datavorous/entrance-exam-dataset huggingface dataset:
An electron moves in a circular arc of radius 10 $\mathrm{m}$ at a contant speed of $2 \times 10^{7} \mathrm{~ms}^{-1}$ with its plane of motion normal to magnetic flux density of $10^{-5} \mathrm{~T}$. What will be the value of specific charge of the electron?

Options:
A $2 \times 10^{4} \mathrm{Ckg}^{-1}$
B $2 \times 10^{5} \mathrm{Ckg}^{-1}$
C $5 \times 10^{6} \mathrm{Ckg}^{-1}$
D $2 \times 10^{11} \mathrm{Ckg}^{-1}$

The correct answer is:
$2 \times 10^{11} \mathrm{Ckg}^{-1}$

Solution:
$\mathrm{Bqv}=\mathrm{mv}^{2} / \mathrm{r}$ or $\mathrm{q} / \mathrm{m}=\mathrm{v} / \mathrm{rB}$.

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2020
An electron revolves in a circular orbit of radius r with angular speed ω. The magnetic field at the centre of electron orbit is

Options:
A μ0eωπr
B μ0eω4πr
C μ0e2ω4πr
D μ0eω2πr

The correct answer is:
μ0eω4πr

Solution:
Magnetic field at the centre of circular loop.

B=μ0l2r ...i

Period of revolution of electron, where ω= angular velocity of electron

T=2πω

where, ω= angular velocity of electron.

Current produced due to movement of electron,

I=eT=e2πω=eω2π

On substituting the value of I in Eq. (i), we get

B=μ02r·eω2π

⇒B=μ0eω4πr

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2019
An elemental crystal has a density of $8570 \mathrm{~kg} / \mathrm{m}^{3}$. The packing efficiency is 0.68 . The closest distance of approach between neighbouring atom is $2.86 Ã…$. What is the mass of one atom approximately?

Options:
A $93 \mathrm{amu}$
B $39 \mathrm{amu}$
C $63 \mathrm{amu}$
D $29 \mathrm{amu}$

The correct answer is:
$93 \mathrm{amu}$

Solution:
The packing efficiency $=0.68$, means the given lattice is $\mathrm{BCC}$. The closest distance of approach $=2 \mathrm{r}$

$$

\begin{array}{l}

2 r=2.86 A=a \sqrt{3} \\

\text { or a }=\frac{2 \times 2.86}{\sqrt{3}}=3.30 Ã…

\end{array}

$$

Let atomic weight of the element $=\mathrm{a}$

$$

\begin{aligned}

\therefore & \frac{2 \times 9}{36 \times 10^{23} \times(3.3)^{3} \times 10^{-24}}=8.57 \\

\mathrm{a} &=8.57 \times 3 \times(3.3)^{3} \times 0.1 \\

&=92.39 \simeq 93 \mathrm{amu}

\end{aligned}

$$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2014
An ellipse has $O B$ as semi minor axis, $F$ and $F^{\prime}$

its focii and the angle $F B F^{\prime}$ is a right angle. Then the eccentricity of the ellipse is

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

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$\begin{aligned} & \because \angle F B F^{\prime}=90^{\circ} \Rightarrow F B^{2}+F^{\prime} B^{2}=F F^{\prime 2} \\ \therefore &\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}=(2 a e)^{2} \\ & \Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2} \Rightarrow e^{2}=\frac{b^{2}}{a^{2}} \ldots(1) \end{aligned}$

Also, $e^{2}=1-b^{2} / a^{2}=1-e^{2}$

(By using equation (i)) $\Rightarrow 2 e^{2}=1 \Rightarrow e=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2013
An ellipse has $\mathrm{OB}$ as semi minor axis, $\mathrm{F}$ and $\mathrm{F}^{\prime}$ its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse is

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

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$\because \angle \mathrm{FBF}^{\prime}=90^{\circ}$

$\Rightarrow \mathrm{FB}^{2}+\mathrm{F}^{\prime} \mathrm{B}^{2}=\mathrm{FF}^{\prime 2}$

$\therefore\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}=(2 a e)^{2}$

$\Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2}$

$\Rightarrow e^{2}=\frac{b^{2}}{a^{2}}$

Also $e^{2}=1-\frac{b^{2}}{a^{2}}=1-e^{2}$

$\begin{array}{l}

\Rightarrow 2 e^{2}=1 \\

\Rightarrow e=\frac{1}{\sqrt{2}}

\end{array}$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2021
An equation of plane parallel to plane x-2y+2z-5=0 and at a unit distance from the origin is

Options:
A x-2y+2z=3
B x-2y+2z+1=0
C x-2y+2z-1=0
D x-2y+2z+5=0

The correct answer is:
x-2y+2z=3

Solution:
Equation of plane parallel to the plane

x-2y+2z-5=0 is x-2y+2z+λ=0

Since, this plane is unit distance from origin

∴ 1=λ1+4+4

|λ|=3⇒λ=±3

∴ Equation of plane is

x-2y+2z=3

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2019
An equilateral prism of mass $\mathrm{m}$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $\mathrm{F}$ is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is -

Options:
A $\frac{\mathrm{mg}}{\sqrt{3}}$
B $\frac{\mathrm{mg}}{4}$
C $\frac{\mu \mathrm{mg}}{\sqrt{3}}$
D $\frac{\mu \mathrm{mg}}{4}$

The correct answer is:
$\frac{\mathrm{mg}}{\sqrt{3}}$

Solution:
The tendency of rotation will be about the point C.

For minimum force, the torque of $\mathrm{F}$ about $\mathrm{C}$ has to be equal to the torque of $\mathrm{mg}$ about $\mathrm{C}$.

$$

\therefore \quad \mathrm{F}\left(\mathrm{a} \frac{\sqrt{3}}{2}\right)=\mathrm{mg}\left(\frac{\mathrm{a}}{2}\right) \Rightarrow \mathrm{F}=\frac{\mathrm{mg}}{\sqrt{3}}

$$

Topic: Physics, Rotational Motion, JEE Main
An equilateral prism of mass $\mathrm{m}$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $\mathrm{F}$ is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is -

Options:
A $\frac{\mathrm{mg}}{\sqrt{3}}$
B $\frac{\mathrm{mg}}{4}$
C $\frac{\mu \mathrm{mg}}{\sqrt{3}}$
D $\frac{\mu \mathrm{mg}}{4}$

The correct answer is:
$\frac{\mathrm{mg}}{\sqrt{3}}$

Solution:
The tendency of rotation will be about the point C.

For minimum force, the torque of $\mathrm{F}$ about $\mathrm{C}$ has to be equal to the torque of $\mathrm{mg}$ about $\mathrm{C}$.

$$

\therefore \quad \mathrm{F}\left(\mathrm{a} \frac{\sqrt{3}}{2}\right)=\mathrm{mg}\left(\frac{\mathrm{a}}{2}\right) \Rightarrow \mathrm{F}=\frac{\mathrm{mg}}{\sqrt{3}}

$$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2014
An equilibrium mixture at 300 K contains N2O4 and NO2 at 0.28 and 1.1 atm pressure, respectively. If the volume of the container is doubled, the new equilibrium pressure of these two gases are respectively.

Options:
A 0.064 atm and 0.095 atm
B 0.64 atm and 0.095 atm
C 0.095 atm and 0.632 atm
D 0.095 atm and 0.64 atm

The correct answer is:
0.095 atm and 0.64 atm

Solution:
N2O4g⇌2NO2g

Pressure at equilibrium
Kp=pNO22pN2O4=1.120.28=4.32 atm

If volume of the container is doubled, the pressure will reduced to half

N2O4⇌2N2O

New pressure,

0.0282-p1.12+2p

Kp=1.12+2p20.282-p=4.32

On solving, we get

p=0.045

∴ pN2O4=0.14-0.045=0.095 atm

pNO2=0.55+2×0.045=0.64 atm

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2018
An extremely hot copper wire reacts with steam to give

Options:
A $\mathrm{CuO}$
B $\mathrm{Cu}_{2} \mathrm{O}$
C $\mathrm{Cu}_{2} \mathrm{O}_{2}$
D $\mathrm{CuO}_{2}$

The correct answer is:
$\mathrm{Cu}_{2} \mathrm{O}$

Solution:
$\underset{\text { Hot }}{2 \mathrm{Cu}}+\underset{\text { Steam }}{\mathrm{H}_{2} \mathrm{O} \longrightarrow} \mathrm{Cu}_{2} \mathrm{O}+\mathrm{H}_{2} \uparrow$

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2012
An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

Options:
A rise
B go down
C remain same
D first rise then go down

The correct answer is:
go down

Solution:
Ice is lighter than water. When ice melts, the volume occupied by water is less than that of ice. Due to which the level of water goes down.

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2012
An ice-berg of density 900 kgm-3 is floating in water of density 1000 kgm-3. The percentage of volume of ice-berg outside the water is

Options:
A 20 %
B 35 %
C 10 %
D 11 %

The correct answer is:
10 %

Solution:
Let the volume of ice-berg is V and its density is ρ. If this ice-berg floats in water with volume Vin inside it, then

Vinσg=Vρg

σ= density of water

⇒ Vin=ρσ·V

⇒ Vout=V-Vin

=σ-ρσ V=1000-9001000 V

=V10

⇒ Vout /V=0.1=10 %

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2017
An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes 32 times of its initial value. If the final pressure of gas is 128 atmospheres, the value of ' $\gamma$ ' of the gas is :

Options:
A $1.5$
B $1.4$
C $1.3$
D $1.6$

The correct answer is:
$1.4$

Solution:
Volume $v=\frac{\mathrm{m}}{d}$ and using $\mathrm{PV}^{\gamma}=$ constant

$\begin{array}{l}

\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{V}}{\mathrm{V}^{\prime}}=\left(\frac{\mathrm{d}^{\prime}}{\mathrm{d}}\right)^{\gamma} \\

\text { or } 128=(32)^{\gamma} \\

\therefore \gamma=\frac{7}{5}=1.4

\end{array}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2021
An ideal gas can't be liquefied because

Options:
A its critical temperature is always above $0^{\circ} \mathrm{C}$
B its molecules are relativelysmaller in size
C it solidifies before becoming a liquid
D forces operated between its molecules are negligible

The correct answer is:
forces operated between its molecules are negligible

Solution:
In an ideal gas, the intermolecular forces of attraction are negligible and hence it cannot be liquefied.

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2012
An ideal gaseous mixture of ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ and ethene $\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)$ occupies 28 litre at 1 atm and $273 \mathrm{~K}$. The mixture reacts completely with $128 \mathrm{~g}$ $\mathrm{O}_{2}$ to produce $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O} .$ Mole fraction at $\mathrm{C}_{2} \mathrm{H}_{6}$ in the mixture is:

Options:
A 0.6
B 0.4
C 0.5
D 0.8

The correct answer is:
0.4

Solution:
$\mathrm{C}_{2} \mathrm{H}_{6}+3.5 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} ;$

$\mathrm{C}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

Let volume of ethane is $\mathrm{x}$ litre, $22.4 \times 4=3.5 \mathrm{x}+3(28-\mathrm{x})$

$\Rightarrow \mathrm{x}=11.2$ litre

at constant $\mathrm{T}$ and $\mathrm{P}, \mathrm{V} \propto \mathrm{n} ;$

$\Rightarrow$ Mole fraction of $\mathrm{C}_{2} \mathrm{H}_{6}$ in mixture $=\frac{11.2}{28}=0.4$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2020
An ideal monoatomic gas at 300 K expands adiabatically to twice its volume. The final temperature of gas is

Options:
A 3002
B 3003
C 3001223
D 300223

The correct answer is:
3001223

Solution:
For an ideal monatomic gas, the ratio of specific heat capacity at constant pressure to specific heat capacity at constant volume is

γ=53

In an adiabatic process,

TVγ-1= constant

i.e. TfVfγ-1=TiViγ-1⇒Tf=TiViVfγ-1=3001253-1

=300122/3

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2019
An ideal refrigerator has a freezer at a temperature of $13^{\circ} \mathrm{C}$. The coefficient of performance of the engine is $5 .$ The temperature of the air (to which heat is rejected) is

Options:
A $320^{\circ} \mathrm{C}$
B $39^{\circ} \mathrm{C}$
C $325 \mathrm{~K}$
D $325^{\circ} \mathrm{C}$

The correct answer is:
$39^{\circ} \mathrm{C}$

Solution:
$\mathrm{T}_{2}=273-13=260$,
$$
\mathrm{K}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}} ; \quad 5=\frac{260}{\mathrm{~T}_{1}-260}
$$
or $\mathrm{T}_{1}-260=52 ; \quad \mathrm{T}_{1}=312 \mathrm{~K}$
$\mathrm{T}_{2}=312-273=39^{\circ} \mathrm{C}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2010
An inductor of inductance $L=400 \mathrm{mH}$ and resistors of resistance $R_{1}=2 \Omega$ and $R_{2}=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0 .$ The potential drop across $L$ as a function of time is

Options:
A $\frac{12}{t} e^{-3 t} \mathrm{~V}$
B $6\left(1-e^{-t / 0.2}\right) \mathrm{V}$
C $12 \mathrm{e}^{-5 t} \mathrm{~V}$
D $6 \mathrm{e}^{-5 t} \mathrm{~V}$

The correct answer is:
$12 \mathrm{e}^{-5 t} \mathrm{~V}$

Solution:
Growth in current in $L R_{2}$ branch when switch is closed is given by

$$

\begin{array}{l}

i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \\

\Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L} e^{-R_{2} t / L}=\frac{E}{L} e^{-\frac{R_{2} t}{L}}

\end{array}

$$

Hence, potential drop across $\mathbf{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}$

$=12 e^{-\frac{2 t}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 t} \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
An inductor of inductance $L=400 \mathrm{mH}$ and resistors of resistance $R_{1}=2 \Omega$ and $R_{2}=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0 .$ The potential drop across $L$ as a function of time is

Options:
A $\frac{12}{t} e^{-3 t} \mathrm{~V}$
B $6\left(1-e^{-t / 0.2}\right) \mathrm{V}$
C $12 \mathrm{e}^{-5 t} \mathrm{~V}$
D $6 \mathrm{e}^{-5 t} \mathrm{~V}$

The correct answer is:
$12 \mathrm{e}^{-5 t} \mathrm{~V}$

Solution:
Growth in current in $L R_{2}$ branch when switch is closed is given by

$$

\begin{array}{l}

i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \\

\Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L} e^{-R_{2} t / L}=\frac{E}{L} e^{-\frac{R_{2} t}{L}}

\end{array}

$$

Hence, potential drop across $\mathbf{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}$

$=12 e^{-\frac{2 t}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 t} \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2016
An instantaneous displacement of a simple harmonic oscillator is $x=A \cos (\omega t+\pi / 4)$. Its speed will be maximum at time

Options:
A $\pi / 4 \omega$
B $\pi / 2 \omega$
C $\pi / \omega$
D $2 \pi \omega$

The correct answer is:
$\pi / 4 \omega$

Solution:
Velocity, $v=\frac{d x}{d t}=-A \omega \sin (\omega t+\pi / 4)$

Velocity will be maximum, when $\omega t+\pi / 4=\pi / 2$ or $\omega t=\pi / 2-\pi / 4=\pi / 4$ or $t=\pi / 4 \omega$

Topic: Physics, Oscillations, BITSAT, BITSAT 2020
An instantaneous displacement of a simple harmonic oscillator is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 4) .$ Its speed will be maximum at time

Options:
A $\pi / 4 \omega$
B $\pi / 2 \omega$
C $\pi / \omega$
D $2 \pi \omega$

The correct answer is:
$\pi / 4 \omega$

Solution:
Velocity, $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{A} \omega \sin (\omega \mathrm{t}+\pi / 4)$

Velocity will be maximum, when $\omega t+\pi / 4=\pi / 2$ or $\omega t=\pi / 2-\pi / 4=\pi / 4$

or $t=\pi / 4 \omega$

Topic: Physics, Oscillations, BITSAT, BITSAT 2013
An inverted conical flask is being filled with water at the rate of 3 cm3 sec-1. The height of the flask is 10 cm and the radius of the base is 5 cm. How fast is the water level rising when the level is 4 cm?

Options:
A 43π cm sec-1
B 34π cm sec-1
C 3π4 cm sec-1
D 43π cm sec-1

The correct answer is:
34π cm sec-1

Solution:
Let depth of water at time t be h and the radius of the base of water level be r.

From similar ∆OMP and ∆ONQ, we have

OMON=PMQN

⇒h10=r5

⇒r=5h10

⇒r=h2

We have, V=13πr2h=13πh22h

⇒V=πh312

On differentiating both sides, we get

dVdt=π12·3h2dhdt=πh24dhdt

Given, dVdt=3 cm3 sec-1 when h=4 cm, so we get

3=π×424dhdt

⇒dhdt=34π cm sec-1

Hence, the water level is rising at 34π cm sec-1

∴ Option (b) is correct.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2018
An iron rod of length $2 \mathrm{~m}$ and cross-sectional area of $50 \mathrm{~mm}^{2}$ stretched by $0.5 \mathrm{~mm}$, when a mass of $250 \mathrm{~kg}$ is hung from its lower end. Young's modulus of iron rod is

Options:
A $19.6 \times 10^{20} \mathrm{~N} / \mathrm{m}^{2}$
B $19.6 \times 10^{18} \mathrm{~N} / \mathrm{m}^{2}$
C $19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
D $19.6 \times 10^{15} \mathrm{~N} / \mathrm{m}^{2}$

The correct answer is:
$19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Solution:
$Y=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}$
$=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2011
An object experiences a net force and accelerates from rest to its final position in $16 \mathrm{~s}$. How long would the object take to reach the same final position from rest if the object's mass was four times larger ?

Options:
A $64 \mathrm{~s}$
B $32 \mathrm{~s}$
C $16 \mathrm{~s}$
D $8 \mathrm{~s}$

The correct answer is:
$32 \mathrm{~s}$

Solution:
When the mass increases by a factor of 4 the acceleration must decrease by a factor of four if the same force is applied. The question asks about position so we need to relate acceleration and time to position. We can do this by the equation $: x_{f}-x_{i}=v_{x i} t+1 / 2 a_{x} t^{2}$ We want the change in position to stay the same. The initial velocity is zero so in order for the change in position to remain constant the term $(1 / 2)$ at $^{2}$ must remain the same. If the acceleration is reduced by a factor of 4 you can see that the time must be increased by a factor of 2 in order for the term to remain the same.

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2011
An object is obseved from the points $A, B$ and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at $\mathrm{B}$ is twice that at $\mathrm{A}$ and at $\mathrm{C}$ three times that at $\mathrm{A} .$ It $\mathrm{AB}=a, \mathrm{BC}=b$, then the height of the object is

Options:
A $\frac{b}{2 a} \sqrt{(a+b)(3 b-a)}$
B $\frac{a}{b} \sqrt{(a+b)(3 b-a)}$
C $\frac{\mathrm{a}}{2 \mathrm{~b}} \sqrt{(\mathrm{a}+\mathrm{b})(3 \mathrm{~b}-\mathrm{a})}$
D $\frac{2 a}{b} \sqrt{(a+b)(3 b-a)}$

The correct answer is:
$\frac{\mathrm{a}}{2 \mathrm{~b}} \sqrt{(\mathrm{a}+\mathrm{b})(3 \mathrm{~b}-\mathrm{a})}$

Solution:
Let $\mathrm{ED}=h, \angle \mathrm{EAB}=\alpha$

$\therefore \quad \angle \mathrm{EBD}=2 \alpha, \angle \mathrm{ECD}=3 \alpha$

Now, $\angle \mathrm{DBE}=\angle \mathrm{EAB}+\angle \mathrm{BEA}$

$\Rightarrow \quad 2 \alpha=\alpha+\angle \mathrm{BEA}$

$\Rightarrow \quad \angle \mathrm{BEA}=\alpha=\angle \mathrm{EAB}$

$\Rightarrow \mathrm{AB}=\mathrm{EB}=a$

Similarly, $\angle \mathrm{EBC}=\alpha$ $\operatorname{In} \Delta \mathrm{EBC}, \frac{\mathrm{BC}}{\sin \alpha}=\frac{\mathrm{EB}}{\sin \left(180^{\circ}-3 \alpha\right)}$

$\Rightarrow \frac{b}{\sin \alpha}=\frac{a}{\sin 3 \alpha} \Rightarrow \frac{a}{b}=\frac{\sin 3 \alpha}{\sin 3 \alpha}$

$\Rightarrow \quad \frac{a}{b}=\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha$

$\Rightarrow 4 \sin ^{2} \alpha=3-\frac{a}{b}=\frac{3 b-a}{b}$

$\Rightarrow \sin \alpha=\sqrt{\frac{3 b-a}{4 b}}$

In $\Delta \mathrm{EBD}, \sin 2 \alpha=\frac{E D}{E B}$

$\Rightarrow \quad \mathrm{ED}=\mathrm{a} \cdot 2 \sin \alpha \cdot \cos \alpha$

$\Rightarrow h=2 a \sqrt{\frac{3 b-a}{4 b}}, \sqrt{1-\frac{3 b-a}{4 b}}$

$=2 a \sqrt{\frac{3 b-a}{4 b}} \sqrt{\frac{b-a}{4 b}}$

$=\frac{a}{2 b} \sqrt{(a+b)(3 b-a)}$

Topic: Mathematics, Trigonometric Equations, JEE Main
An object is obseved from the points $A, B$ and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at $\mathrm{B}$ is twice that at $\mathrm{A}$ and at $\mathrm{C}$ three times that at $\mathrm{A} .$ It $\mathrm{AB}=a, \mathrm{BC}=b$, then the height of the object is

Options:
A $\frac{b}{2 a} \sqrt{(a+b)(3 b-a)}$
B $\frac{a}{b} \sqrt{(a+b)(3 b-a)}$
C $\frac{\mathrm{a}}{2 \mathrm{~b}} \sqrt{(\mathrm{a}+\mathrm{b})(3 \mathrm{~b}-\mathrm{a})}$
D $\frac{2 a}{b} \sqrt{(a+b)(3 b-a)}$

The correct answer is:
$\frac{\mathrm{a}}{2 \mathrm{~b}} \sqrt{(\mathrm{a}+\mathrm{b})(3 \mathrm{~b}-\mathrm{a})}$

Solution:
Let $\mathrm{ED}=h, \angle \mathrm{EAB}=\alpha$

$\therefore \quad \angle \mathrm{EBD}=2 \alpha, \angle \mathrm{ECD}=3 \alpha$

Now, $\angle \mathrm{DBE}=\angle \mathrm{EAB}+\angle \mathrm{BEA}$

$\Rightarrow \quad 2 \alpha=\alpha+\angle \mathrm{BEA}$

$\Rightarrow \quad \angle \mathrm{BEA}=\alpha=\angle \mathrm{EAB}$

$\Rightarrow \mathrm{AB}=\mathrm{EB}=a$

Similarly, $\angle \mathrm{EBC}=\alpha$ $\operatorname{In} \Delta \mathrm{EBC}, \frac{\mathrm{BC}}{\sin \alpha}=\frac{\mathrm{EB}}{\sin \left(180^{\circ}-3 \alpha\right)}$

$\Rightarrow \frac{b}{\sin \alpha}=\frac{a}{\sin 3 \alpha} \Rightarrow \frac{a}{b}=\frac{\sin 3 \alpha}{\sin 3 \alpha}$

$\Rightarrow \quad \frac{a}{b}=\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha$

$\Rightarrow 4 \sin ^{2} \alpha=3-\frac{a}{b}=\frac{3 b-a}{b}$

$\Rightarrow \sin \alpha=\sqrt{\frac{3 b-a}{4 b}}$

In $\Delta \mathrm{EBD}, \sin 2 \alpha=\frac{E D}{E B}$

$\Rightarrow \quad \mathrm{ED}=\mathrm{a} \cdot 2 \sin \alpha \cdot \cos \alpha$

$\Rightarrow h=2 a \sqrt{\frac{3 b-a}{4 b}}, \sqrt{1-\frac{3 b-a}{4 b}}$

$=2 a \sqrt{\frac{3 b-a}{4 b}} \sqrt{\frac{b-a}{4 b}}$

$=\frac{a}{2 b} \sqrt{(a+b)(3 b-a)}$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2013
An object moves with speed $v_1, v_2$ and $v_3$ along a line segment $A B, B C$ and $C D$ respectively as shown in figure. Where $A B=B C$ and $A D=3 A B$, then average speed of the object will be:

Options:
A $\frac{v_1 v_2 v_3}{3\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$
B $\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$
C $\frac{\left(u_1+z_2+v_3\right)}{3}$
D $\frac{\left(v_1+v_2+v_3\right)}{3 v_1 v_2 v_3}$

The correct answer is:
$\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$

Solution:
Consider,
$\mathrm{AB}=x$
$\mathrm{BC}=x$
$2 x+\mathrm{CD}=3 x \Rightarrow \mathrm{CD}=3 x-2 x=x$
Average speed of the object $ < \mathrm{v}>$
$=\frac{\text { Total distance }}{\text { Total time }}$
$ < \mathrm{v}>=\frac{3 x}{\frac{x}{v_1}+\frac{x}{v_2}+\frac{x}{v_3}}=\frac{3 v_1 v_2 v_3}{v_2 v_3+v_1 v_3+v_1 v_2}$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2023 (Memory Based Paper 1)
An object of mass 5 kg is projected with a velocity 20 m s-1 at an angle 60°, to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses 1 kg and 4 kg. The fragments separate horizontally after the explosion, which releases internal energy such that the KE of the system af the highest point is doubled. The separation between the two fragments when they reach the ground is

Options:
A 52.25 m
B 44.25 m
C 65.32 m
D 78.76 m

The correct answer is:
44.25 m

Solution:
Given, m=5 kg,v=20 m s-1,θ=60°

Vertical component of velocity, vy=v sin 60°

=20×32=103 m s-1

Time taken to reach the highest point = Time taken to reach the ground from highest point.

t=v sin θg=vyg=1039.8=1.77 s

At the highest point, m splits up into two parts of masses m1=1 kg and m2=4 kg

If their velocities are v1 and v2 respectively, then applying the principle of conservation of linear momentum, we get

m1v1+m2v2=mv cos θ

v1+4v2=5×20×12 ∵ θ=60°

v1+4v2=5×10=50 ....i

Initial KE=12 m v cos θ2

=12×5×102=250 J

Final KE=2 (initial KE)=2×250=500 J

∴ 12m1v12+12m2v22=500

or 12×1×v12+12×4×v22=500

or v12+4v22=1000 ....ii

Solving Eq. (i) and (ii), we get

v1=30 m s-1, v2=5 m s-1

Hence, the separation between the two fragments

=v1-v2×t=30-5×1.77 m=44.25 m

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2018