datavorous/entrance-exam-dataset huggingface dataset:
Phosphate pollution is caused by:

Options:
A Sewage and agricultural fertilizers.
B Weathering of phosphate rocks only.
C $\quad$ Agriculutral fertilizers only.
D Phosphate rocks and sewage.

The correct answer is:
Sewage and agricultural fertilizers.

Solution:
Phosphate pollution is caused by sewage and agricultural fertilizers.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2010
Phosphine is not obtained by the reaction

Options:
A White $P$ is heated with $\mathrm{NaOH}$
B $\operatorname{Red} P$ is heated with $\mathrm{NaOH}$
C $\mathrm{Ca}_{3} \mathrm{P}_{2}$ reacts with water
D Phosphorus trioxide is boiled with water

The correct answer is:
$\operatorname{Red} P$ is heated with $\mathrm{NaOH}$

Solution:
Red P does not react with $\mathrm{NaOH}$ to give
$\mathrm{PH}_{3}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2011
Phosphine is not obtained by which of the following reaction?

Options:
A White $\mathrm{P}$ is heated with $\mathrm{NaOH}$.
B Red P is heated with $\mathrm{NaOH}$.
C $\mathrm{Ca}_{3} \mathrm{P}_{2}$ reacts with water.
D Both (a) and (c).

The correct answer is:
Red P is heated with $\mathrm{NaOH}$.

Solution:
Red P doesnot react with $\mathrm{NaOH}$ to give $\mathrm{PH}_{3}$.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2020
Photochemical smog formed in congested metropolitan cities mainly consists of

Options:
A
ozone, peroxyacetyl nitrate and $\mathrm{NO}_x$
B
smoke, peroxyacetyl nitrate and $\mathrm{SO}_2$
C
hydrocarbons, $\mathrm{SO}_2$ and $\mathrm{CO}_2$
D
hydrocarbons, ozone and $\mathrm{SO}_x$

The correct answer is:
ozone, peroxyacetyl nitrate and $\mathrm{NO}_x$

Solution:
Photochemical smog $\Rightarrow \mathrm{O}_3, \mathrm{PAN}, \mathrm{NO}_x$

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Plots showing the variation of the rate constant $(k)$ with temperature $(T)$ are given below. The plot that follows Arrhenius equation is

Options:
A

B

C

D

The correct answer is:

Solution:
As per Arrhenius equation $\left(k=A \mathrm{e}^{-E_a / R T}\right)$, the rate constant increases exponentially with temperature.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Point out the false statement :

Options:
A
The colloidal solution of a liquid in liquid is called gel
B
Hardy Schulze rule is related with coagulation
C
Brownian movement and Tyndall effect are shown by colloidal system
D
Gold number is a measure of the protective power of lyophilic colloid

The correct answer is:
The colloidal solution of a liquid in liquid is called gel

Solution:
Colloid of liquid in liquid is called emulsion. Colloid of liquid in solid is gel.

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Polymer used in bullet proof glass is

Options:
A Lexan
B PMMA
C Nomex
D Kevlar

The correct answer is:
PMMA

Solution:
PMMA is used in bullet proof glass

Topic: Chemistry, Polymers, BITSAT, BITSAT 2015
Position of non-polar and polar part in micelle is

Options:
A polar at outer surface and non-polar at inner surface.
B polar at inner surface and non-polar at outer surface.
C distributed all over the surface.
D present in the surface only.

The correct answer is:
polar at outer surface and non-polar at inner surface.

Solution:

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2011
Prabhat wants to invest the total amount of $Rs 15,000$ in saving certificates and national saving bonds. According to rules, he has to invest at least $Rs 2000$ in saving certificates and $Rs 2500$ in national saving bonds. The interest rate is $8 \%$ on saving certificate and $10 \%$ on national saving bonds per annum. He invest $Rs$ x in saving certificate and $Rs$ y in national saving bonds. Then the objective function for this problem is

Options:
A $0.08 \mathrm{x}+0.10 \mathrm{y}$
B $\frac{x}{2000}+\frac{y}{2500}$
C $2000 \mathrm{x}+2500 \mathrm{y}$
D $\frac{x}{8}+\frac{y}{10}$

The correct answer is:
$0.08 \mathrm{x}+0.10 \mathrm{y}$

Solution:
The function is given by profit function
$$
=x \cdot \frac{8}{100}+y \times \frac{10}{100}=0.08+0.10 y .
$$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2012
Primary amines can be distinguished from secondary and tertiary amines by reacting with

Options:
A
Chloroform and alcoholic $\mathrm{KOH}$
B
Methyl iodide
C
Chloroform alone
D
Zinc Dust

The correct answer is:
Chloroform and alcoholic $\mathrm{KOH}$

Solution:
$1^{\circ}$ amines (aliphatic and aromatic) react with $\mathrm{CHCl}_3 / \mathrm{KOH}$ to yield isocyanide (foul smelling) This is known as carbylamine test which is not given by $2^{\circ}$ and $3^{\circ}$ amines.

Topic: Chemistry, Amines, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Product (D) in above reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
(A)

(B)

(C)

(D)

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2020
Proton (p) and electron (e) will have same deBroglie wavelength when the ratio of their momentum is (assume, $m_p=1849 \mathrm{~m}_{\mathrm{e}}$ )

Options:
A $1: 43$
B $43: 1$
C $1: 1849$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
De Broglie wavelength is $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}$
$\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}} \Rightarrow \mathrm{p}_{\mathrm{e}}=\mathrm{p}_{\mathrm{p}}$
$\therefore \frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\mathrm{e}}}=\frac{1}{1}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Radius of the largest circle which passes through the focus of the parabola y2=4x and contained in it, is

Options:
A 8
B 4
C 2
D 5

The correct answer is:
4

Solution:
Let r be the radius of the largest circle passing through the focus 1,0 of y2=4x

Clearly, centre of the circle will be on X-axis and its coordinates are 1+r,0.

The equation of the circle, is

x-1-r2+y2=r2.

It touches y2=4x. Therefore, the equation x-r-12+4x=r2 must have equal roots.

∴ 41-r2-42r+1=0

⇒r=4

Hence, option (b) is correct.

Topic: Mathematics, Circle, JEE Main
Radius of the largest circle which passes through the focus of the parabola y2=4x and contained in it, is

Options:
A 8
B 4
C 2
D 5

The correct answer is:
4

Solution:
Let r be the radius of the largest circle passing through the focus 1,0 of y2=4x

Clearly, centre of the circle will be on X-axis and its coordinates are 1+r,0.

The equation of the circle, is

x-1-r2+y2=r2.

It touches y2=4x. Therefore, the equation x-r-12+4x=r2 must have equal roots.

∴ 41-r2-42r+1=0

⇒r=4

Hence, option (b) is correct.

Topic: Mathematics, Circle, BITSAT, BITSAT 2018
Ratio of thermal energy released in two resistor $R$ and $3 R$ connected in parallel in an electric circuit is :

Options:
A
3 : 1
B
1 : 1
C
1 : 3
D
1 : 27

The correct answer is:
3 : 1

Solution:
Thermal energy is given by
$\mathrm{H}=\mathrm{P} \times \mathrm{t}=\frac{\mathrm{V}^2}{\mathrm{R}} \times \mathrm{t}$
Here, voltage $\mathrm{V}$ is same.
$\therefore \quad \frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}}}{\frac{\mathrm{V}^2 \mathrm{t}}{3 \mathrm{R}}}=3: 1$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Reductive ozonolysis of CH32C=CCH32 followed by hydrolysis gives

Options:
A only one type of ketone
B only one type of aldehyde
C two types of ketone
D two types of aldehyde

The correct answer is:
only one type of ketone

Solution:
Reductive ozonolysis of CH32C=CCH32 followed by hydrolysis gives only one type of ketones because of symmetry across the C=C and bonding of two same alkyl groups CH3 with both the carbons bonded through double bond.

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2017
Reflection of the line x-1-1=y-23=z-41 in the plane x+y+z=7 is

Options:
A x-13=y-21=z-41
B x-1-3=y-2-1=z-41
C x-1-3=y-21=z-4-1
D x-13=y-21=z-4-1

The correct answer is:
x-1-3=y-21=z-4-1

Solution:
Given line x-1-1=y-23=z-41 is passes through P(1,2,4).

To find the reflection of line we need one more point on the line clearly M(0,5,5) also lie on line.

Let N(α,β,γ) be the reflection M in the x+y+z=7

∴ α2+β+52+γ+52=7

⇒ α+β+γ=4

Also, MN is perpendicular, i.e. parallel to the normal of the plane

∴ α1=β-51=γ-51=λ

⇒ α=λ,β=λ+5,γ=λ+5

∴ λ+λ+5+λ+5=4

⇒ λ=-2

∴ N=(-2,3,5)

Equation PN is x-1-3=y-21=z-4-1

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2019
Resolving power of the telescope will be more, if the diameter of the objective is

Options:
A larger
B smaller
C it does not depends on diameter
D None of these

The correct answer is:
larger

Solution:
No Solution

Topic: Physics, Ray Optics, BITSAT, BITSAT 2014
Rolle's theorem is not applicable for the function f(x)=|x| in the interval [-1,1] because

Options:
A f'(1) does not exist
B f'(-1) does not exist
C f(x) is discontinuous at x=0
D f'(0) does not exist

The correct answer is:
f'(0) does not exist

Solution:
We have,

f(x)=|x|

We know that the function is not differentiable at the sharp edges. Hence, fx is not differentiable at x=0. Therefore, Rolle's theorem is not applicable for the function

f(x)=|x| in [-1,1]

∵ f'(0) does not exist.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
$\mathrm{S}$ and $\mathrm{T}$ are the foci of an ellipse and $\mathrm{B}$ is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{2}{3}$

The correct answer is:
$\frac{1}{2}$

Solution:
Let eq. of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $\mathrm{S}$ is $(-\mathrm{ae}, 0), \mathrm{T}$ is $(\mathrm{ae}, 0)$ and $\mathrm{B}$ is $(0, \mathrm{~b})$

$\Rightarrow \mathrm{SB}=\sqrt{(0+\mathrm{ae})^{2}+\mathrm{b}^{2}}$

Also $\mathrm{SB}^{2}=\mathrm{ST}^{2} \Rightarrow 4 \mathrm{a}^{2} \mathrm{e}^{2}=\mathrm{a}^{2} \mathrm{e}^{2}+\mathrm{b}^{2}$

$\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)=a^{2}-a^{2} e^{2}$

$\Rightarrow 4 a^{2} e^{2}=a^{2} \Rightarrow e^{2}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$

Topic: Mathematics, Ellipse, JEE Main
$\mathrm{S}$ and $\mathrm{T}$ are the foci of an ellipse and $\mathrm{B}$ is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{2}{3}$

The correct answer is:
$\frac{1}{2}$

Solution:
Let eq. of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $\mathrm{S}$ is $(-\mathrm{ae}, 0), \mathrm{T}$ is $(\mathrm{ae}, 0)$ and $\mathrm{B}$ is $(0, \mathrm{~b})$

$\Rightarrow \mathrm{SB}=\sqrt{(0+\mathrm{ae})^{2}+\mathrm{b}^{2}}$

Also $\mathrm{SB}^{2}=\mathrm{ST}^{2} \Rightarrow 4 \mathrm{a}^{2} \mathrm{e}^{2}=\mathrm{a}^{2} \mathrm{e}^{2}+\mathrm{b}^{2}$

$\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)=a^{2}-a^{2} e^{2}$

$\Rightarrow 4 a^{2} e^{2}=a^{2} \Rightarrow e^{2}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2013
$\mathrm{S}-\mathrm{S}$ bond is not present in

Options:
A $\mathrm{S}_2 \mathrm{O}_4^{2-}$
B $\mathrm{S}_2 \mathrm{O}_5^{2-}$
C $\mathrm{S}_2 \mathrm{O}_3^{2-}$
D $\mathrm{S}_2 \mathrm{O}_7^{2-}$

The correct answer is:
$\mathrm{S}_2 \mathrm{O}_7^{2-}$

Solution:
$\mathrm{S}_2 \mathrm{O}_7^{2-}$ has $\mathrm{S}-\mathrm{O}-\mathrm{S}$ bond.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2023 (Memory Based Paper 1)
Schotten-Baumann reaction is a reaction of phenols with

Options:
A benzoyl chloride and sodium hydroxide.
B acetyl chloride and sodium hydroxide.
C salicylic acid and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$.
D acetyl chloride and conc $\mathrm{H}_{2} \mathrm{SO}_{4}$.

The correct answer is:
benzoyl chloride and sodium hydroxide.

Solution:

Topic: Chemistry, Amines, JEE Main
Schotten-Baumann reaction is a reaction of phenols with

Options:
A benzoyl chloride and sodium hydroxide.
B acetyl chloride and sodium hydroxide.
C salicylic acid and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$.
D acetyl chloride and conc $\mathrm{H}_{2} \mathrm{SO}_{4}$.

The correct answer is:
benzoyl chloride and sodium hydroxide.

Solution:

Topic: Chemistry, Amines, BITSAT, BITSAT 2011
Select the correct statement about the complex
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Br}$

Options:
A
Its ionisation isomer is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Br}\right] \mathrm{SO}_4$.

B
It gives yellow precipitate with $\mathrm{AgNO}_3$.

C
Its ionisation isomer give while precipitate with $\mathrm{BaCl}_2$

D
All the above are correct statements.

The correct answer is:

All the above are correct statements.

Solution:

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2022
Select the incorrect statement.

Options:
A Equanil is used to control depression and hypertension.
B Mifepristone is a synthetic steroid used as "moming after pill".
C 0.2 per cent solution of phenol is an antiseptic while its 1.0 per cent solution is a disinfectant.
D A drug which kills the organism in the body is called bacteriostatic.

The correct answer is:
A drug which kills the organism in the body is called bacteriostatic.

Solution:
Bacteriostatic drugs inhibit the growth of organisms while bactericidal drugs kill the microorganisms.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2020
Semiconductor materials like Si and Ge are usually purified by

Options:
A distillation
B zone refining
C liquation
D electrolytic refining

The correct answer is:
zone refining

Solution:
Semiconductor materials like Si and Ge are usually purified by zone refining. Zone refining is based on the principle of fractional crystallisation i.e. difference in

states of metal, so that the zones of impurities are formed and finally removed.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2015
Sodium nitroprusside, when added to an alkaline solution of sulphide ions, produce a

Options:
A red colouration
B blue colouration
C purple colouration
D brown colouration

The correct answer is:
purple colouration

Solution:
This test is used for the presence of sulphur in any compound. When sodium nitroprusside, react with sulphide ion, e.g.Na2 S the following reaction takes place

Na2S+Na2FeCN5NO⟶Na4FeCN5NO5Sod. nitroprusside purple coloration sod. thionitroprusside

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2019
Sodium peroxide in contact with moist air turns white due to formation of:

Options:
A $\mathrm{Na}_{2} \mathrm{CO}_{3}$
B $\mathrm{Na}_{2} \mathrm{O}$
C $\mathrm{NaOH}$
D $\mathrm{NaHCO}_{3}$

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

Solution:
$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\frac{1}{2} \mathrm{O}_{2}$

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2010
Sodium sulphate is soluble in water but barium sulphate is insoluble because

Options:
A hydration energy of Na2SO4 is more than of its lattice energy
B lattice energy of BaSO4 is more than its hydration energy
C Both (a) and (b)
D None of the above

The correct answer is:
Both (a) and (b)

Solution:
For any solution to become soluble in water hydration energy of the solution should me more than its lattice energy. Here the hydration energy of sodium sulphate is more than lattice energy but in case of barium sulphate lattice energy of barium sulphate is more than its hydration energy.

For dissolution hydration energy must be more higher (-ve) value than of lattice energy. Thus, (c) is the correct answer.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2017
Solution of $2^x+2^{|x|} \geq 2 \sqrt{2}$ is

Options:
A $\left(-\infty, \log _2(\sqrt{2}+1)\right)$
B $(0, \infty)$
C $\left(\frac{1}{2}, \log _2(\sqrt{2}-1)\right)$
D $\left(-\infty, \log _2(\sqrt{2}-1)\right] \cup\left[\frac{1}{2}, \infty\right)$

The correct answer is:
$\left(-\infty, \log _2(\sqrt{2}-1)\right] \cup\left[\frac{1}{2}, \infty\right)$

Solution:
$2^x+2^{|x|} \geq 2 \sqrt{2}$ ...(i)
Case I: $x \geq 0$, then Eq. (i) becomes
$2^x+2^x \geq 2 \sqrt{2}$
$\Rightarrow 2^x \geq \sqrt{2} \Rightarrow x \geq \frac{1}{2}$
Case II: $x < 0$, then Eq. (i) becomes
$2^x+2^{-x} \geq 2 \sqrt{2}$
$\Rightarrow t+\frac{1}{t} \geq 2 \sqrt{2}$, where $2^x=t$
$\Rightarrow t^2-2 \sqrt{2} t+1 \geq 0$
$\Rightarrow x \leq \log _2(\sqrt{2}-1)$
Also, $0 < \sqrt{2}-1 < 1, \log _2(\sqrt{2}-1) < 0$.
$\therefore$ The solution is
$\left.\left(-\infty, \log _2(\sqrt{2}-1)\right)\right] \cup\left[\frac{1}{2}, \infty\right)$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Solution of differential equation

$$

x^{2}=1+\left(\frac{x}{y}\right)^{-1} \frac{d y}{d x}+\frac{\left(\frac{x}{y}\right)^{-2}\left(\frac{d y}{d x}\right)^{2}}{2 !}+\frac{\left(\frac{x}{y}\right)^{-3}\left(\frac{d y}{d x}\right)^{3}}{3 !}+\ldots \text { is }

$$

Options:
A $y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$
B $y=x^{2}(\ln x-1)+C$
C $y^{2}=x(\ln x-1)+\mathrm{C}$
D $y=x^{2} e^{x^{2}}+\mathrm{C}$

The correct answer is:
$y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$

Solution:
$x^{2}=e^{\left(\frac{x}{y}\right)^{-1}\left(\frac{d y}{d x}\right)} \Rightarrow x^{2}=e^{\left(\frac{y}{x}\right)\left(\frac{d y}{d x}\right)}$

$\Rightarrow \ln x^{2}=\frac{y}{x} \frac{d y}{d x}$ or $\int x \ln x^{2} d x=\int y d y$

Put $x^{2}=t \Rightarrow 2 x d x=d t \therefore \frac{1}{2} \int \ln t d t=\frac{y^{2}}{2}$

$C+t \ln t-t=y^{2}$ or $y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2014
Solution of the differential equation dydx=sinx+y+cosx+y is equal to

Options:
A log2+secx+y2=x+C
B log1+tanx+y=x+C
C log1+tanx+y2=y+C
D log1+tanx+y2=x+C

The correct answer is:
log1+tanx+y2=x+C

Solution:
We have,

dydx=sinx+y+cosx+y

Put x+y=v⇒1+dydx=dvdx

dvdx-1=sinv+cosv

⇒dvdx=sinv+cosv+1

⇒dvsinv+cosv+1=dx

⇒∫1sinv+cosv+1dv=∫dx

⇒∫12tanv21+tan2v2+1-tan2v21+tan2v2+1dv=∫dx

⇒∫1+tan2v22tanv2+1-tan2v2+1+tan2v2dv=x+C

⇒12∫sec2v2tanv2+1dv=x+C

⇒logtanv2+1=x+C

⇒logtanx+y2+1=x+C

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2018
Solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}=\sin x\) is

Options:
A \(x(y+\cos x)=\cos x+C\)
B \(x(y-\cos x)=\sin x+C\)
C \(x(y+\cos x)=\sin x+C\)
D None of these

The correct answer is:
\(x(y+\cos x)=\sin x+C\)

Solution:
\(\begin{aligned}
& \frac{d y}{d x}+\frac{1}{x} \cdot y=\sin x\left[\text { Type } \frac{d y}{d x}+P y=Q\right] \\
& e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x
\end{aligned}\)
\(\therefore\) Sol. is \(\mathrm{yx}=\int \mathrm{x} \sin \mathrm{xdx}+\mathrm{C}\)
\(\begin{aligned}
& =x(-\cos x)-\int 1 \cdot(-\cos x) d x+C \\
& =-x \cos x+\sin x+C \Rightarrow x(y+\cos x)=\sin x+C
\end{aligned}\)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2009
Solution of the equation dydx=ex-yex-ey is

Options:
A ey=ex-1+ce-ex
B ey-x=-1+ce-ex
C ex+ey=ce-ex
D None of these

The correct answer is:
ey=ex-1+ce-ex

Solution:
We have,

dydx=ex-yex-ey

⇒ eydydx+eyex=e2x

Put ey=v

∴ eydydx=dvdx

∴ dvdx+vex=e2x

This is linear equation

I.F==e∫exdx=eex

∴ Required solution is

v·eex=∫eex·e2xdx+C

v·eex=eexex-1+C

⇒ ey·eex=eexex-1+C

⇒ ey=ex-1+ce-ex

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2019
Solve $\sqrt{5} x^2+x+\sqrt{5}=0$

Options:
A $\pm \frac{\sqrt{19}}{5} \mathrm{i}$
B $\pm \frac{\sqrt{19} i}{2}$
C $\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}}$
D $\frac{-1 \pm \sqrt{19} \mathrm{i}}{\sqrt{5}}$

The correct answer is:
$\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}}$

Solution:
Here, $\mathrm{b}^2-4 \mathrm{ac}=1^2-4 \times \sqrt{5} \times \sqrt{5}$ $=1-20=-19$
Therefore, the solutions are
$\frac{-1 \pm \sqrt{-19}}{2 \sqrt{5}}=\frac{-1 \pm \sqrt{19 i}}{2 \sqrt{5}}$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Some amount of a radioactive substance (half-life =10 days ) is spread inside a room and consequently, the level of radiation becomes 50 times the permissible level for normal occupancy of the room. After how many days will the room be safe for occupation?

Options:
A 20 days
B 34.8 days
C 56.4 days
D 62.9 days

The correct answer is:
56.4 days

Solution:
since, the initial activity is 50 times the activity for safe occupancy, therefore R0=50R, where R=λN.
since, R∝N

RR0=NN0=12n=12t/T

Or 12t/10=150

2t/10-50⇒t10log102=log1050

⇒ t=10log1050log102=10×1.6990.301=56.4 days

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2018
Some salts containing two different metallic elements give test for only one of them in solution, such salts are

Options:
A double salts
B normal salts
C complex salts
D None of these

The correct answer is:
complex salts

Solution:
Complex compounds contains two different metallic elements but give test only for one of them. Because complex ions such as [Fe $\left.(\mathrm{CN})_{6}\right]^{4-}$ of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right],$ do not dissociate

into Fe $^{2+}$ and $\mathrm{CN}$ -ions.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2015
Sometimes, the colour observed in Lassaigne's test for nitrogen is green. It is because

Options:
A of green colour of ferrous sulphate
B ferric ferrocyanide is also green
C of green colour of copper sulphate
D of excess of Fe $^{3+}$ ions whose yellow colour makes the blue colour of ferric ferrocyanide to appear green.

The correct answer is:
of excess of Fe $^{3+}$ ions whose yellow colour makes the blue colour of ferric ferrocyanide to appear green.

Solution:
Although blue coloured ferric ferrocyanide is formed but due to the presence of yellow coloured $\mathrm{Fe}^{3+}$ salts, the blue colour gives the shade of green.

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2016
Spin only magnetic moment of Mn4+ ion is nearly

Options:
A 3 BM
B 6 BM
C 4 BM
D 5 BM

The correct answer is:
4 BM

Solution:
Electronic configuration of Mn is Z=25=Ar3d54s2

∴ Electronic configuration of Mn4+ is =Ar3d3

Thus, Mn4+ has 3-unpaired electron and its spin only magnetic moment μ is, μ=nn+2 BM

where, n= number of unpaired electrons.

μ=3(3+2). BM⇒μ=15=3.87BM≈4BM

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2019
$Statement- 1$: If the general equation $x^{2}+y^{2}+2 x y+2 g x+2 f y+4=0$ represents a pair of real lines then $|\mathrm{g}| \geq 2$.

$Statement- 2$: The equation $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ represents pair

of real lines if $a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}=0$

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 true, Statement- 2 is false

Solution:
The equation represents a pair of lines if $1 \cdot 1 \cdot 4+2 \cdot \mathrm{f} \cdot \mathrm{g} \cdot 1-1 \cdot \mathrm{f}^{2}-1 \cdot \mathrm{g}^{2}-4 \cdot 1^{2}=0$

$\Rightarrow(f-g)^{2}=0 \Rightarrow f=g$

The equation becomes $(x+y)^{2}+2 g(x+y)+4=0$ Which represents pair of parallel lines, which are real provided $(2 \mathrm{~g})^{2}-4 \cdot 4 \cdot 1 \geq 0 \Rightarrow|\mathrm{g}| \geq 2$

Topic: Mathematics, Pair of Lines, BITSAT, BITSAT 2020
Standard entropy of $X_2, Y_2$ and $X Y_2$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} X_2+\frac{3}{2} Y_2 \longrightarrow X Y_3, \Delta H=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be

Options:
A
$1250 \mathrm{~K}$

B
$500 \mathrm{~K}$

C
$750 \mathrm{~K}$

D
$1000 \mathrm{~K}$

The correct answer is:

$750 \mathrm{~K}$

Solution:
$\Delta S=S\left(X Y_3\right)-\frac{1}{2} S\left(X_2\right)-\frac{3}{2} S\left(Y_2\right)$
$=50-30-60=-40 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
$\Delta H=-30 \mathrm{~kJ}=-3000 \mathrm{~J}$
$\Delta G=\Delta H-T \Delta S$
At equilibrium,
$\Delta G=0$
$T=\frac{\Delta H}{\Delta S}=\frac{-3000}{-40}=750 \mathrm{~K}$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2022