datavorous/entrance-exam-dataset huggingface dataset:
The heat of reaction for

C10H8s+12O2g⟶10CO2g+4H2O(l) at constant volume is -1228.2 kcal at 250C. The heat of reaction at constant pressure and same temperature is

Options:
A -1228.2 kcal
B -1229.3 kcal
C -1232.9 kcal
D -1242.6 kcal

The correct answer is:
-1229.3 kcal

Solution:
Heat of reaction at constant pressure is represented by ΔH

Heat of reaction at constant volume is represented by ΔU

C10H8s+12O2g⟶10CO2g+4H2O

We known that ΔH=ΔE+ΔngRT

Δng=10-12=-2

ΔH=-1228.2×103+-2×2×298

=-1229393 Cal

=-1229.39 kcal

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2019
The heats of neutralisation of $\mathrm{CH}_{3} \mathrm{COOH}$, $\mathrm{HCOOH}, \mathrm{HCN}$ and $\mathrm{H}_{2} \mathrm{~S}$ are -13.2,-13.4,-2.9

and $-3.8 \mathrm{kCal}$ per equivalent respectively. Arrange the acids in increasing order of acidic strength.

Options:
A $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$
B $\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCOOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$
C $\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCN}$
D $\mathrm{HCOOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCN}$

The correct answer is:
$\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$

Solution:
The greater the (negative value) of heat of neutralisation, the more is the strength of the acid. Hence,

$$

\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}

$$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2014
The helical structure of protein is stabilised by

Options:
A peptide bonds
B dipeptide bonds
C hydrogen bonds
D vander waals forces

The correct answer is:
hydrogen bonds

Solution:
Fibrous proteins have thread like molecules which lie side by side to form fibres. The various molecules are held together by hydrogen bonds.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2012
The helical structure of protein is stabilized by

Options:
A dipeptide bonds
B hydrogen bonds
C ether bonds
D peptide bonds

The correct answer is:
hydrogen bonds

Solution:
The $\alpha$-helix structure is formed when the chain of $\alpha$-amino acids coils as a right handed screw (called $\alpha$-helix) because of the formation of hydrogen bonds between amide groups of the same peptide chain, i.e., NH group in one unit is linked to carbonyl oxygen of the third unit by hydrogen bonding. This hydrogen bonding between different units is responsible for holding helix in a position.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The horizontal range and maximum height attained by a projectile are R and H respectively. If a constant horizontal acceleration a=g4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be

Options:
A R+H,H2
B R+H2,2H
C R+2H,H
D R+H,H

The correct answer is:
R+H,H

Solution:
T=2Uyg

H=Uy22g

Here maximum height will be same as H because there is no acceleration change in vertical direction.

and R=Ux×T

When a horizontal acceleration is also given to the projectile Uy,T and H will remain unchanged while range will become

R'=Ux×T+12aT2

=R+12g44Uy2g2=R+H ∵a=g4

Therefore horizontal range & maximum height will be R+H & H.

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2017
The hybridisation of Fe is K3[Fe(CN)6] is

Options:
A sp3
B dsp3
C sp3d2
D d2sp3

The correct answer is:
d2sp3

Solution:
Electronic configuration of Fe=[Ar]4 s23 d6 Fe2+=[Ar]3d6

CN- are strong field ligands

d2sp3 hybridisation

∵ CN- is a strong ligand and from six coordinate bonds with Fe3+ . Thus Fe show d2sp3 hybridisation in K3Fe(CN)6.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2017
The hybridisation of phosphorous in PO43- is

Options:
A sp
B sp2
C sp3
D sp3d

The correct answer is:
sp3

Solution:
The structure of PO43- is as follows

In the given molecule, The P atom have 5 valence electrons. The hybridization can be determined by the number of sigma bonds and the number of lone pair.

As phosphorous central atom forms four 4 sigma bonds and has no lone pair of electrons. It uses four hybrid orbital and thus, hybridisation of p-atom is sp3.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The hybridisation of phosphorous in PO43- is

Options:
A sp
B sp2
C sp3
D sp3d

The correct answer is:
sp3

Solution:
The structure of PO43- is as follows

In the given molecule, The P atom have 5 valence electrons. The hybridization can be determined by the number of sigma bonds and the number of lone pair.

As phosphorous central atom forms four 4 sigma bonds and has no lone pair of electrons. It uses four hybrid orbital and thus, hybridisation of p-atom is sp3.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2019
The hypothetical complex chlorodiaquatriamminecobalt (III) chloride can be represented as

Options:
A $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}$
B $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]$
C $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}\right]$
D $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right] \mathrm{Cl}_{3}$

The correct answer is:
$\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}$

Solution:
The complex chlorodiaquatriammine cobalt (III) chloride can have the structure $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}$

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2016
The I - V characteristics shown in figure represents

Options:
A ohmic conductors
B non-ohmic conductors
C insulators
D superconductors

The correct answer is:
non-ohmic conductors

Solution:
The figure is showing I - V characteristics of non ohmic or non linear conductors.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2021
The image of the point (1,6,3) in the line x1=y-12=z-23 is

Options:
A (-1,0,7)
B (-1,0,-7)
C (1,0,7)
D (2,0,7)

The correct answer is:
(1,0,7)

Solution:
Let P(1,6,3) be the given point and L be the foot of perpendicular from P to the given line. The coordinates of a general point on the given line are

x-01=y-12=z-23=λ (say)

i.e. x=λ

y=2λ+1, z=3λ+2

If the coordinates of L are (λ,2λ+1,3λ+2), then the direction ratios of PL are (λ-1,2λ-5,3λ-1).

Since, the direction ratios of given line which is perpendicular to PL, are 1,2 and 3. Therefore, (λ-1)1+(2λ-5)2+(3λ-1)3=0,which gives λ=1.

Hence, coordinates of L are (1,3,5).

Let Qx1,y1,z1 be the image of P(1,6,3) in the given line. Then, L is the mid-point of PQ.

Therefore, x1+12=1,y1+62=3,z1+32=5

⇒x1=1,y=0,z=7

Hence, the image of (1,6,3) in the given line is (1,0,7).

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2017
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is

Options:
A 0.4
B 0.6
C 0.8
D 1

The correct answer is:
0.6

Solution:
$\tan \phi=\frac{\mathrm{X}}{\mathrm{R}} \times \frac{4}{3}$
Power factor $=\cos \phi=\frac{3}{5}=0.6$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2012
The important antioxidant used in food is

Options:
A $\mathrm{BHT}$
B $\mathrm{BHC}$
C $\mathrm{BTX}$
D All the three

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

Solution:
BHT is the important antioxidant used in food.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2016
The incorrect statement about carbonate CO32- ion is,

Options:
A It has planar structure
B It has one coordinate bond
C It has three resonating structure
D Hydrolysis of CO32- ion gives basic solution

The correct answer is:
It has one coordinate bond

Solution:
Resonating structure of carbonate CO32- ion are

In resonating structures bonds are not fixed and therefore, all bond lengths are equal.

The hybridisation of central atom is sp2.

Thus, there is no coordinate bond present in CO32-. Hence, the second statement is incorrect.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2018
The instantaneous current from an a.c. source is $\mathrm{I}=6 \sin 314 \mathrm{t}$. What is the rms value of the current?

Options:
A $3 \sqrt{2} \mathrm{amp}$
B $2 \sqrt{2}$ amp
C $\sqrt{2} \mathrm{amp}$
D $2 \mathrm{amp}$

The correct answer is:
$3 \sqrt{2} \mathrm{amp}$

Solution:
$\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3 \sqrt{2} \mathrm{amp} .$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2009
The integral ∫1+x-1xex+1xdx is equal to

Options:
A x-1ex+1x+C
B xex+1x+C
C x+1ex+1x+C
D -xex+1x+C

The correct answer is:
xex+1x+C

Solution:
We have,

I=∫1+x-1xex+1xdx

I=∫ex+1xdx+∫x-1xex+1xdx

I=∫ex+1xdx+∫x1-1x2ex+1xdx

I=∫ex+1xdx+xex+1x-∫ex+1xdx

I=xex+1x+C

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2019
The integrating factor of $\frac{x d y}{d x}-y=x^4-3 x$ is

Options:
A
x
B $\log x$
C $\frac{1}{x}$
D $-\mathbf{X}$

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

Solution:
$\begin{aligned}
& \text { (c) Since } x \frac{d y}{d x}-y=x^4-3 x \\
& \therefore \frac{d y}{d x}-\frac{y}{x}=x^3-3
\end{aligned}$
Hence $I F=e^{\int P d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The intermediate formed during the addition of $\mathrm{HCl}$ to propene in the presence of peroxide is

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

The correct answer is:
$\mathrm{CH}_{2} \mathrm{CHCH}_{3}$

Solution:
The addition of HCl to propene proceeds by ionic mechanism and not by free radical mechanism. Hence it forms intermediate carbonium ion.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2009
The interval in which the function $2 \mathrm{x}^{3}+15$ increases less rapidly than the function $9 x^{2}-12 x,$ is $-$

Options:
A $(-\infty, 1)$
B (1,2)
C $(2, \infty)$
D None of these

The correct answer is:
(1,2)

Solution:
Let $f(x)=2 x^{3}+15$ and $g(x)=9 x^{2}-12 x$ then $\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}$

$\therefore \mathrm{f}(\mathrm{x})$ is increasing function $\forall \mathrm{x} \in \mathrm{R}$

$$

\text { Also, } g^{\prime}(x)>0 \Rightarrow 18 x-12>0 \Rightarrow x>\frac{2}{3}

$$

Thus, $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ both increases for $\mathrm{x}>\frac{2}{3}$

$$

\text { Let } F(x)=f(x)-g(x), F^{\prime}(x)<0

$$

$(\because \mathrm{f}(\mathrm{x})$ increases less rapidly than the function $\mathrm{g}(\mathrm{x}))$

$$

\Rightarrow 6 \mathrm{x}^{2}-18 \mathrm{x}+12<0 \Rightarrow 1<\mathrm{x}<2

$$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2014
The interval in which the function $f(x)=$ $\frac{4 x^2+1}{x}$ is decreasing is :

Options:
A $\left(-\frac{1}{2}, \frac{1}{2}\right)$
B $\left[-\frac{1}{2}, \frac{1}{2}\right]$
C $(-1,1)$
D $[-1,1]$

The correct answer is:
$\left(-\frac{1}{2}, \frac{1}{2}\right)$

Solution:
Given $f(x)=\frac{4 x^2+1}{x}$ Thus $f^{\prime}(x)=4-\frac{1}{x^2}$
$\mathrm{f}(\mathrm{x})$ will be decreasing if $\mathrm{f}^{\prime}(\mathrm{x}) < 0$
Thus $4-\frac{1}{x^2} < 0 \Rightarrow \frac{1}{x^2}>4 \Rightarrow \frac{-1}{2} < x < \frac{1}{2}$
Thus interval in which $f(x)$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The inverse of the function fx=logax+x2+1 (where, a<0, a≠1) is

Options:
A 12ax-a-x
B not defined for all x
C defined for x>0
D None of these

The correct answer is:
12ax-a-x

Solution:
Let fx=y, then

ay=x+x2+1

⇒a-y=1x+x2+1

⇒a-y=x-x2+1-1 [rationalising]

∴ ay-a-y=2x

⇒x=12ay-a-y

∴ f-1y=12ay-a-y ∵ fx=y⇒x=f-1y

⇒f-1x=12ax-a-x [replacing y by x]

Hence, option (a) is correct.

Topic: Mathematics, Functions, BITSAT, BITSAT 2018
The ionic conductance of $\mathrm{Ba}^{2+}$ and $\mathrm{Cl}^{-}$ respectively are 127 and $76 \Omega^{-1} \mathrm{~cm}^{2}$ at infinite dilution. The equivalent conductance of $\mathrm{BaCl}_{2}$, at infinite dilution will be

Options:
A $330 \Omega^{-1} \mathrm{~cm}^{2}$
B $203 \Omega^{-1} \mathrm{~cm}^{2}$
C $139 \Omega^{-1} \mathrm{~cm}^{2}$
D $51 \Omega^{-1} \mathrm{~cm}^{2}$

The correct answer is:
$139 \Omega^{-1} \mathrm{~cm}^{2}$

Solution:
$\begin{aligned} & \lambda_{\mathrm{m}}^{\circ} \text { for } \mathrm{BaCl}_{2}=\lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}^{2+}+\lambda_{\mathrm{m}}^{\circ} \mathrm{Cl}^{-} \\ &=\frac{1}{2} \times 127+76=139.5 \Omega^{-1} \mathrm{~cm}^{2} \end{aligned}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2012
The ionization constant of phenol is higher than that of ethanol because :

Options:
A Phenoxide ion is bulkier than ethoxide
B Phenoxide ion is stronger base than ethoxide
C Phenoxide ion is stabilized through delocalization
D Phenoxide ion is less stable than ethoxide

The correct answer is:
Phenoxide ion is stabilized through delocalization

Solution:
The acidic nature of phenol is due to the formation of stable phenoxide ion in solution

$$

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{\text {Phenoxide ion }}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}

$$

The phenoxide ion is stable due to

resonance.

The negative charge is delocalized in the benzene ring which is a stabilizing factor in the phenoxide ion and increase acidity of phenol. wheras no resonance is possible in alkoxide ions $\left(\mathrm{RO}^{-}\right)$ derived from alcohol. The negative charge is localized on oxygen atom. Thus, alcohols are not acidic.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2014
The ions $\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}$ and $\mathrm{Al}^{3+}$ are

isoelectronic. Their ionic radii show:

Options:
A A decrease from $\mathrm{O}^{2-}$ to $\mathrm{F}^{-}$ and then increase from $\mathrm{Na}^{+}$ to $\mathrm{Al}^{3+}$
B A significant increase from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$
C A significant decrease from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$
D An increase from $\mathrm{O}^{2-}$ to $\mathrm{F}^{-}$ and then decrease from $\mathrm{Na}^{+}$ to $\mathrm{Al}^{3+}$

The correct answer is:
A significant decrease from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$

Solution:
Amongst isoelectronic species, ionic radii of anion is more than that of cations. Further size of anion increase with increase in -ve charge and size of cation decrease with increase in $+$ ve charge. Hence ionic radii decreases from $\mathrm{O}^{-}$ to $\mathrm{Al}^{3+}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2013
The isoelectric-point of a colloidially dispersed material is the $\mathrm{pH}$ value at which

Options:
A the dispersed phase migrate in an electric field.
B the dispersed phase does not migrate in an electric field.
C the dispersed phase has $\mathrm{pH}$ equal to 7 .
D the dispersed phase has pH equal to zero.

The correct answer is:
the dispersed phase does not migrate in an electric field.

Solution:
At isoelectric point there is no migration of dispersed phase in an electric field.

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2014
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_{2} \mathrm{~S}$ and $\mathrm{HgS}$ are $10^{-31}, 10^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :

Options:
A $\quad \mathrm{Ag}_{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{HgS}$
B $\quad \mathrm{AgS}>\mathrm{HgS}>\mathrm{CuS}$
C $\mathrm{HgS}>\mathrm{Ag}_{2} \mathrm{~S}>\mathrm{CuS}$
D $\quad\mathrm{CuS}>\mathrm{Ag}_{2} \mathrm{~S}>\mathrm{HgS}$

The correct answer is:
$\quad \mathrm{Ag}_{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{HgS}$

Solution:
For CuS, solubility $=\left(10^{-31}\right)^{1 / 2}$;
$$
\begin{array}{l}
\text { For } \mathrm{Ag}_{2} \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}=\left(\frac{10^{-44}}{4}\right)^{\frac{1}{3}} \text { and } \\
\text { for } \mathrm{HgS}=\left(10^{-54}\right)^{\frac{1}{2}}
\end{array}
$$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2010
The larger of two angles made with the X-axis of a straight line drawn through (1,2) so that it intersects the line x+y=4 at a point distant 6/3 from the point (1,2) is

Options:
A 60°
B 75°
C 105°
D None of these

The correct answer is:
75°

Solution:
Let A=(1,2) equation of line is

x-1cosθ=y-2sinθ=r=63

∴ x=63cosθ+1,y=63sinθ+2

This point lies on line x+y=4

∴ 63cosθ+63sinθ=1

⇒ sinθ+cosθ=32

⇒ sinπ4+θ=32

⇒ π4+θ=60° or 120°

⇒ θ=15° or 75°

∴ Larger angle =75°

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2019
The Lassaigne's extract is boiled with dil. $\mathrm{HNO}_{3}$ before testing for halogens because

Options:
A silver halides are soluble in $\mathrm{HNO}_{3}$
B $\mathrm{Na}_{2} \mathrm{~S}$ and $\mathrm{NaCN}$ are decomposed by $\mathrm{HNO}_{3}$
C $\mathrm{Ag}_{2} \mathrm{~S}$ is soluble in $\mathrm{HNO}_{3}$
D $\mathrm{AgCN}$ is soluble in $\mathrm{HNO}_{3}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{~S}$ and $\mathrm{NaCN}$ are decomposed by $\mathrm{HNO}_{3}$

Solution:
$\mathrm{Na}_{2} \mathrm{~S}$ and $\mathrm{NaCN},$ formed during fusion with metallic sodium, must be removed before adding $\mathrm{AgNO}_{3},$ otherwise black ppt. due to $\mathrm{Na}_{2} \mathrm{~S}$ or white precipitate due to $\mathrm{AgCN}$ will be formed and thus white precipitate of $\mathrm{AgCl}$ will not be identified easily.

$\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{AgNO}_{3} \longrightarrow 2 \mathrm{NaNO}_{3}+\underset{\text { Black }}{\mathrm{Ag}_{2} \mathrm{~S}} \downarrow$

$\mathrm{NaCN}+\mathrm{AgNO}_{3} \longrightarrow \mathrm{NaNO}_{3} + \underset{\text { White }}{\mathrm{Ag} \mathrm{CN}} \downarrow$

$\mathrm{NaCl}+\mathrm{AgNO}_{3} \longrightarrow \mathrm{NaNO}_{3}+\underset{\text { White }}{\mathrm{Ag} \mathrm{Cl}} \downarrow$

$\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{HNO}_{3} \stackrel{\text { boil }}{\longrightarrow} 2 \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{~S} \uparrow$

$\mathrm{NaCN}+\mathrm{HNO}_{3} \stackrel{\text { boil }}{\longrightarrow} \mathrm{NaNO}_{3}+\mathrm{HCN} \uparrow$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2020
The laws of electromagnetic induction have been used in the construction of a

Options:
A galvanometer
B voltmeter
C electric motor
D generator

The correct answer is:
generator

Solution:
Hint:
The laws of Electromagnetic Induction give us idea about the various ways in which we can produce Electric Current using Electric or Magnetic field.
Correct Option is D.
Explanation for correct answer:
- Galvanometer is a device which detects current in the circuit.
- Voltmeter is a device which measures potential difference between two points in the circuit.
- Electric motor is a device which converts Electrical Energy into Mechanical Energy.
- Electric Generator is an device which produces Electric Current from Mechanical energy, Chemical energy, or by using Magnetic or Electric Fields. The laws if Electromagnetic Induction are used in Electric Generator which produces Electric Current from Magnetic or Electric Fields.

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2012
The least positive non-integral solution of the equation $\sin \pi\left(x^{2}+x\right)=\sin \pi x^{2}$ is

Options:
A rational
B irrational of the form $\sqrt{\mathrm{p}}$
C irrational of the form $\frac{\sqrt{\mathrm{p}}-1}{4},$ where $\mathrm{p}$ is an odd integer
D irrational of the form $\frac{\sqrt{\mathrm{p}}+1}{4},$ where $\mathrm{p}$ is an even integer

The correct answer is:
rational

Solution:
We have, $\sin \pi\left(x^{2}+x\right)=\sin \pi x^{2}$

$\Rightarrow \quad \pi\left(\mathrm{x}^{2}+\mathrm{x}\right)=\mathrm{n} \pi+(-1)^{\mathrm{n}} \pi \mathrm{x}^{2}$

$\therefore$ Either $\mathrm{x}^{2}+\mathrm{x}=2 \mathrm{~m}+\mathrm{x}^{2} \Rightarrow \mathrm{x}=2 \mathrm{~m} \in \mathrm{I}$

or $\quad \mathrm{x}^{2}+\mathrm{x}=\mathrm{k}-\mathrm{x}^{2},$ where $\mathrm{k}$ is an odd integer

$\Rightarrow \quad 2 \mathrm{x}^{2}+\mathrm{x}-\mathrm{k}=0 \Rightarrow \mathrm{x}=\frac{-1 \pm \sqrt{1+8 \mathrm{k}}}{4}$

For least positive non-integral solution

is $x=\frac{1}{2},$ when $k=1$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2020
The least value of the function f(x)=∫0x3sinx+4cosxdx in the interval 5π4,4π3 is

Options:
A 3-32
B 5-432
C 7-432
D 9-432

The correct answer is:
9-432

Solution:
We have, fx=∫0x3sinx+4cosxdx

f'x=3sinx+4cosx=5sinx+tan-143

5π4<x<4π3

π<x+tan-14/3<2π

∵ f'x<0, i.e. fx is decreasing.

∵ least value of fx is 5π4,4π3=f4π3

least fx=∫04π/33sinx+4cosxdx

=-3cosx+4sinx04π/3

=-3-12-1+4-32

=9-432

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2019
The length of a potentiometer wire is $\ell$. A cell of emf $E$ is balanced at a length $\ell / 3$ from the positive end of the wire. If the length of the wire is increased by $\ell / 2 .$ At what distance will be the same cell give a balance point.

Options:
A $2 \ell / 3$
B $\ell / 2$
C $\quad \ell / 6$
D $4 \ell / 3$

The correct answer is:
$\ell / 2$

Solution:
Potential gradient in the first case $=\frac{\mathrm{E}_{0}}{\ell}$ $$
\therefore \quad \mathrm{E}=\left(\frac{\ell}{3}\right) \cdot\left(\frac{\mathrm{E}_{0}}{\ell}\right)=\frac{\mathrm{E}_{0}}{3}
$$
Potential gradient in second case
$$
=\frac{E_{0}}{3 \ell / 2} \text { (x) } \frac{2 E_{0}}{3 \ell}
$$
From equations (i) and (ii),
$$
\frac{\mathrm{E}_{0}}{3}-\left(\frac{2 \mathrm{E}_{0}}{3 \ell}\right) \mathrm{x} \Rightarrow \mathrm{x}=\frac{\ell}{2}
$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2009
The length of perpendicular drawn from the point (2,3,4) to line 4-x2=y6=1-z3, is

Options:
A 37101
B 27101
C 27103
D 37103

The correct answer is:
37101

Solution:
Let P be the foot of the perpendicular drawn from A(2,3,4) to the given line l.

Then, x-4-2=y6=z-1-3

Now, any point on the line l is given by

x=4-2λ,y=6λ,z=1-3λ

The coordinates of P are (4-2λ,6λ,1-3λ)

The direction ratios of AP are

(4-2λ-2,6λ-3,1-3λ-4)

i.e. (2-2λ,6λ-3,-3-3λ)

And the direction ratios of l are-2,6 and -3.

Given, AP⊥l

∴ -2(2-2λ)+6(6λ-3)-3(-3-3λ)=0

⇒ λ=1349

∴ AP2=(4-2λ-2)2+(6λ-3)2+(1-3λ-4)2

=22-26λ+49λ2

Put λ=1349, we get

AP2=90949⇒AP=37101

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2017
The length of the chord $x+y=3$ intercepted by the circle $x^{2}+y^{2}-2 x-2 y-2=0$ is

Options:
A $\frac{7}{2}$
B $\frac{3 \sqrt{3}}{2}$
C $\sqrt{14}$
D $\frac{\sqrt{7}}{2}$

The correct answer is:
$\sqrt{14}$

Solution:
The centre of the circle is \(\mathrm{C}(1,1)\) and radius of the circle is 2 , perpendicular distance from \(\mathrm{C}\) on \(\mathrm{AB}\), the chord \(\mathrm{x}+\mathrm{y}=3\)

\(\begin{aligned} &\mathrm{CD}=\left|\frac{1+1-3}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}} \\ &\Rightarrow \mathrm{AD}=\sqrt{4-\frac{1}{2}}=\sqrt{\frac{7}{2}}\left[\mathrm{AD}=\sqrt{\mathrm{AC}^{2}-\mathrm{CD}^{2}}\right] \end{aligned}\)

Hence, the length of the chord \(\mathrm{AB}=2 \mathrm{AD}=2 \sqrt{\frac{7}{2}}=\sqrt{14}\).

Topic: Mathematics, Circle, BITSAT, BITSAT 2016
The length of the latus rectum of the parabola which has focus at (-1,1) and the directrix is $4 \mathrm{x}$ $+3 y-24=0$ is

Options:
A 4
B 6
C 8
D 10

The correct answer is:
10

Solution:
Length of latus rectum $=2 \times$ distance of focus from directrix $=2 \times \frac{|1-4+3-24|}{5}=10$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2020