datavorous/entrance-exam-dataset huggingface dataset:
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state

Options:
A $\quad V_{A B}=V_{B C}=100 \mathrm{~V}$
B $\quad V_{A B}=75 V, V_{B C}=25 V$
C $\quad V_{A B}=25 V, V_{B C}=75 V$
D $\quad V_{A B}=V_{B C}=50 \mathrm{~V}$

The correct answer is:
$\quad V_{A B}=25 V, V_{B C}=75 V$

Solution:
The equivalent circuit is shown in figure.

$$

\begin{array}{l}

V_{1}+V_{2}=100 \\

\text { and } 2 V_{1}=6 V_{2}

\end{array}

$$

On solving above equations, we get $V_{1}=75 V, V_{2}=25 V$

Topic: Physics, Capacitance, BITSAT, BITSAT 2016
In the following circuit diagram, the current through the battery is

Options:
A 4 A
B 3 A
C 1 A
D 2 A

The correct answer is:
2 A

Solution:
From the figure, It is clear that diodes D1 and D3 are reverse biased, therefore they will not conduct any current and the resistance of 11 Ω and 5 Ω will be not in use. The diode D2 is forward biased, therefore the current flowing through the diode D2 is same as the current through the battery.

On redrawing the circuit as shown, we can obtain the value of current as

i=207+3=2 A

Topic: Physics, Semiconductors, BITSAT, BITSAT 2019
In the following circuit the equivalent resistance between $X$ and $Y$ is ...... $\Omega$

Options:
A
$5$

B
$12$

C
$16$

D
$20$

The correct answer is:

$16$

Solution:
According to the given figure $X$ is at lower potential w.r.t $Y$. Hence, both diodes are in reverse biasing, so, equivalent circuit can be redrawn as follows

Equivalent resistance between $X$ and $Y$
$$
R_{\text {eq }}=8+2+6=16 \Omega
$$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2022
In the following figures, which figure represents the variation of the electric field with distance from the centre of a uniformly charged non-conducting spheres of radius R ?

Options:
A
B
C
D

The correct answer is:

Solution:
The electric field intensity at a point lying outside the sphere (non-conducting) is

E=14πε0·qr2

i.e. E∝1r2 …i

The electric field intensity at the surface (r=R),

E=14πε0qR2

⇒E∝1R2

where, R being the radius of sphere,

The electric field intensity inside the sphere is

E=qr4πε0R3

⇒E∝r

At the centre of sphere, r=0, ∴E=0

Topic: Physics, Electrostatics, JEE Main
In the following figures, which figure represents the variation of the electric field with distance from the centre of a uniformly charged non-conducting spheres of radius R ?

Options:
A
B
C
D

The correct answer is:

Solution:
The electric field intensity at a point lying outside the sphere (non-conducting) is

E=14πε0·qr2

i.e. E∝1r2 …i

The electric field intensity at the surface (r=R),

E=14πε0qR2

⇒E∝1R2

where, R being the radius of sphere,

The electric field intensity inside the sphere is

E=qr4πε0R3

⇒E∝r

At the centre of sphere, r=0, ∴E=0

Topic: Physics, Electrostatics, BITSAT, BITSAT 2019
In the given circuit diagram,

E=5 V, r=1 Ω, R2=4 Ω, R1=R3=1 Ω and C=3 μF

Then, what will be the numerical value of charge on each plates of the capacitor.

Options:
A 24 μC
B 12 μC
C 6 μC
D 3 μC

The correct answer is:
6 μC

Solution:
Current will flow only through the branch containing resistance R2

∴ i=ER2+r=54+1=1 A

Potential difference across R2

=1×4=4 V

If q be the charge on each plate of capacitor, then

qc+qc=4

or 2qc=4

or 2q3×10-6=4

q=6 μC

Topic: Physics, Capacitance, BITSAT, BITSAT 2017
In the given circuit, it is observed that the current I is independent of the value of resistance R5. Then, the resistance value must satisfy

Options:
A 1R5+1R6=1R1+R2·1R3+R4
B R1R4=R2R3
C R1R2R5=R3R4R6
D R1R3=R2R4=R5R6

The correct answer is:
R1R4=R2R3

Solution:
From the given circuit, we can say that I is independent of resistance R5, so no current flows through R5. This required that the junction of R1 and R2 is at the same potential as the junction of R3 and R4 (Wheatstone's bridge balancing condition)

So, R1R2=R3R4

or R1R4=R3R2

Topic: Physics, Current Electricity, BITSAT, BITSAT 2018
In the given circuit, rms value of current (Irms) through the resistor $R$ is :

Options:
A $2 \mathrm{~A}$
B $\frac{1}{2} \mathrm{~A}$
C $20 \mathrm{~A}$
D $2 \sqrt{2} \mathrm{~A}$

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

Solution:
(a) Given,
Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=100 \Omega$
Inductive reactance, $\mathrm{X}_{\mathrm{L}}=200 \Omega$
Resistance, $R=100 \Omega$
Impedance, $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}$
$Z=\sqrt{100^2+(200-100)^2}=100 \sqrt{2} \Omega$
RMS value of current,
$i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{200 \sqrt{2}}{100 \sqrt{2}}=2 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main
In the given circuit, rms value of current (Irms) through the resistor $R$ is :

Options:
A $2 \mathrm{~A}$
B $\frac{1}{2} \mathrm{~A}$
C $20 \mathrm{~A}$
D $2 \sqrt{2} \mathrm{~A}$

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

Solution:
(a) Given,
Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=100 \Omega$
Inductive reactance, $\mathrm{X}_{\mathrm{L}}=200 \Omega$
Resistance, $R=100 \Omega$
Impedance, $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}$
$Z=\sqrt{100^2+(200-100)^2}=100 \sqrt{2} \Omega$
RMS value of current,
$i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{200 \sqrt{2}}{100 \sqrt{2}}=2 \mathrm{~A}$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2023 (Memory Based Paper 2)
In the given circuit, The current through battery is

Options:
A 0.5 A
B 1 A
C 1.5 A
D 2.5 A

The correct answer is:
1.5 A

Solution:
In the given circuit, diode D1 is reverse biased, so it will not conduct but D2,D3 are forward biased, so they will conduct, hence corresponding equivalent circuit. For the given circuit is

Now, the equivalent resistance of the circuit is

Req=515×205+5+20+203 Ω

Current through battery, I=10 V203 Ω=1.5 A

Topic: Physics, Semiconductors, BITSAT, BITSAT 2018
In the given figure, the equation of the larger circle is $x^{2}+y^{2}+4 y-5=0$ and the distance between centres is 4 . Then the equation of smaller circle is

Options:
A $(x-\sqrt{7})^{2}+(y-1)^{2}=1$
B $\quad(x+\sqrt{7})^{2}+(y-1)^{2}=1$
C $x^{2}+y^{2}=2 \sqrt{7} x+2 y$
D None of these

The correct answer is:
$(x-\sqrt{7})^{2}+(y-1)^{2}=1$

Solution:
We have $x^{2}+y^{2}+4 y-5=0$. Its centre is $\mathrm{C}_{1}(0,-2)$

$\mathrm{r}_{1}=\sqrt{4+5}=3$. Let $\mathrm{C}_{2}(\mathrm{~h}, \mathrm{k})$ be the centre of the

smaller circle and its radius $\mathrm{r}_{2}$. Then $\mathrm{C}_{1} \mathrm{C}_{2}=4$ $\Rightarrow \sqrt{\mathrm{h}^{2}+(\mathrm{k}+2)^{2}}=3+\mathrm{r}_{2}=4$

$\Rightarrow \mathrm{r}_{2}=1$

But $\mathrm{k}=\mathrm{r}_{2}=1 \quad$ [it touches $\mathrm{x}$ -axis

$\therefore$ From eq $(1), 4=\sqrt{\mathrm{h}^{2}+(1+2)^{2}}$

$\Rightarrow 16=\mathrm{h}^{2}+9 \Rightarrow \mathrm{h}^{2}=7 \Rightarrow \mathrm{h}=\pm \sqrt{7}$

Since $\mathrm{h}>0 \quad \therefore \mathrm{h}=\sqrt{7}$

Hence, required circle is $(x-\sqrt{7})^{2}+(y-1)^{2}=1$

Topic: Mathematics, Circle, BITSAT, BITSAT 2020
In the kinetic theory of gases, which of these statements is/are true?

(i) The pressure of a gas is proportional to the mean speed of the molecules.

(ii) The root mean square speed of the molecules is proportional to the pressure.

(iii) The rate of diffusion is proportional to the mean speed of the molecules.

(iv) The mean translational kinetic energy of a gas is proportional to its kelvin temperature.

Options:
A (ii) and (iii) only
B (i),(ii) and (iv) only
C (i) and (iii) only
D (iii) and (iv) only

The correct answer is:
(iii) and (iv) only

Solution:
No Solution

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2014
In the laboratory, manganese (II) salt is oxidise to permanganate ion in aqueous solution by

Options:
A hydrogen peroxide
B conc. nitric acid
C peroxy disulphate
D dichromate

The correct answer is:
peroxy disulphate

Solution:
In laboratory, manganese (II) ion salt is oxidised to permanganate ion in aqueous solution by peroxodisulphate.
$2 \mathrm{Mn}^{2+}+\underset{\text{peroxodisulphate ion}}{5 \mathrm{~S}_2 \mathrm{O}_8^{2-}}+8 \mathrm{H}_2 \mathrm{O} \rightarrow$ $2 \mathrm{MnO}_4^{-}+10 \mathrm{SO}_4^{2-}+16 \mathrm{H}^{+}$

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2023 (Memory Based Paper 1)
In the manufacture of iron from haematite, limestone is added to act as :

Options:
A Flux
B Slag
C A reducing agent
D An oxidizing agent

The correct answer is:
Flux

Solution:
Limestone $\left(\mathrm{CaCO}_{3}\right)$ is mixed with $\mathrm{Fe}_{2} \mathrm{O}_{3}$ and it acts as flux to form slag $\left(\mathrm{CaSiO}_{3}\right)$.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2010
In the molecules $\mathrm{CH}_4, \mathrm{NF}_3, \mathrm{NH}_4^{+}$and $\mathrm{H}_2 \mathrm{O}$

Options:
A
number of lone pairs are same

B
all have same hybridisation of centre of atom

C
the bond angles are same

D
number of bond pairs are same

The correct answer is:

all have same hybridisation of centre of atom

Solution:
$\mathrm{CH}_4 ; \mathrm{bp}=4 ; s p^3$-hybridisation
$\mathrm{NF}_3 ; \mathrm{bp}=3 ; l p=1 ; s p^3$-hybridisation
$\mathrm{NH}_4^{+} ; \mathrm{bp}=4 ; l p=0 ; s p^3$-hybridisation
$\mathrm{H}_2 \mathrm{O} ; \mathrm{bp}=2 ; l p=2 ; s p^3$-hybridisation
Hence, all have the same slope.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2022
In the periodic table, with the increase in atomic number, the metallic character of an element

Options:
A decreases in a period and increases in a group
B increases in a period and decreases in a group
C increases both in a period and the group
D decreases in a period and the group

The correct answer is:
decreases in a period and increases in a group

Solution:
No Solution available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2010
In the reaction $2 \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{4}^{+}+\mathrm{PCl}_{6}^{-}$, the change in hybridisation is from

Options:
A $s p^{3} d$ to $s p^{3}$ and $s p^{3} d^{2}$
B $s p^{3} d$ to $s p^{2}$ and $s p^{3}$
C $s p^{3} d$ to $s p^{3} d^{2}$ and $s p^{3} d^{3}$
D $s p^{3} d^{2}$ to $s p^{3}$ and $s p^{3} d$

The correct answer is:
$s p^{3} d$ to $s p^{3}$ and $s p^{3} d^{2}$

Solution:
$2 \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{4}^{+}+\mathrm{PCl}_{6}^{-}$

$\uparrow s p^{3} d \quad \uparrow s p^{3} \quad \uparrow s p^{3} d^{2}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2020
In the reaction
$3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 5 \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}^{-}$

Options:
A Bromine is oxidised and carbonate is reduced.
B Bromine is reduced and water is oxidised.
C Bromine is neither reduced nor oxidised.
D Bromine is both reduced and oxidised.

The correct answer is:
Bromine is both reduced and oxidised.

Solution:
$\begin{aligned}
3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} & \rightarrow \\
5 \mathrm{Br}^{-} &+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}^{-}
\end{aligned}$
O.N. of $\mathrm{Br}_{2}$ changes from 0 to -1 and +5 , hence it is reduced as well as oxidised.

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2012
In the reaction,

CH3·CH2·Cl⟶KCNA⟶LiAlH4P end product P is

Options:
A CH3CH2·NO2
B CH3CH2CH2NO2
C CH3CH2NH2
D CH3·CH2·CH2NH2

The correct answer is:
CH3·CH2·CH2NH2

Solution:
When the reducing agent is lithium aluminium hydride, the aldehyde group is reduced to primary alcohol but the carbon-carbon double bond is unaffected.

In the given reaction the end product is (d). It is obtained
as follows.

CH3CH2·Cl⟶KCNCH3CH2CN(A)→(Reduction)LiAlH4CH3CH2CH2·NH2

Topic: Chemistry, Amines, BITSAT, BITSAT 2017
In the reaction of $\mathrm{KMnO}_{4}$ with an oxalate in acidic medium, $\mathrm{MnO}_{4}^{-}$ is reduced to $\mathrm{Mn}^{2+}$ and $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is oxidised to $\mathrm{CO}_{2}$. Hence, $50 \mathrm{~mL}$ of 0.02 $\mathrm{M} \mathrm{KMnO}_{4}$ is equivalent to

Options:
A $100 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
B $50 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
C $25 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
D $50 \mathrm{~mL}$ of $0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$

The correct answer is:
$50 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$

Solution:
eg. of \(\mathrm{KMnO}_{4}\)
\(\begin{aligned}
&=5 \times 0.02 \times 50 \\
&=250 \times 0.02 \\
&=509 .
\end{aligned}\)

eq. of \(\mathrm{KAnO}_{4} \equiv \mathrm{eq}_{1}\) of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }_{2}-\)

\(\begin{aligned} 5=\operatorname{option} 2 &=50 \times 0.05 \times 2 \\ &=589 . \end{aligned}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2012
In the reaction, $\mathrm{P}+\mathrm{Q} \longrightarrow ? \mathrm{R}+\mathrm{S}$

The time taken for $75 \%$ reaction of $P$ is twice the time taken for $50 \%$ reaction of $\mathrm{P}$. The concentration of $\mathrm{Q}$ varies with reaction time as shown in the figure. The overall order of the reaction is

Options:
A 2
B 3
C 0
D 1

The correct answer is:
1

Solution:
For $\mathrm{P}$, if $\mathrm{t}_{50 \%}=\mathrm{x}$ then $\mathrm{t}_{75 \%}=2 \mathrm{x}$

This is true only for first order reaction. So, order with respect to $\mathrm{P}$ is $1 .$ Further the graph shows that concentration of $\mathrm{Q}$ decreases with time. So rate, with respect to $\mathrm{Q},$ remains constant. Hence, it is zero order wrt Q. So, overall order is $1+0=1$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2016
In the reaction, $\mathrm{P}+\mathrm{Q} \longrightarrow ? \mathrm{R}+\mathrm{S}$

The time taken for $75 \%$ reaction of $P$ is twice the time taken for $50 \%$ reaction of $\mathrm{P}$. The concentration of $\mathrm{Q}$ varies with reaction time as shown in the figure. The overall order of the reaction is

Options:
A 2
B 3
C 0
D 1

The correct answer is:
1

Solution:
For $\mathrm{P}$, if $\mathrm{t}_{50 \%}=\mathrm{x}$ then $\mathrm{t}_{75 \%}=2 \mathrm{x}$

This is true only for first order reaction. So, order with respect to $\mathrm{P}$ is $1 .$ Further the graph shows that concentration of $\mathrm{Q}$ decreases with time. So rate, with respect to $\mathrm{Q},$ remains constant. Hence, it is zero order wrt Q. So, overall order is $1+0=1$

Topic: Chemistry, Chemical Kinetics, JEE Main
In the reaction:

Which of the following compounds will be formed?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2021
In the reaction X and Y are respectively.

Options:
A
B
C
D

The correct answer is:

Solution:
The given reaction is for preparation of DDT (dichlorodiphenyl trichloroethane). Complete reaction is given as follows.

(i)

(ii)

Dichloro-diphenyl-trichloroethane (DDT) is prepared by heating chloral and chlorobenzene in 1:2 ratio in the presence of conc. sulphuric acid.

DDT is used as an insecticide. It is also used as a mosquito repellant.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In the reaction X and Y are respectively.

Options:
A
B
C
D

The correct answer is:

Solution:
The given reaction is for preparation of DDT (dichlorodiphenyl trichloroethane). Complete reaction is given as follows.

(i)

(ii)

Dichloro-diphenyl-trichloroethane (DDT) is prepared by heating chloral and chlorobenzene in 1:2 ratio in the presence of conc. sulphuric acid.

DDT is used as an insecticide. It is also used as a mosquito repellant.

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2019
In the real number system, the equation x+3-4x-1+x+8-6x-1=1 has

Options:
A No solution
B Exactly two distinct solution
C Exactly four distinct solution
D Infinitely many solution

The correct answer is:
Infinitely many solution

Solution:
We have,x+3-4x-1+x+8-6x-1=1

(x-1)2-2×2x-1+4+(x-1)2-2×3x-1+9=1

⇒ (x-1-2)2+(x-1-3)2=1

⇒ |x-1-2|+|x-1-3|=1

⇒ |x-1-2|+|x-1-3|=(x-1-2)-(x-1-3)

We know that,

If |x-a|+|x-b|=(x-a)-(x-b)

then (x-a)(x-b)<0

∴ (x-1-2)(x-1-3)<0

⇒ 2<x-1<3⇒5<x<10

∴ Equation has infinite many solutions.

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2019
In the relation:
$\mathrm{P}$ is pressure, $\mathrm{Z}$ is distance, $\mathrm{k}$ is Boltzmann constant and $\theta$ is the temperature. The
dimensional formula of $\beta$ will be

Options:
A $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]$
B $\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{1}\right]$
C $\left[\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]$
D $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]$

The correct answer is:
$\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]$

Solution:
No Solution available

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2009
In the shown arrangement of the experiment of the meter bridge if $A C$ corresponding to null deflection of galvanometer is $\mathrm{x},$ what would be its value if the radius of the wire $\mathrm{AB}$ is doubled?

Options:
A $x$
B $x / 4$
C $4 x$
D $2 x$

The correct answer is:
$x$

Solution:
At null point $\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}=\frac{x}{100-x}$

If radius of the wire is doubled, then the resistance of $A C$ will change and also the resistance of $C B$ will change. But since

$\frac{R_{1}}{R_{2}}$ does not change so, $\frac{R_{3}}{R_{4}}$ should also not change at null point. Therefore the point $C$ does not change.

Topic: Physics, Experimental Physics, BITSAT, BITSAT 2016
In the shown situation, if middle portion of the lens is painted black, then

Options:
A complete image of the object will form with low intensity
B complete image of the object will form with high intensity
C incomplete image of the object will form with low intensity
D incomplete image of the object will form with high intensity

The correct answer is:
complete image of the object will form with low intensity

Solution:
If middle portion of lens is painted black, then less number of refracted rays will intersect, therefore intensity of the corresponding image of object will be less. Though, the rays coming from object refracted with small portion of lens, but complete image of object will be formed but of lesser intensity.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2019
In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $\mathrm{K},(\lambda$ being the wave length of light used). The intensity at a point where the path difference is $\lambda / 4,$ will be :

Options:
A $\mathrm{K}$
B $\mathrm{K} / 4$
C $\mathrm{K} / 2$
D Zero

The correct answer is:
$\mathrm{K} / 2$

Solution:
For path difference $\lambda,$ phase difference $=2 \pi \mathrm{rad}$

For path difference $\frac{\lambda}{4},$ phase difference

$$

=\frac{\pi}{2} \mathrm{rad}

$$

As $\mathrm{K}=4 \mathrm{I}_{0}$ so intensity at given point where

path difference is $\frac{\lambda}{4}$

$\mathrm{K}^{\prime}=4 \mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{4}\right)\left(\cos \frac{\pi}{4}=\cos 45^{\circ}\right)$

$=2 \mathrm{I}_{0}=\frac{\mathrm{K}}{2}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2016
In van der Waal's equation of state for a nonideal gas, the term that accounts for intermolecular forces is

Options:
A $(V-b)$
B $R T$
C $\left(P+\frac{a}{V^2}\right)$
D $(R T)^{-1}$

The correct answer is:
$\left(P+\frac{a}{V^2}\right)$

Solution:
$\left(P+\frac{a}{V^2}\right)(V-b)=R T$; Here $\left(P+\frac{a}{V^2}\right)$ represents the intermolecular forces.

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 2)
In which case the $\mathrm{NO}_{2}$ will attack at the meta position

Options:
A I, II, III
B II, IV
C II and III only
D II only

The correct answer is:
I, II, III

Solution:
$-\mathrm{CCl}_{3},-\mathrm{NO}_{2}$ and $-\mathrm{NH}_{3}$ are
$m$ -directing in nature.

Topic: Chemistry, Hydrocarbons, JEE Main
In which case the $\mathrm{NO}_{2}$ will attack at the meta position

Options:
A I, II, III
B II, IV
C II and III only
D II only

The correct answer is:
I, II, III

Solution:
$-\mathrm{CCl}_{3},-\mathrm{NO}_{2}$ and $-\mathrm{NH}_{3}$ are
$m$ -directing in nature.

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2012
In which of the following conversions, phosphorus pentachloride is used as the reagent?

Options:
A $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$
C $\mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{CH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}$
D $\mathrm{CH} \equiv \mathrm{CH} \longrightarrow \mathrm{CH}_{2}=\mathrm{CHCl}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$

Solution:
When ethyl alcohol is treated with $\mathrm{PCl}_{5}$, then ethyl chloride is formed.
$$
\begin{aligned}
\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{OH} &+\mathrm{PCl}_{5} \longrightarrow_{\Delta} \\
& \mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{Cl}+\mathrm{HCl}+\mathrm{POCl}_{3}
\end{aligned}
$$

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2010
In which of the following reactions, $\mathrm{H}_{2} \mathrm{O}_{2}$ is acting as a reducing agent?

Options:
A $\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{SO}_{2} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}$
B $2 \mathrm{KI}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{KOH}+\mathrm{I}_{2}$
C $\quad \mathrm{PbS}+4 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4}+4 \mathrm{H}_{2} \mathrm{O}$
D $\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Ag}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

The correct answer is:
$\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Ag}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

Solution:
$\mathrm{SO}_{2}$ changes to $\mathrm{H}_{2} \mathrm{SO}_{4}$ (O.N. changes from +4 to +6 oxidation $)$ $2 \mathrm{KI} \rightarrow \mathrm{I}_{2}(\mathrm{O} . \mathrm{S} .$ changes from -1 to 0
oxidation) $\mathrm{PbS} \rightarrow \mathrm{PbSO}_{4}(\mathrm{O} . \mathrm{S}$. changes from -2 to +6
oxidation) $\mathrm{Ag}_{2} \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{O} . \mathrm{S} .$ changes from +1 to 0
reduction)

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2010
In which of the following reactions, standard entropy change $\left(\Delta \mathrm{S}^{\circ}\right)$ is positive and standard Gibb's energy change $\left(\Delta \mathrm{G}^{\circ}\right)$ decreases sharply with increasing temperature?

Options:
A $\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})$
B $\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$
C $\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{MgO}(\mathrm{s})$
D $\frac{1}{2} \mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{CO}_{2}(\mathrm{~g})$

The correct answer is:
$\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})$

Solution:
Since, in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. $\Delta \mathrm{S}=+\mathrm{ve}$.
$\mathrm{C}(\mathrm{gr} .)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{S}^{\circ}=+\mathrm{ve}$
Since, $\Delta \overrightarrow{\mathrm{G}^{\circ}}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}$ hence the value of
$\Delta$ G decrease on increasing temperature.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2012
In which pair or pairs is the stronger bond found in the first species?
I. $\mathrm{O}_2^{2-}, \mathrm{O}_2$; II. $\mathrm{N}_2, \mathrm{~N}_2^{+}$ III. $\mathrm{NO}^{+}, \mathrm{NO}^{-}$

Options:
A
I only

B
II only

C
I and II only

D
II and III only

The correct answer is:

II and III only

Solution:
Larger the bond order, stronger the bond formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2022