datavorous/entrance-exam-dataset huggingface dataset:
The number of all three elements subsets of the set $\left\{a_{1}, a_{2}, a_{3} \ldots a_{n}\right\}$ which contain $a_{3}$ is

Options:
A ${ }^{\mathrm{n}} \mathrm{C}_{3}$
B ${ }^{\mathrm{n}-1} \mathrm{C}_{3}$
C ${ }^{\mathrm{n}-1} \mathrm{C}_{2}$
D None of these

The correct answer is:
${ }^{\mathrm{n}-1} \mathrm{C}_{2}$

Solution:
The number of three elements subsets containing $\mathrm{a}_{3}$ is equal to the number of ways of selecting 2 elements out of $\mathrm{n}-1$ elements. So, the required number of subsets is ${ }^{\mathrm{n}-1} \mathrm{C}_{2}$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2014
The number of coulombs required for the deposition of $108 \mathrm{~g}$ of silver is

Options:
A 96500
B 48250
C 193000
D 10000

The correct answer is:
96500

Solution:
Amt. deposited $=\frac{\text { E. wt } \times \text { Q }}{96500}$;
$$
107.870=\frac{107.870}{96500} \times \mathrm{Q} ; \quad \therefore \mathrm{Q}=96500 \mathrm{C}
$$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2011
The number of different seven-digit numbers that can be written using only the three digit 1 , 2 and 3 with the condition that the digit 2 occurs twice in each number is

Options:
A
${ }^7 C_2 2^5$

B
${ }^7 p_2 2^5$

C
${ }^7 C_2 5^2$

D
None of these

The correct answer is:

${ }^7 C_2 2^5$

Solution:
Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is $2 \times 2 \times 2 \times 2 \times 2=2^5$. For 2 , selecting 2 places out of 7 is ${ }^7 C_2$. Hence, the required number of ways is ${ }^7 C_2 \times 2^5$.

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2022
The number of double bonds in gammexane is:

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Gammexane is $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{Cl}_{6}$ or (6,6,6) . It is a saturated compound so no double bond is there in it.

Topic: Chemistry, Hydrocarbons, JEE Main
The number of double bonds in gammexane is:

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Gammexane is $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{Cl}_{6}$ or (6,6,6) . It is a saturated compound so no double bond is there in it.

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2014
The number of elements present in the d-block of the periodic table is

Options:
A 40
B 41
C 45
D 46

The correct answer is:
40

Solution:
40 elements are present in d-block.

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2015
The number of enantiomers of the compound $\mathrm{CH}_{3} \mathrm{CHBrCHBrCOOH}$ is

Options:
A 0
B 1
C 3
D 4

The correct answer is:
4

Solution:
The structure $\mathrm{CH}_{3}{ }^{*} \mathrm{CH} \mathrm{Br} \mathrm{C}^{*} \mathrm{H} \mathrm{Br} \mathrm{COOH}$

has two different chiral carbon atoms, hence number of enantiomers (optically active forms) is $2^{n}=2^{2}=4$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2013
The number of geometrical isomers of $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CHCl}$ is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
The given structure has three double bonds whose each carbon atom is differently substituted hence number of geometrical isomers will be $2^{n}=2^{3^{\prime}}=8$, where $n$ is the number of double bonds whose each carbon atom is differently substituted.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2010
The number of integral values of $\lambda$ for which $\mathrm{x}^{2}+\mathrm{y}^{2}+\lambda \mathrm{x}+(1-\lambda) \mathrm{y}+5=0$ is the equation of a ircle whose radius cannot exceed $5,$ is

Options:
A 14
B 18
C 16
D None

The correct answer is:
16

Solution:
Radius $\leq 5$

$$

\begin{array}{l}

\sqrt{\frac{\lambda^{2}}{4}+\frac{(1-\lambda)^{2}}{4}-5} \leq 5 \Rightarrow \lambda^{2}+(1-\lambda)^{2}-20 \leq 100 \\

\Rightarrow 2 \lambda^{2}-2 \lambda-119 \leq 0 \\

\therefore \frac{1-\sqrt{239}}{2} \leq \lambda \leq \frac{1+\sqrt{239}}{2} \Rightarrow-7.2 \leq \lambda \leq 8.2 \\

\quad(\text { approx. }) \\

\therefore \lambda=-7,-6,-5, \ldots \ldots .7,8, \text { in all } 16

\end{array}

$$ values

Topic: Mathematics, Circle, BITSAT, BITSAT 2016
The number of points at which the function $\mathrm{f}(\mathrm{x})=\frac{1}{\log |\mathrm{x}|}$ is discontinuous is:

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
The function $\log |x|$ is not defined at $x=0$, so, $\mathrm{x}=0$ is a point of discontinuity. Also for $\mathrm{f}(\mathrm{x})$ to be defined, $\log |\mathrm{x}| \neq 0 \Rightarrow \mathrm{x} \neq\pm 1$

Hence, 0,1,-1 are three points of discontinuity.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2016
The number of positive integral solution of $\mathrm{abc}=30 \mathrm{is}$

Options:
A 30
B 27
C 8
D None of these

The correct answer is:
27

Solution:
We have : $30=2 \times 3 \times 5 .$ So, 2 can be assigned to either a or b or ci.e. 2 can be assigned in 3 ways. Similarly, each of 3 and 5 can be assigned in 3 ways. Thus, the number of solutions is $3 \times 3 \times 3=27$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2012
The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is

Options:
A $6^{n}-1$
B $3^{n}-1$
C $6^{n}-3^{n}$
D None of these

The correct answer is:
$6^{n}-3^{n}$

Solution:
Total number of ways $=6 \times 6 \times \ldots \ldots$ to $n$ times $=6^{n}$ Total number of ways to show only even number $=3 \times 3 \times \ldots \ldots$ to $n$ times $=3^{n}$

$\therefore$ required number of ways $=6^{\mathrm{n}}-3^{\mathrm{n}}$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2020
The number of real numbers λ for which the equality sin(λα)sinα-cos(λα)cosα=λ-1, holds for all real α which are not integral multiple of π/2 is

Options:
A 1
B 2
C 3
D Infinite

The correct answer is:
2

Solution:
We have, sinλαsinα-cosλαcosα=λ-1

sinλαcosα-cosλαsinα=λ-1sinαcosα

sinλ-1α=λ-1sinαcosα

sinλ-1α=λ-12sin2α

It is possible, when λ-1=0⇒λ=1

and λ-12=1⇒λ=3

∴ λ has only two solution.

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2019
The number of real roots of the equation e $\mathrm{e}^{(x-1)}+\mathrm{x}$ $-2=0$ is

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

The correct answer is:
1

Solution:
Let $f(x)=e^{x-1}+x-2$

check for $x=1$ Then, $\mathrm{f}(1)=\mathrm{e}^{0}+1-2=0$

So, $x=1$ is a real root of the equation $f(x)=0$ Let $x=\alpha$ be the other root such that $\alpha>1$ or

$\alpha<1$. Consider the interval $[1, \alpha]$ or $[\alpha, 1]$

Clearly $\mathrm{f}(1)=\mathrm{f}(\alpha)=0$

By Rolle's theorem $\mathrm{f}(\mathrm{x})=0$ has a root in $(1, \alpha)$ or in $(\alpha, 1)$. But $f^{\prime}(x)=e^{x-1}+1>0,$ for all $x$. Thus

$\mathrm{f}^{\prime}(\mathrm{x}) \neq 0,$ for any $\mathrm{x} \in(1, \alpha)$ or $\mathrm{x} \in(\alpha, 1)$

which is a contradiction. Hence, $\mathrm{f}(\mathrm{x})=0$ has no real root other than 1 .

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
The number of real solutions of the equation

$(x-1)^{2}+(x-2)^{2}+(x-3)^{2}=0$ is

Options:
A 2
B 1
C 0
D 3

The correct answer is:
0

Solution:
Equation holds if and only if $x-1=0, x-2=0$ and $x-3=0$ simultanesouly, which is not possible.

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2020
The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$ is

Options:
A none
B one
C two
D infinite

The correct answer is:
one

Solution:
Since, $\frac{d y}{d x}=\frac{y+1}{x-1}$
$\Rightarrow \quad \frac{d y}{y+1}=\frac{d x}{x-1}$
After integrating on both sides, we have
$\log (y+1)=\log (x-1)-\log C$
$C(y+1)=(x-1)$
$C=\frac{x-1}{y+1}$
If $x=1$, then $y=2$, so $C=0$
Therefore, $x-1=0$
Hence, there is only one solution.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The number of terms in the expansion of $(1+5 \sqrt{2} x)^9+(1-5 \sqrt{2} x)^9$, is

Options:
A
5

B
7

C
9

D
10

The correct answer is:

5

Solution:
We have,
$(1+5 \sqrt{2} x)^9+(1-5 \sqrt{2} x)^9$
$=2\left\{{ }^2 C_0+{ }^9 C_2(5 \sqrt{2} x)^2+\ldots+{ }^9 C_8(5 \sqrt{2} x)^8\right\}$
Clearly, it has 5 terms.

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2022
The number of times the digit 5 will be written when listing the integers from 1 to 1000, is

Options:
A 271
B 272
C 300
D None of these

The correct answer is:
300

Solution:
Any number between 1 to 999 is of the form abc when 0≤a,b,c≤9. Let us first count the number in which 5 occurs exactly once.

Since, 5 can occur at one place in 1×C13×9×9=243 ways, next 5 can occur in exactly two places in C23×9=27.

Lastly, 5 can occur in all three digits in only one way.

Hence, the number of times 5 occurs

=1×243+27×2+1×3

=243+54+3=300

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2017
The number of values of satisfying the equation ${ }^{39} \mathrm{C}_{3 \mathrm{r}-1}-{ }^{39} \mathrm{C}_{\mathrm{r}^{2}}={ }^{39} \mathrm{C}_{\mathrm{r}^{2}-1}-{ }^{39} \mathrm{C}_{3 \mathrm{r}}$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
${ }^{39} \mathrm{C}_{3 \mathrm{r}-1}-{ }^{39} \mathrm{C}_{\mathrm{r}^{2}}={ }^{39} \mathrm{C}_{\mathrm{r}^{2}-1}-{ }^{39} \mathrm{C}_{3 \mathrm{r}}$

$\Rightarrow{ }^{39} \mathrm{C}_{3 \mathrm{r}-1}+{ }^{39} \mathrm{C}_{3 \mathrm{r}}={ }^{39} \mathrm{C}_{\mathrm{r}^{2}-1}+{ }^{39} \mathrm{C}_{\mathrm{r}^{2}}$

$\Rightarrow{ }^{40} \mathrm{C}_{3 \mathrm{r}}={ }^{40} \mathrm{C}_{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{r}^{2}=3 \mathrm{r}$ or $\mathrm{r}^{2}=40-3 \mathrm{r}$

$\Rightarrow \mathrm{r}=0,3$ or -8,5

3 and 5 are the values as the given equation is not defined by $r=0$ and $r=-8$. Hence, the number of values of $\mathrm{r}$ is 2 .

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2016
The number of ways of arranging letters of the word HAVANA so that $\mathrm{V}$ and $\mathrm{N}$ do not appear together is

Options:
A
40

B
60

C
80

D
100

The correct answer is:

80

Solution:
We can arrange the letters $\mathrm{H}, \mathrm{A}, \mathrm{A}, \mathrm{A}$ in $\frac{4 !}{3 !}=4$ ways. If one possible arrangement is
$$
X X X X
$$
Then, we can arrange $\mathrm{V}, \mathrm{N}$ at any of the two places marked with $\mathrm{O}$ in the following arrangement
$$
\mathrm{O} \times \mathrm{O} \times \mathrm{O} \times \mathrm{O} \times \mathrm{O}
$$
Thus, we can arrange $V$ and $N$ in ${ }^5 P_2=20$ ways. Thus, the number of ways in which letters can be arranged is $4 \times 20=80$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2022
The numbers 32sin2α-1,14 and 34-2sin2α form first three terms of an A.P., then its fifth term is

Options:
A -25
B -12
C 40
D 53

The correct answer is:
53

Solution:
Since, 32sin2α-1,14 and 34-2sin2α are in AP. Therefore,

2×14=32sin2α-1+34-2sin2α

⇒ 28=a3+34a, where a=32sin2α

⇒ a2-84a+243=0

⇒ a-81a-3=0

⇒ a=81, a=3

⇒ 32sin2α=34 or 32sin2α=3

⇒ 2sin2α=1 ∵ 2sin2α≠4

⇒ sin2α=12⇒2α=30° ∵ sin30°=1/2

Thus, the first three terms of the AP are 1,14,27.

Hence, its fifth term a5=a1+5-1d

=1+4×13=1+52=53

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2017
The one which is least basic is

Options:
A $\mathrm{NH}_{3}$
B $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
C $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{~N}$
D $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{NH}$

The correct answer is:
$\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{~N}$

Solution:
More the electron density on $\mathrm{N}$, higher will be the basicity. Density on $\mathrm{N}$ is influenced by the (i) nature of the group (+I or -I) present in alkyl group or benzene nucleus and (ii) resonance (delocalisation of the electron present on $\mathrm{N}$ ). In $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{~N}$ :
electron pair is delocalised to the maximum extent due to three benzene rings and hence least available for protonation, thus it will be least basic)

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2009
The orbital angular momentum of an electron is 2s orbital is

Options:
A h4π
B zero
C h2π
D 2·h2π

The correct answer is:
zero

Solution:
For s-subshell l=0

Orbital angular momentum (mvr)=h2πl(l+1)

For 2s orbital, l (azimuthal quantum number) =0
∴ orbital angular momentum =h2π0(0+1)

=h2π0=0

Thus, the orbital angular momentum of an electron in 2s -orbital is Zero.

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2017
The order of acidic strength of boron trihalides

Options:
A BF3<BCl3<BBr3<BI3
B BI3<BBr3<BCl3<BF3
C BCl3<BBr3<BI3<BF3
D BBr3<BCl3<BF3<BI3

The correct answer is:
BF3<BCl3<BBr3<BI3

Solution:
The strength of acidic character of boron trihalides depends upon pπ-pπ back bonding.

In boron trihalides, pπ-pπ -back bonding occurs due to empty orbital of boron and filled orbitals of halogens.

The pπ-pπ back bonding is shown maximum by BF3, as the size of B and F are small and comparatively same. Due to this effect tendency of accepting lone-pair of electrons of boron decreases as size of halogen increases. The order of size of halogens are F<Cl<Br<I. Thus, acidic nature is in order

BF3<BCl3<BBr3<BI3

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The order of acidic strength of boron trihalides

Options:
A BF3<BCl3<BBr3<BI3
B BI3<BBr3<BCl3<BF3
C BCl3<BBr3<BI3<BF3
D BBr3<BCl3<BF3<BI3

The correct answer is:
BF3<BCl3<BBr3<BI3

Solution:
The strength of acidic character of boron trihalides depends upon pπ-pπ back bonding.

In boron trihalides, pπ-pπ -back bonding occurs due to empty orbital of boron and filled orbitals of halogens.

The pπ-pπ back bonding is shown maximum by BF3, as the size of B and F are small and comparatively same. Due to this effect tendency of accepting lone-pair of electrons of boron decreases as size of halogen increases. The order of size of halogens are F<Cl<Br<I. Thus, acidic nature is in order

BF3<BCl3<BBr3<BI3

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2019
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a:

Options:
A NOT gate
B NOR gate
C AND gate
D OR gate

The correct answer is:
NOR gate

Solution:
$(\overline{A+B})=$ NOR gate When both inputs of NAND gate are connected, it behaves as NOT gate $\mathrm{OR}+\mathrm{NOT}=\mathrm{NOR}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2011
The oxidation state of sulphur in $\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$ is

Options:
A 6
B $\frac{+3}{2}$
C $\frac{+5}{2}$
D -2

The correct answer is:
$\frac{+5}{2}$

Solution:
$\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$

$2+4 x-12=0$

$4 x-10=0$

$x=\frac{10}{4}=\frac{+5}{2}$

Oxidation state of $\mathrm{S}$ is $=\frac{+5}{2}$

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2013
The oxidation states of sulphur in the anions $\mathrm{SO}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ follow the order

Options:
A $\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$
B $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{SO}_{3}^{2-}$
C $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}$
D $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$

The correct answer is:
$\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$

Solution:
According to chemical bond method:

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation states of sulphur in the anions $\mathrm{SO}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ follow the order

Options:
A $\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$
B $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{SO}_{3}^{2-}$
C $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}$
D $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$

The correct answer is:
$\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$

Solution:
According to chemical bond method:

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2009
The $\mathrm{P}-\mathrm{P}-\mathrm{P}$ angle in $\mathrm{P}_{4}$ molecule and $\mathrm{S}-\mathrm{S}-\mathrm{S}$

angle in $\mathrm{S}_{8}$ molecule is(in degree) respectively

Options:
A $60^{\circ}, 107^{\circ}$
B $107^{\circ}, 60^{\circ}$
C $40^{\circ}, 60^{\circ}$
D $60^{\circ}, 40^{\circ}$

The correct answer is:
$60^{\circ}, 107^{\circ}$

Solution:
In $P_{4}$ molecule, the four sp $^{3}$ -hybridised phosphorous atoms lie at the corners of a regular tetrahedron with $\angle \mathrm{PPP}=60^{\circ}$.

In $\mathrm{S}_{8}$ molecule S-S-S angle is $107^{\circ}$ rings.

In $\mathrm{S}_{8}$ molecule $\mathrm{S}-\mathrm{S}-\mathrm{S}$ angle is $107^{\circ} \mathrm{rings} .$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2015
The pair in which both species have same magnetic moment (spin onlyvalue) is:

Options:
A $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{CoCI}_{4}\right]^{2-}$
B $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\right.$
C $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right)^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
D $\left[\mathrm{CoCl}_{4}\right)^{2-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\right.$

Solution:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \mathrm{Cr}$ is in $\mathrm{Cr}^{2+}$ form

In $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+} \mathrm{Fe}^{2+}$ form. Both will have 4

unpaired electrons.

Topic: Chemistry, Coordination Compounds, JEE Main
The pair in which both species have same magnetic moment (spin onlyvalue) is:

Options:
A $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{CoCI}_{4}\right]^{2-}$
B $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\right.$
C $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right)^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
D $\left[\mathrm{CoCl}_{4}\right)^{2-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\right.$

Solution:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \mathrm{Cr}$ is in $\mathrm{Cr}^{2+}$ form

In $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+} \mathrm{Fe}^{2+}$ form. Both will have 4

unpaired electrons.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2013
The parabola having its focus at (3,2) and directrix along the $y$ -axis has its vertex at

Options:
A (2,2)
B $\left(\frac{3}{2}, 2\right)$
C $\left(\frac{1}{2}, 2\right)$
D $\left(\frac{2}{3}, 2\right)$

The correct answer is:
$\left(\frac{3}{2}, 2\right)$

Solution:
Vertex of the parabola is a point which lies on the axis of the parabola, which is a line $\perp$ to the directrix through the focus, i.e., $\mathrm{y}=2$ and equidistant from the focus and directrix $\mathrm{x}=0$, so that the vertex is $\left(\frac{3}{2}, 2\right)$.

Topic: Mathematics, Parabola, JEE Main
The parabola having its focus at (3,2) and directrix along the $y$ -axis has its vertex at

Options:
A (2,2)
B $\left(\frac{3}{2}, 2\right)$
C $\left(\frac{1}{2}, 2\right)$
D $\left(\frac{2}{3}, 2\right)$

The correct answer is:
$\left(\frac{3}{2}, 2\right)$

Solution:
Vertex of the parabola is a point which lies on the axis of the parabola, which is a line $\perp$ to the directrix through the focus, i.e., $\mathrm{y}=2$ and equidistant from the focus and directrix $\mathrm{x}=0$, so that the vertex is $\left(\frac{3}{2}, 2\right)$.

Topic: Mathematics, Parabola, BITSAT, BITSAT 2016
The percentage hydrolysis of $0.15 \mathrm{M}$ solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$

and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $1.8 \times 10^{-5}$

Options:
A 0.556
B 4.72
C 9.38
D 5.56

The correct answer is:
0.556

Solution:
$\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}}=0.55$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2016
The period of $\tan 3 \theta$ is

Options:
A $\pi$
B $3 \pi 4$
C $\pi / 2$
D None of these

The correct answer is:
None of these

Solution:
$\tan \theta$ is of period $\pi$ so that $\tan 3 \theta$ is of period $\pi / 3$

Topic: Mathematics, Functions, BITSAT, BITSAT 2015
The perpendicular distance of \(P(1,2,3)\) from the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\) is

Options:
A 7
B 5
C 0
D 6

The correct answer is:
7

Solution:
The point \(\mathrm{A}(6,7,7)\) is on the line. Let the perpendicular from \(\mathrm{P}\) meet the line in \(\mathrm{L}\). Then
\(\mathrm{AP}^2=(6-1)^2+(7-2)^2+(7-3)^2=66\)
Also \(\mathrm{AL}=\) projection of AP on line
\(\begin{aligned}
& \left(\text {actual d.c.'s } \frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right) \\
\Rightarrow & (6-1) \cdot \frac{3}{\sqrt{17}}+(7-2) \cdot \frac{2}{\sqrt{17}}+(7-3) \frac{-2}{\sqrt{17}} \\
& =\sqrt{17}
\end{aligned}\)
\(\therefore \perp\) distance \(\mathrm{d}\) of \(\mathrm{P}\) from the line is given by
\(\mathrm{d}^2=\mathrm{AP}^2-\mathrm{AL}^2=66-17=49 \text { so that } \mathrm{d}=7\)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2010
The pH of a solution is increased from 3 to 6 ; its $\mathrm{H}^{+}$ ion concentration will be

Options:
A reduced to half
B doubled
C reduced by 1000 times
D increased by 1000 times

The correct answer is:
reduced by 1000 times

Solution:
$\mathrm{pH}=3 \ldots\left[\mathrm{H}^{+}\right]=10^{-3} ; \quad \mathrm{pH}=6$
$\therefore\left[\mathrm{H}^{+}\right]=10^{-6}$. Hence $\left[\mathrm{H}^{+}\right]$ is reduced by
$10^{-3}$ times.

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2011
The photochemical smog does not generally contain:

Options:
A $\mathrm{NO}$
B $\mathrm{NO}_2$
C $\mathrm{SO}_2$
D $\mathrm{HCHO}$

The correct answer is:
$\mathrm{SO}_2$

Solution:
Photochemical smog contains nitrogen dioxide $\left(\mathrm{NO}_2\right)$, Ozone $\left(\mathrm{O}_3\right)$, PAN (peroxyacetylnitrate), and compounds containing $-\mathrm{CHO}$ group.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The photoelectric threshold of metal is $2000 \AA$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500 \AA$ is

Options:
A $2.0 \mathrm{eV}$
B $1.5 \mathrm{eV}$
C $15 \mathrm{eV}$
D $150 \mathrm{eV}$

The correct answer is:
$2.0 \mathrm{eV}$

Solution:
$\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=2.0 \mathrm{eV}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2010
The pOH value of a solution whose hydroxide ion concentration is $6.2 \times 10^{-9} \mathrm{~mol} /$ litre is

Options:
A 8.21
B 6.21
C 7.75
D 7.21

The correct answer is:
8.21

Solution:
$-\log (\mathrm{OH})=\mathrm{pOH} ;-\log 6.2 \times 10^{-9}=\mathrm{pOH}$
$\therefore \mathrm{pOH}=8.21$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2012