datavorous/entrance-exam-dataset huggingface dataset:
In a sample initially, there are an equal number of atoms of two radioactive isotopes A and B. 3 days later, the number of atoms of A is twice that of B. The half-life of B is 1.5 days. What is the half-life (in days) of isotope A?

Options:

The correct answer is:
3

Solution:
ABt=0N0N0t0=3 days2NN
2N=N0(0.5)t0/τ1
N=N0(0.5)t0/τ2
2=(0.5)t0(1τ1−1τ2)
0.5−1=(0.5)3τ1−2
⇒-1=3τ1-2                  ∴τ1=3 days

Topic: Physics, Nuclear Physics, JEE Main
In a sample initially, there are an equal number of atoms of two radioactive isotopes A and B. 3 days later, the number of atoms of A is twice that of B. The half-life of B is 1.5 days. What is the half-life (in days) of isotope A?

Options:

The correct answer is:
3

Solution:
ABt=0 N0N0t0=3âdays 2NN
2N=N0(0.5)t0/Ï1
N=N0(0.5)t0/Ï2
2=(0.5)t0(1Ï1â1Ï2)
0.5â1=(0.5)3Ï1â2
â-1=3Ï1-2âââââââââââââââââââ´Ï1=3 daysâ

Topic: Physics, Nuclear Physics, JEE Main
In a sample of excited hydrogen atoms, electrons make transition from n=2 to n=1 . Emitted quanta strike on a metal of work function ϕ 4.2 ev. Calculate the wavelength (in Å ) associated with ejected electrons having maximum kinetic energy.

Options:

The correct answer is:
5

Solution:
E=Ï+KE
10.2=4.2+KE
λ e − = 150 6 =5 A o

Topic: Chemistry, Structure of Atom, JEE Main
In a sarcomere, the H-zone is paler than the rest of A-band because it consists of

Options:
A only myosin filaments
B only actin filaments
C both actin and myosin filaments
D filaments are absent

The correct answer is:
only myosin filaments

Solution:
Skeletal muscles are different from visceral muscles as these muscles show striations and are voluntary in nature. These are made up of many bundles known as Fasciculi. Each bundle has many muscle fibres which in turn has many myofibrils. Each myofibril consists of repeating units called as sarcomere. This is the main functional unit of contraction and relaxation of muscles. One sarcomere is the portion between two Z lines. Z lines bisect the 'I' or Isotrophic or light bands. In the centre of Sarcomere or between two Z lines there are present Dark bands or 'A'orAnisotrophic bands. Hensons zone or H zone is also seen in the centre which is demarcated by M line in the middle. Dark band has both actin and myosin fibres and hence appears dark. Hensons zone has only myosin filaments and hence appears paler.

Topic: Biology, Locomotion and Movement, NEET
In a sarcomere, the thin filaments are bisected by:

Options:
A Z-line
B M-line
C H-zone
D F-line

The correct answer is:
Z-line

Solution:
Skeletal muscles are different from visceral muscles as these muscles show striations and are voluntary in nature. These are made up of many bundles known as fasciculi. Each bundle has many muscle fibres which in turn has many myofibrils. Each myofibril consists of repeating units called as sarcomere. This is the main functional unit of contraction and relaxation of muscles. One sarcomere is the portion between two Z lines. Z lines bisect the 'I' or Isotrophic or light bands. In the centre of Sarcomere or between two Z lines there are present Dark bands or 'A'or Anisotrophic bands. Hensons zone or H zone is also seen in the centre which is demarcated by M line in the middle.

Topic: Biology, Structural Organisation in Animals, NEET
In a satellite if the time of revolution is $T$, then K.E. is proportional to

Options:
A $\frac{1}{T}$
B $\frac{1}{T^2}$
C $\frac{1}{T^3}$
D $T^{-2 / 3}$

The correct answer is:
$T^{-2 / 3}$

Solution:
$V=\sqrt{\frac{G M}{r}} \quad \therefore \quad K . E . \propto v^2 \propto \frac{1}{r}$ and $T^2 \propto r^3$
$\therefore K . E . \propto T^{-2 / 3}$

Topic: Physics, Gravitation, JEE Main
In a satellite, if the time of revolution is \( T \), then kinetic energy is proportional to

Options:
A \( \frac{1}{T} \)
B \( \frac{1}{T^{2}} \)
C \( \frac{1}{T^{3}} \)
D \( T^{-\frac{2}{3}} \)

The correct answer is:
\( T^{-\frac{2}{3}} \)

Solution:
Suppose mass of satellite is M and radius of satellite is r.

Velocity of satellite, v=GMr, KE∝v2∝1r
and time period is
T=2πr3GM
∴T2∝r3,

So, KE∝T-23.

Topic: Physics, Gravitation, JEE Main
In a sautrated solution of the sparingly soluble strong electrolyte $\mathrm{AgIO}_3$ (Molecular mass $=283$) the equilibrium which sets in is
$\mathrm{AgIO}_{3(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{IO}_{3(\mathrm{aq})}^{-}$
If the solubility product constant $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgIO}_3$ at a given temperature is $1.0 \times 10^{-8}$, what is the mass of $\mathrm{AgIO}_3$ contained in $100 \mathrm{~ml}$ of its saturated solution?

Options:
A
$28.3 \times 10^{-2} \mathrm{~g}$

B
$2.83 \times 10^{-3} \mathrm{~g}$

C
$1.0 \times 10^{-7} \mathrm{~g}$

D
$1.0 \times 10^{-4} \mathrm{~g}$

The correct answer is:
$2.83 \times 10^{-3} \mathrm{~g}$

Solution:
$\mathrm{AgIO}_3(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{IO}_3^{-}(\mathrm{aq})$
Let the solubility of $\mathrm{AgIO}_3$ be $\mathrm{s}$
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{IO}_3^{-}\right]$
$1.0 \times 10^{-8}=\mathrm{s}^2$
$\mathrm{s}=10^{-4} \mathrm{~mol} / \mathrm{litre}$
$=\frac{10^{-4} \times 283}{1000} \times 100$
$=283 \times 10^{-5}$
$=2.83 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~ml}$
Hence, (B) is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2007
In a school there are $40 \%$ science students and the remaining $60 \%$ are arts students. It is known that $5 \%$ of the science students are girls and $10 \%$ of the arts students are girls. One student selected at random is a girl. What is the probability that she is an arts student?

Options:
A $\frac{1}{3}$
B $\frac{3}{4}$
C $\frac{1}{5}$
D $\frac{3}{5}$

The correct answer is:
$\frac{3}{4}$

Solution:
Let there be 100 students. So, there are 40 students of science and 60 students of
arts.
$5 \%$ of $40=2$ science students (girls) $10 \%$ of $60=6$ science students (girls) Total girls students $=8$ If a girl is chosed then
$\mathrm{P}(\mathrm{arts})=\frac{6}{8}=\frac{3}{4}$

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 1)
In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?

Options:
A P and Q
B P and R
C Q and R
D None of these

The correct answer is:
None of these

Solution:
In P, the shaded region is lying in all the circles. It means some students play all the games.

In Q, the shaded region is lying in all the circles. It means some students play all the games.

In R, the shaded region is lying in all the circles. It means some students play all the games.

Hence, none of the Venn diagrams justify the statement.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2021 (17 Mar Shift 1)
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible error, the thickness of the wire is

Options:
A 0.4300 cm
B 0.3150 cm
C 0.0430 cm
D 0.2150 cm

The correct answer is:
0.2150 cm

Solution:
The Least count=0.255×100 cm=5×10-4 cm.

Reading =4×0.05 cm+30×5×10-4 cm
=0.2+0.0150 cm=0.2150 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2018 (15 Apr)
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, what is the thickness (in mm) of the wire?

Options:

The correct answer is:
21.5

Solution:
Least count =0.255Ã100 cm=5Ã10-4 cm
Reading =4Ã0.05 cm+30Ã5Ã10-4 cm
=0.2+0.0150 cm=2.15 mm

Topic: Physics, Units and Dimensions, JEE Main
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of $0.25$ $\mathrm{cm}$. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

Options:
A
$0.0430 \mathrm{~cm}$

B
$0.3150 \mathrm{~cm}$

C
$0.4300 \mathrm{~cm}$

D
$0.2150 \mathrm{~cm}$

The correct answer is:
$0.2150 \mathrm{~cm}$

Solution:
Least count
$$
=\frac{\text { Value of } 1 \text { part on main scale }}{\text { Number of parts on vernier scale }}
$$
$$
=\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm}
$$
Reading $=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4} \mathrm{~cm}$ $=(0.2+0.0150) \mathrm{cm}=0.2150 \mathrm{~cm}$ (Thickness of wire)

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

Options:
A 5.20 mm
B 5.00 mm
C 5.15 mm
D 5.25 mm

The correct answer is:
5.15 mm

Solution:
L.C. of screw Gauge=0.550=0.01 mm
Zero error of screw Gauge =5×L.C.=5×0.01=0.05 mm
Reading of screw gauge = main scale reading + circular scale reading ×L.C.-Zero error =5+20×0.01-0.05=5.15 mm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
In a scries LCR resonance circuit, if we change the resistance only, from a lower to higher value :

Options:
A The bandwidth of resonance circuit will increase.
B The resonance frequency will increase.
C The quality factor will increase.
D The quality factor and the resonance frequency will remain constant.

The correct answer is:
The bandwidth of resonance circuit will increase.

Solution:
The bandwidth of resonance circuit will increase.

As we know that Bandwidth =R/L.

Bandwidth ∝R.

So, bandwidth will increase.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (18 Mar Shift 1)
In a semiconductor, mobility of electron, i.e., drift velocity per unit applied electric field is \( 1.6 \) (S.I. unit). The density of electron is \( 10^{19} \mathrm{~m}^{-3} \) (Neglect holes concentration). The resistivity of semiconductor is \( n \Omega \mathrm{m} \).

Options:

The correct answer is:
0.4

Solution:
ρ=EJ=Eneυd=1neμ
μ=mobility
ρ=11019×1.6×10-19×1.6=12.56 Ω-m
=0.4 Ω-m .

Topic: Physics, Current Electricity, JEE Main
In a semiconductor, the number density of intrinsic charge carriers at 27°C is 1.5×1016 m-3. If the semiconductor is doped with an impurity atom, the hole density increases to 4.5×1022 m-3. The electron density in the doped semiconductor is _______×109 m-3

Options:

The correct answer is:
5

Solution:
nenh=ni2

ne=ni2nh=1.5×101624.5×1022

=1.5×1.5×10324.5×1022

5×109 m-3

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (25 Jul Shift 2)
In a sensitive meter bridge apparatus the bridge wire should possess

Options:
A
high resistivity and low temperature coefficient.

B
low resistivity and high temperature coefficient.

C
low resistivity and low temperature coefficient.

D
high resistivity and high temperature coefficient.

The correct answer is:
high resistivity and low temperature coefficient.

Solution:
Bridge wire in a sensitive meter bridge wire should be of high resistivity and low temperature coefficient.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (12 May Online)
In a sequence of \( 21 \) terms, the first \( 11 \) terms are in AP with common difference \( 2 \) and the last \( 11 \) terms are in GP with common ratio \( 2 \). If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is

Options:
A \( -\frac{10}{31} \)
B \( \frac{10}{31} \)
C \( \frac{32}{31} \)
D \( -\frac{31}{32} \)

The correct answer is:
\( -\frac{10}{31} \)

Solution:
Since, the first 11 terms in AP
d=2
∴ a11=a+10d=a+20
The middle term of AP is T6=a+5d=a+10
For the next 11 terms in GP, r=2
∴ The middle term of GP is b25 where b is the first term of a GP which is the last term of AP.
∴ b25=a+20 32
According to the given condition,
a+10=a+20 32
⇒ 31a=10-640 ⇒ a= -63031
∴ Middle term of entire sequence is 11th term.
∴ T11= -63031+10×d
= -63031+10 ×2= -1031

Topic: Mathematics, Sequences and Series, JEE Main
In a series circuit $R=300 \Omega, L=0.9 \mathrm{H}, C=2.0 \mu \mathrm{F}$ and $\omega=1000 \mathrm{rad} / \mathrm{sec}$. The impedance of the circuit is

Options:
A $1300 \Omega$
B $900 \Omega$
C $500 \Omega$
D $400 \Omega$

The correct answer is:
$500 \Omega$

Solution:
For series $R-L-C$ circuit, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
$=\sqrt{(300)^2+\left(1000 \times 0.9-\frac{10^6}{1000 \times 2}\right)^2}=500 \Omega$

Topic: Physics, Alternating Current, JEE Main
In a series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit, $\mathrm{C}=10^{-11}$ Farad, $\mathrm{L}=10^{-}$ ${ }^5$ Henry and $\mathrm{R}=100 \mathrm{Ohm}$, when a constant D.C. voltage $\mathrm{E}$ is applied to the circuit, the capacitor acquires a charge $10^{-9}$ C. The D.C. source is replaced by a sinusoidal voltage source in which the peak voltage $\mathrm{E}_0$ is equal to the constant D.C. voltage E. At resonance the peak value of the charge acquired by the capacitor will be :

Options:
A
$10^{-15} \mathrm{C}$

B
$10^{-6} \mathrm{C}$

C
$10^{-10} \mathrm{C}$

D
$10^{-8} \mathrm{C}$

The correct answer is:
$10^{-8} \mathrm{C}$

Solution:
No Solution Available

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (22 Apr Online)
In a series $L C R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=20 \Omega$. The phase difference between the current and the voltage is

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

The correct answer is:
$60^{\circ}$

Solution:
$\cos \phi=\frac{R}{Z}=\frac{10}{20}=\frac{1}{2} \Rightarrow \phi=60^{\circ}$

Topic: Physics, Alternating Current, JEE Main
In a series LCR circuit $R=200 \Omega$ and the voltage and the frequency of the main supply is $220 \mathrm{~V}$ and $50 \mathrm{~Hz}$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $30^{\circ}$. On taking out the inductor from the circuit the current leads the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is

Options:
A
$305 \mathrm{~W}$

B
$210 \mathrm{~W}$

C
Zero W

D
$242 \mathrm{~W}$

The correct answer is:
$242 \mathrm{~W}$

Solution:
The given circuit is under resonance as $X_L=X_C$ Hence power dissipated in the circuit is
$$
\mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=242 \mathrm{~W}
$$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2010
In a series LCR circuit, the inductance, capacitance and resistance are L=100 mH, C=100 μFand R=10 Ω respectively. They are connected to an AC source of voltage 220 V and frequency of 50 Hz. The approximate value of current in the circuit will be _____ A.

Options:

The correct answer is:
22

Solution:
Angular frequency, ω=2πf=100π rad s-1

Here, inductive reactance XL=ωL=100π100×10-3=10π Ω

And capacitive reactanceXC=1ωC=1100π×100×10-6=100π Ω=10π Ω

Now, impedance of the series circuit is Z=R2+XL-XC2=R=10 Ω

Therefore,

IRMS=VRMSZ,

=22010=22 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (25 Jun Shift 2)
In a series LCR circuit, the inductive reactance XL is 10 Ω and the capacitive reactance XC is 4 Ω. The resistance (R) in the circuit is 6 Ω. The power factor of the circuit is :

Options:
A 12
B 122
C 12
D 32

The correct answer is:
12

Solution:
We know that power factor is cosϕ,

cosϕ=RZ ...1

Z=R2+XL-XC2 ...2

⇒Z=62+(10-4)2

⇒Z=62∣cosϕ=662

cosϕ=12

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In a series LCR circuit, the voltage drop across the resistance is 8 V, across the inductor is 6 V and across the capacitor is 12 V. Then,

Options:
A the voltage of the source will be leading the current in the circuit
B the voltage drop across each element will be less than the applied voltage
C the power factor of the circuit will be 43
D none of these

The correct answer is:
none of these

Solution:
Since, cosâ¡Ï=RZ=IRIZ=810=45
(Also cosâ¡Ï can never be greater than 1)
Hence, (c) is wrong
Also IXC>IXL â XC>XL
â´ Current will be leading.
In an LCR circuit,
V= VL-VC2+VR2=6-122+82
V=10 V, which is less than the voltage drop across the capacitor.

Topic: Physics, Alternating Current, JEE Main
In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be _______

Options:

The correct answer is:
282.84

Solution:
Q=XLR=ωLR=1LCLR=LRC

Q'=2 LR2C=22Q=22100

=282.84

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as n3π μF, then value of n is

Options:

The correct answer is:
400

Solution:
P=Vm. in cosϕ

400=250×1m×0.8

irms=2A

1m2.R=P

4×R=400

⇒ R=100Ω.

cosϕ=RR2+XL2

1002+XL2=1000.82

1002+XL2=1000.82

XL=75Ω

Power factor is unity

XC=XL=75

1ω=75

⇒ C=175×2H×50=17500πF

3π×2500

=13π×4×102 mF

=4003πμF

N=400

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (06 Sep Shift 2)
In a series of 2n observations, half of them are equal to a2 and the remaining half are equal to âa2. If the standard deviation of the observations is 2, then a is equal to

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

The correct answer is:
2

Solution:
xi=a2 for i=1,2,..,n and xi=-a2 for i=n+1,...2n
S.D=12nâi=12nxi-x-2
â2=12nâi=12nxi2Sinceâi=12nxi=0
â2= 1 2n â2n a 4 â a = 2

Topic: Mathematics, Statistics, JEE Main
In a series of $2 n$ observations, half of them equal a and remaining half equal $-a$. If the standard deviation of the observations is 2 , then $|a|$ equals

Options:
A
$\frac{1}{n}$

B
$\sqrt{2}$

C
2

D
$\frac{\sqrt{2}}{n}$

The correct answer is:
2

Solution:
$x_i=a$ for $i=1,2, \ldots, n$ and $x_i=-a$ for $i=n, \ldots ., 2 n$
S.D. $=\sqrt{\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_i-\bar{x}\right)^2} \Rightarrow 2=\sqrt{\frac{1}{2 n} \sum_{i=1}^{2 n} x_i^2} \quad\left(\right.$ Since $\left.\sum_{i=1}^{2 n} x_i=0\right) \Rightarrow 2=\sqrt{\frac{1}{2 n} \cdot 2 n a^2} \Rightarrow|a|=2$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2004
In a series resonant LCR circuit, the voltage across $R$ is 100 volts and $R=1 \mathrm{k} \Omega$ with $C=2 \mu \mathrm{F}$. The resonant frequency $\omega$ is $200 \mathrm{rad} / \mathrm{s}$. At resonance the voltage across $L$ is

Options:
A
$4 \times 10^{-3} \mathrm{~V}$

B
$2.5 \times 10^{-2} \mathrm{~V}$

C
$40 \mathrm{~V}$

D
$250 \mathrm{~V}$

The correct answer is:
$250 \mathrm{~V}$

Solution:
$\mathrm{i}=\frac{100}{1000}=0.1 \mathrm{~A}$
$V_L=V_C=\frac{0.1}{200 \times 2 \times 10^{-6}}=250 \mathrm{~V}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2006
In a set of 2n distinct observations, each of the observation below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then, the mean of the new set of observations :

Options:
A Increases by 2 .
B Increases by 1 .
C Decreases by 2 .
D Decreases by 1 .

The correct answer is:
Increases by 1 .

Solution:
New mean=nΣxi+5n+nΣyi-3n2n
=nΣxi+nΣyi2n+2n2n
=old mean +2n2n
=old mean+1

Topic: Mathematics, Statistics, JEE Main, JEE Main 2014 (09 Apr Online)
In a set of $2 n$ observations, half of them are equal to ' $\mathrm{a}$ ' and the remaining hall are equal to ' $-\mathrm{a}^{\prime}$ '. If the standard deviation of all the observations is 2 ; then the value of $|a|$ is :

Options:
A
2

B
$\sqrt{2}$

C
4

D
$2 \sqrt{2}$

The correct answer is:
2

Solution:
Clearly mean $\mathrm{A}=0$
Now, standard deviation $\sigma=\sqrt{\frac{\Sigma(x-\mathrm{A})^2}{2 n}}$
$$
\begin{aligned}
2=& \sqrt{\frac{(a-0)^2+(a-0)^2+\ldots \ldots+(0-a)^2+\ldots . .}{2 n}} \\
&=\sqrt{\frac{a^2 \cdot 2 n}{2 n}}=|a|
\end{aligned}
$$
Hence, $|a|=2$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2013 (25 Apr Online)
In a set of reaction acetic acid yielded a product 'd'.

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In a set of reactions acetic acid yielded a product
D.
$$
\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{A} \stackrel{\text { Berzene }}{\text { anhy. AlCl }_{3}} \mathrm{~B} \stackrel{\text { HCN }}{\longrightarrow}\mathrm{C} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{D}
$$
The structure of D would be

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COCl} \frac{\text { Bennen }}{\text { Arhy.A.Cl }_{3}}$

Topic: Chemistry, Hydrocarbons, JEE Main
In a set of reactions p - nitrotoluene yielded a product E

The product E would be :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
In a set of reactions p - nitrotoluene yielded a product E

The product E would be :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2014 (19 Apr Online)
In a set of the given reactions, acetic acid yielded a product $C$
\(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{PCl}_5 \rightarrow A \xrightarrow[\text { Anh. } \mathrm{AlCl}_3]{\mathrm{C}_6 \mathrm{H}_6} B \xrightarrow[\text { ether }]{\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}} C\)
Product $C$ would be

Options:
A

B $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{C}_2 \mathrm{H}_3$
C $\mathrm{CH}_3 \mathrm{COC}_6 \mathrm{H}_5$
D $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{C}_6 \mathrm{H}_5$

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In a set reactions acid yielded a product $D$
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{SOCl}_2} A \xrightarrow[\text { Anhr } A \mathrm{AlCl}_3]{\text { Benzene }} B \xrightarrow{\mathrm{HCN}} C \xrightarrow{\mathrm{HOH}} D$.

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In a shooting competition a man can score 5, 4, 3, 2, 1 or 0 points for each shot. Then the number of different ways in which he can score 10 in seven shots is

Options:
A 6538
B 6648
C 6468
D 6236

The correct answer is:
6538

Solution:
Required number of ways = Coefficient of x10 in x0+x1+...+x57
= coefficient of x10 in 1-x671-x-7
= coefficient of x10 in 1-7x61-x-7 (Neglecting the terms higher than x10 )
= coefficient of x10 in {1-x-7-7x61-x-7}
=16C6-7.10C4=6538

Topic: Mathematics, Permutation Combination, JEE Main
In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement -1: The number of different ways the child can buy the six ice-creams is ${ }^{10} \mathrm{C}_5$.
Statement $-2$ : The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging $6 \mathrm{~A}^{\text {'s }}$ and 4 B's in a row.

Options:
A
Statement $-1$ is false, Statement $-2$ is true

B
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

D
Statement $-1$ is true, Statement $-2$ is false.

The correct answer is:
Statement $-1$ is false, Statement $-2$ is true

Solution:
$$
\begin{aligned}
& x_1+x_2+x_3+x_4+x_5=6 \\
& 5+6-1 C_{5-1}={ }^{10} C_4
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2008