datavorous/entrance-exam-dataset huggingface dataset:
A plano-convex lens of focal length $30 \mathrm{~cm}$ has its plane surface silvered An object is placed $40 \mathrm{~cm}$ from the lens on the convex side. The distance of the image from the lens is

Options:
A $18 \mathrm{~cm}$
B $24 \mathrm{~cm}$
C $30 \mathrm{~cm}$
D $40 \mathrm{~cm}$

The correct answer is:
$24 \mathrm{~cm}$

Solution:
No Solution available

Topic: Physics, Ray Optics, BITSAT, BITSAT 2009
A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre C. When released, it reaches $\mathrm{C}$ under the gravitational attraction of the ring. Its speed at $C$ will be

Options:
A $\sqrt{\frac{2 \mathrm{G} m}{r}(\sqrt{2}-1)}$
B $\sqrt{\frac{\mathrm{G} m}{r}}$
C $\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
D $\sqrt{\frac{2 \mathrm{G} m}{r}}$

The correct answer is:
$\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

Solution:
Let ' $\mathrm{M}$ ' be the mass of the particle

Now, $E_{\text {initial }}=E_{\text {final }}$

i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} \mathrm{MV}^{2}$

or, $\frac{1}{2} \mathrm{MV}^{2}=\frac{\mathrm{GM} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$

$\Rightarrow \frac{1}{2} \mathrm{~V}^{2}=\frac{\mathrm{G} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$ or,

$\mathrm{V}=\sqrt{\frac{2 \mathrm{Gm}}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2021
A point particle of mass $0.1 \mathrm{~kg}$ is executing S.H.M. of amplitude of $0.1 \mathrm{~m}$. When the particle passes through the mean position, its kinetic energy is 8 $\times 10^{-3}$ Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is $45^{\circ} .$

Options:
A $y=0.1 \sin \left(\pm 4 t+\frac{\pi}{4}\right)$
B $\quad y=0.2 \sin \left(\pm 4 t+\frac{\pi}{4}\right)$
C $y=0.1 \sin \left(\pm 2 t+\frac{\pi}{4}\right)$
D $y=0.2 \sin \left(\pm 2 t+\frac{\pi}{4}\right)$

The correct answer is:
$y=0.1 \sin \left(\pm 4 t+\frac{\pi}{4}\right)$

Solution:
No Solution

Topic: Physics, Oscillations, BITSAT, BITSAT 2016
A positive charge q is projected in magnetic field of width mv2qB with velocity v. Then, the time taken by charged particle to emerge from the magnetic field is

Options:
A m2qB
B πm4qB
C πm2qB
D πm2·qB

The correct answer is:
πm4qB

Solution:
Referring the diagram,

width of magnetic field w=mv2qB

Also we know that when a charge is projected in perpendicular magnetic field then radius of path is

R=mvqB

Now in triangle OAB,

sinθ=wR=mv2qBmvqB

=12

⇒ sinθ=sin45°

=sinπ4

Time period for angle 2π is T=2πmqB

∴ Time by charged particle for θ=π4in order to emerge from magnetic field is

T=πm4qB.

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2017
A projectile A is thrown at an angle of $30^{\circ}$ to the horizontal from point P. At the same time, another projectile $\mathrm{B}$ is thrown with velocity $\mathrm{v}_{2}$ upwards from the point $Q$ vertically below the highest point. For $\mathrm{B}$ to collide with $\mathrm{A}, \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ should be

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

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

Solution:
This happen when vertical velocity of both are same.

$\therefore v_{2}=v_{1} \sin 30^{\circ} \text { or } \frac{v_{2}}{v_{1}}=\frac{1}{2}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2020
A projectile can have the same range $\mathrm{R}$ for two angles of projection. If $t_{1}$ and $t_{2}$ be the times of flight in two cases, then what is the product of two times of flight?

Options:
A $\mathrm{t}_{1} \mathrm{t}_{2} \propto \mathrm{R}$
B $\mathrm{t}_{1} \mathrm{t}_{2} \propto \mathrm{R}^{2}$
C $\mathrm{t}_{1} \mathrm{t}_{2} \propto \mathrm{l} / \mathrm{R}$
D $\mathrm{t}_{1} \mathrm{t}_{2} \propto 1 / \mathrm{R}^{2}$

The correct answer is:
$\mathrm{t}_{1} \mathrm{t}_{2} \propto \mathrm{R}$

Solution:
$\mathrm{t}_{1}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$ and
$\mathrm{t}_{2}=\frac{2 \mathrm{u} \sin (90-\theta)}{\mathrm{g}}=\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}$
$\begin{aligned} \therefore \mathrm{t}_{1} \mathrm{t}_{2} &=\frac{4 \mathrm{u}^{2} \cos \theta \sin \theta}{\mathrm{g}^{2}}=\frac{2}{\mathrm{~g}}\left[\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\right] \\ &=\frac{2}{\mathrm{~g}} \mathrm{R}, \end{aligned}$
where $\mathrm{R}$ is the range.
Hence $t_{1} t_{2} \propto R$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2012
A projectile is fired with a velocity u making an angle $\theta$ with the horizontal. What is the magnitude of change in velocity when it is at the highest point -

Options:
A $u \cos \theta$
B $\mathrm{u}$
C $\mathrm{u} \sin \theta$
D $\mathrm{u} \cos \theta-\mathrm{u}$

The correct answer is:
$\mathrm{u} \sin \theta$

Solution:
Initially $\mathrm{u}=\cos \theta \hat{\mathrm{i}}+\mathrm{u} \sin \theta \hat{\mathrm{j}}$.

At highest point $\mathrm{v}=\mathrm{u} \cos \theta \hat{\mathrm{i}}$

$\therefore$ difference is $\mathrm{u} \sin \theta$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2014
A projectile is given an initial velocity of i^+3 j^ m s-1, where i^ is along the ground and j^ is along the vertical. Then, the equation of the path of projectile is [Take, g=10 m s-2

Options:
A y=3x-5x2
B y=3x+5x2
C x=3y+5x2
D x2=y2+3

The correct answer is:
y=3x-5x2

Solution:
Velocity of projectile at time t is given by

v=u-g t j^

v=i^+3 j^-10 t j^

∴ r=∫v dt

=∫ i^+3 j^-10 t j^ dt

=t i^+3 t j^-10t22j^

=t i^+3 t j^-5 t2 j^

r=t i^+3 t-5 t2 j^

or x i^+y j^=t i^+3 t-5 t2 j^

On comparing, x=t ...(i)

and y=3 t-5 t2 ...(ii)

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

y=3 x-5 x2

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2019
A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40 \mathrm{~ms}^{-1}$. The velocity of the projectile at $\mathrm{t}=2 \mathrm{~s}$ from the start will be: (Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $20 \sqrt{3} \mathrm{~ms}^{-1}$
B $40 \sqrt{3} \mathrm{~ms}^{-1}$
C $20 \mathrm{~ms}^{-1}$
D
Zero

The correct answer is:
$20 \sqrt{3} \mathrm{~ms}^{-1}$

Solution:
Given,
Initial velocity of projectile, $u=40 \mathrm{~m} / \mathrm{s}$
Angle, $\theta=30^{\circ}$
Time of flight
$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 40 \times 1}{10 \times 2}=45\left(\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
It means projectile is at maximum height at $t=$ $2 \mathrm{~s}$. At maximum height vertical component of velocity is zero.
$\begin{aligned}
& \text { Velocity at } \mathrm{t}=2 \mathrm{~s}=\mathrm{V}_{\mathrm{x}}=\mathrm{u} \cos \theta=40 \cos 30^{\circ} \\
& =20 \sqrt{3} \mathrm{~ms}^{-1} \text {. } \\
&
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A projectile is projected with the velocity of $(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The horizontal range of the projectile will be

Options:
A
$1.2 \mathrm{~m}$

B
$2.4 \mathrm{~m}$

C
$3.6 \mathrm{~m}$

D
$4.5 \mathrm{~m}$

The correct answer is:

$2.4 \mathrm{~m}$

Solution:
Given, $\mathbf{v}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
$$
\therefore \quad v=|\mathrm{v}|=\sqrt{3^2+4^2} \Rightarrow v=5 \mathrm{~m} / \mathrm{s}
$$
From figure,

$\sin \theta=\frac{4}{5}$ and $\cos \theta=\frac{3}{5}$
$\because \quad R=\frac{v^2 \sin 2 \theta}{g}$
$=\frac{v^2 \cdot 2 \sin \theta \cdot \cos \theta}{g}$
$=\frac{5 \times 5 \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10}=24 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2022
A projectile projected at an angle $30^{\circ}$ from the horizontal has a range $\mathrm{R}$. If the angle of projection at the same initial velocity be $60^{\circ}$, then the range will be-

Options:
A $\mathrm{R}$
B $\mathrm{R} / 2$
C $2 \mathrm{R}$
D $\mathrm{R}^{2}$

The correct answer is:
$\mathrm{R}$

Solution:
If sum of angle of projection $=90^{\circ}$ for given speed then range for that angle of projection is same.

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2009
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:

Options:
A
4 : 1
B
2 : 1
C
8 : 1
D
16 : 1

The correct answer is:
4 : 1

Solution:
De-Broglie wavelength,
$\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}} \\
& \therefore \frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{2 \mathrm{~m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{2 \mathrm{~m}_\alpha \mathrm{V}_\alpha \mathrm{q}_\alpha}} \\
& \because \mathrm{V}_{\mathrm{p}}=2 \mathrm{~V} \text { and } \mathrm{V}_\alpha=4 \mathrm{~V} \text { (given) } \\
& \mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}} \text { and } \mathrm{q}_\alpha=2 \mathrm{q}_{\mathrm{p}}
\end{aligned}$
$\therefore \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4} \Rightarrow \lambda_p: \lambda_\alpha=4: 1$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A proton is revolving on a circular path of radius 2 mm with frequency 10 Hz. Magnetic dipole moment associated with proton is

Options:
A 2×10-24 A-m2
B 4×10-24 A-m2
C 3×10-20 A-m2
D 6×10-20 A-m2

The correct answer is:
2×10-24 A-m2

Solution:
Given, charge on proton, q=e
=1.6×10-19 C

Radius, r=2 mm=2×10-3 cm

Frequency, f=10 Hz

Current associated by proton is given by

I=qt [T→ time period ]

=qf ∵T=1f

=1.6×10-19×10=1.6×10-18 A

∴ Magnetic dipole moment, M=IA

=I·πr2=1.6×10-18×3.14×2×10-32
=2×10-24 A-m2

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2019
A proton moving with a velocity $3 \times 10^{5} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of $0.3$ tesla at an angle of $30^{\circ}$ with the field. The radius of curvature of its path will be $(\mathrm{e} / \mathrm{m}$ for proton $=10^{8} \mathrm{C} / \mathrm{kg}$ )

Options:
A $2 \mathrm{~cm}$
B $0.5 \mathrm{~cm}$
C $0.02 \mathrm{~cm}$
D $1.25 \mathrm{~cm}$

The correct answer is:
$0.5 \mathrm{~cm}$

Solution:
$r=\frac{m v \sin \theta}{\mathrm{B} e}=\frac{3 \times 10^{5} \sin 30^{\circ}}{0.3 \times 10^{8}}$

$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}}=0.5 \times 10^{-2}$

$m=0.5 \mathrm{~cm}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2021
A pulley of radius 2 m is rotated about its axis by a force =20t-5t2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotation made by the pulley before its direction of motion is reversed is

Options:
A more than 3 but less than 6
B more than 6 but less than 9
C more than 9
D less than 3

The correct answer is:
more than 3 but less than 6

Solution:
We have,

Torque =r×F=lα

or 220t-5t2=10α

α=4t-t2

But α=dωdt, so

dωdt=4t-t2

dω=4t-t2dt

Integrating ω=2t2-t33

ω will be zero at t=6 s

So, ω=dθdt=2t2-t33

or dθ=2t2-t33dt

Again, integrating both sides
θ=2t33-t412

since. t=6 s

so, θ=2×633-6412=36

Number of turns, n=θ2π=362π=5.73

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2018
A radioactive element X converts into another stable element Y. Half life of $\mathrm{X}$ is $2 \mathrm{hrs}$. Initially only $\mathrm{X}$ is present. After time $\mathrm{t},$ the ratio of atoms of $X$ and $Y$ is found to be $1: 4,$ then $t$ in hours is

Options:
A 2
B 4
C between 4 and 6
D 6

The correct answer is:
between 4 and 6

Solution:
Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}=0$

Then at $\mathrm{t}=4$ hrs (two half lives)

$\mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_{0}}{4}$ and $\mathrm{N}_{\mathrm{y}}=\frac{3 \mathrm{~N}_{0}}{4}$

$\therefore \mathrm{N}_{\mathrm{x}} / \mathrm{N}_{\mathrm{y}}=1 / 3$

and at $\mathrm{t}=6 \mathrm{hrs}$ (three half lives)

$\mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_{0}}{8}$ and $\mathrm{N}_{\mathrm{y}}=\frac{7 \mathrm{~N}_{0}}{8}$

or $\frac{\mathrm{N}_{\mathrm{x}}}{\mathrm{N}_{\mathrm{y}}}=\frac{1}{7}$

The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$ Therefore, t lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$.

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2016
A radioactive isotope having a half - life period of 3 days was received after 12 days. If $3 \mathrm{~g}$ of the isotope is left in the container, what would be the initial mass of the isotope?

Options:
A $12 \mathrm{~g}$
B $36 \mathrm{~g}$
C $48 \mathrm{~g}$
D $24 \mathrm{~g}$

The correct answer is:
$48 \mathrm{~g}$

Solution:
Given $t_{1 / 2}=3$ Total time $\mathrm{T}=12$

No. of half lives $(n)=\frac{12}{3}=4$ $\left(\frac{1}{2}\right)^{n}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}

\therefore\left(\frac{1}{2}\right)^{4}=\frac{3}{\mathrm{~N}}$

$\frac{3}{\mathrm{~N}}=\frac{1}{16}$

$N=48$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2020
A radio-active material is reduced to $1 / 8$ of its original amount in 3 days. If $8 \times 10^{-3} \mathrm{~kg}$ of the material is left after 5 days. The initial amount of the material is

Options:
A $700 \mathrm{gm}$
B $900 \mathrm{gm}$
C $475 \mathrm{gm}$
D $256 \mathrm{gm}$

The correct answer is:
$256 \mathrm{gm}$

Solution:
$\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\mathrm{N}=\frac{\mathrm{N}_0}{8}$
$\frac{\mathrm{N}_0}{8}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\mathrm{n}=3$
3 half lives $=3$ days
1 half life $=1$ day
5 days $=5$ half life
$\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 8 \times 10^{-3}=\mathrm{N}_0\left(\frac{1}{2}\right)^5$
$\Rightarrow \quad N_0=2^5 \times 8 \times 10^{-3}=256 \mathrm{gm}$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A radioactive sample of half life 23.1 days is disintegrating continuously. The percentage decay of the sample between the 15th and 16th day will be [Take, e0.03=1.03]

Options:
A 5 %
B 1 %
C 2.9 %
D 3.5 %

The correct answer is:
2.9 %

Solution:
Given, half life T1/2=23.1 days

If N0 be the initial amount, then active nuclei at t=15 days

N1=N0e-15λ ...i

Active nuclei at t=16 days

N2=N0e-16λ ...ii

Disintegration constant,

λ=0.693T1/2=0.69323.1=0.03

∴ % decay in 15th to 16th days

=N1-N2N1×100=N0e-15λ-N0e-16λN0e-15λ×100

=1-e-λ×100=1-e-0.03×100

=1-1e0.03×100=1-11.03×100=2.9 %

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2019
A ray of light coming from the point (1,2) is reflected at a point A on the $x$ -axis and then passes through the point $(5,3) .$ The co-ordinates of the point $A$ is

Options:
A $\left(\frac{13}{5}, 0\right)$
B $\left(\frac{5}{13}, 0\right)$
C (-7,0)
D None of these

The correct answer is:
$\left(\frac{13}{5}, 0\right)$

Solution:
No Solution

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2016
A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of $\mu_{\mathrm{g}}$ would be

Options:
A $(4 / 3) \sin i$
B $1 / \sin i$
C $4 / 3$
D 1

The correct answer is:
$1 / \sin i$

Solution:
$\mu_{g} \sin i=\mu_{\text {air }} \sin 90^{\circ} \Rightarrow \mu_{g}=\frac{1}{\sin i}$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2012
A ray of light of intensity I is incident on a parallel glass slab at point $A$ as shown in diagram. It undergoes partial reflection and refraction. At each reflection, $25 \%$ of incident energy is reflected. The rays $\mathrm{AB}$ and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ undergo interference. The ratio of $\mathrm{I}_{\max }$ and $\mathrm{I}_{\min }$ is :

Options:
A 49: 1
B 7: 1
C 4: 1
D 8: 1

The correct answer is:
49: 1

Solution:
From figure $\mathrm{I}_{1}=\frac{\mathrm{I}}{4}$ and $\mathrm{I}_{2}=\frac{9 \mathrm{I}}{64}$

$\begin{aligned} \text { By using } \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }} &=\left(\frac{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}+1}{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}-1}\right)^{2} \\ &=\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)^{2}=\frac{49}{1} \end{aligned}$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2016
A ray parallel to principal axis is incident at $30^{\circ}$ from normal on concave mirror having radius of curvature $R$. The point on principal axis where rays are focussed is $Q$ such that $\mathrm{PQ}$ is

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

The correct answer is:
$R\left(1-\frac{1}{\sqrt{3}}\right)$

Solution:
From similar triangles,

$$

\frac{Q C}{\sin 30^{\circ}}=\frac{R}{\sin 120^{\circ}}

$$

or $\quad Q C=R \times \frac{\sin 30^{\circ}}{\sin 120^{\circ}}=\frac{R}{\sqrt{3}}$

Thus $P Q=P C-Q C=R-\frac{R}{\sqrt{3}}=R\left(1-\frac{1}{\sqrt{3}}\right)$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2015
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then

Options:
A T∝ρ
B T∝1A
C T∝1ρ
D T∝1m

The correct answer is:
T∝1A

Solution:
Let block is displaced through x m, then weight of displaced water or upthrust (upwards)

=-Axρg

Where, A is area of a cross-section of the block and $\rho$ is density. This must be equal to force =ma applied where m is mass of the block and a is acceleration.

∴ ma=-Axρg

or a=-Aρgmx=-ω2x

This is equation of SHM.

The time period of oscillation,

T=2πω=2πmAρg

⇒ T∝1A

Topic: Physics, Oscillations, BITSAT, BITSAT 2017
$[\mathrm{A}] \stackrel{\text { reduction }}{\longrightarrow}[\mathrm{B}] \stackrel{\mathrm{CHCl}_{3}+\mathrm{KOH}}{\longrightarrow}[\mathrm{C}] \stackrel{\text { reduction }}{\longrightarrow} \mathrm{N}-$ Methylaniline, A is?

Options:
A Formaldehyde
B Trichloromethane
C Nitrobenzene
D Toluene

The correct answer is:
Nitrobenzene

Solution:

Topic: Chemistry, Amines, BITSAT, BITSAT 2021
A reservoir is at 827 °C and Carnot's engine takes a thousand kilocalorie of heat from it and exhausts it to a sink at 27 °C. What is the amount of work and the efficiency of the engine?

Options:
A 2.7×105 cal, 70.70 %
B 7.27×105 cal, 72.72 %
C 2.70×105 cal, 80.70 %
D 3.70×105 cal, 70.70 %

The correct answer is:
7.27×105 cal, 72.72 %

Solution:
Given, Q1=106 cal

T1=827 °C=(827+273)=1100 K,

T2=27 °C=(27+273)=300 K.

As, Q1T1=Q2T2

∴ Q2=T2T1Q1=3001100×106=2.72×105 cal
work done, W=Q1-Q2=106-2.72×105=7.27×105 cal

The efficiency of the engine,
η=1-T2T1×100

⇒η=1-3001100×100

=72.72 %

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2018
A resistance $\mathrm{R}$ and inductance $\mathrm{L}$ and a capacitor $\mathrm{C}$ all are connected in series with an AC supply. The resistance of $\mathrm{R}$ is 16 ohm and for a given frequency, the inductive reactance of $L$ is 24 ohm and capacitive reactance of $\mathrm{C}$ is $12 \mathrm{ohm} .$ If the current in the circuit is 5 amp., find the potential difference across $\mathrm{R}, \mathrm{L}$ and $\mathrm{C}$.

Options:
A 30,20,50 volt
B 40,100,60 volt
C 70,110,60 volt
D 80,120,60 volt

The correct answer is:
80,120,60 volt

Solution:
$\mathrm{V}_{\mathrm{R}}=\mathrm{i} \mathrm{R}=5 \times 16=80 \mathrm{Volt}$
$$
\begin{array}{l}
\mathrm{V}_{\mathrm{L}}=\mathrm{i} \times(\omega \mathrm{L})=5 \times 24=120 \mathrm{Volt} \\
\mathrm{V}_{\mathrm{C}}=\mathrm{i} \times(\mathrm{l} / \omega \mathrm{C})=5 \times 12=60 \mathrm{Volt}
\end{array}
$$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2010
A resistor of resistance $R$, capacitor of capacitance $C$ and inductor of inductance $L$ are connected in parallel to AC power source of voltage $\varepsilon_{0} \sin \omega t$. The maximum current through the resistance is half of the maximum current through the power source. Then value of $\mathrm{R}$ is

Options:
A $\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}$
B $\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right|$
C $\sqrt{5}\left|\frac{1}{\omega C}-\omega L\right|$
D None of these

The correct answer is:
$\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}$

Solution:
$\frac{i_{R_{0}}}{\sqrt{\left(i_{R_{0}}\right)^{2}+\left(i_{c_{0}}-i_{L_{0}}\right)^{2}}}=\frac{1}{2}$

$\Rightarrow \frac{\varepsilon_{0} / R}{\sqrt{\left(\varepsilon_{0} / R\right)^{2}+\left(\varepsilon_{0} \omega C-\frac{\varepsilon_{0}}{\omega L}\right)^{2}}}=\frac{1}{2}$

$\Rightarrow R=\frac{\sqrt{3}}{\left(\omega C-\frac{1}{\omega L}\right)}$

Topic: Physics, Alternating Current, JEE Main
A resistor of resistance $R$, capacitor of capacitance $C$ and inductor of inductance $L$ are connected in parallel to AC power source of voltage $\varepsilon_{0} \sin \omega t$. The maximum current through the resistance is half of the maximum current through the power source. Then value of $\mathrm{R}$ is

Options:
A $\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}$
B $\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right|$
C $\sqrt{5}\left|\frac{1}{\omega C}-\omega L\right|$
D None of these

The correct answer is:
$\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}$

Solution:
$\frac{i_{R_{0}}}{\sqrt{\left(i_{R_{0}}\right)^{2}+\left(i_{c_{0}}-i_{L_{0}}\right)^{2}}}=\frac{1}{2}$

$\Rightarrow \frac{\varepsilon_{0} / R}{\sqrt{\left(\varepsilon_{0} / R\right)^{2}+\left(\varepsilon_{0} \omega C-\frac{\varepsilon_{0}}{\omega L}\right)^{2}}}=\frac{1}{2}$

$\Rightarrow R=\frac{\sqrt{3}}{\left(\omega C-\frac{1}{\omega L}\right)}$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2016
A resistor R, an inductor L, a capacitor C and voltmeters V1,V2 and V3 are connected to an oscillator in the circuit as shown in the adjoining diagram. When the frequency of the oscillation is increased, then at the resonant frequency, the voltmeter reading is zero in the case of

Options:
A voltmeter V1 only
B voltmeter V2 only
C voltmeter V3 only
D All the three voltmeters

The correct answer is:
voltmeter V2 only

Solution:
At resonance, XL=XC or ωL=1ωC

Voltage across the series L-C combination,

V2=iXL-XC=0

Topic: Physics, Alternating Current, BITSAT, BITSAT 2018
A rifle man, who together with his rifle has a mass of $100 \mathrm{~kg}$, stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass $10 \mathrm{~g}$ and a muzzle velocity of $800 \mathrm{~ms}^{-1}$. The velocity which the rifle man attains after firing 10 shots is

Options:
A $8 \mathrm{~ms}^{-1}$
B $0.8 \mathrm{~ms}^{-1}$
C $0.08 \mathrm{~ms}^{-1}$
D $-0.8 \mathrm{~ms}^{-1}$

The correct answer is:
$0.8 \mathrm{~ms}^{-1}$

Solution:
According to law of conservation of momentum,

$$

\begin{array}{l}

100 \mathrm{v}=-\frac{10}{1000} \times 10 \times 800 \\

\text { i.e., } \mathrm{v}=0.8 \mathrm{~ms}^{-1}

\end{array}

$$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2014
A round disc of moment of inertia $\mathrm{I}_{2}$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $\mathrm{I}_{1}$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is

Options:
A $\frac{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega}{\mathrm{I}_{1}}$
B $\frac{\mathrm{I}_{2} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$
C $\omega$
D $\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$

The correct answer is:
$\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$

Solution:
Total angular momentum of the system initially \(\mathrm{L}_{\mathrm{i}}=\mathrm{I}_{1} \mathrm{w}+\mathrm{I}_{2}(0)=\mathrm{I}_{1} \mathrm{w}\)

Total angular momentum of the system finally \(L_{f}=\left(I_{1}+I_{2}\right) w_{2}\)

According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)

\(\begin{aligned}

&\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2} \\

&\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}

\end{aligned}\)

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2021