datavorous/entrance-exam-dataset huggingface dataset:
What is the value of $n$ so that the angle between the lines having direction ratios (1,1,1) and $(1,-1, n)$ is $60^{\circ} ?$

Options:
A $\sqrt{3}$
B $\sqrt{6}$
C 3
D None of these

The correct answer is:
$\sqrt{6}$

Solution:
If $\left(l_{1}, m_{1}, n_{1}\right)$ and $\left(l_{2}, m_{2}, n_{2}\right)$ are the direction ratios then angle between the lines is
$\cos \theta=\frac{l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}$
Here $l_{1}=1, m_{1}=1, n_{1}=1$ and
$l_{2}=1, m_{2}=-1, n_{2}=n$ and $\theta=60^{\circ}$
$\therefore \cos 60^{\circ}=\frac{1 \times 1+1 \times(-1)+1 \times n}{\sqrt{h^{2}+1^{2}+1^{2} \times} \sqrt{1^{2}+1^{2}+y^{2}}}$
$\Rightarrow \frac{1}{2}=\frac{1}{\sqrt{3} \sqrt{2+n^{2}}} \Rightarrow 3\left(2+n^{2}\right)=4 n^{2}$
$\Rightarrow n^{2}=6 \Rightarrow n=\pm \sqrt{6}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2012
What is the $x$ -coordinate of the point on the curve $f(x)=\sqrt{x}(7 x-6),$ where the tangent is parallel to $\mathrm{x}$ -axis?

Options:
A $-\frac{1}{3}$
B $\frac{2}{7}$
C $\frac{6}{7}$
D $\frac{1}{2}$

The correct answer is:
$\frac{2}{7}$

Solution:
$f(x)=\sqrt{x}(7 x-6)=7 x^{3 / 2}-6 x^{1 / 2}$

$f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}$

When tangent is parallel to $x$ axis $f^{\prime}(x)=0$

or, $\quad \frac{21}{2} \mathrm{x}^{1 / 2}-3 \mathrm{x}^{-1 / 2}=0$

$\frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}$

or, $7 x=2 \Rightarrow x=\frac{2}{7}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2016
What is $\mathrm{X}$ in the following conversion?

Options:
A $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{H}_{2} \mathrm{O}$
B $\mathrm{HIO}_{4}$
C $\mathrm{PCC} / \mathrm{CH}_{2} \mathrm{Cl}_{2}$
D $\mathrm{OsO}_{4},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C.COOH},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}, \mathrm{OH}^{-}$

The correct answer is:
$\mathrm{PCC} / \mathrm{CH}_{2} \mathrm{Cl}_{2}$

Solution:
Pyridinium chloro chromate selectively oxidises a primary alcohol to an aldehyde

e.g.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2020
What is $x$ in the following reaction?
$\mathrm{Al}(\mathrm{s})+\mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow x+\mathrm{H}_2(\mathrm{~g})$

Options:
A $\mathrm{Na}_2\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}$
B $\mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}$
C $\mathrm{Na}_2\left[\mathrm{Al}(\mathrm{OH})_6\right]^{-}$
D $\mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_6\right]^{-}$

The correct answer is:
$\mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}$

Solution:
$2 \mathrm{Al}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow$ $2 \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}(\mathrm{aq})+3 \mathrm{H}_2(\mathrm{~g})$

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2023 (Memory Based Paper 1)
What should be the characteristic of fuse wire?

Options:
A High melting point, high specific resistance.
B Low melting point, low specific resistance.
C High melting point, low specific resistance.
D Low melting point, high specific resistance.

The correct answer is:
Low melting point, high specific resistance.

Solution:
Fuse wire should be such that it melts immediatley when strong current flows through the circuit. The same is possible if its melting point is low and resistivity is high.

Topic: Physics, Current Electricity, BITSAT, BITSAT 2013
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400 \mathrm{~km}$.

Options:
A $8.7 \times 10^{-7} \mathrm{rad} / \mathrm{s}$
B $7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s}$
C $6.7 \times 10^{-4} \mathrm{rad} / \mathrm{s}$
D $7.4 \times 10^{-3} \mathrm{rad} / \mathrm{s}$

The correct answer is:
$7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s}$

Solution:
True weight at equator, $W=m g$ Observed weight at equator,

$$

\mathrm{W}^{\prime}=\mathrm{mg}^{\prime}=\frac{3}{5} \mathrm{mg}

$$

At equator, latitude $\lambda=0$ Using the formula,

$$

\begin{array}{l}

\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{mR} \omega^{2} \cos ^{2} \lambda \\

\frac{3}{5} \mathrm{mg}=\mathrm{mg}-\mathrm{mR} \omega^{2} \cos ^{2} 0=\mathrm{mg}-\mathrm{mR} \omega^{2} \\

\Rightarrow \mathrm{mR} \omega^{2}=\mathrm{mg}-\frac{3}{5} \mathrm{mg}=\frac{2}{5} \mathrm{mg} \\

\therefore \quad \omega=\left(\frac{2}{5} \frac{\mathrm{g}}{\mathrm{R}}\right)^{1 / 2} \\

=\left(\frac{2 \times 9.8}{5 \times 6.4 \times 10^{6}}\right)^{1 / 2}=7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s}

\end{array}

$$

Topic: Physics, Gravitation, BITSAT, BITSAT 2016
What time in sec is required for depositing all the silver present in 125 ml of 1 M AgNO3 solution by passing a current of 241.25 A (Given 1 F= 96500 C)

Options:
A 10
B 50
C 100
D 1000

The correct answer is:
50

Solution:
Given, 125 ml of 1 M AgNO3

which is equal to 13.5 g of Ag

∵ 1000 ml of AgNO3=108 g of Ag

∴ 125 ml of AgNO3=108×1251000=13.5 g

Also, ∵ 108 g of Ag is deposited by =96500 C

∴ 13.5 g of Ag is deposited by =96500×13.5108=12062.5 C

∵ Q=it

∴ t (time) =Q/i=12062.50241.25, t=50 sec

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2019
What will be the acceleration due to gravity at a depth $d$, where $g$ is acceleration due to gravity on the surface of earth?

Options:
A
$\frac{g}{\left[1+\frac{d}{R}\right]^2}$

B
$g\left[1-\frac{2 d}{R}\right]$

C
$\frac{g}{\left[1-\frac{d}{R}\right]^2}$

D
$g\left[1-\frac{d}{R}\right]$

The correct answer is:

$g\left[1-\frac{d}{R}\right]$

Solution:
Acceleration due to gravity at the surface of earth,
$$
g=\frac{G M}{R^2}=\frac{4}{3} \pi \rho G R
$$
$$
\left[\because M=\frac{4}{3} \pi R^3 \cdot \rho\right]
$$
Where, $\rho=$ density of earth.
Acceleration due to gravity at depth $d$ from the surface of earth,
$$
g^{\prime}=\frac{4}{3} \pi \rho G(R-d)
$$
From Eqs. (i) and (ii), we get
$$
g^{\prime}=g\left[1-\frac{d}{R}\right]
$$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2022
What will be the correct structural formula of product for the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What will be the correct structural formula of product for the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2013
What will be the main product when acetylene reacts with hypochlorous acid?

Options:
A Trichloroacetaldehyde
B Acetaldehyde
C Dichloroacetaldehyde
D Chloro acetaldehyde

The correct answer is:
Dichloroacetaldehyde

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2013
What will be the product in the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
What will be the product in the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2011
What will happen when D-(+)-glucose is treated with methanolic - HCl followed by Tollens' reagent?

Options:
A A black ppt. will be formed
B A red ppt. will be formed
C A green colour will appear
D No characteristic colour or ppt. will be formed.

The correct answer is:
No characteristic colour or ppt. will be formed.

Solution:
Reaction of $\mathrm{D}-(+)$ -glucose with methanolic $-\mathrm{HCl}$ leads to formation of methyl glucoside $\left(\mathrm{C}_{1}-\mathrm{OH}\right.$ group is methylated) which, being acetal, is not hydrolysable by base, so it will not respond Tollens'reagent.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2016
When $0.50 Ã$ X-rays strike a material, the photoelectrons from the $\mathrm{k}$ shell are observed to move in a circle of radius $23 \mathrm{~mm}$ in a magnetic field of $2 \times 10^{-2}$ tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?

Options:
A $3.5 \mathrm{keV}$
B $6.2 \mathrm{keV}$
C $2.9 \mathrm{keV}$
D $5.5 \mathrm{keV}$

The correct answer is:
$6.2 \mathrm{keV}$

Solution:
As we know,

$$

\mathrm{F}=\mathrm{qvB}=\mathrm{m} \frac{\mathrm{v}^{2}}{\mathrm{R}} \Rightarrow \mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{BR}

$$

The kinetic energy of the photoelectron

$$

\begin{array}{l}

=\frac{1}{2} m v^{2}=\frac{1}{2} \frac{e^{2} B^{2} R^{2}}{m} \\

=\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^{2}\left(2 \times 10^{-2}\right)^{2}\left(23 \times 10^{-3}\right)^{2}}{\left(9.1 \times 10^{-31}\right)} \\

=2.97 \times 10^{-15} \mathrm{~J} \\

=\frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}}=18.36 \mathrm{keV}

\end{array}

$$

Fnergy of the incident $\mathrm{n}=\frac{\mathrm{hc}}{\lambda}$

$$

=\frac{12.4}{0.50}=24.8 \mathrm{keV}

$$

Therefore, Binding energy $=24.8-18.6=$ $6.2 \mathrm{keV}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2015
When a capacitor is fully charged as shown in the following figure, then current drawn from the cell is

Options:
A 1 mA
B 0.5 mA
C 2 mA
D 5 mA

The correct answer is:
0.5 mA

Solution:
When capacitor is fully charged, then it acts like a open switch, therefore 8 kΩ resistor has no use. So, rest of the resistances 4 kΩ, 7 kΩ and 9 kΩ will be in series. Hence, current drawn from cell

I=104+7+9×103=1020×103=0.5 mA

Topic: Physics, Capacitance, JEE Main
When a capacitor is fully charged as shown in the following figure, then current drawn from the cell is

Options:
A 1 mA
B 0.5 mA
C 2 mA
D 5 mA

The correct answer is:
0.5 mA

Solution:
When capacitor is fully charged, then it acts like a open switch, therefore 8 kΩ resistor has no use. So, rest of the resistances 4 kΩ, 7 kΩ and 9 kΩ will be in series. Hence, current drawn from cell

I=104+7+9×103=1020×103=0.5 mA

Topic: Physics, Capacitance, BITSAT, BITSAT 2019
When a crystal of caustic soda is exposed to air, a liquid layer is deposited because :

Options:
A Crystal loses water
B Crystal absorbs moisture and $\mathrm{CO}_{2}$
C Crystal melts
D Crystal sublimes

The correct answer is:
Crystal absorbs moisture and $\mathrm{CO}_{2}$

Solution:
It is hygroscopic and deliquescent and hence absorbs moisture and $\mathrm{CO}_{2}$ to form $\mathrm{Na}_{2} \mathrm{CO}_{3}$
$$
2 \mathrm{NaOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}
$$

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2011
When a mica sheet of thickness 7 microns and $\mu=1.6$ is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe. The wavelength of light used will be

Options:
A $4000 \mathrm{~A}$
B $5000 \AA$
C $6000 \mathrm{~A}$
D $7000 \AA$

The correct answer is:
$6000 \mathrm{~A}$

Solution:
$\lambda=\frac{(\mu-1) \mathrm{t}}{\mathrm{n}}$
According to question
$$
\mathrm{n}=7, \mu=1.6, \mathrm{t}=7 \times 10^{-6} \mathrm{~meter}
$$
From eqs. (1) and $(2), \lambda=6 \times 10^{-7}$ meter

Topic: Physics, Wave Optics, BITSAT, BITSAT 2009
When a string is divided into three segments of length $l_{1}, l_{2},$ and $l_{3}$, the fundamental frequencies of these three segments are $v_{1}, v_{2}$ and $v_{3}$ respectively. The original fundamental frequency

(v) of the string is

Options:
A $\quad \sqrt{v}=\sqrt{v_{1}}+\sqrt{v_{2}}+\sqrt{v_{3}}$
B $v=v_{1}+v_{2}+v_{3}$
C $\frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$.
D $\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{v_{1}}}+\frac{1}{\sqrt{v_{2}}}+\frac{1}{\sqrt{v_{3}}}$

The correct answer is:
$\frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$.

Solution:
Fundamental frequency is given by

$$

v=\frac{1}{2 l} \sqrt{\frac{T}{\mu}} \Rightarrow v \propto \frac{1}{l} \Rightarrow P \propto \frac{1}{v}

$$

Since, $P$ divided into $l_{1}, l_{2}$ and $l_{3}$ segments Here $l=l_{1}+l_{2}+l_{3}$

So $\quad \frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2014
When aqueous solutions of borax is acidified with HCl, we get

Options:
A B2H6
B H3BO3
C B2O3
D All of these

The correct answer is:
H3BO3

Solution:
When aqueous solution of borax Na2B4O7 is acidified
with HCl, we get H3BO3 and NaCl as follows

Na2B4O7Borox+2HCl+5H2O⟶2NaCl+4H3BO3Boric acid

Boric acid is not a proticacid it does not give H+ ion accepts a pair of electrons and acts as a Lewis acid.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2017
When $\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{CH}_{4}$ and $\mathrm{C}_{2} \mathrm{H}_{4}$ pass through a test tube which has ammonical $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$, find out which gas comes out unaffected from test tube?

Options:
A $\mathrm{C}_{2} \mathrm{H}_{2}$ and $\mathrm{CH}_{4}$
B $\mathrm{C}_{2} \mathrm{H}_{2}$ and $\mathrm{C}_{2} \mathrm{H}_{4}$
C $\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{CH}_{4}$
D $\mathrm{C}_{2} \mathrm{H}_{2}$

The correct answer is:
$\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{CH}_{4}$

Solution:
$\mathrm{C}_{2} \mathrm{H}_{2}$ forms copper acetylide with ammonical $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$

$\begin{array}{l}

\mathrm{Cu}_{2} \mathrm{Cl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{Cu}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{NH}_{4} \mathrm{Cl} \\

\mathrm{Cu}_{2} \mathrm{O}+\mathrm{C}_{2} \mathrm{H}_{2} \rightarrow \underset{\text { (copper acetylide) }}{\mathrm{Cu}_{2} \mathrm{C}_{2}}+\mathrm{H}_{2} \mathrm{O}

\end{array}$

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2021
When chlorine water is exposed to sunlight, $\mathrm{O}_{2}$ is liberated. Hence,

Options:
A hydrogen has little affinity to $\mathrm{O}_{2}$
B hydrogen has more affinity to $\mathrm{O}_{2}$
C hydrogen has more affinity to chlorine
D it is a reducing agent

The correct answer is:
hydrogen has more affinity to chlorine

Solution:
$\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{HCl}+\frac{1}{2} \mathrm{O}_{2}$
Hydrogen has more affinity for chlorine.

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2012
When dil. sulphuric acid reacts with aqueous solution of potassium chromate, the colour changes from yellow to orange. This shows that

Options:
A chromate ions reduced
B chromate ions are oxidised
C monocentric complex is converted into dicentric complex
D oxygen gets removed from chromate ions

The correct answer is:
monocentric complex is converted into dicentric complex

Solution:
On acidification of potassium chromate solution, yellow colour changes to orange colour due to formation of dichromate which suggests that monocentric complex is converted into dicentric complex.

2K2CrO4+H2SO4Dil.⟶K2Cr2O7+ K2SO4+H2O

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2018
When esters are hydrolysed the product gives hydrogen ions. The product which gives hydrogen ion is

Options:
A acid
B alcohol
C both
D either acid or alcohol

The correct answer is:
acid

Solution:
When esters are hydrolysed, then acid and alcohol are formed, where acid gives hydrogen ion.

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2009
When ethylamine react with sodium metal, the gas evolved is

Options:
A $\mathrm{H}_{2}$
B $\mathrm{C}_{2} \mathrm{H}_{5}$
C $\mathrm{N}_{2}$
D $\mathrm{NH}_{3}$

The correct answer is:
$\mathrm{H}_{2}$

Solution:
When ethylamine is heated with sodium metal, then hydrogen gas is evolved. $2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}+2 \mathrm{Na} \longrightarrow$
$$
2 \mathrm{C}_{2} \mathrm{H}_{5} \overline{\mathrm{N}} \mathrm{HNa}^{+}+\mathrm{H}_{2} \uparrow
$$

Topic: Chemistry, Amines, BITSAT, BITSAT 2010
When H2S gas is passed into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ ion in an acidified aqueous solution, the precipitates formed are

Options:
A CuS and HgS
B

MnS and CuS

C MnS and NiS
D NiS and HgS

The correct answer is:
CuS and HgS

Solution:
In presence of acidic medium, ionisation of H2S is supressed so less number of S2- ions are produced. So only those sulphides are precipitated which have low solubility product Ksp value.

Thus, in acidic medium, H2S is very feebly ionised giving very small concentration of sulphide ion for precipitation. Therefore, the most insoluble salts CuS and HgS are precipitated.

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2018
When $\mathrm{H}_{2} \mathrm{~S}$ gas is passed through the $\mathrm{HCl}$ containing aqueous solution of $\mathrm{CuCl}_{2}, \mathrm{HgCl}_{2}$, $\mathrm{BiCl}_{3}$ and $\mathrm{CoCl}_{2}$ it does not precipitate out

Options:
A $\mathrm{CuS}$
B $\mathrm{HgS}$
C $\mathrm{Bi}_{2} \mathrm{~S}_{3}$
D $\mathrm{CoS}$

The correct answer is:
$\mathrm{CoS}$

Solution:
$\operatorname{CoS}$ is not precipitated in acidic medium.

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2010
When light is refracted, which of the following does not change ?

Options:
A Wavelength
B Frequency
C Velocity
D Amplitude

The correct answer is:
Frequency

Solution:
Frequency does not change on refraction.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2013
When $\mathrm{NH}_{3}$ is passed over heated metal $\mathrm{A}$, its amide is formed. The metal is

Options:
A $\mathrm{Mg}$
B $\mathrm{K}$
C $\mathrm{Al}$
D $\mathrm{Pb}$

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

Solution:
When Potassium is treated with ammonia, then potassium amide is obtained

$\mathrm{K}+\mathrm{NH}_{3} \longrightarrow \underset{\text { Pot. amide }}{\mathrm{KNH}_{2}}+\frac{1}{2} \mathrm{H}_{2}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2013
When photon of energy 4.0 eV strikes the surface of a metal A, the elected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectron liberated from another metal B by photon of energy 4.50 eV is TB=TA-1.50 eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then choose the correct statement(s).

Options:
A The work function of A is 1.50 eV
B The work function of B is 4.0 eV
C TA=3.2 eV
D All of the above

The correct answer is:
The work function of B is 4.0 eV

Solution:
From Einstein photoelectric equation,
E=ϕ0+KEmax

For metal A 4=ϕA+TA....i
For metal B 4.5=ϕB+TA-1.5....ii

From Eq. (i) and (ii), wo got
ϕB-ϕA=2

Now, according to de-Broglie hypothesis,
λA=hmv=h2mTA

Similarly, λB=h2mTB

∴ λAλB=TBTA=TA-1.5TA=1-1.5TA1/2

122=1-1.5TA

On solving, TA=2.0eV

or ϕA=4-TA=4-2=2.0eV

ϕB=6-TA=6-2=4.0eV

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2018
When rain is accompanied by a thunderstorm, the collected rain water will have a $\mathrm{pH}$ value:

Options:
A Slightly lower than that of rain water without thunderstorm.
B Slightly higher than that when the thunderstorm is not there.
C Uninfluenced by occurrence of thunderstorm.
D Which depends upon the amount of dust in air.

The correct answer is:
Slightly lower than that of rain water without thunderstorm.

Solution:
Normal rain water has pH 5.6. Thunderstorm results in the formation $\mathrm{NO}$ and $\mathrm{HNO}_{3}$ which lowers the $\mathrm{pH}$

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2014
When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution, the ratio of volumes of hydrogen evolved is:

Options:
A 1: 1
B 1: 2
C 2: 1
D 9: 4

The correct answer is:
1: 1

Solution:
$\mathrm{Zn}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{H}_{2}$

$\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}$

$\therefore$ Ratio of volumes of $\mathrm{H}_{2}$ evolved is 1: 1

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2013
When the first ionisation energies are plotted against atomic number, the peaks are occupied by

Options:
A Alkali metals
B Rare gases
C Halogens
D Transition elements

The correct answer is:
Rare gases

Solution:
Rare gases; as the $\mathrm{e}^{-}$ is to be removed from stable configuration.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2013
When the heat of a reaction at constant pressure is -25×10-3 cal and entropy change for the reaction is 7.4 cal deg-1, it is predicted that the reaction at 250C is

Options:
A revesible
B spontaneous
C non-spontaneous
D irreversible

The correct answer is:
spontaneous

Solution:
Heat at constant pressure means enthalpy. i.e., ΔG=ΔH-TΔS

ΔH=2.5×103cal

ΔS=7.4caldeg-1

T=298 K

ΔG=ΔH-TΔS

=-25×103-298×7.4

=-4705.2 cal

As the value of ΔG is −4705.2 cal, which is a negative value, so this reaction will be spontaneous.

Hence, the process is spontaneous.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2018
When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(b)

Solution:
In electrolytic purification cathode is of pure metal and anode is of impure metal. At anode: $\mathrm{M} \rightarrow \mathrm{M}^{\mathrm{n}+}+\mathrm{n} \overline{\mathrm{e}}$
At cathode: $\mathrm{M}^{\mathrm{n}+}+\mathrm{n} \overline{\mathrm{e}} \rightarrow \mathrm{M}$.
The pure metal is thus deposited at cathode.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2009
When the temperature is raised, the viscosity of liquid decreases this is because

Options:
A decreased volume of the solution.
B increased attraction between molecules.
C decreased covalent and hydrogen bond forces.
D increase in temperature increase the average kinetic energy of molecules which overcomes the attractive force between them.

The correct answer is:
increased attraction between molecules.

Solution:
No Solution available

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2010
Which alkene on ozonolysis gives $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$ and

Options:
A
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{3}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}$
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2012
Which among the following actinoids does not have stable electronic configuration

Options:
A Protactinium
B Nobelium
C Americium
D Lawrencium

The correct answer is:
Protactinium

Solution:
Protactinium Pa is the element that does not have stable electronic configuration.

Protactinium Pa=Rn5f26d17s2 (Atomic number =91)

Nobelium No=Rn5f146d07s2 (Atomic number =102)

Lawrencium Lr=Rn5f146d17s2 (Atomic number =103)

Americium Am=Rn5f76d07s2 (Atomic number =95)

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2018
Which among the following is not an antibiotic?

Options:
A Penicillin
B Oxytocin
C $\quad$ Ofloxacin
D Tetracycline

The correct answer is:
Oxytocin

Solution:
Oxytocin is a hormone (nanopetide) which contracts uterus after the child birth and produces lactation in the mammary glands.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2014
Which compound has antifluorite structure?

Options:
A
$\mathrm{MnO}_4$

B
$\mathrm{Na}_2 \mathrm{O}$

C
$\mathrm{Na}_2 \mathrm{O}_2$

D
$\mathrm{Li}_2 \mathrm{O}_2$

The correct answer is:

$\mathrm{Na}_2 \mathrm{O}$

Solution:
Anti-fluorite structure refers to an anion array with tetrahedral cations and the compound having $A_2 B$ formula have anti-fluorite structure. So, $\mathrm{Na}_2 \mathrm{O}$ has an anti-fluorite structure in which $\mathrm{Na}^{+}$ions occupy all the tetrahedral voids and $\mathrm{O}^{2-}$occupy half of the cubic holes.

Topic: Chemistry, Solid State, BITSAT, BITSAT 2022
Which form coloured salts:

Options:
A Non-metals
B Metals
C p-block elements
D Transitional elements

The correct answer is:
Transitional elements

Solution:
Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid state and in aqueous solution in contrast to the compounds of $s$ and $p$ -block elements due to the presence of incomplete d-subshell.

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2014