datavorous/entrance-exam-dataset huggingface dataset:
The ratio of intensity at the centre of a bright fringe to the intensity at a point distance one-fourth of the distance between two successive bright fringes will be

Options:
A 4
B 3
C 2
D 1

The correct answer is:
2

Solution:
Intensity at centre of bright fringe,

I0=I+I+2II·cos0

I0=2I+2I ∵ cos0°=1

=4I

Similarly Intensity at point P4

(with phase difference =2π4=π2)

I'=I+I+2IIcosπ2

=2I+2II×0

=2I

∴ I0I'=4I2I=2

Topic: Physics, Wave Optics, BITSAT, BITSAT 2017
The ratio of oxidation states of $\mathrm{Cl}$ in potassium chloride to that in potassium chlorate is

Options:
A $\frac{+1}{5}$
B $\frac{-1}{5}$
C $\frac{-2}{5}$
D $\frac{+3}{5}$

The correct answer is:
$\frac{-1}{5}$

Solution:
\mathrm{KCl} x & =-1 & \mathrm{KClO}_{3}

potassium chloride $+1+x-6=0$

$$

x=+5

$$

potassium chlorate.

$\therefore$ Ratio of oxidation state of $\mathrm{Cl}=\frac{-1}{5}$

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2014
The ratio of secondary and primary turns of step-up transformer is $4: 1 .$ If a current of 4 A is applied to the primary, the induced current in secondary will be

Options:
A $8 \mathrm{~A}$
B $2 \mathrm{~A}$
C $1 \mathrm{~A}$
D $0.5 \mathrm{~A}$

The correct answer is:
$1 \mathrm{~A}$

Solution:
$\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}=\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{1}{4} \Rightarrow \mathrm{I}_{\mathrm{S}}=\frac{1}{4} \times 4=1 \mathrm{~A}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2013
The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:

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

The correct answer is:
$4: 1$

Solution:
Given $\mathrm{M}_{\mathrm{H}_2}=2 ; \mathrm{M}_{\mathrm{O}_2}=32$
Speed of sound, $\mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}}$
$\Rightarrow \quad v \propto \frac{1}{\sqrt{M}}$
$\therefore \quad \frac{\mathrm{v}_{\mathrm{H}_2}}{\mathrm{v}_{\mathrm{O}_2}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{H}_2}}}=\sqrt{\frac{32}{2}}=4: 1$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The ratio of the accelerations for a solid sphere (mass 'm' and radius 'R') rolling down an incline of angle ' $\theta$ ' without slipping and slipping down the incline without rolling is:

Options:
A 5: 7
B 2: 3
C 2: 5
D 7: 5

The correct answer is:
5: 7

Solution:
For solid sphere rolling without slipping on inclined plane, acceleration
$$
a_{1}=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}
$$
For solid sphere slipping on inclined plane without rolling, acceleration
$$
a_{2}=g \sin \theta
$$
Therefore required ratio $=\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}$
$$
=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}
$$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2010
The reaction

Options:
A
Rosenmund's reaction
B
Stephen's reaction
C
Cannizzaro's reaction
D
Gatterman-Koch reaction

The correct answer is:
Gatterman-Koch reaction

Solution:
This reaction proceeds in the presence of anhydrous $\mathrm{AlCl}_3$ or $\mathrm{CuCl}$.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reaction
is called:

Options:
A Williamson continuous etherification process
B Etard reaction
C Gatterman - Koch reaction
D Williamson Synthesis

The correct answer is:
Williamson Synthesis

Solution:
Williamson synthesis is one of the best methods for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2011
The reaction
is called:

Options:
A Williamson continuous etherification process
B Etard reaction
C Gatterman - Koch reaction
D Williamson Synthesis

The correct answer is:
Williamson Synthesis

Solution:
Williamson synthesis is one of the best methods for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reaction,

is known as:

Options:
A Wurtz reaction
B Koch reaction
C Clemmensen reduction
D Kolbe's reaction

The correct answer is:
Koch reaction

Solution:
No Solution

Topic: Chemistry, Carboxylic Acid Derivatives, BITSAT, BITSAT 2014
The reaction,

is known as:

Options:
A Wurtz reaction
B Koch reaction
C Clemmensen reduction
D Kolbe's reaction

The correct answer is:
Koch reaction

Solution:
No Solution

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCH}_3$ with $\mathrm{HBr}$ produces

Options:
A

B

C
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$

D

The correct answer is:

Solution:
Electrophile addition reaction takes place via more stable carbocation.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCH}_3$ with $\mathrm{HBr}$ produces

Options:
A

B

C
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$

D

The correct answer is:

Solution:
Electrophile addition reaction takes place via more stable carbocation.

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2022
The reaction, $\mathrm{SO}_{2}+\mathrm{Cl}_{2} \longrightarrow \mathrm{SO}_{2} \mathrm{Cl}_{2}$ is
exothermic and reversible. A mixture of $\mathrm{SO}_{2}(\mathrm{~g})$, $\mathrm{Cl}_{2}(\mathrm{~g})$ and $\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{l})$ is at equilibrium in a closed container. Now a certain quantity of extra $\mathrm{SO}_{2}$ is introduced into the container, the volume remaining the same. Which of the following is/are true?

Options:
A The pressure inside the container will not change.
B The temperature will not change.
C The temperature will increase.
D The temperature will decrease.

The correct answer is:
The temperature will increase.

Solution:
Byaddition of $\mathrm{SO}_{2}$, equilibrium will shift to RHS which is exothermic. Hence temp, will increase.

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2012
The reaction that does NOT take place in a blast furnace between $900 \mathrm{~K}$ to $1500 \mathrm{~K}$ temperature range during extraction of iron is :

Options:
A $\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2$
B $\mathrm{FeO}+\mathrm{CO} \rightarrow \mathrm{Fe}+\mathrm{CO}_2$
C $\mathrm{C}+\mathrm{CO}_2 \rightarrow 2 \mathrm{CO}$
D $\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3$

The correct answer is:
$\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2$

Solution:
At $900-1500 \mathrm{~K}$ (higher temperature range in the blast furnace)
Reaction which take place are:
$\mathrm{C}+\mathrm{CO}_2 \longrightarrow 2 \mathrm{CO}$
$\mathrm{FeO}+\mathrm{CO} \longrightarrow \mathrm{Fe}+\mathrm{CO}_2$
$\mathrm{CaO}+\mathrm{SiO}_2 \longrightarrow \mathrm{CaSiO}_3$ (Slag formation)
$\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \longrightarrow 2 \mathrm{FeO}+\mathrm{CO}_2$ at $500-800 \mathrm{~K}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The reaction

Options:
A
Rosenmund's reaction
B
Stephen's reaction
C
Cannizzaro's reaction
D
Gatterman-Koch reaction

The correct answer is:
Gatterman-Koch reaction

Solution:
This reaction proceeds in the presence of anhydrous $\mathrm{AlCl}_3$ or $\mathrm{CuCl}$.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The reagent (s) which can be used to distinguish acetophenone from benzophenone is (are)

Options:
A 2,4 - dinitrophenylhydrazine
B aqueous solution of $\mathrm{NaHSO}_{3}$
C benedict reagent
D $\mathrm{I}_{2}$ and $\mathrm{NaOH}$

The correct answer is:
$\mathrm{I}_{2}$ and $\mathrm{NaOH}$

Solution:
$\mathrm{I}_{2}$ and $\mathrm{NaOH}$ react with acetophenone $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}\right)$ to give yellow ppt. of $\mathrm{CHI}_{3}$
but benzophenone $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}\right)$ does not
give haloform test.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2011
The reduction of peroxydisulphate ion by I ion

is expressed by $\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+3 \mathrm{I}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{3}^{-}$. If

rate of disappearance of $\mathrm{I}^{-}$ is $9 / 2 \times 10^{-3} \mathrm{~mol}$ lit $^{-1}$

$\mathrm{s}^{-1},$ what is the rate of formation of $2 \mathrm{SO}_{4}^{2-}$ during same time?

Options:
A $3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
B $2 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
C $10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
D $4 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

The correct answer is:
$3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

Solution:
$\frac{1 d\left[\mathrm{SO}_{4}^{2-}\right]}{2}=\frac{1}{3}\left(-\frac{\mathrm{d}\left(\mathrm{I}^{-}\right)}{d t}\right)$

$\frac{1}{2} \times \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=\frac{1}{3} \times \frac{9}{2} \times 10^{-3}$

$\therefore \quad \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2014
The reflection of the point \((4,-13)\) in the line \(5 x+y+6=0\), is

Options:
A \((-1,-14)\)
B \((3,4)\)
C \((1,2)\)
D \((-4,13)\)

The correct answer is:
\((-1,-14)\)

Solution:
Let \(\mathrm{Q}(a, b)\) be the reflection of \(\mathrm{P}(4,-13)\) in the line \(5 x+y+6=0\)
Then the mid-point \(R\left(\frac{a+4}{2}, \frac{b-13}{2}\right)\) lies on
\(\begin{aligned}
& 5 x+y+6=0 \\
& \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0 \\
& \Rightarrow 5 a+b+19=0 \quad ...(i)
\end{aligned}\)
Also \(\mathrm{PQ}\) is perpendicular to \(5 x+y+6=0\)
Therefore \(\frac{b+13}{a-4} \times\left(-\frac{5}{1}\right)=-1\)
\(\Rightarrow a-5 b-69=0 \quad ...(ii)\)
Solving (i) and (ii), we get \(a=-1, b=-14\)

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2010
The resistance of a wire at 20 °C is 20 Ω and 500 °C is 60 Ω. At which temperature, its resistance will be 25 Ω ?

Options:
A 50 °C
B 60 °C
C 70 °C
D 80 °C

The correct answer is:
80 °C

Solution:
We have, Rt=R0 1+αt

Here, 20=R0 1+20α

60=R0 1+500α

⇒ 6020=R0 1+500αR0 1+20α

3=1+500 α1+20 α

⇒ 3 1+20 α=1+500 α

⇒ 3+60α=1+500α

⇒ 440 α=2

⇒ α=2440=1220I°C

Also given Rt=25 °C, t= ?

Now, 2025=R0 1+20220R0 1+t220

4 1+t220=5 1+20220

On solving,

t=80 °C

Topic: Physics, Current Electricity, BITSAT, BITSAT 2017
The resistance of a wire is $5 \Omega$. It's new resistance in ohm if stretched to 5 times of its original length will be :

Options:
A 625
B 5
C 125
D 25

The correct answer is:
125

Solution:
Let resistance of a wire $R$ and length 1 .
$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=5 \Omega$
$\therefore \quad$ Volume of wire is constant in stretching
$V_i=V_f \Rightarrow A_i \ell_i=A_f \ell_f$
$\mathrm{A} \ell=\mathrm{A}^{\prime}(5 \ell) \Rightarrow \mathrm{A}^{\prime}=\frac{\mathrm{A}}{5}$
$R_f=\frac{\rho \ell_f}{A_f}=\frac{\rho(5 \ell)}{\left(\frac{A}{5}\right)}=25\left(\frac{\rho \ell}{A}\right)=25 \times 5=125 \Omega$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The resistance of a wire is $R$. It is bent at the middle by $180^{\circ}$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is

Options:
A $2 R$
B $R / 2$
C $R / 4$
D $R / 8$

The correct answer is:
$R / 4$

Solution:
Resistance of wire $(R)=\rho \frac{l}{A}$ If wire is bent in the middle then

$$

l^{\prime}=\frac{l}{2}, A^{\prime}=2 A

$$

$\therefore$ New resistance, $\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}=\frac{\rho \frac{l}{2}}{2 A}=$

$\frac{\rho l}{4 A}=\frac{R}{4}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2015
The r.m.s. value of potential difference $\mathrm{V}$ shown in the figure is

Options:
A $\mathrm{V}_{0}$
B $\mathrm{V}_{0} / \sqrt{2}$
C $\mathrm{V}_{0} / 2$
D $\mathrm{V}_{0} / \sqrt{3}$

The correct answer is:
$\mathrm{V}_{0} / \sqrt{2}$

Solution:
$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{(\mathrm{T} / 2) \mathrm{V}_{0}{ }^{2}+0}{\mathrm{~T}}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}$.

Topic: Physics, Alternating Current, BITSAT, BITSAT 2012
The r.m.s. value of potential difference $\mathrm{V}$ shown in the figure is

Options:
A $\mathrm{V}_{0}$
B $\mathrm{V}_{0} / \sqrt{2}$
C $\mathrm{V}_{0} / 2$
D $\mathrm{V}_{0} / \sqrt{3}$

The correct answer is:
$\mathrm{V}_{0} / \sqrt{2}$

Solution:
$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{(\mathrm{T} / 2) \mathrm{V}_{0}{ }^{2}+0}{\mathrm{~T}}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}$.

Topic: Physics, Alternating Current, JEE Main
The root of the equation

$2(1+i) x^{2}-4(2-i) x-5-3 i=0$ which has

greater modulus is

Options:
A $\frac{3-5 i}{2}$
B $\frac{5-3 \mathrm{i}}{2}$
C $\frac{3-i}{2}$
D none

The correct answer is:
$\frac{3-5 i}{2}$

Solution:
$\begin{aligned} \text { Roots } &=\frac{4(2-i) \pm \sqrt{16(2-i)^{2}+8(1+i)(5+3 i)}}{4(1+i)} \\ &=\frac{4-i}{1+i} \text { or } \frac{-i}{1+i}=\frac{3-5 i}{2} \text { or } \frac{-1-i}{2} \end{aligned}$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2015
The roots of the equation \(x^2-2 \sqrt{2} x+1=0\) are

Options:
A Real and different
B Imaginary and different
C Real and equal
D Rational and different

The correct answer is:
Real and different

Solution:
The discriminant of the equation \((-2 \sqrt{2})^2-4(1)(1)=8-4=4 > 0\) and \(a\) perfect square, so roots are real and different but we can't say that roots are rational because coefficients are not rational therefore.
\(\frac{\sqrt{2 \sqrt{2} \pm \sqrt{(2 \sqrt{2})^2-4}}}{2}=\frac{2 \sqrt{2} \pm 2}{2}=\sqrt{2} \pm 1\)
this is irrational \(\therefore\) the roots are real and different.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2010
The roots of the given equation $(p-q) x^2+(q-r) x+(r-p)=0$ are :

Options:
A $\frac{p-q}{r-p}, 1$
B $\frac{q-r}{p-q}, 1$
C $\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1$
D
None of these

The correct answer is:
$\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1$

Solution:
Given equation is
$(p-q) x^2+(q-r) x+(r-p)=0$
By using formula for finding the roots viz: $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$, we get
$\begin{aligned} & x=\frac{(r-q) \pm \sqrt{(q-r)^2-4(r-p)(p-q)}}{2(p-q)} \\ & \Rightarrow x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1\end{aligned}$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The roots of the given equation $(p-q) x^2+(q-r) x+(r-p)=0$ are :

Options:
A $\frac{\mathrm{p}-\mathrm{q}}{\mathrm{r}-\mathrm{p}}, 1$
B $\frac{\mathrm{q}-\mathrm{r}}{\mathrm{p}-\mathrm{q}}, 1$
C $\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1$
D None of these

The correct answer is:
$\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1$

Solution:
Given equation is
$(p-q) x^2+(q-r) x+(r-p)=0$
By using formula for finding the root viz: $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$, we get
$x=\frac{(r-q) \pm \sqrt{(q-r)^2-4(r-p)(p-q)}}{2(p-q)}$
$\Rightarrow \quad x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The rotation of the earth having radius $\mathrm{R}$ about its axis speeds upto a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:

Options:
A $8 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
B $8 \pi \sqrt{\frac{g}{R}}$
C $\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
D $4 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$

The correct answer is:
$\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

Solution:
No Solution available

Topic: Physics, Gravitation, BITSAT, BITSAT 2009
The screening effect of $d$ -electrons is

Options:
A Equal to $p$ -electrons
B Much more than $p$ -electrons
C Same as $f$ -electrons
D Less than $p$ -electrons.

The correct answer is:
Less than $p$ -electrons.

Solution:
The screening effect of inner electron of the nucleus causes the decrease in ionization potential, therefore the order of the screening effect is

$$

f
$$

Hence, the screening effect of $d$ -electron is less than $p$ -electron.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2013
The secondary structure of a protein refers to

Options:
A fixed configuration of the polypeptide backbone
B $\alpha$ -helical backbone
C hydrophobic interactions
D sequence of $\alpha$ -amino acids.

The correct answer is:
$\alpha$ -helical backbone

Solution:
The secondary structure of a protein refers to the shape in which a long peptide chain can exist. There are two different conformations of the peptide linkage present in protein are $\alpha$ -helix and $\beta$ -conformation. The $\alpha$ -helix always has a right handed arrangement. In $\beta$ -conformation all peptide chains are streched out to nearly maximum extension and then laid side by side held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as $\beta$ -pleated shelt.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2010
The set $(A \backslash B) \cup(B \backslash A)$ is equal to]

Options:
A $[A \backslash(A \cap B)] \cap[B \backslash(A \cap B)]$
B $(A \cup B) \backslash(A \cap B)$
C $A \backslash(A \cap B)$
D $\overrightarrow{A \cap B} \backslash A \cup B$

The correct answer is:
$(A \cup B) \backslash(A \cap B)$

Solution:
Given set can be written as $(A-B) \cup(B-A)=(A \cup B)-(A \cap B)$

(By definition of symmetric difference) Hence, $(\mathrm{A} \backslash \mathrm{B}) \cup(\mathrm{B} \backslash \mathrm{A})=(\mathrm{A} \cup \mathrm{B}) \backslash(\mathrm{A} \cap \mathrm{B})$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2014
The set of all real $x$ satisfying the inequality

$\frac{3-|x|}{4-|x|} \geq 0,$ is

Options:
A [-3,3]$\cup(-\infty,-4) \cup(4, \infty)$
B $(-\infty,-4) \cup(4, \infty)$
C $(-\infty,-3) \cup(4, \infty)$
D $(-\infty,-3) \cup(3, \infty)$

The correct answer is:
[-3,3]$\cup(-\infty,-4) \cup(4, \infty)$

Solution:
Given, $\frac{3-|x|}{4-|x|} \geq 0$

$\Rightarrow 3-|x| \leq 0$ and $4-|x|<0$

or $3-|x| \geq 0$ and $4-|x|<0$

$\Rightarrow|x| \geq 3$ and $|x|>4$

or $|x| \leq 3$ and $|x|<4$

$\Rightarrow|x|>4$ or $|x| \leq 3$

$\Rightarrow x \in(-\infty,-4) \cup[-3,3] \cup(4, \infty)$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2013
The set of all values of a for which the function $f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+$

$\sin 1$ does not possess critical points is

Options:
A $[1, \infty)$
B (0,1)$\cup(1,4)$
C (-2,4)
D (1,3)$\cup(3,5)$

The correct answer is:
(0,1)$\cup(1,4)$

Solution:
\(f(x)=\left(a^{2}=-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=(\mathrm{a}-1)(\mathrm{a}-2) \cos \mathrm{x} / 2+(\mathrm{a}-1) \mathrm{x}+\sin 1\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{2}(\mathrm{a}-1)(\mathrm{a}-2) \sin \frac{\mathrm{x}}{2}+(\mathrm{a}-1)\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{a}-1)\left[1-\frac{(\mathrm{a}-2)}{2} \sin \frac{\mathrm{x}}{2}\right]\)

If \(f(x)\) does not possess critical points, then \(f^{\prime}(x)\) = 0 for any \(x \in R\)

\(\Rightarrow(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) = 0 for any \(x \in R\)

\(\Rightarrow \mathrm{a}=1\) and \(1-\left(\frac{\mathrm{a}-2}{2}\right) \sin \frac{\mathrm{x}}{2}=0\)

must not have any solution in \(\mathrm{R}\).

\(\Rightarrow \mathrm{a}=1\) and \(\sin \frac{\mathrm{x}}{2}=\frac{2}{\mathrm{a}-2}\) is not solvable in \(\mathrm{R}\).

\(\Rightarrow \mathrm{a} = 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\left[\right.\) For \(\mathrm{a}=2, \mathrm{f}(\mathrm{x})=\mathrm{x}+\sin 1 \therefore \mathrm{f}^{\prime}(\mathrm{x})=1\)

\(\Rightarrow \mathrm{a}=1\) and \(0<\mathrm{a}<4 \Rightarrow \mathrm{a} \in(0,1) \cup(1,4)\).

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2016
The set of points of discontinuity of the function \(1 / \log |\mathrm{x}|\) is -

Options:
A \(\{-1,0,1\}\)
B \(\{0\}\)
C \(\{0,1\}\)
D None of these

The correct answer is:
\(\{-1,0,1\}\)

Solution:
Let \(f(x)=\frac{1}{\log |x|}\)
The points of disconinuity of \(f(x)\) are those points where \(f(x)\) is undefined or infinite. It is undefined when \(x=0\) and is infinite when \(\log |x|=0,|x|=1\), i.e. \(x= \pm 1\).
\(\therefore \quad\) Set of points of discontinuity \(=\{-1,0,1\}\).

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2009
The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is :

Options:
A $\frac{18}{\sqrt{5}}$
B $\frac{22}{3 \sqrt{5}}$
C $\frac{46}{3 \sqrt{5}}$
D $6 \sqrt{3}$

The correct answer is:
$\frac{18}{\sqrt{5}}$

Solution:
Give lines are $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$
Lines passes through the points
$\vec{a}_1=(3,2,1) \text { and } \vec{a}_2=(-3,6,5) \text {, }$
$\vec{b}_1=2 \hat{i}+3 \hat{j}-\hat{k}$
$\vec{b}_1=2 \hat{i}+\hat{j}-3 \hat{k}, \vec{a}_2-\vec{a}_1=6 \hat{i}-4 j-4 \hat{k}$
Shortest distance $=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right)\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}$
$\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}$
$\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=60+32+16=108$
$\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{100+64+16}=\sqrt{180}$
$S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The shortest distance between the lines $\mathrm{x}=\mathrm{y}+$ $2=6 z-6$ and $x+1=2 y=-12 z$ is

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

The correct answer is:
2

Solution:
The lines are $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$
and $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
Here,
$\begin{aligned}
& \vec{a}_1=-2 \hat{j}+\hat{k}, b_1+6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_2=-\hat{i}, \\
& \vec{b}_2=12 \hat{i}+6 \hat{j}-\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & k \\
6 & 6 & 1 \\
12 & 6 & -1
\end{array}\right|=-12 \hat{i}+18 \hat{j}-36 \hat{k} \\
& \text { Shortest distance }=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1-\vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}
\end{aligned}$
$\begin{aligned} & =\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i+18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^2+(18)^2+(-36)^2}} \\ & =\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The shortest distance between the parabolas y2=4x and y2=2x-6 is

Options:
A 2
B 5
C 3
D None

The correct answer is:
5

Solution:
Shortest distance between two curves occured along the common normal.

∴ Normal to y2=4x at m2,2m is

y+mx-2m-m3=0

Normal to y2=2x-3 at m22+3,m is

y+mx-3-m-m32=0

Both normals are same, if

-2m-m3=-4m-12m3⇒m=0,2

So, points will be (4,4) and (5,2) or (4,-4) and (5,-2).

Hence, shortest distance will be

(1+4)=5 unit.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
The significant result deduced from the Rutherford's scattering experiment is that

Options:
A whole of the positive charge is concentrated at the centre of atom
B there are neutrons inside the nucleus
C $\alpha$ - particles are helium nuclei
D electrons are embedded in the atom

The correct answer is:
whole of the positive charge is concentrated at the centre of atom

Solution:
The significant result deduced from the Rutherford's scatter ing is that whole of the positive charge is concentrated at the centre of atom i.e. nucleus.

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2021