datavorous/entrance-exam-dataset huggingface dataset:
A graph is plotted between log K versus 1T for calculation of activation energy Ea. The correct plot is

Options:
A

B

C

D

The correct answer is:

Solution:
According to Arrhenius equation,

k∝T and k=Ae−Ea​/RT

logk=logA-Ea2.303×R​​×1T​

y=mx+c, the equation of straight line

The value of log k should increase uniformily with T or decrease with 1T.

From this, that the slope is negative, which means that angle with x - axis is greater than 90° also that the point on which the straight line of graph intercepts the y - axis is log A which is positive.

Hence all the conclusions point out towards option b as it has negative slope and a positive y intercept.

Topic: Chemistry, Chemical Kinetics, JEE Main
A graph is plotted between log K versus 1T for calculation of activation energy Ea. The correct plot is

Options:
A

B

C

D

The correct answer is:

Solution:
According to Arrhenius equation,

k∝T and k=Ae−Ea​/RT

logk=logA-Ea2.303×R​​×1T​

y=mx+c, the equation of straight line

The value of log k should increase uniformily with T or decrease with 1T.

From this, that the slope is negative, which means that angle with x - axis is greater than 90° also that the point on which the straight line of graph intercepts the y - axis is log A which is positive.

Hence all the conclusions point out towards option b as it has negative slope and a positive y intercept.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2017
A gun fires two bullets at $60^{\circ}$ and $30^{\circ}$ with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio of

Options:
A $2: 1$
B $3: 1$
C $4: 1$
D $1: 1$

The correct answer is:
$3: 1$

Solution:
The bullets are fired at the same initial speed

$\begin{array}{l}

\frac{\mathrm{H}}{\mathrm{H}^{\prime}}=\frac{\mathrm{u}^{2} \sin ^{2} 60^{\circ}}{2 \mathrm{~g}} \times \frac{2 \mathrm{~g}}{\mathrm{u}^{2} \sin ^{2} 30^{\circ}}=\frac{\sin ^{2} 60^{\circ}}{\sin ^{2} 30^{\circ}} \\

=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3}{1}

\end{array}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2021
A hollow metal sphere of radius $5 \mathrm{~cm}$ is charged such that the potential on its surface is $10 \mathrm{~V}$. The potential at a distance of $2 \mathrm{~cm}$ from the centre of the sphere is

Options:
A zero
B $10 \mathrm{~V}$
C $4 \mathrm{~V}$
D $10 / 3 \mathrm{~V}$

The correct answer is:
$10 \mathrm{~V}$

Solution:
Potential at any point inside the sphere $=$ potential at the surface of the sphere $=10 \mathrm{~V}$.

Topic: Physics, Electrostatics, BITSAT, BITSAT 2011
A horizontal overhead powerline is at height of $4 \mathrm{~m}$ from the ground and carries a current of $100 \mathrm{~A}$ from east to west. The magnetic field directly below it on the ground is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right)$

Options:
A $2.5 \times 10^{-7} \mathrm{~T}$ southward
B $5 \times 10^{-6}$ T northward
C $5 \times 10^{-6} \mathrm{~T}$ southward
D $2.5 \times 10^{-7}$ T northward

The correct answer is:
$5 \times 10^{-6} \mathrm{~T}$ southward

Solution:
The magnetic field is
$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}=10^{-7} \times \frac{2 \times 100}{4}=5 \times 10^{-6} \mathrm{~T}$

According to right hand palm rule, the magnetic field is directed towards south.

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2012
A house subtends a right angle at the window of a opposite house and the angle of elevation of the window from the bottom of the first house is $60$. If the distance between two houses be $6 \mathrm{~m}$, then the height of the first house is

Options:
A
$8 \sqrt{3} \mathrm{~m}$

B
$6 \sqrt{3} \mathrm{~m}$

C
$4 \sqrt{3} \mathrm{~m}$

D
None of these

The correct answer is:

$8 \sqrt{3} \mathrm{~m}$

Solution:
Let $P Q$ be the house subtending a right angle at the window $B$ of opposite house $A B$

In $\triangle P A B$, we have
$\tan 60^{\circ}=\frac{A B}{6}=A B=6 \sqrt{3} \mathrm{~m}$
In $\triangle C B Q$, we have
$\tan 30^{\circ}=\frac{h-C P}{B C}$
$\begin{array}{ll}\Rightarrow & \frac{1}{\sqrt{3}}=\frac{h-6 \sqrt{3}}{6} \quad[\because A B=C P, B C=A P] \\ \Rightarrow & h=6\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \\ \Rightarrow & h=8 \sqrt{3} \mathrm{~m}\end{array}$

Topic: Mathematics, Heights and Distances, BITSAT, BITSAT 2022
A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$. Then the ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$ will be.

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

The correct answer is:
$\frac{4}{3}$

Solution:
In adiabatic process
$\begin{aligned}
& \mathrm{PV}^\gamma=\text { constant } \\
& \Rightarrow \quad \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \Rightarrow \quad \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^\gamma \Rightarrow \frac{81}{16}=\left(\frac{27}{8}\right)^\gamma \\
& \Rightarrow\left(\frac{9}{4}\right)^2=\left(\frac{9 \times 3}{4 \times 2}\right)^\gamma \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{3}{2}\right)^\gamma \\
& \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{9}{4}\right)^{\gamma / 2} \Rightarrow 2-\gamma=\frac{\gamma}{2} \\
& \Rightarrow 2=\frac{3}{2} \gamma \Rightarrow \gamma=\frac{4}{3}
\end{aligned}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A is an optically inactive alkyl chloride which on reaction with aqueous $\mathrm{KOH}$ gives $\mathrm{B}$. $\mathrm{B}$ on heating with $\mathrm{Cu}$ at $300^{\circ} \mathrm{C}$ gives an alkene $\mathrm{C}$, what are $\mathrm{A}$ and $\mathrm{C}$

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{CH}_{2}=\mathrm{CH}_{2}$
B $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{MeCH}=\mathrm{CH} \cdot \mathrm{Me}$
C $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$
D $\mathrm{Me}_{2} \mathrm{CH} . \mathrm{CH}_{2} \mathrm{Cl} \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$

The correct answer is:
$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$

Solution:
OR

Topic: Chemistry, Hydrocarbons, JEE Main
A is an optically inactive alkyl chloride which on reaction with aqueous $\mathrm{KOH}$ gives $\mathrm{B}$. $\mathrm{B}$ on heating with $\mathrm{Cu}$ at $300^{\circ} \mathrm{C}$ gives an alkene $\mathrm{C}$, what are $\mathrm{A}$ and $\mathrm{C}$

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{CH}_{2}=\mathrm{CH}_{2}$
B $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{MeCH}=\mathrm{CH} \cdot \mathrm{Me}$
C $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$
D $\mathrm{Me}_{2} \mathrm{CH} . \mathrm{CH}_{2} \mathrm{Cl} \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$

The correct answer is:
$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$

Solution:
OR

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2011
A ladder rests against a wall so that its top touches the roof of the house. If the ladder makes an angle of \(60^{\circ}\) with the horizontal and height of the house be \(6 \sqrt{3}\) meters, then the length of the ladder in meters is

Options:
A \(12 \sqrt{3}\)
B 12
C \(12 / \sqrt{3}\)
D None of these

The correct answer is:
12

Solution:
Length of ladder \(=\frac{6 \sqrt{3}}{\sin 60^{\circ}}=12 \mathrm{~m}\)

Topic: Mathematics, Heights and Distances, BITSAT, BITSAT 2011
A large drop of oil (density $0.8 \mathrm{~g} / \mathrm{cm}^{3}$ and viscosity $\eta_{0}$ ) floats up through a column of another liquid (density $1.2 \mathrm{~g} / \mathrm{cm}^{3}$ and viscosity $\eta_{\mathrm{L}}$ ). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:

Options:
A $\eta_{0}$ only
B $\eta_{L}$ only
C both on $\eta_{0}$ and $\eta_{\mathrm{L}}$
D neither $\eta_{0}$ nor $\eta_{L}$

The correct answer is:
$\eta_{L}$ only

Solution:
Forces of viscosity arise in a fluid when different sections of the fluid move with different velocities. Here, there is no such motion inside the oil drop and hence the viscosity of oil does not come into play.

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2012
A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Options:
A $f$ and $\frac{I}{4}$
B $\frac{3 f}{4}$ and $\frac{I}{2}$
C $f$ and $\frac{3 I}{4}$
D $\frac{f}{2}$ and $\frac{I}{2}$

The correct answer is:
$f$ and $\frac{3 I}{4}$

Solution:
Bycovering aperture, focal length does not change. But intensity is reduced by $\frac{1}{4}$ times

as aperture diameter $\frac{d}{2}$ is covered.

$$

\begin{array}{l}

\therefore \quad I^{\prime}=I-\frac{I}{4}=\frac{3 I}{4} \\

\therefore \quad \text { New focal length }=\mathrm{f} \text { and intensity }=\frac{3 I}{4} .

\end{array}

$$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2016
A letter'A' is constructed of a uniform wire with resistance $1.0 \Omega$ per $\mathrm{cm}$, The sides of the letter are $20 \mathrm{~cm}$ and the cross piece in the middle is 10 $\mathrm{cm}$ long. The apex angle is $60 .$ The resistance between the ends of the legs is close to:

Options:
A $50.0 \Omega$
B $10 \Omega$
C $36.7 \Omega$
D $26.7 \Omega$

The correct answer is:
$26.7 \Omega$

Solution:
For ADE $\frac{1}{R^{\prime}}=\frac{1}{2 x}+\frac{1}{10} \quad$ or $\quad R^{\prime}=\frac{20 x}{10+2 x}$
$\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} \quad \ldots$(i)
or $\quad \frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$
Solving we get $\mathrm{x}=10 \Omega$
Putting the value of $x=10 \Omega$ in equation (i) We get $\mathrm{R}_{\mathrm{BC}}=\frac{20 \times 10}{10+2 \times 10}+20-10+20-10$
$=\frac{80}{3}=26.7 \Omega$

Topic: Physics, Current Electricity, JEE Main
A letter'A' is constructed of a uniform wire with resistance $1.0 \Omega$ per $\mathrm{cm}$, The sides of the letter are $20 \mathrm{~cm}$ and the cross piece in the middle is 10 $\mathrm{cm}$ long. The apex angle is $60 .$ The resistance between the ends of the legs is close to:

Options:
A $50.0 \Omega$
B $10 \Omega$
C $36.7 \Omega$
D $26.7 \Omega$

The correct answer is:
$26.7 \Omega$

Solution:
For ADE $\frac{1}{R^{\prime}}=\frac{1}{2 x}+\frac{1}{10} \quad$ or $\quad R^{\prime}=\frac{20 x}{10+2 x}$
$\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} \quad \ldots$(i)
or $\quad \frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$
Solving we get $\mathrm{x}=10 \Omega$
Putting the value of $x=10 \Omega$ in equation (i) We get $\mathrm{R}_{\mathrm{BC}}=\frac{20 \times 10}{10+2 \times 10}+20-10+20-10$
$=\frac{80}{3}=26.7 \Omega$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2012
A light having wavelength $300 \mathrm{nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?

Options:
A $2.3 \mathrm{~V}$
B $2.59 \mathrm{~V}$
C $1.59 \mathrm{~V}$
D $1.29 \mathrm{~V}$

The correct answer is:
$1.59 \mathrm{~V}$

Solution:
Energy of incident radiation is given by:
\(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.64 * 10^{-34} * 3 * 10^{8}}{300 * 10^{-9}}=6.64 * 10^{-19}=4.05 \mathrm{eV}\)
The work function of the metal is equal to \(2.54 \mathrm{eV}\)
Hence the maximum kinetic energy of an ejected photoelectron would be \(1.59 \mathrm{eV}\)
And the stopping potential needed to stop this would be \(1.59 \mathrm{~V}\).

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2012
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2},$ find the work done $(\mathrm{in}$

joules) by the string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.

Options:
A $4 \mathrm{~J}$
B $2 \mathrm{~J}$
C $8 \mathrm{~J}$
D $10 \mathrm{~J}$

The correct answer is:
$8 \mathrm{~J}$

Solution:
Given $m=0.36 \mathrm{~kg}, M=0.72 \mathrm{~kg}$. The figure shows the forces on $m$ and $M$. When the system is released, let the acceleration be $a$. Then

$T-m g=m a$

$M g-T=M a$

$$

a=\frac{(M-m) g}{M+m}=g / 3

$$

and $T=4 \mathrm{mg} / 3$

For block m:

$$

u=0, a=g / 3, t=1, s=?

$$

$$

s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times \frac{g}{3} \times 1^{2}=g / 6

$$

Work done by the string on $m$ is

$$

\bar{T} \bar{s}=T s=4 \frac{m g}{3} \times \frac{g}{6}=\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}=8 \mathrm{~J}

$$

Topic: Physics, Work Power Energy, JEE Main
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2},$ find the work done $(\mathrm{in}$

joules) by the string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.

Options:
A $4 \mathrm{~J}$
B $2 \mathrm{~J}$
C $8 \mathrm{~J}$
D $10 \mathrm{~J}$

The correct answer is:
$8 \mathrm{~J}$

Solution:
Given $m=0.36 \mathrm{~kg}, M=0.72 \mathrm{~kg}$. The figure shows the forces on $m$ and $M$. When the system is released, let the acceleration be $a$. Then

$T-m g=m a$

$M g-T=M a$

$$

a=\frac{(M-m) g}{M+m}=g / 3

$$

and $T=4 \mathrm{mg} / 3$

For block m:

$$

u=0, a=g / 3, t=1, s=?

$$

$$

s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times \frac{g}{3} \times 1^{2}=g / 6

$$

Work done by the string on $m$ is

$$

\bar{T} \bar{s}=T s=4 \frac{m g}{3} \times \frac{g}{6}=\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}=8 \mathrm{~J}

$$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2013
A line makes angles of $45^{\circ}$ and $60^{\circ}$ with the positive axes of $X$ and $Y$ respectively. The angle made by the same line with the positive axis of $Z$, is

Options:
A $30^{\circ}$ or $60^{\circ}$
B $60^{\circ}$ or $90^{\circ}$
C $90^{\circ}$ or $120^{\circ}$
D $60^{\circ}$ or $120^{\circ}$

The correct answer is:
$60^{\circ}$ or $120^{\circ}$

Solution:
Given $\alpha=45^{\circ}, \beta=60^{\circ}, \gamma=?$

$\begin{array}{l}

\because \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\

\therefore \cos ^{2} \gamma=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\

\Rightarrow \mathrm{y}=60^{\circ} \text { or } 120^{\circ}

\end{array}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2021
A line makes angles α, β, γ with the coordinate axes. If α+β=π2, then cosα+cosβ+cosγ2 is equal to

Options:
A 1+sin2α
B 1-sin2α
C 1+cos2α
D None of these

The correct answer is:
1+sin2α

Solution:
We have, cos2α+cos2β+cos2γ=1

⇒cos2α+cos2π2-α+cos2γ=1 [given α+β=π2]

⇒cos2α+sin2α+cos2γ=1

⇒1+cos2γ=1

⇒cos2γ=0

⇒cosγ=0

∴ cosα+cosβ+cosγ2=cosα+sinα2

=1+2sinα·cosα=1+sin2α

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2018
A line makes the same angle $\theta$ with each of the $\mathrm{X}$ and $\mathrm{Z}$ -axes. If the angle $\beta$, which it makes with Y-axis, is such that $\sin ^{2} \beta=3 \sin ^{2} \theta,$ then $\cos ^{2} \theta$ equals

Options:
A $2 / 5$
B $1 / 5$
C $3 / 5$
D $2 / 3$

The correct answer is:
$3 / 5$

Solution:
Let $l, \mathrm{~m}$ and $\mathrm{n}$ be the direction cosines. Then, $l=\cos \theta, \mathrm{m}=\cos \beta, \mathrm{n}=\cos \theta$

$$

\begin{array}{l}

\text { we have } l^{2}+m^{2}+n^{2}=1 \\

\Rightarrow \cos ^{2} \theta+\cos ^{2} \beta+\cos ^{2} \theta=1 \\

\Rightarrow 2 \cos ^{2} \theta+1-\sin ^{2} \beta=1 \\

\Rightarrow 2 \cos ^{2} \theta-\sin ^{2} \beta=0 \\

\Rightarrow 2 \cos ^{2} \theta-3 \sin ^{2} \beta=0 \\

\quad\left[\because \sin ^{2} \beta=3 \sin ^{2} 0(\text { given })\right] \\

\Rightarrow \tan ^{2} \theta=2 / 3 \\

\therefore \cos ^{2} \theta=\frac{1}{1+\tan ^{2} \theta}=\frac{1}{1+2 / 3}=\frac{3}{5}

\end{array}

$$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2015
A line segment has length 63 and direction ratios are \(3,-2,6\). If the line makes an obtuse angle with \(\mathrm{x}\)-axis, the components of the line vector are

Options:
A \(27,-18,54\)
B \(-27,18,54\)
C \(-27,18,-54\)
D \(27,-18,-54\)

The correct answer is:
\(-27,18,-54\)

Solution:
Let the components of the line vector be a, b, c. Then \(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=(63)^2\) ...(i)
Also \(\frac{\mathrm{a}}{3}=\frac{\mathrm{b}}{-2}=\frac{\mathrm{c}}{6}=\lambda\) (say), then \(\mathrm{a}=3 \lambda\), \(\mathrm{b}=-2 \lambda\) and \(\mathrm{c}=6 \lambda\) and from (i) we have \(9 \lambda^2+4 \lambda^2+36 \lambda^2=(63)^2\)
\(\Rightarrow 49 \lambda^2=(63)^2\)
\(\Rightarrow \lambda= \pm \frac{63}{7}= \pm 9\)
Since \(a=3 \lambda < 0\) as the line makes an obtuse angle with \(x\)-axis, \(\lambda=-9\) and the required components are \(-27,18,-54\).

Topic: Mathematics, Probability, BITSAT, BITSAT 2009
A liquid is flowing through a non-sectional tube with its axis horizontally. If two points $\mathrm{X}$ and $\mathrm{Y}$ on the axis of tube has a sectional area $2.0 \mathrm{~cm}^{3}$ and $25 \mathrm{~mm}^{2}$ respectively then find the flow velocity at $Y$ when the flow velocity at $X$ is $10 \mathrm{~m} / \mathrm{s}$.

Options:
A $20 \mathrm{~m} / \mathrm{s}$
B $40 \mathrm{~m} / \mathrm{s}$
C $80 \mathrm{~m} / \mathrm{s}$
D $60 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$80 \mathrm{~m} / \mathrm{s}$

Solution:
According to principle of continuity
$$
v_{y}=\frac{v_{x} A_{x}}{A_{y}}=\frac{10(m / s) \times 2\left(c m^{2}\right)}{25 \times 10^{-2}\left(c m^{2}\right)}=80 \mathrm{~m} / \mathrm{s}
$$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2009
A load of mass m falls from a height h on the scale pan hung from a spring as shown. If the spring constant is k and the mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is

Options:
A mgk
B mgk1+2hkmg
C mgk+mgk1+2hkmg
D None of these

The correct answer is:
mgk1+2hkmg

Solution:
From the energy conservation principle,

mgh=12kX02-mgX0

where, X0 is maximum elongation in spring.

⇒ 12kX02-mgX0-mgh=0

⇒ X02-2mgkX0-2mgkh=0

X0=2mg2±2mgk2+4×2mgkh2=mg2+mgk2+2mgkh

Amplitude = elongation in spring for lowest extreme positionX0 - elongation in spring for equilibrium positionX1

=X0-X1=mgk1+2hkmg , ∵X1=mgk

Topic: Physics, Oscillations, BITSAT, BITSAT 2017
A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is $\mathrm{k}$ and mass of the scale pan is zero and the mass $\mathrm{m}$ does not bounce relative to the pan, then the amplitude of vibration is

Options:
A $\mathrm{mg} / \mathrm{d}$
B $\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$
C $\frac{\mathrm{mg}}{\mathrm{k}}+\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$
D $\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}-\frac{\mathrm{mg}}{\mathrm{k}}\right)}$

The correct answer is:
$\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$

Solution:
According to energy conservation principle, If, $\mathrm{x}_{1}$ is maximum elongation in the spring when the particle is in its lowest extreme position. Then,

$$

\begin{array}{l}

\mathrm{mgh}=\frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1} \\

\Rightarrow \frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1}-\mathrm{mgh}=0 \\

\text { or, } \mathrm{x}_{1}^{2}-\frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{x}_{1}-\frac{2 \mathrm{mg}}{\mathrm{k}} \cdot \mathrm{h}=0 \\

\therefore \mathrm{x}_{1}=\frac{\frac{2 \mathrm{mg}}{\mathrm{k}} \pm \sqrt{\left.\left(\frac{2 \mathrm{mg}}{\mathrm{k}}\right)^{2}+4 \times \frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{h}\right]}}{2}

\end{array}

$$

Amplitude $\mathrm{A}=\mathrm{X}_{1}-\mathrm{X}_{0}$ (elongation in spring for equilibrium position)

$$

\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(1+\frac{2 \mathrm{hk}}{\mathrm{mg}}\right)}

$$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2015
A long straight wire along the $Z$ -axis carries a current I in the negative $Z$ -direction. The

magnetic vector field $\vec{B}$ at a point having coordinates $(x, y)$ in the $Z=0$ plane is

Options:
A $\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$
B $\frac{\mu_{0} I(x \hat{i}+y \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$
C $\frac{\mu_{0} I(x \hat{j}-y \hat{i})}{2 \pi\left(x^{2}+y^{2}\right)}$
D $\frac{\mu_{0} I(x \hat{i}-y \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$

The correct answer is:
$\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$

Solution:
The wire carries a current $I$ in the negative $z$ -direction. We have to consider the

magnetic vector field $\vec{B}$ at $(x, y)$ in the $z=0$ plane.

Magnetic field $\vec{B}$ is perpendicular to $O P$ $\therefore \vec{B}=B \sin \theta \hat{i}-B \cos \theta \hat{j}$

$$

\begin{array}{l}

\sin \theta=\frac{y}{r}, \cos \theta=\frac{x}{r} B=\frac{\mu_{0} I}{2 \pi r} \\

\therefore \quad \vec{B}=\frac{\mu_{0} I}{2 \pi r^{2}}(y \hat{i}-x \hat{j})

\end{array}

$$

or $\quad \vec{B}=\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2015
A long string is stretched by $2 \mathrm{~cm}$ and the potential energy is $\mathrm{V}$. If the spring is stretched by $10 \mathrm{~cm}$, its potential energy will be

Options:
A $\mathrm{V} / 25$
B $\mathrm{V} / 5$
C $5 \mathrm{~V}$
D $25 \mathrm{~V}$

The correct answer is:
$25 \mathrm{~V}$

Solution:
$\quad V=\frac{1}{2} k(x)^{2}=\frac{1}{2} k(2)^{2}$ or $k=\frac{2 V}{4}=\frac{V}{2}$
$$
\mathrm{V}^{\prime}=\frac{1}{2} \mathrm{k}(10)^{2}=\frac{1}{2} \times\left(\frac{\mathrm{V}}{2}\right)(10)^{2}=25 \mathrm{~V}
$$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2010
A machine gun fires 300 bullets per $\mathrm{min}$ if the mass of each bullet is $10 \mathrm{~g}$ and the velocity of the bullets is $600 \mathrm{~ms}^{-1}$, the power (in $\mathrm{kW}$ ) of the gun is

Options:
A
$43200$

B
$9$

C
$72$

D
$7.2$

The correct answer is:

$9$

Solution:
Work done by the gun $=$ total kinetic energy of the bullets
$$
\begin{aligned}
& =n \frac{1}{2} m v^2=300 \times \frac{1}{2} \times 10 \times 10^{-3} \times(600)^2 \\
& =150 \times 10 \times 10^{-3} \times 600 \times 600
\end{aligned}
$$
$$
\begin{aligned}
\text { Power of gun } & =\frac{\text { work done }}{\text { Time taken }} \\
& =\frac{150 \times 10 \times 10^{-3} \times 600 \times 600}{60 \mathrm{~s}} \\
& =9 \mathrm{~kW} \quad(\because 1 \text { minute }=60 \text { seconds })
\end{aligned}
$$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2022
A man of mass $100 \mathrm{~kg}$. is standing on a platform of mass $200 \mathrm{~kg}$. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed $30 \mathrm{~m} / \mathrm{sec}$ relative to the platform then calculate with what velocity relative to the ice the platform will recoil?

Options:
A $5 \mathrm{~m} / \mathrm{sec}$
B $10 \mathrm{~m} / \mathrm{sec}$
C $15 \mathrm{~m} / \mathrm{sec}$
D $20 \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$10 \mathrm{~m} / \mathrm{sec}$

Solution:
Consider the situation shown in figure.Suppose the man moves at a speed w towards right and the the platform recoils at a speed \(\mathrm{V}\) towards left, both relative to the ice. Hence, the speed of the man relative to the platform is
\(\mathrm{V}+\mathrm{w}\). By the question,
\(\mathrm{V}+\mathrm{w}=\mathrm{v}\), or \(\mathrm{w}=\mathrm{v}-\mathrm{V} \ldots\) (i)
Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Thus,
\(0=\mathrm{MV}-\underset{m v}{\operatorname{mor}} \mathrm{MV}=\mathrm{m}(\mathrm{v}-\mathrm{V})[\mathrm{Using}(\mathrm{i})]\)
or, \(V=\frac{m v}{M+m}\)
Hence, the velocity relative to the ice is \(V=\frac{m v}{M+m}\).

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2012