. Then horizontal velocity which must be imparted to end C of the rod to deflect it to the horizontal position is

Options:
A 125gl
B 3gl
C 165gl
D 2gl

The correct answer is:
125gl

Solution:
Loss in kinetic energy = gain in potential energy
∴12Iω2=mgl2+mgl ...(i)
Here, I=ml24+ml2=54ml2
From (i),
1254ml2ω2=mgl2+mgl=32mgl
ml2ω2=125mgl
∴vc=12gl5

Topic: Physics, Rotational Motion, JEE Main
A weightless string can support a tension up to $30 \mathrm{~N}$. A stone of mass $0.5 \mathrm{~kg}$ is tied to its one end and is revolved in a circular path of radius $2 \mathrm{~m}$ in a vertical plane. Then the maximum angular velocity of the stone will be (acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $10 \mathrm{rad} / \mathrm{s}$
B $\sqrt{60} \mathrm{rad} / \mathrm{s}$
C $\sqrt{30} \mathrm{rad} / \mathrm{s}$
D $5 \mathrm{rad} / \mathrm{s}$

The correct answer is:
$5 \mathrm{rad} / \mathrm{s}$

Solution:
Considering force balance:
$\begin{aligned} & T_{\max }=m \omega_{\max }^2 r+m g \\ & \Rightarrow \frac{T_{\text {max }}}{m}=\omega_{\max }^2 r+g \\ & \Rightarrow \frac{30 \mathrm{~N}}{0.5 \mathrm{~kg}}-10 \mathrm{~m} / \mathrm{s}^2=\omega_{\max }^2 r \\ & \Rightarrow \omega_{\max }=\sqrt{\frac{50}{2}}=5 \mathrm{rad} / \mathrm{s}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A well stoppered thermos flask contains some ice cubes. This is an example of a

Options:
A Closed system
B Open system
C Isolated system
D Non-thermodynamic system

The correct answer is:
Isolated system

Solution:
Isolated system

Topic: Chemistry, Thermodynamics (C), JEE Main
(a) What is the largest average velocity of blood flow in an artery of radius $2 \times 10^{-3} \mathrm{~m}$ if the flow must remain laminar?
(b) What is the corresponding flow rate? Take viscosity of blood to be $2.084 \times 10^{-3}$ pa-s. Density of blood is $1.06 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$.

Options:

None

Solution:
Given, $r=2 \times 10^{-3} \mathrm{~m}, \mathrm{D}=2 \mathrm{r}=2 \times 2 \times 10^{-3} \mathrm{~m}$ $=4 \times 10 \mathrm{~m}^{-3}$
$\eta=2.084 \times 10^{-3} \mathrm{~Pa}-\mathrm{s}, \rho=1.06 \times 10^3 \mathrm{kgm}^{-3}$
(a) Now, $\mathrm{V}_{\mathrm{e}}=\mathrm{N}_{\mathrm{R}} \eta / \rho \mathrm{D}$
$$
\begin{array}{r}
=\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^{-3}\right) \times\left(4 \times 10^{-3}\right)} \\
=0.98 \mathrm{~m} / \mathrm{s}
\end{array}
$$
(b) Volume flowing per second
$$
\begin{aligned}
&=\pi \mathrm{r}^2 \mathrm{~V}_{\mathrm{e}} \\
&=\frac{22}{7} \times\left(2 \times 10^{-3}\right)^2 \times 0.98=1.23 \times 10^{-5}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids
(a) What would happen if a large volume of untreated sewage is discharged into a river?
(b) In what way anaerobic sludge diagestion is important in sewage treatments?

Options:

None

Solution:
(a) Sewage, if untreated, and discharged directly intorivers it will lead to severe pollution of the waters with organic matter and pathogenic bacteria protozoa and many other disease causing organisms. This water will cause outbreaks of water borne diseases upon use.
(b) Anaerobic bacteria, digest the aerobic bacteria and the fungi in the sludge and the remaining organic matter in anaerobic sludge digestion. Bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide during this digestion. These gases (biogas) is inflammable and can be used as source of energy.

Topic: Biology, Microbes in Human Welfare
A Wheatstone bridge has the resistances $10 \Omega$. $10 \Omega, 10 \Omega$ and $30 \Omega$ in its four arms. What resistance joined in parallel to the $30 \Omega$ resistance will bring it to the balanced condition?

Options:
A $2 \Omega$
B $5 \Omega$
C $10 \Omega$
D $15 \Omega$

The correct answer is:
$15 \Omega$

Solution:
In balanced Wheatstone bridge,
$$
\begin{array}{l}
\frac{1}{10}=\frac{1}{x}+\frac{1}{30} \\
x=15 \Omega
\end{array}
$$

Topic: Physics, Current Electricity, JEE Main
A wheel has angular acceleration of 3.0 $\mathrm{rad} / \mathrm{sec}^2$ and an initial angular speed of $2~ \mathrm{rad}/ \mathrm{sec}$. In a time of $2 \mathrm{sec}$ it has rotated through an angle (in radian) of:

Options:
A 10
B 12
C 4
D 6

The correct answer is:
10

Solution:
Here $\theta=\omega t+\frac{1}{2} t^2$
$\begin{aligned}
& =(2)(2)+\frac{1}{2}(3)(2)^2 \\
& =4+6=10 \mathrm{rad}
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
âA wheel having mass m has charges +q and -q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of a uniform vertical electric field E. The value of E is

Options:
A mgq
B mg2q
C mgtanθ2q
D none

The correct answer is:
mg2q

Solution:
For rotational equilibrium, â Ï P =0
âmgacosÎ± âqE(2acos ) = 0

â E= mg 2q

Topic: Physics, Electrostatics, JEE Main
A wheel having moment of inertia $2 \mathrm{~kg}$ $\mathrm{m}^2$ about its vertical axis, rotates at the rate of $60 \mathrm{rpm}$ about the axis. The torque which can stop the wheel's rotation in one minute would be:

Options:
A $\frac{2 \pi}{15} \mathrm{~N}-\mathrm{m}$
B $\frac{\pi}{12} \mathrm{~N}-\mathrm{m}$
C $\frac{\pi 5}{15} \mathrm{~N}-\mathrm{m}$
D $\frac{\pi}{18} \mathrm{~N}-\mathrm{m}$

The correct answer is:
$\frac{\pi 5}{15} \mathrm{~N}-\mathrm{m}$

Solution:
Given, $\mathrm{I}=2 \mathrm{~kg}-m^2$,
$\omega_0=\frac{60}{60} \times 2 \pi \mathrm{rad} / \mathrm{s}$
$\omega_0$ and $t=60 \mathrm{~s}$
The torque required to stop the wheel's rotation is
$\begin{aligned}
\tau & =\mathrm{I} \propto l\left(\frac{\omega_0-\omega}{t}\right) \\
\tau & =\frac{2 \times 2 \pi \times 60}{60 \times 60} \\
& =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal the acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Options:

None

Solution:
Wheel is a rigid body. It is in uniform motion about axis passing through its centre and perpendicular to the plane of wheel.
The particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. That acceleration arises due to internal elastic forces, which cancel out in pairs.
In a half wheel, the distribution of mass (of half wheel) about its centre of mass (through which axis of rotation passes) is not symmetrical about the axis of wheel. So, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Thus, an external torque is required to maintain the motion of the half wheel.

Topic: Physics, System of Particles and Rotational Motion
A wheel is at rest. Its angular velocity increases uniformly and become 80 radian second after 5 seconds. The total angular displacement is:

Options:
A 800 rad
B 400 rad
C 200 rad
D 100 rad

The correct answer is:
200 rad

Solution:
Initial angular speed, $\omega_0=0 \mathrm{rad} / \mathrm{s}$
Final angular speed, $\omega=80 \mathrm{rad} / \mathrm{s}$
Time taken, $\mathrm{t}=5 \mathrm{~s}$
From kinematics:
$\begin{aligned} & \Rightarrow 80=\alpha \times 5 \\ & \Rightarrow \alpha=\frac{80}{5}=16 \mathrm{rad} / \mathrm{s}^2\end{aligned}$
where, $\alpha$ is angular acceleration.
Also,
$\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \Rightarrow \theta=0+\frac{1}{2} \times 16 \times 25 \\
& \Rightarrow \theta=200 \mathrm{rad}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A wheel is rotating freely with an angular speed ω on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:

Options:
A 56
B 14
C 0
D 34

The correct answer is:
34

Solution:
From angular momentum conservation
Iω+0=IωC+3IωC
ωc=ω4

Loss of kinetic energy =12Iω2-12(I+3I)ω42

=12Iω2-12Iω24

=38Iω2

Fractional loss =34

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 seconds, it rotates through an angle $\theta_1$. In the next 2 seconds, it rotates through an additional angle $\theta_2$. Find the ratio of $\theta_2 / \theta_1$.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Using relation $\theta=w_0 t+\frac{1}{2} a^2$
here we know that $\mathrm{w}_{\mathrm{o}}=\mathrm{o}, \mathrm{t}=2 \mathrm{sec}$
$\therefore \quad \theta_1=\frac{1}{2}(\alpha)(2)^2=(2 \alpha)$
Now using same equation for $t=4 \mathrm{sec}, w_0=0$
$\begin{aligned}
& \theta_1+\theta_2=\frac{1}{2} a(4)^2=8 \alpha \\
& \theta_2=8 \alpha-2 a=6 a
\end{aligned}$
From(i)and(ii),
$\begin{aligned}
& \theta_1=2 a, \quad \theta_2=6 a \\
& \therefore \frac{\theta_2}{\theta_1}=\frac{6 a}{2 a}=3
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A Wheel of moment of inertia 2 kg m2 is rotating about an axis passing through centre and perpendicular to its plane at a speed 60 rad s-1. Due to friction, it comes to rest in 5 minutes. The angular momentum of the wheel three minutes before it stops rotating is

Options:
A 24Â kgÂ m2/s
B 48 kgÂ m2/s
C 72 kg m2/s
D 96 kgÂ m2/s

The correct answer is:
72 kg m2/s

Solution:
I=2 kg m2, ωo=60 rad s-1
t=5 min=5×60=300 s
∝ =0-60300=-60300=-15 rad s-2
for 2min (from starting) (2 min = 120 sec)
ω=ω0+∝t
=60-15×120=60-24
ω=36 rad s-1 and L=Iω=2×36=72 kg m2 s-1

Topic: Physics, Rotational Motion, JEE Main
A wheel of radius R is trapped in a mud pit and spinning. As the wheel is spinning, it splashes mud blobs with initial speed u from various points on its circumference. The maximum height from the centre of the wheel, to which a mud blob can reach is

Options:
A u2/2g
B u22g+gR22u2
C 0
D R+u22g

The correct answer is:
u22g+gR22u2

Solution:
Let a mud blob is detached from the circumference of wheel with initial speed u at angle θ as shown below.

Height upto which mud blob can be thrown is

h= Maximum height of projectile+Height at which mud blob is thrown

h=u2sin2θ2g+Rcosθ....i
h is maximum when dhdθ=0

⇒ddθu2sin2θ2g+Rcosθ

⇒u22g.2sinθcosθ-Rsinθ=0

⇒u2g·cosθ=R or cosθ=Rgu2

⇒sin2θ=1-R2g2u2

Substituting these values in Eq (i), we get

λmax=u22g1-R2g2u4+RRgu2

=u22g-R2g2u2+R2gu2=u22g+R2g2u2

Topic: Physics, Motion In Two Dimensions, JEE Main
A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of the topmost point of the wheel with respect to the bottommost point is

Options:
A v2R
B 2v2R
C v22R
D 4v2R

The correct answer is:
2v2R

Solution:
As it is rolling
â Ï=VR
V Q =0
âVP/Q=Ï(2R)=2V
â a P/Q = V P/Q 2 2R = ( 2V ) 2 2R = 2 V 2 R

Topic: Physics, Rotational Motion, JEE Main
A wheel of the radius 0.4 m can rotate freely about its axis. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad s-2 is produced in it due to the torque. Then, the moment of inertia of the wheel is (g=10 m s-2 )

Options:
A 2 kg m2

B 1 kg m2
C 4 kg m2
D 8 kg m2

The correct answer is:
2 kg m2

Solution:
Given, r=0.4 m,

α = 8 rad s - 2
m=4 kg, I =?

Torque, τ=Iα
⇒mgr=Iα

or 4 × 1 0 × 0.4 = I × 8

⇒ I=168=2 kg m2

Or I=2 kg m2

Topic: Physics, Rotational Motion, JEE Main
A whistle of frequency 500 Hz, tied to the end of a string of length 1.2 m, revolves at 400 rev/min. A listener standing some distance away in the plane of rotation of the whistle, hears frequency in the range of (speed of sound = 340 ms-1)

Options:
A 386 Hz-436 Hz
B 426 Hz-474 Hz
C 426 Hz-586 Hz
D 436 Hz-586 Hz

The correct answer is:
436 Hz-586 Hz

Solution:
vs=rω
vs=1.2×2×3.14×40060=50ms-1
vmin⁡ =vv+vsν
=340340+50×500=436 Hz
vmax⁡ =vv-vsv
=340340-50×500

=586 Hz

Topic: Physics, Waves and Sound, JEE Main
A whistle producing sound waves of frequencies $9500 \mathrm{~Hz}$ and above is approaching a stationary person with speed $v \mathrm{~ms}^{-1}$. The velocity of sound in air is $300 \mathrm{~ms}^{-1}$. If the person can hear frequencies upto a maximum of $10,000 \mathrm{~Hz}$, the maximum value of $\mathrm{v}$ upto which he can hear the whistle is

Options:
A
$30 \mathrm{~ms}^{-1}$

B
$15 \sqrt{2} \mathrm{~ms}^{-1}$

C
$15 / \sqrt{2} \mathrm{~ms}^{-1}$

D
$15 \mathrm{~ms}^{-1}$

The correct answer is:
$15 \mathrm{~ms}^{-1}$

Solution:
$f_{a p p}=\frac{f(300)}{300-v} \Rightarrow v=15 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2006
$A$ whistle whose air column is open at both ends has a fundamental frequency of $5100 \mathrm{Hz}$. If the speed of sound in air is $340 \mathrm{ms}^{-1}$, the length of the whistle, in $\mathrm{cm}$, is

Options:
A $5 / 3$
B $10 / 3$
C 5
D $20 / 3$

The correct answer is:
$10 / 3$

Solution:
For an open pipe, the frequency
$$
\begin{array}{l}
f=\frac{v}{2 l} \Rightarrow 5100=\frac{340}{2 \times 1} \\
l=\frac{340}{5100 \times 2} \Rightarrow l=\frac{2}{30 \times 2}=\frac{1}{30} \mathrm{m} \\
l=\frac{100}{30}=\frac{10}{3} \mathrm{cm}
\end{array}
$$

Topic: Physics, Waves and Sound, JEE Main
A white precipitate was formed when BaCl2 was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCI. The anion present in the inorganic salt is:

Options:
A I-
B SO32-
C S2-
D NO2-

The correct answer is:
SO32-

Solution:
A white precipitate was formed when BaCl2 when added to water extract of SO3-2 salt. which is an inorganic salt.white pracipitate of BaSO3 is formed. Further, a gas SO2 with burning sulphur like odour will be released when theBaSO3 is dissolved in dilute HCI.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the aforementioned solution, a white precipitate is obtained which does not dissolve in dil. nitric acid. The anion is :

Options:
A
$\mathrm{CO}_3^{2-}$

B
$\mathrm{SO}_4^{2-}$

C
$\mathrm{S}^{2-}$

D
$\mathrm{Cl}^{-}$

The correct answer is:
$\mathrm{Cl}^{-}$

Solution:
Therefore, the required locus is $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the before mentioned solution, a white precipitate is obtained which does not dissolve in dilute nitric acid. The anion is:

Options:
A CO32-
B SO42-
C S2-
D Cl-

The correct answer is:
Cl-

Solution:
Na(aq)++ Cl-(aq)+ AgNO3→AgCl↓White+NaNO3

AgCl  + HNO3(dil)→Does not dissolve

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2018 (15 Apr)
A wide cylindrical vessel 50 cm in height is filled with water and rests on a table. Assuming the viscosity to be negligible, find at what height (in cm) from the bottom of the vessel a small hole should be perforated for the water jet coming out of it to hit the surface of the table at the maximum distance from the vessel?

Options:

The correct answer is:
25

Solution:
Applying Bernoulli's theorem just inside (at point (1)) and just outside of hole (at point(2)).
âµ Pâ¡ 1 Ï g + vâ¡ 1 2 2 g + y = Pâ¡ 2 Ï g + vâ¡ 2 2 2 g + y
Here v1 = 0, P2 = P0 = atmospheric pressure.
P1 = P0 +Ïg (H â y)
v2 = v

â

Putting the values, we obtain
1 2 Ï vâ¡ 2 = Hâ¡ - y Ï g
â´ vâ¡ = 2 g Hâ¡ - y In horizontal direction
Horizontal range, l = vt
For vertical motion,
y = 1 2 gt 2
â´ t = 2 y g
â´ lâ¡ = 2 g Hâ¡ - y 2 y g = 2 y H - y
For the maximum value of l, dlâ¡ dâ¡ hâ¡ = 0
After solving, y=Hâ¡2=25 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A wind with speed 40 m s-1 blows parallel to the roof of a house. The area of the roof is 250 m2 . Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the direction of the force will be : ρair =1.2 kg m-3

Options:
A 4.8×105 N, downwards
B 4.8×105 N, upwards
C 2.4×105 N, upwards
D 2.4×105 N, downwards

The correct answer is:
2.4×105 N, upwards

Solution:
By Bernoulli's equation,
P+12ρv2=P0+0
P0-P=12ρv2
F=12ρv2A
=12×1.2×402×250
=12×1.2×1600×250
=2.4×105 N, upwards

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed $v$, the electrical power output will be most likely proportional to

Options:
A
$v^4$

B
$v^2$

C
$\mathrm{v}$

D
$v^3$

The correct answer is:
$v^3$

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2013 (25 Apr Online)
A wire, 3 m in length and 1 mm in diameter at 30Â°C is kept in low temperature at -170Â°C and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is [Y=2Ã1011 Nm-2, g=10 ms-2 and Î±=1.2Ã10-5/Â°C]

Options:
A 5.2 mm
B 2.5 mm
C 7.2 mm
D 3.5 mm

The correct answer is:
5.2 mm

Solution:
The change in length of the wire due to change in temperature

=αLT=1.2×10-5×3×-170-30
=-7.2×10-3m
The change in the length of the wire due to the stretching force
=F LA Y= 10×10×30.75×10-6×2×1011
=2×10-3 m
The total change in the length of the wire
=-7.2×10-3+2×10-3
=-5.2×10-3 m=-5.2 mm

The negative sign indicates contraction of the wire

Topic: Physics, Thermal Properties of Matter, JEE Main
A wire $34 \mathrm{~cm}$ long is to be bent in the form of a quadrilateral of which each angle is $90^{\circ}$. What is the maximum area which can be enclosed inside the quadrilateral?

Options:
A $68 \mathrm{~cm}^{2}$
B $70 \mathrm{~cm}^{2}$
C $71.25 \mathrm{~cm}^{2}$
D $\begin{array}{lll}\text { } & 72.25 \mathrm{~cm}^{2} & {}\end{array}$

The correct answer is:
$\begin{array}{lll}\text { } & 72.25 \mathrm{~cm}^{2} & {}\end{array}$

Solution:
Let one side of quadrilateral be $x$ and another side be $y$ so, $2(x+y)=34$
or, $(\mathrm{x}+\mathrm{y})=17$
We know from the basic principle that for a given perimeter square has the maximum area, so, $x=y$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$

Topic: Mathematics, Application of Derivatives, JEE Main