datavorous/entrance-exam-dataset huggingface dataset:
Starch can be used as an indicator for the detection of traces of

Options:
A glucose in aqueous solution
B proteins in blood
C iodine in aqueous solution
D urea in blood

The correct answer is:
iodine in aqueous solution

Solution:
$\mathrm{I}_{2}$ gives blue colour with starch.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2014
Steam at 100 °C is passed into 1.1 kg of water contained in a calorimoter of water equivalent to 0.2 kg at 15 °C till the temperature of the calorimeter and its contents rises to 80 °C. The mass of steam condensed (in kg ) is

Options:
A 0.130
B 0.065
C 0.260
D 0.135

The correct answer is:
0.130

Solution:
(a) According to principle of calorimetry,
heat gained = heat lost
Heat is lost by steam in two stages

(i) Change of state from steam at 100 °C to water at 100 °C is m×540.

(ii) To change water at 100 °C to water at 80 °C is m×1×100-80, where m is the mass of the steam condensed.

Total heat lost by steam is
m×540+m×20=m 540+20=560 m

Heat gained by calorimeter and its contents is
1.1+0.2×80-15=1.12×65 cal

⇒ 560 m=1.12×65

⇒ m=1.12×65560=0.130 kg

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2018
Straight lines 3x+4y=5 and 4x-3y=15 intersect at the point A. If point B and C are chosen on these two lines such that AB=AC, then the possible equation of the line BC passing through the point 1,2 is

Options:
A x+7y+13=0 or 7x+y+9=0
B x+7y+13=0 or 7x+2y+7=0
C x-7y+13=0 or 7x+y-9=0
D None of the above

The correct answer is:
x-7y+13=0 or 7x+y-9=0

Solution:
The given straight lines are 3x+4y=5 and 4x-3y=15. Clearly, these straight lines are perpendicular to each other m1m2=-1 and intersect at A. Now, B and C are points on these lines such that AB=AC and BC passes through 1,2.

From figure it is clear that ∠B=∠C=45°

Let slope of BC be m. Then,

tan45°=m+341-34m

⇒±1=4m+34-3m

4m+3=±4-3m

4m+3=4-3m

or 4m+3=-4+3m

m=17

or m=-7

Hence, equation of BC is

y-2=17x-1

or y-2=-7x-1

⇒7y-14=x-1

or y-2=-7x+7

⇒x-7y+13=0

or 7x+y-9=0

Hence, option (c) is correct.

Topic: Mathematics, Straight Lines, JEE Main
Straight lines 3x+4y=5 and 4x-3y=15 intersect at the point A. If point B and C are chosen on these two lines such that AB=AC, then the possible equation of the line BC passing through the point 1,2 is

Options:
A x+7y+13=0 or 7x+y+9=0
B x+7y+13=0 or 7x+2y+7=0
C x-7y+13=0 or 7x+y-9=0
D None of the above

The correct answer is:
x-7y+13=0 or 7x+y-9=0

Solution:
The given straight lines are 3x+4y=5 and 4x-3y=15. Clearly, these straight lines are perpendicular to each other m1m2=-1 and intersect at A. Now, B and C are points on these lines such that AB=AC and BC passes through 1,2.

From figure it is clear that ∠B=∠C=45°

Let slope of BC be m. Then,

tan45°=m+341-34m

⇒±1=4m+34-3m

4m+3=±4-3m

4m+3=4-3m

or 4m+3=-4+3m

m=17

or m=-7

Hence, equation of BC is

y-2=17x-1

or y-2=-7x-1

⇒7y-14=x-1

or y-2=-7x+7

⇒x-7y+13=0

or 7x+y-9=0

Hence, option (c) is correct.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2018
Strong reducing behaviour of $\mathrm{H}_{3} \mathrm{PO}_{2}$ is due to

Options:
A presence of one $-\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ bonds
B high electron gain enthalpy of phosphorus
C high oxidation state of phosphorus
D presence of two -OH groups and one $\mathrm{P}-\mathrm{H}$ bond

The correct answer is:
presence of one $-\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ bonds

Solution:
The acids which contain $\mathrm{P}-\mathrm{H}$ bond have strong reducing properties. Thus $\mathrm{H}_{3} \mathrm{PO}_{2}$ acid is good reducing agent as it contains two P-H bonds. For example, it reduces $\mathrm{AgNO}_{3}$ to metallic silver.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Strong reducing behaviour of $\mathrm{H}_{3} \mathrm{PO}_{2}$ is due to

Options:
A presence of one $-\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ bonds
B high electron gain enthalpy of phosphorus
C high oxidation state of phosphorus
D presence of two -OH groups and one $\mathrm{P}-\mathrm{H}$ bond

The correct answer is:
presence of one $-\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ bonds

Solution:
The acids which contain $\mathrm{P}-\mathrm{H}$ bond have strong reducing properties. Thus $\mathrm{H}_{3} \mathrm{PO}_{2}$ acid is good reducing agent as it contains two P-H bonds. For example, it reduces $\mathrm{AgNO}_{3}$ to metallic silver.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2013
Suppose a parabola y=ax2+bx+c has two x intercepts, one positive and one negative, and its vertex is (2,-2), then which of the following is true?

Options:
A ab>0
B bc>0
C ac>0
D a+b+c>0

The correct answer is:
bc>0

Solution:
Given, y=ax2+bx+c has two x-intercepts one is positive and one is negative and vertex is (2,-2). The graph of y=ax2+bx+c is

Clearly from graph

c>0

-ba>0⇒-b>0

⇒ b<0

f(1)=a+b+c<0

ab<0,ac<0,bc>0

Topic: Mathematics, Parabola, BITSAT, BITSAT 2019
Suppose $\mathrm{p}, \mathrm{q}, \mathrm{r} \neq 0$ and system of equation $(\mathrm{p}+\mathrm{a}) \mathrm{x}+\mathrm{by}+\mathrm{cz}=0$

$a x+(q+b) y+c z=0$

$a x+b y+(r+c) z=0$ has a non-trivial solution,

then value of $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}$ is

Options:
A -1
B 0
C 1
D 2

The correct answer is:
-1

Solution:
Since the system is homogeneous and has a non-trivial solution

$0=\left|\begin{array}{ccc}p+a & b & c \\ a & q+b & c \\ a & b & r+c\end{array}\right|=\left|\begin{array}{ccc}p & -q & 0 \\ 0 & q & -r \\ a & b & r+c\end{array}\right|$

using $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$

Expanding along $\mathrm{C}$, we get $\Rightarrow \mathrm{p}[\mathrm{q}(\mathrm{r}+\mathrm{c})+\mathrm{br}]+\mathrm{aqr}=0$

$\Rightarrow \mathrm{pqr}+\mathrm{pqc}+\mathrm{prb}+\mathrm{qra}=0$

$\Rightarrow \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2020
Suppose that $f(0)=-3$ and $f^{\prime}(x) \leq 5$ for all values of $x$ Then, the largest value which $f(2)$ can attain is

Options:
A 7
B 10
C 2
D 9

The correct answer is:
7

Solution:
Using LMVT in [0,2] $\frac{f(2)-f(0)}{2-0}=f^{\prime}(c),$ where $c \in(0,2)$

$\frac{f(2)+3}{2} \leq 5 \Rightarrow f(2) \leq 7$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2020
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $\mathrm{R}$ ' around the sun will be proportional to

Options:
A $\mathrm{R}^{\mathrm{n}}$
B $\mathrm{R}^{\left(\frac{\mathrm{n}-1}{2}\right)}$
C $\mathrm{R}\left(\frac{\mathrm{n}+1}{2}\right)$
D $\mathrm{R}^{\left(\frac{\mathrm{n}-2}{2}\right)}$

The correct answer is:
$\mathrm{R}\left(\frac{\mathrm{n}+1}{2}\right)$

Solution:
$$
\mathrm{F}=\mathrm{KR}^{-\mathrm{n}}=\mathrm{MR} \omega^{2} \Rightarrow \omega^{2}=\mathrm{KR}^{-(\mathrm{n}+1)}
$$
or $\omega=\mathrm{K}^{\prime} \mathrm{R}^{\frac{-(\mathrm{n}+1)}{2}}$
[where $\mathrm{K}^{\prime}=\mathrm{K}^{1 / 2}$, a constant $]$
$\frac{2 \pi}{T} \alpha R^{\frac{-(n+1)}{2}} \quad \therefore T \alpha R^{\frac{(n+1)}{2}}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2010
Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. F $\propto \frac{1}{r^{3}}$, but still remaining a central force. Then

Options:
A Kepler's law of area still holds
B Kepler's law of period still holds
C Kepler's law of area and period still holds
D neither the law of area, nor the law of period still holds

The correct answer is:
Kepler's law of area still holds

Solution:
According to kepler's law of area $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{L}}{2 \mathrm{~m}}$

For central forces, torque $=0$

$\therefore \mathrm{L}=$ constant

$\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=$ constant

Topic: Physics, Gravitation, BITSAT, BITSAT 2021
Suppose the limit L=limn→∞n∫0111+x2ndx exists and is larger than 12, then

Options:
A 12<L<2
B 2<L<3
C 3<L<4
D L≥4

The correct answer is:
12<L<2

Solution:
We have,

L=limn→∞n∫01dx1+x2n

We know that, 1+x2n>1+nx2

⇒ 11+x2n<11+nx2

⇒ ∫01dx1+x2n<∫0111+nx2dx

⇒ ∫0111+x2n<1ntan-1nx01

=1ntan-1n

L=limn→∞n∫0111+x2ndx<limn→∞n1ntan-1n

∴ L=π2

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2019
Taking Rydberg's constant $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^{7} \mathrm{~m}$, first and second wavelength of Balmer series in hydrogen spectrum is

Options:
A $2000 \mathrm{~A}, 3000 \mathrm{~A}$
B $1575 \mathrm{~A}, 2960 \mathrm{~A}$
C $6529 \mathrm{~A}, 4280 \mathrm{~A}$
D $6552 \mathrm{~A}, 4863 \mathrm{~A}$

The correct answer is:
$6552 \mathrm{~A}, 4863 \mathrm{~A}$

Solution:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$. For first wavelength, $\mathrm{n}_{1}$
$$
=2, \mathrm{n}_{2}=3
$$
$\Rightarrow \lambda_{1}=6563 \AA$. For second wavelength, $\mathrm{n}_{1}$
$=2, \mathrm{n}_{2}=4 \Rightarrow \lambda_{2}=4861 \AA$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2009
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ equal to-

Options:
A $\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
B $\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
C $\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
D $\tan ^{-1}\left(\frac{1}{2}\right)$

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

Solution:
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$
$\quad=\tan ^{-1}\left[\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right]=\tan ^{-1}\left[\frac{1}{2}\right]$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2012
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ is equal to

Options:
A $\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
B $\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
C $\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
D $\tan ^{-1}\left(\frac{1}{2}\right)$

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

Solution:
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left\{\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right\}$
$=\tan ^{-1}\left\{\frac{9+8}{36-2}\right\}=\tan ^{-1}\left(\frac{1}{2}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2023 (Memory Based Paper 1)
$\int \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$ is equal to

Options:
A $(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C$
B $x \tan ^{-1} \sqrt{x}-\sqrt{x}+C$
C $\sqrt{x}-x \tan ^{-1} \sqrt{x}+C$
D $\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C$

The correct answer is:
$(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C$

Solution:
We have, $I=\int 1 \cdot \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$
Using by parts,
$\begin{aligned}
I & =\tan ^{-1} \sqrt{x} \cdot(x)-\int \frac{1}{1+x} \times \frac{1}{2 \sqrt{x}} \times x d x \\
& =x \tan ^{-1} \sqrt{x} \cdot-\int \frac{x}{(1+x) 2 \sqrt{x}} d x \\
& =x \tan ^{-1} \sqrt{x}-\int\left(\frac{1+x}{(1+x) 2 \sqrt{x}}-\frac{1}{(1+x) 2 \sqrt{x}}\right) d x \\
& =x \tan ^{-1} \sqrt{x}-\int \frac{d x}{2 \sqrt{x}}+\int \frac{d x}{2 \sqrt{x}(1+x)} \\
& =x \tan ^{-1} \sqrt{x}-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\
& =(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C
\end{aligned}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2023 (Memory Based Paper 2)
\(\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}\) is equal to

Options:
A \(\sqrt{3} / 2\)
B \(\sqrt{3} / 4\)
C \(\sqrt{3}\)
D 1

The correct answer is:
\(\sqrt{3}\)

Solution:
\(\begin{aligned}
\sqrt{3} & =\tan 60^{\circ}=\tan \left(40^{\circ}+20^{\circ}\right) \\
& =\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\tan 40^{\circ} \tan 20^{\circ}} \\
\therefore \quad & \sqrt{3}-\sqrt{3} \tan 40^{\circ} \tan 20^{\circ} \\
& =\tan 40^{\circ}+\tan 20^{\circ}
\end{aligned}\)
Hence \(\tan 40^{\circ}+\tan 20^{\circ}+\sqrt{3} \tan 40^{\circ} \tan 20^{\circ}\) \(=\sqrt{3}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2010
Tangents are drawn from the origin to the curve $\mathrm{y}=\cos \mathrm{x} .$ Their points of contact lie on

Options:
A $x^{2} y^{2}=y^{2}-x^{2}$
B $x^{2} y^{2}=x^{2}+y^{2}$
C $x^{2} y^{2}=x^{2}-y^{2}$
D None of these

The correct answer is:
$x^{2} y^{2}=x^{2}-y^{2}$

Solution:
Let $\left(x_{1}, y_{1}\right)$ be one of the points of contact. Given curve is $y=\cos x$

$$

\begin{array}{l}

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x} \\

\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}=-\sin \mathrm{x}_{1}

\end{array}

$$

Now the equation of the tangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is

$$

\begin{aligned}

& y-y_{1}\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}\left(x-x_{1}\right) \\

\Rightarrow & y-y_{1}=-\sin x_{1}\left(0-x_{1}\right)

\end{aligned}

$$

Since, it is given that equation of tangent passes through origin.

$$

\begin{array}{lr}

\therefore & 0-y_{1}=-\sin x_{1}\left(0-x_{1}\right) \\

\Rightarrow & y_{1}=-x_{1} \sin x_{1}

\end{array}

$$

Also, point $\left(x_{1}, y_{1}\right)$ lies on $y=\cos x$

$$

\therefore \quad y_{1}=\cos x_{1}

$$

From Eqs. (i), (ii), we get

$$

\begin{aligned}

& \sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1 \\

\Rightarrow & x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}

\end{aligned}

$$

Hence, the locus of $\left(x_{1}, y_{1}\right)$ is

$$

x^{2}=y^{2}+y^{2} x^{2} \Rightarrow x^{2} y^{2}=x^{2}-y^{2}

$$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
The \(100^{\text {th }}\) term of the sequence \(1,2,2,3,3,3,4,4\), \(4,4, \ldots\) is

Options:
A 12
B 13
C 14
D 15

The correct answer is:
14

Solution:
\(\begin{aligned}
& 1^{\text {st }} \text { term } \rightarrow 1,2^{\text {nd }} \text { term }=2,4^{\text {th }} \text { term } \rightarrow 3, \\
& 7^{\text {th }} \text { term } \rightarrow 4,11^{\text {th }} \text { term } \rightarrow 5, \ldots
\end{aligned}\)
Series is \(1,2,4,7,11, \ldots\)
\(a_n=1+\frac{n(n-1)}{2}=\frac{n^2-n+2}{2}\)
If \(n=14\), then \(a_n=92, \mathrm{If} n=15\), then \(a_n=106\).

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2009
The activation energy for a simple chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$ is $E_{a}$ in forward direction. The activation energy for reverse reaction

Options:
A is always less than $E_{a}$
B can be less than or more than $E_{a}$
C is always double of $E_{a}$
D is negative of $E_{a}$

The correct answer is:
can be less than or more than $E_{a}$

Solution:
Since the nature of reaction (i.e. exothermic or endothermic) is not given, $\mathrm{E}_{\mathrm{a}}$ for reverse reaction can be more or less.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2010
The activity of a radioactive sample is measured as N0 counts per minute at t=0 and N0e counts per minute at t=5 min. The time (in minutes) at which the activity reduces to half its value is

Options:
A loge25
B 5loge2
C 5log102
D 5loge2

The correct answer is:
5loge2

Solution:
Fraction remains after n half-lives

NN0=12n=12tT

Given, N=N0e⇒N0eN0=125T

⇒ 1e=125T

Taking log on both sides, we get

log1-loge=5Tlog12

⇒ -1=5T-log2

⇒ T=5loge2

Now, let t be the time after which activity reduces to half

12=12t/5loge2

⇒ t'=5loge2

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2017
The addition of HBr to 1-butene gives a mixture of products x, y and z

The mixture consists of

Options:
A x and y as major and z as minor products
B y as major, x and z as minor products
C y as minor, x and z as major products
D x and y as minor and z as major products

The correct answer is:
x and y as major and z as minor products

Solution:
The alkene is unsymmetrical, hence will follow Markownikoff's rule to give major product.

since, y contains, a chiral carbon, it exists in two enantiomers (x and y) which are mirror images of each other.

Topic: Chemistry, General Organic Chemistry, JEE Main
The addition of HBr to 1-butene gives a mixture of products x, y and z

The mixture consists of

Options:
A x and y as major and z as minor products
B y as major, x and z as minor products
C y as minor, x and z as major products
D x and y as minor and z as major products

The correct answer is:
x and y as major and z as minor products

Solution:
The alkene is unsymmetrical, hence will follow Markownikoff's rule to give major product.

since, y contains, a chiral carbon, it exists in two enantiomers (x and y) which are mirror images of each other.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2018
The alcohol product(s) of the reduction of 2 -methyl-3-pentanone with $\mathrm{LiAlH}_{4}$ is (are)

Options:
A a single enantiomer
B racemic mixture
C two diastereoisomers
D two structural isomers

The correct answer is:
racemic mixture

Solution:
The resultingcompound is $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{2} \mathrm{H}_{5}$

which is optically inactive and reaction leads the racemisation

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The alcohol product(s) of the reduction of 2 -methyl-3-pentanone with $\mathrm{LiAlH}_{4}$ is (are)

Options:
A a single enantiomer
B racemic mixture
C two diastereoisomers
D two structural isomers

The correct answer is:
racemic mixture

Solution:
The resultingcompound is $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{2} \mathrm{H}_{5}$

which is optically inactive and reaction leads the racemisation

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2020
The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders?

Options:
A $\mathrm{CsH}>\mathrm{RbH}>\mathrm{KH}>\mathrm{NaH}>\mathrm{LiH}$
B $\mathrm{KH}>\mathrm{NaH}>\mathrm{LiH}>\mathrm{CsH}>\mathrm{RbH}$
C $\mathrm{NaH}>\mathrm{LiH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$
D $\mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$

The correct answer is:
$\mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$

Solution:
The stability of alkali metal hydrides decreases from Li to Cs. It is due to the fact that $\mathrm{M}-\mathrm{H}$ bonds becomes weaker with increase in size of alkali metals as we move down the group from $\mathrm{Li}$ to $\mathrm{Cs}$. Thus the order of stability of hydrides is $\mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2013
The amount of $\mathrm{H}_{2} \mathrm{O}_{2}$ present in 1 litre of $1.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{O}_{2}$ solution, is:

Options:
A $25.5 \mathrm{~g}$
B $3.0 \mathrm{~g}$
C $8.0 \mathrm{~g}$
D $2.5 \mathrm{~g}$

The correct answer is:
$25.5 \mathrm{~g}$

Solution:
Molecular weight of $\mathrm{H}_{2} \mathrm{O}_{2}=34$ Equivalent weight of $\mathrm{H}_{2} \mathrm{O}_{2}=17$

$\therefore 1 \mathrm{~L}$ of $1 \mathrm{~N} \mathrm{H}_{2} \mathrm{O}_{2} \mathrm{has}=17 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$

$\therefore 1$ Lof $1.5 \mathrm{NH}_{2} \mathrm{O}_{2}$ has $=1.5 \times 17=25.5 \mathrm{~g} \mathrm{of} \mathrm{H}_{2} \mathrm{O}_{2}$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2020
The amplitude of $\sin \frac{\pi}{5}+i\left(1-\cos \frac{\pi}{5}\right)$

Options:
A $\pi / 5$
B $2 \pi / 5$
C $\pi / 10$
D $\pi / 15$

The correct answer is:
$\pi / 10$

Solution:
$\sin \frac{\pi}{5}+\mathrm{i}\left(1-\cos \frac{\pi}{5}\right)$
$=2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}+\mathrm{i} 2 \sin ^{2} \frac{\pi}{10}$
$=2 \sin \frac{\pi}{10}\left(\cos \frac{\pi}{10}+\mathrm{i} \sin \frac{\pi}{10}\right)$
For amplitude, $\tan \theta=\frac{\pi}{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}=\tan \frac{\pi}{10}$
$\Rightarrow \theta=\frac{\pi}{10}$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2012
The angle between any two diagonal of a cube is

Options:
A $45^{\circ}$
B $60^{\circ}$
C $30^{\circ}$
D $\tan ^{-1}(2 \sqrt{2})$

The correct answer is:
$\tan ^{-1}(2 \sqrt{2})$

Solution:
for a unit cube unit vector along the diagonal

$$

\mathrm{OP}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})

$$

unit vector along the diagonal

$\mathrm{CD}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\therefore \cos \theta=\frac{1}{3}(1+1-1)=\frac{1}{3} \quad \therefore \tan \theta=2 \sqrt{2}$

Topic: Mathematics, Vector Algebra, JEE Main
The angle between any two diagonal of a cube is

Options:
A $45^{\circ}$
B $60^{\circ}$
C $30^{\circ}$
D $\tan ^{-1}(2 \sqrt{2})$

The correct answer is:
$\tan ^{-1}(2 \sqrt{2})$

Solution:
for a unit cube unit vector along the diagonal

$$

\mathrm{OP}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})

$$

unit vector along the diagonal

$\mathrm{CD}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\therefore \cos \theta=\frac{1}{3}(1+1-1)=\frac{1}{3} \quad \therefore \tan \theta=2 \sqrt{2}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2014
The angle between the lines whose direction cosine satisfy the equations l+m+n=0 and l2=m2+n2 is

Options:
A π3
B π4
C π6
D π2

The correct answer is:
π3

Solution:
We have,

l+m+n=0 ...(i)

and l2=m2+n2 ...(ii)

Putting the value of l in Eq. (ii), we get

(m+n)2=m2+n2

⇒ m2+n2+2mn=m2+n2

⇒ mn=0

⇒ m=0,n=0

when m=0,l=-n

∴ Direction cosines are -12,0,12.

When n=0,l=-m

∴ Direction cosines are -12,12,0.

Angle between lines are

cosθ=-12×-12+0×12+12×0

cosθ=12,θ=π/3

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2019
The angle of dip at a place is $37^{\circ}$ and the vertical component of the earth's magnetic field is $6 \times 10^{-5} \mathrm{~T}$. The earth's magnetic field at this place is $\left(\tan 37^{\circ}=\frac{3}{4}\right)$

Options:
A $7 \times 10^{-5} \mathrm{~T}$
B $6 \times 10^{-5} \mathrm{~T}$
C $5 \times 10^{-5} \mathrm{~T}$
D $10^{-4} \mathrm{~T}$

The correct answer is:
$10^{-4} \mathrm{~T}$

Solution:
$\tan \theta=\frac{\mathrm{V}}{\mathrm{H}}=\frac{3}{4}\left[\because \tan 37^{\circ}=\frac{3}{4}\right]$

$\therefore \mathrm{V}=\frac{3}{4} \mathrm{H}$

$\mathrm{V}=6 \times 10^{-5} \mathrm{~T}$

$\mathrm{H}=\frac{4}{3} \times 6 \times 10^{-5} \mathrm{~T}=8 \times 10^{-5} \mathrm{~T}$

$\therefore \mathrm{B}_{\text {total }}=\sqrt{\mathrm{V}^{2}+\mathrm{H}^{2}}=\sqrt{(36+64)} \times 10^{-5}$

$=10 \times 10^{-5}=10^{-4} \mathrm{~T} .$

Topic: Physics, Magnetic Properties of Matter, BITSAT, BITSAT 2021
The angle of intersection of the two circles $x^{2}+y^{2}-2 x-2 y=0$ and $x^{2}+y^{2}=4,$ is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $45^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Here circles are

$$

\begin{array}{l}

x^{2}+y^{2}-2 x-2 y=0...(1) \\

x^{2}+y^{2}=4...(2) \\

\text { Now, } \mathrm{C}_{1}(1,1), r_{1}=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\

C_{2}(0,0), r_{2}=2

\end{array}

$$

If $\theta$ is the angle of intersection then

$$

\begin{array}{l}

\cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-\left(c_{1} c_{2}\right)^{2}}{2 r_{1} r_{2}} \\

=\frac{2+4-(\sqrt{2})^{2}}{2 \cdot \sqrt{2.2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}

\end{array}

$$

Topic: Mathematics, Circle, BITSAT, BITSAT 2014
The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

Options:
A $\frac{\mathrm{L}}{2}$
B
zero
C
L
D
2L

The correct answer is:
L

Solution:
From the Bohr's quantization rule
Angular momentum $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}, \mathrm{L}_1=\frac{1 . \mathrm{h}}{2 \pi}=\mathrm{L}$ ( $n=1$ for first orbit)
in ( $n=2$ for second orbit) $L_2=\frac{2 h}{2 \pi}=2 L$
Hence, change in angular momentum $\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{~L}-$ $\mathrm{L}=\mathrm{L}$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The angular size of the central maxima due to a single slit diffraction is $\mathrm{a} \rightarrow$ slit width $)$

Options:
A $\frac{\lambda}{\mathrm{a}}$
B $\frac{2 \lambda}{\text { a }}$
C $\frac{3 \lambda}{2 \mathrm{a}}$
D $\frac{\lambda}{2 \mathrm{a}}$

The correct answer is:
$\frac{2 \lambda}{\text { a }}$

Solution:
Angular size of central maxima is

$$

2 \times\left(\frac{\lambda}{a}\right)=\frac{2 \lambda}{a}

$$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2014