datavorous/entrance-exam-dataset huggingface dataset:
Adsorption of a gas follows Freundlich adsorption isotherm. $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. The plot of $\frac{x}{m}$ versus $\log p$ is shown in the given graph.

$\frac{x}{m}$ is proportional to :

Options:
A $p^{2 / 3}$
B $p^{3 / 2}$
C $p^{3}$
D $p^{2}$

The correct answer is:
$p^{2 / 3}$

Solution:
According to Freundlich adsorption isotherm $\frac{x}{m} \propto p^{\frac{1}{n}}, \frac{x}{m}=k p^{\frac{1}{n}}$

Slope $=\frac{2}{3}$

$\log \frac{x}{m}=\log k+\frac{1}{n} \log p$

Slope $=\frac{1}{n}=\frac{2}{3}$

$\frac{x}{m} \propto p^{\frac{2}{3}}$

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2021
$\mathrm{Al}$ (at. wt 27 ) crystallizes in the cubic system with a cell edge of $4.05 Ã…$. Its density is $2.7 \mathrm{~g}$ per $\mathrm{cm}^{3}$. Determine the unit cell type calculate the radius of the $\mathrm{Al}$ atom

Options:
A $\mathrm{fcc}, 2.432 Ã$
B bcc, $2.432 Ã$
C bcc, $1.432 Ã$
D $\mathrm{fcc}, 1.432 Ã$

The correct answer is:
$\mathrm{fcc}, 1.432 Ã$

Solution:
$\rho=\frac{Z \times M}{N_{0} \times a^{3}}$,

$$

2.7=\frac{Z \times 27}{6.02 \times 10^{23} \times(4.05)^{3} \times 10^{-24}}

$$

$\therefore Z=4$

Hence it is face centred cubic unit lattice.

Again $4 r=a \sqrt{2}=5.727 Ã…$

$\therefore \mathrm{r}=1.432 Ã…$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2016
Alizarin is an example of

Options:
A Triayl dye
B Azo dye
C Vat dye
D Anthraquinone dye

The correct answer is:
Anthraquinone dye

Solution:
Alizarin is an anthraquinone dye. It gives a bright red colour with aluminium and a blue colour with barium.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2012
All the words that can be formed using alphabets A, $H, L, U$ and $R$ are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is

Options:
A 71
B 72
C 73
D 74

The correct answer is:
74

Solution:
No. of words starting with $\mathrm{A}$ are $4 !=24$

No. of words starting with $\mathrm{H}$ are $4 !=24$

No. of words starting with $L$ are $4 !=24$

These account for 72 words Next word is RAHLU and the $74^{\text {th }}$ word RAHUL

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2016
Ammonia forms the complex ion $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it?

Options:
A In acidic solutions protons coordinate with ammonia molecules forming $\mathrm{NH}_{4}^{+}$ ions and $\mathrm{NH}_{3}$ molecules are not available.
B In alkaline solutions insoluble $\mathrm{Cu}(\mathrm{OH})_{2}$ is precipitated which is soluble in excess of any alkali.
C Copper hydroxide is an amphoteric substance.
D In acidic solutions hydration protects copper ions.

The correct answer is:
In acidic solutions protons coordinate with ammonia molecules forming $\mathrm{NH}_{4}^{+}$ ions and $\mathrm{NH}_{3}$ molecules are not available.

Solution:
$\quad \mathrm{N} \mathrm{H}_{3}+\mathrm{H}^{+}$ (acidic medium) $\rightleftharpoons \mathrm{NH}_{4}^{+}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2009
Among $\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SiO}_{2}, \mathrm{P}_{2} \mathrm{O}_{3}$ and $\mathrm{SO}_{2}$, the correct
order of acid strength is:

Options:
A $\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}$
B $\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{P}_{2} \mathrm{O}_{3}$
C $\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}$
D $\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SO}_{2}$

The correct answer is:
$\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SO}_{2}$

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2011
Among $\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SiO}_{2}, \mathrm{P}_{2} \mathrm{O}_{3}$ and $\mathrm{SO}_{2}$, the correct
order of acid strength is:

Options:
A $\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}$
B $\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{P}_{2} \mathrm{O}_{3}$
C $\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}$
D $\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SO}_{2}$

The correct answer is:
$\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SO}_{2}$

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Among the following four structures I to IV,

it is true that

Options:
A only I and II are chiral compounds.
B only III is a chiral compound.
C only II and IV are chiral compounds.
D all four are chiral compounds.

The correct answer is:
only I and II are chiral compounds.

Solution:
A chiral object or structure has four different groups attached to the carbon.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2013
Among the following statements the incorrect one is:

Options:
A Calamine and siderite are carbonates.
B Argentite and cuprite are oxides.
C Zinc blende and iron pyrites are sulphides.
D Malachite and azurite are ores of copper.

The correct answer is:
Argentite and cuprite are oxides.

Solution:
Cuprite is $\mathrm{Cu}_{2} \mathrm{O}$ and Argentite is $\mathrm{Ag}_{2} \mathrm{~S}$.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2009
Among the following substituted silanes, the one which will give rise to cross linked silicone polymer on hydrolysis is?

Options:
A R4 Si
B R SiCl3
C R2 SiCl2
D R3 SiCl

The correct answer is:
R SiCl3

Solution:
Among the following substituted silanes, the one which will give rise to cross-linked silicone polymer on hydrolysis is trichloroalkylsilane RSiCl3.

RSCl3 gives cross-linked silicon polymer on hydrolysis.

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2019
Among the following the lowest degree of paramagnetism per mole of the compound at 298 $\mathrm{K}$ will be shown by

Options:
A $\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}$
B $\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}$
C $\mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$
D $\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}$

Solution:
$\begin{array}{lllll}\text { Ion } & \mathrm{Mn}^{2+} & \mathrm{Cu}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Ni}^{2+} \\ \text { EC } & 3 d^{5} & 3 d^{9} & 3 d^{6} & 3 d^{8} \\ \text { Number of } & 5 & 1 & 4 & 2\end{array}$
unpaired electron
Hence lowest paramagnetism is shown by $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2012
Among the following, which have highest melting point?

Options:
A lonic solids
B Pseudo solids
C Molecular solids
D Amorphous solids

The correct answer is:
lonic solids

Solution:
Ionic solids are formed by three dimensional arrangements of positive charged ions i.e. cations and negative charged ions i.e. anions. These ions are held together by coulombic(electrostatic) forces and hence, ionic solids are hard and brittle.

Ionic solids have highest melting point due to strong electrostatic force of attraction.

Topic: Chemistry, Solid State, BITSAT, BITSAT 2017
Among the given halides, which one will give same product in both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions.

Options:
A (III) only
B (I) and (II)
C (III) and (IV)
D (I), (III) and (IV)

The correct answer is:
(III) and (IV)

Solution:
$S_{\mathbb{N}}^2$ and $S_N 1$ same, if $C^{\oplus}$ not rearrange.

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Among the ligands $\mathrm{NH}_3$, en, $\mathrm{CN}^{-}$and $\mathrm{CO}$ the correct order of their increasing field strength, is :

Options:
A $\mathrm{NH}_3 < $ en $ < \mathrm{CN}^{-} < \mathrm{CO}$
B $\mathrm{CN}^{-} < \mathrm{NH}_3 < \mathrm{NH}_3 < $ en
C en $ < \mathrm{CN}^{-} < \mathrm{NH}_3 < \mathrm{CO}$
D $\mathrm{CO} < \mathrm{NH}_3 < $ en $ < \mathrm{CN}^{-}$

The correct answer is:
$\mathrm{NH}_3 < $ en $ < \mathrm{CN}^{-} < \mathrm{CO}$

Solution:
Ligands can be arranged in a series in the orders of increasing field strength as given below:
Weak field ligands:
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{S}^{2-} < \mathrm{SCN} < \mathrm{Cl}^{-} < \mathrm{N}_3^{-}, \mathrm{F}^{-}$ $ < $ Urea, OH $ < $ oxalate
Strong field ligands
$\mathrm{O}^{--} < \mathrm{H}_2 \mathrm{O} < \mathrm{NCS}^{-} < \text {EDTA } < \mathrm{Py}, \mathrm{NH}_3 < $ en $=\mathrm{SO}_3^{-} < $bipy, Phen $ < \mathrm{NO}_2^{-} < \mathrm{CH}_3^{-}$ $ < \mathrm{C}_6 \mathrm{H}_5^{-} < \mathrm{CN}^{-} < \mathrm{CO}$
Such a series is termed as spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2023 (Memory Based Paper 1)
An achromatic convergent doublet of two lense in contact has a power of $+5 \mathrm{D}$. The power of converging lens is $+6 \mathrm{D}$. The ratio of the dispersive power of the convergent and divergent lenses is

Options:
A
$3: 7$

B
$2: 3$

C
$1: 5$

D
$5: 3$

The correct answer is:

$1: 5$

Solution:
- The condition of achromatism is $W_1 P_1+W_2 P_2=0$
$$
\begin{array}{rlrl}
\Rightarrow & W_1 P_1 & =-W_2 P_2 \\
\Rightarrow & & \frac{W_1}{W_2} & =-\frac{P_2}{P_1}
\end{array}
$$
Now, $\quad P_1+P_2=4 \mathrm{D}$
but, Power of converging lens,
$$
P_1=5 \mathrm{D}
$$
$\because$ Power of diverging lens
$$
\begin{aligned}
P_2 & =4 \mathrm{D}-P_1 \\
& =4 \mathrm{D}-5 \mathrm{D}=-\mathrm{D}
\end{aligned}
$$
[From ii]
$\therefore$ From Eq. (i), we have
$$
\frac{W_1}{W_2}=-\frac{P_2}{P_1}=\frac{-(-D)}{5 D}=\frac{1}{5} \Rightarrow \frac{W_1}{W_2}=\frac{1}{5}
$$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2022
An aeroplane is flying in a horizontal direction with a velocity u and at a height of 2000 m. When it is vertically below a point A on the ground a food packet is released from it. The packet strikes the ground at point B. If AB=3 km and g=10 m s-2, then the value of u is

Options:
A 54 km h-1
B 540 km h-1
C 150 km h-1
D 300 km h-1

The correct answer is:
540 km h-1

Solution:
Given R=3 km=3000 m

Range, R=u2hg

⇒ u=Rg2hor u=3000×102×2000

=3000×120

=150 m s-1

=150×185 km h-1

=540 km h-1

Topic: Physics, Motion In One Dimension, JEE Main
An aeroplane is flying in a horizontal direction with a velocity u and at a height of 2000 m. When it is vertically below a point A on the ground a food packet is released from it. The packet strikes the ground at point B. If AB=3 km and g=10 m s-2, then the value of u is

Options:
A 54 km h-1
B 540 km h-1
C 150 km h-1
D 300 km h-1

The correct answer is:
540 km h-1

Solution:
Given R=3 km=3000 m

Range, R=u2hg

⇒ u=Rg2hor u=3000×102×2000

=3000×120

=150 m s-1

=150×185 km h-1

=540 km h-1

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2017
An air bubble of volume $\mathrm{v}_{0}$ is released by a fish at a depth $\mathrm{h}$ in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake. The volume of the bubble just before touching the surface will be (density) of water is $\rho$

Options:
A $\mathrm{v}_{0}$
B $\quad v_{0}(\rho g h / p)$
C $\frac{\mathrm{v}_{0}}{\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)}$
D $\mathrm{v}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)$

The correct answer is:
$\mathrm{v}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)$

Solution:
As the bubble rises the pressure gets reduced for constant temperature, if $\mathrm{P}$ is the standard atmospheric pressure, then $(\mathrm{P}+\rho \mathrm{gh}) \mathrm{V}_{0}=\mathrm{PV}$
$$
\text { or } \mathrm{V}=\mathrm{V}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{P}}\right)
$$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2011
An alternating voltage $V=V_{0}$ sin ot is appliedacross a circuit. As a result, a current $I=I_{0} \sin (\omega t-\pi / 2)$ flows in it. The power consumed per cycle is

Options:
A zero
B $0.5 \mathrm{~V}_{0} \mathrm{I}_{0}$
C $0.707 \mathrm{~V}_{0} \mathrm{I}_{0}$
D $1.414 \mathrm{~V}_{0} \mathrm{I}_{0}$

The correct answer is:
zero

Solution:
The phase angle between voltage $\mathrm{V}$ and current I is $\pi / 2$.

Topic: Physics, Alternating Current, BITSAT, BITSAT 2010
An alternating voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ is applied across a circuit. As a result, a current $\mathrm{I}=\mathrm{I}_{0} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ flows in it. The power consumed per cycle is

Options:
A zero
B $0.5 \mathrm{~V}_{0} \mathrm{I}_{0}$
C $0.707 \mathrm{~V}_{0} \mathrm{I}_{0}$
D $1.414 \mathrm{~V}_{0} \mathrm{I}_{0}$

The correct answer is:
zero

Solution:
The phase angle between voltage $V$ and current $I$ is $\frac{\pi}{2}$.

Therefore, power factor $\cos \phi=\cos \left(\frac{\pi}{2}\right)=0$.

Hence the power consumed is zero.

Topic: Physics, Alternating Current, BITSAT, BITSAT 2021
An ammeter of resistance 1.5 Ω can measure currents upto 1 A. The value of shunt resistance to measure current upto 4 A is

Options:
A 0.5 Ω
B 1 Ω
C 1.5 Ω
D 0.2 Ω

The correct answer is:
0.5 Ω

Solution:
Given, for an ammeter, Ig=1 A, Rg=1.5 Ω

Total current, I=4 A

If Rs be the shunt resistance, then according to the question circuit diagram is given, Rs‖Rg, hence have equal potential difference)

where, IgRg=I-IgRs

Rs=IgI-Ig×Rg=14-1×1.5=0.5 Ω

Topic: Physics, Current Electricity, BITSAT, BITSAT 2019
An ammeter reads upto 1 ampere. Its internal resistance is $0.81 \mathrm{ohm} .$ To increase the range to 10 A the value of the required shunt is

Options:
A $0.03 \Omega$
B $0.3 \Omega$
C $0.9 \Omega$
D $0.09 \Omega$

The correct answer is:
$0.09 \Omega$

Solution:
No Solution available

Topic: Physics, Current Electricity, BITSAT, BITSAT 2009
An aqueous solution freezes at 272.4 K, while pure water freezes at 273 K, given Kf=1.86 K kg mol-1, and Kb=0.512 K kg mol-1; the molality of solution and boiling point of solution respectively will be -

Options:
A 0.322 and 373.16 K
B 0.222 and 273.15 K
C 0.413 and 400 K
D 0.5 and 300.73 K

The correct answer is:
0.322 and 373.16 K

Solution:
Depression in freezing point is given as ΔTf=Kfm

Molality of solution m=ΔTfKf

or m=0.61.86=0.322 m

Elevation in boiling point of solution is given as

ΔTb=kb×m

ΔTb=0.512×0.322

ΔTb=0.165

∴ Boiling point of solution =373+0.16

=373.16 K

Topic: Chemistry, Solutions, BITSAT, BITSAT 2019
An aqueous solution of a substance gives a white precipitate on treatment with dil. HCl which dissolves on heating. When $\mathrm{H}_{2} \mathrm{~S}$ is passed through the hot acidic solution, a black precipitate is obtained. The substance is a

Options:
A $\mathrm{Hg}_{2}^{2+}$ salt
B $\mathrm{Cu}^{2+}$ salt
C $\mathrm{Ag}^{+}$ salt
D $\mathrm{Pb}^{2+}$ salt

The correct answer is:
$\mathrm{Pb}^{2+}$ salt

Solution:
$\mathrm{PbCl}_{2}$ is insoluble in cold water, soluble in hot water and $\mathrm{PbS}$ is black ppt in acidic medium.

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2009
An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is

Options:
A $11 / 4 \mathrm{~m}$
B $8 / 3 \mathrm{~m}$
C $7 / 2 \mathrm{~m}$
D $2 \mathrm{~m}$

The correct answer is:
$8 / 3 \mathrm{~m}$

Solution:
The equation of the ellipse is

$$

\frac{x^{2}}{\left(\frac{9}{2}\right)^{2}}+\frac{y^{2}}{9}=1

$$

Where centre is assumed as origin and base as $x$ -axis. Put $x=2,$ we get

$$

\frac{16}{81}+\frac{\mathrm{y}^{2}}{9}=1 \Rightarrow \mathrm{y}=\frac{\sqrt{65}}{3} \approx \frac{8}{3} \mathrm{~m}

$$

(approximately)

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2014
An aromatic compound has molecular formula $\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{Br}$. Give the possible isomers and the appropriate method to distinguish them.

Options:
A 3 isomers; by heating with $\mathrm{AgNO}_{3}$ solution
B 4 isomers; by treating with $\mathrm{AgNO}_{3}$ solution
C 4 isomers; by oxidation
D 5 isomers; by oxidation

The correct answer is:
4 isomers; by oxidation

Solution:

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2016
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height $(h)$ of the satellite above the earth's surface is (Take radius of earth as $R_{e}$ )

Options:
A $h=R_{e}^{2}$
B $h=R_{e}$
C $h=2 R_{e}$
D $h=4 R_{e}$

The correct answer is:
$h=R_{e}$

Solution:
The escape velocity from earth is given by $v_{e}=\sqrt{2 \mathrm{~g} R_{e}} \ldots$ (i)

The orbital velocity of a satellite revolving around earth is given by

$$

v_{0}=\frac{\sqrt{\mathrm{G} M_{e}}}{\left(R_{e}+h\right)}

$$

where, $M_{e}=$ mass of earth, $R_{e}=$ radius of earth, $h=$ height of satellite from surface of earth.

By the relation $\mathrm{G} M_{e}=\mathrm{g} R_{e}^{2}$

$$

\text { So, } \quad v_{0}=\frac{\sqrt{g R_{e}^{2}}}{\left(R_{e}+h\right)} \ldots

$$

Dividing equation (i) by (ii), we get

$$

\frac{v_{e}}{v_{0}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{\left(R_{e}\right)}

$$

Given, $v_{0}=\frac{v_{e}}{2}$

$$

\frac{2 v_{e}}{v_{e}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{R_{e}}

$$

Squaring on both side, we get

$$

4=\frac{2\left(R_{e}+h\right)}{R_{e}}

$$

or $R_{e}+h=2 R_{e} \quad$ i.e., $h=R_{e}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2015
An asteroid of mass m is approaching earth, initially at a distance 10Re with speed vi. It hite oarth with a speed vf , Re and Me are radius and mass of earth), then

Options:
A vf2=vi2+2GmRe1+110
B vf2=vi2+2GMeRe1+110
C vf2=vi2+2GMeRθ1-110
D vf2=vi2+2GmRe1-110

The correct answer is:
vf2=vi2+2GMeRθ1-110

Solution:
Initial energy of the asteroid is
Ei=Ki+Ui=12mvi2-GMθm10Re

Final energy of the asteroid,
Ef=12mvf2-GMθmRθ

According to law of conservation of energy, Ei=Ef

12mvi2-GMθm10Rθ=12mvf2-GMθmRθ

vf2-2GMθRθ=vi2-2GMθ10Rθ

⇒ vf2=vi2+2GMθRθ1-110

Topic: Physics, Gravitation, BITSAT, BITSAT 2018
An automobile moving with a speed of 36 km h-1 reaches an upward inclined road of angle 30°, its engine becomes switch off. If the coefficient of friction is 0.1, then how much distance will automobile move before coming to rest?

Options:
A 12.53 m
B 21.42 m
C 15.43 m
D 8.53 m

The correct answer is:
8.53 m

Solution:
Given, initial speed, u=36 km h-1

=36×100060×60=10 m s-1

θ=30°, μ=0.1, s=?

Here, work done in moving up the inclined road= KE of the vehicle

mg sin θ+F s=12mu2

mg sin θ+μR×s=12mu2

mg sin θ+μ mg cos θ×s=12mu2

s=12mu2mg sin θ+μ cos θ=u22 g sin θ+μ cos θ

=10×102×10× sin 30°+0.1 cos 30°=8.53 m

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2018
An e $^{-}$ has magnetic quantum number as $-3,$ what is its principal quantum number?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
When $m=-3, l=3, \therefore \mathrm{n}=4$.

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2016
An elastic string of unstretched length $\mathrm{L}$ and force constant $\mathrm{k}$ is stretched by a small length $\mathrm{x}$. It is further stretched by another small length y. The work done in the second stretching is

Options:
A $1 / 2 \mathrm{Ky}^{2}$
B $1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})$
C $1 / 2 \mathrm{~K}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$
D $1 / 2 \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$

The correct answer is:
$1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})$

Solution:
In the string elastic force is conservative in nature. $\therefore \mathrm{W}=-\Delta \mathrm{U}$

Work done by elastic force of string,

$$

\begin{aligned}

\mathrm{W} &=-\left(\mathrm{U}_{\mathrm{F}}-\mathrm{U}_{\mathrm{i}}\right)=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{F}} \\

\mathrm{W} &=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{\mathrm{k}}{2}(\mathrm{x}+\mathrm{y})^{2} \\

&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right) \\

&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy}) \\

&=-\mathrm{k} \mathrm{x} \mathrm{y}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}

\end{aligned}

$$

Therefore, the work done against elastic force

$\mathrm{W}_{\text {external }}=-\mathrm{W}=\frac{\mathrm{K} y}{2}(2 \mathrm{x}+\mathrm{y})$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2015
An electric current $I$ enters and leaves a uniform circular wire of radius $r$ through diametrically opposite points. A charged particle $q$ moves along the axis of circular wire passes through its centre at speed $v$.
The magnetic force on the particle when it passes through the centre has a magnitude.

Options:
A
$\frac{q v \mu_0 I}{2 \pi r}$

B
$q v \frac{\mu_o I}{\pi r}$

C
$q v \mu_0 I$

D
0

The correct answer is:

0

Solution:
Force on a moving charged particle in uniform magnetic field
$$
F=B q v \sin \theta
$$
Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^{\circ}$ or $180^{\circ}$
When $\theta=0^{\circ} ; F=B q v \sin 0^{\circ}$
$$
F=0
$$
When $\theta=180^{\circ}, F=B q u \sin 180^{\circ}$
$$
F=0
$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2022
An electromagnetic wave is propagating along $X$-axis. At $x=1 \mathrm{~cm}$ and $t=18 \mathrm{~s}$, its electric vector $|E|=8 \mathrm{~V} / \mathrm{m}$, then the magnitude of its magnetic vector is

Options:
A
$266 \times 10^{-8}$

B
$3 \times 10^{-7}$

C
$3.14 \times 10^{-8}$

D
$3.16 \times 10^{-7}$

The correct answer is:

$266 \times 10^{-8}$

Solution:
As we know that, $\frac{E}{B}=c$
$$
\text { so, } B=\frac{E}{c}=\frac{8}{3 \times 10^8}=2.66 \times 10^{-8}
$$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2022
An electromagnetic wave passes through space and its equation is given by $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$
where $\mathrm{E}$ is electric field. Energy density of electromagnetic wave in space is

Options:
A $\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
B $\frac{1}{4} \varepsilon_{0} \mathrm{E}_{0}^{2}$
C $\varepsilon_{0} \mathrm{E}_{0}^{2}$
D $2 \varepsilon_{0} \mathrm{E}_{0}^{2}$

The correct answer is:
$\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$

Solution:
Energy density
$$
=\varepsilon_{0} \mathrm{E}_{\mathrm{rm}}^{2}=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}
$$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2011
An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of $-3.4 \mathrm{eV}$. The kinetic energy and the de-Broglie wavelength of the electron are respectively

Options:
A $+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$
B $-3.4 \mathrm{eV}, 1.99 \times 10^{-9} \mathrm{~m}$
C $2.8 \mathrm{eV}, 2.38 \times 10^{-10} \mathrm{~m}$
D $1.1 \mathrm{eV}, 1.28 \times 10^{-9} \mathrm{~m}$

The correct answer is:
$+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$

Solution:
$E_{n}=-3.4 \mathrm{eV}$
The kinetic energyis equal to the magnitude of total energy in this case. $\therefore \mathrm{K.E}=+3.4 \mathrm{eV}$
The de Broglie wavelength of electron
$\begin{aligned} \lambda=\frac{h}{\sqrt{2 m K}} &=\frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \mathrm{eV} \\ &=0.66 \times 10^{-9} \mathrm{~m} \end{aligned}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2011