datavorous/entrance-exam-dataset huggingface dataset:
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

Options:
A $\frac{3 K \varepsilon_{0} A}{4 d}$
B $\frac{4 K \varepsilon_{0} A}{3 d}$
C $\frac{(K+1) \varepsilon_{0} A}{2 d}$
D $\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$

The correct answer is:
$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$

Solution:
$$

c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d}

$$

$$

\therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)}

$$

$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )

Topic: Physics, Capacitance, BITSAT, BITSAT 2016
A parallel plate capacitor of capacitance $\mathrm{C}$ is connected to a battery and is charged to a potential difference $\mathrm{V}$. Another capacitor of capacitance $2 \mathrm{C}$ is similarly charge to a potential difference $2 \mathrm{~V}$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Options:
A Zero
B $\frac{3}{2} \mathrm{CV}^{2}$
C $\frac{25}{6} \mathrm{CV}^{2}$
D $\frac{9}{2} \mathrm{CV}^{2}$

The correct answer is:
$\frac{3}{2} \mathrm{CV}^{2}$

Solution:
From the figure. The net charge shared between the capacitors

$$

Q^{\prime}=Q_{2}-Q_{2}=4 C V-C V=3 C V

$$

The two capacitors will have some potential, say $\mathrm{V}^{\prime}$. The net capacitance of the parallel combination of the two capacitors

$$

C^{\prime}=C_{1}+C_{2}=C+2 C+3 C

$$

The potential of the capacitors

$$

\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{3 \mathrm{CV}}{3 \mathrm{C}}=\mathrm{V}

$$

The electrostatic energy of the capacitors

$$

\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{\prime 2}=\frac{1}{2}(3 \mathrm{C}) \mathrm{V}^{2}=\frac{3}{2} \mathrm{CV}^{2}

$$

Topic: Physics, Capacitance, BITSAT, BITSAT 2015
A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between the plates is d. The space between the plates is now filled with two dielectrics constant K1=3 and thickness d3 while the other one has dielectric constant K2=6 and thickness 2d3. Capacitance of the capacitor is now

Options:
A 1.8 pF
B 45 pF
C 40.5 pF
D 20.25 pF

The correct answer is:
40.5 pF

Solution:
The two capacitors formed are in parallel, hence capacitance of the combination

C=C1C2C1+C2 ...(i)

where, C1=K1ε0Ad3 ∵C=ε0Ad ...(ii)

C2=K2ε0A2d3 ...(iii)

Ceq=3K1ε0Ad×K2ε0A·32d3K1ε0Ad+3K2ε0A2d

=9K1K2ε02A22d2×dε0A×18

It is given

ε0Ad=9 pF

Using given values

Ceq=40.5 pF

Topic: Physics, Capacitance, BITSAT, BITSAT 2017
A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is

Options:
A $\sqrt{2 \mathrm{~A}}$
B $2 \mathrm{~A}$
C $\frac{1}{\sqrt{2}} \mathrm{~A}$
D $\frac{1}{2} \mathrm{~A}$

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

Solution:
Let the distance from the mean position is $\mathrm{X}$.
Given $\mathrm{KE}=\mathrm{PE}$
So, $\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2$
$A^2-x^2=x^2 \Rightarrow A^2=2 \times 2$
$\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A particle executing simple harmonic motion along $y$ -axis has its motion described by the equation $y=A \sin (\omega t)+B$. The amplitude of the simple harmonic motion is

Options:
A $\mathrm{A}$
B $\mathrm{B}$
C $\mathrm{A}+\mathrm{B}$
D $\sqrt{A+B}$

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

Solution:
The amplitude is a maximum displacement from the mean position.

Topic: Physics, Oscillations, BITSAT, BITSAT 2020
A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in $t^{\text {th }}$ and $(t+1)^{\text {th }}$ seconds is $100 \mathrm{~cm}$, then its velocity after $t$ seconds, in $\mathrm{cm} / \mathrm{s}, \mathrm{is}$

Options:
A 80
B 50
C 20
D 30

The correct answer is:
50

Solution:
No Solution available

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2011
A particle moving along X-axis has acceleration f, at time t given byf=f01-tT, where f0 and T are constants. The particle at t=0 and the instant when f=0, the particle's velocity vx is

Options:
A f0T
B 12f0T2
C f0T2
D 12f0T

The correct answer is:
12f0T

Solution:
Acceleration

f=dvdt=f01+tT

or dv=f01-tT·dt ......... Eq(i)

Integrating Eq. (i) on both sides, we get

v=f0t-f0T·t22+C

After applying boundary conditions

v=0 at t=0

We get, C=0

⇒ v=f0t-f0T·t22 .......... Eq(ii)

As f=f01-tT

When f0=0,t=T

Substituting t=T in Eq. (ii), then velocity

vx=f0T-f0T·T22=12f0T

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2017
A particle of mass $2 \mathrm{~m}$ is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $20 \sqrt{2} \mathrm{~m} / \mathrm{s}$. After $1 \mathrm{~s}$, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is

Options:
A $50 \mathrm{~m}$
B $25 \mathrm{~m}$
C $40 \mathrm{~m}$
D $35 \mathrm{~m}$

The correct answer is:
$35 \mathrm{~m}$

Solution:
Given : Initial velocity $u_{0}=20 \sqrt{2} \mathrm{~m} / \mathrm{s} ;$ angle of projection $\theta=45^{\circ}$

Therefore horizontal and vertical components of initial velocity are $\mathrm{u}_{\mathrm{x}}=20 \sqrt{2} \cos 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$

and $\mathrm{u}_{\mathrm{y}}=20 \sqrt{2} \sin 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$

After $1 \mathrm{~s}$, horizontal component remains unchanged while the vertical component becomes $v_{y}=u_{y}-g t$

Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second

part will become $v_{y}^{\prime}=20 m / s$ Therefore, maximum height attained by the second part will be $\mathrm{H}=\mathrm{h}_{1}+\mathrm{h}_{2}$

Using, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$\Rightarrow \mathrm{h}_{1}=(20 \times 1)-\frac{1}{2} \times 10 \times(1)^{2}=15 \mathrm{~m}$

$\mathrm{a}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$

$\mathrm{h}_{2}=\frac{\mathrm{v}_{\mathrm{y}}^{\prime 2}}{2 \mathrm{~g}}=\frac{(20)^{2}}{2 \times 10}=20 \mathrm{~m}$

$\mathrm{H}=20+15=35 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2015
A particle of mass $10 \mathrm{~kg}$ moving eastwards with a speed $5 \mathrm{~ms}^{-1}$ collides with another particle of the same mass moving north-wards with the same speed $5 \mathrm{~ms}^{-1}$. The two particles coalesce on collision. The new particle of mass $20 \mathrm{~kg}$ will move in the north-east direction with velocity

Options:
A $10 \mathrm{~ms}^{-1}$
B $5 \mathrm{~ms}^{-1}$
C $\quad(5 / \sqrt{2}) \mathrm{ms}^{-1}$
D none of these

The correct answer is:
$\quad(5 / \sqrt{2}) \mathrm{ms}^{-1}$

Solution:
Here $\hat{i} m v+\hat{j} m v=2 m \vec{V}$
That is $\overrightarrow{\mathrm{V}}=\frac{\mathrm{v}}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
Hence $V=\frac{v}{2} \times \sqrt{2}=\frac{v}{\sqrt{2}} \cdot\left[\right.$ Here $\left.v=5 \mathrm{~ms}^{-1}\right]$
$$
\text { So, } V=\frac{5}{\sqrt{2}} \mathrm{~ms}^{-1}
$$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2009
A particle of mass ' $m$ ' is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' $h$ ' is

Options:
A $\frac{\sqrt{3}}{2} \frac{m v^{2}}{g}$
B zero
C $\frac{m v^{3}}{\sqrt{2 g}}$
D $\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$

The correct answer is:
$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$

Solution:
Angular momentum $L_{0}=p r_{\perp}$ $\left(\because\right.$ linear momentum $p=m v \cos \theta$ and $\left.\mathrm{r}_{1}=H\right)$

$\Rightarrow L_{0}=m v \cos \theta H$

$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$

$=\frac{\sqrt{3} m v^{3}}{16 g}$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass ' $m$ ' is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' $h$ ' is

Options:
A $\frac{\sqrt{3}}{2} \frac{m v^{2}}{g}$
B zero
C $\frac{m v^{3}}{\sqrt{2 g}}$
D $\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$

The correct answer is:
$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$

Solution:
Angular momentum $L_{0}=p r_{\perp}$ $\left(\because\right.$ linear momentum $p=m v \cos \theta$ and $\left.\mathrm{r}_{1}=H\right)$

$\Rightarrow L_{0}=m v \cos \theta H$

$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$

$=\frac{\sqrt{3} m v^{3}}{16 g}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2013
A particle of mass $m$ is projected with velocity $v$ at an angle $\theta$ with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original trajectory, then the velocity of second piece is

Options:
A
$2 v \cos \theta$

B
$v \cos \theta$

C
$3 v \cos \theta$

D
$\frac{v}{2} \cos \theta$

The correct answer is:

$v \cos \theta$

Solution:
According to the law of conservation of linear momentum at the highest point,
$$
m v \cos \theta=\frac{m}{2}(v \cos \theta)+\frac{m}{2} v^{\prime}
$$
$\Rightarrow \quad m v \cos \theta-\frac{m}{2} v \cos \theta=\frac{m}{2} v^{\prime}$
$\Rightarrow \quad v^{\prime}=2\left(v \cos \theta-\frac{v \cos \theta}{2}\right)=v \cos \theta$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2022
A particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2 \mathrm{~m}$. After collision, they stick together and continue to move together with velocity

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

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

Solution:
Applying conservation of linear momentum
$\Rightarrow \overrightarrow{\mathrm{P}}_{\mathrm{i}}=\overrightarrow{\mathrm{P}}_{\mathrm{f}}(\because \mathrm{P}=\mathrm{mv})$
$m v_1+2 m v_2=(m+2 m) v^{\prime}$
$m v^1+2 m \times 0=(3 m) v^{\prime}$
$\Rightarrow \mathrm{mv}=3 \mathrm{mv}^{\prime} \Rightarrow \mathrm{v}^{\prime}=\frac{\mathrm{v}}{3}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A particle starts moving rectilinearly at time $\mathrm{t}=0$ such that its velocity $\mathrm{v}$ changes with time $\mathrm{t}$ according to the equation $v=t^{2}-t$ where $t$ is in seconds and $v$ is in $\mathrm{m} / \mathrm{s}$. Find the time interval for which the particle retards.

Options:
A $\frac{1}{2}<\mathrm{t}<1$
B $\frac{1}{2}>t>1$
C $\frac{1}{4}<\mathrm{t}<1$
D $\frac{1}{2}

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

Solution:
Acceleration of the particle $a=2 t-1$ The particle retards when acceleration is opposite to velocity.
$$
\Rightarrow \text { a } \cdot v<0 \Rightarrow(2 t-1)\left(t^{2}-t\right)<0 \Rightarrow t(2 t-1)
$$
$(\mathrm{t}-1)<0$
Now t is always positive
$$
\therefore(2 t-1)(t-1)<0
$$
or $2 \mathrm{t}-1<0$ and $\mathrm{t}-1>0 \Rightarrow \mathrm{t}<\frac{1}{2}$ and $\mathrm{t}>1$.
This is not possible or $2 \mathrm{t}-1>0 \& \mathrm{t}-1<0 \Rightarrow 1 / 2<\mathrm{t}<1$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2009
A particle X of mass m and initial velocity u collides with another particle Y of mass 3m4, which is at rest. The collision is head-on and perfectly elastic. The ratio of de-Broglie wavelengths λY and λX after the collision is,

Options:
A 4:3
B 2:32
C 3:4
D 1:6

The correct answer is:
1:6

Solution:
For perfectly elastic collision,

total momentum before collision = total momentum after collision

i.e. mu+3m4×0=mvx+3m4vy

⇒4u=4vx+3vy ...i

Coefficient of restitution,

e=Velocity of separation after collision Velocity of approach before collision = vy-vxux-uy

⇒1=vy-vxu-0 [∵ ux=u and uy=0]

u=vy-vx ...ii

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

vx=u7 and vy=8u7

∴ λyλx=h34mvyhm×vx=43×vxvy=43u78u7=16

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2019
A passenger in a open car travelling at $30 \mathrm{~m} / \mathrm{s}$ throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is $20 \mathrm{~m} / \mathrm{s}$ at $60^{\circ}$ to the horizontal. The angle of projection of the ball with respect to the horizontal road will be

Options:
A $\tan ^{-1}\left(\frac{2}{3}\right)$
B $\tan ^{-1}\left(\frac{\sqrt{3}}{4}\right)$
C $\tan ^{-1}\left(\frac{4}{\sqrt{3}}\right)$
D $\tan ^{-1}\left(\frac{3}{4}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Solution:
$$\begin{array}{l}
\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{x}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{x}} \\
20 \cos 60^{\circ}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-30 \\
\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}=40 ;\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{y}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{y}} \\
20 \sin 60^{\circ}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-0 \\
\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}=10 \sqrt{3} ; \tan \theta=\frac{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}}{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}}=\frac{10 \sqrt{3}}{40}=\frac{\sqrt{3}}{4}
\end{array}
$$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2011
A passenger sitting in a train A moving at $90 \mathrm{~km} /$ $h$ observes another train B moving in the opposite direction for $8 \mathrm{~s}$. If the velocity of the train B is $54 \mathrm{~km} / \mathrm{h}$, then length of train B is :

Options:
A $80 \mathrm{~m}$
B $200 \mathrm{~m}$
C $120 \mathrm{~m}$
D $320 \mathrm{~m}$

The correct answer is:
$320 \mathrm{~m}$

Solution:
Velocity of train A
\(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)
Velocity of train B
\(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)
Velocity of train B w.r.t. train
\(\begin{aligned}
A=\vec{V}_B & -\vec{V}_A \\
& =15-(-25) \mathrm{m} / \mathrm{s} \\
& =40 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Time of crossing \(=\frac{\text { length of train }}{\text { relative velocity }}\)
\((8)=\frac{\ell}{40}\)
\(\ell=8 \times 40=320\) meter.

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A pendulum bob carries a +ve charge $+\mathrm{q} . \mathrm{A}$ positive charge $+\mathrm{q}$ is held at the point of support. Then the time period of the bob is [where, $\mathrm{L}=$ length of pendulum, $\mathrm{g}_{\mathrm{eff}}=$ effective value of $\mathrm{g}]$

Options:
A greater than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$
B less than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$
C equal to $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$
D equal to $2 \pi \sqrt{2 \mathrm{~L} / \mathrm{g}_{\text {eff }}}$

The correct answer is:
greater than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$

Solution:
Effective $g^{\prime}=g-\frac{k q^{2}}{d^{2} m}, \quad T \propto \frac{1}{\sqrt{g_{\text {eff }}}}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2009
A penicillin is a member of a family of drugs that have a

Options:
A four membered cyclic amide fused to a five membered thiazole ring.
B three membered cyclic amide fused to a five-membered thiazole ring.
C four-membered cyclic amide fused to have a four membered thiazole ring.
D five-membered cyclic amide fused to have a five membered thiazole ring.

The correct answer is:
four membered cyclic amide fused to a five membered thiazole ring.

Solution:
A penicillin is a member of a family of drugs that have a four membered cyclic amide fused to a five membered thiazole ring. It is narrow spectrum antibiotic.

Penicillin is an antibiotic or group of antibiotics produced naturally by certain blue moulds, now usually prepared synthetically.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2018
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Options:
A $\frac{(10) !}{6 !}$
B $\frac{(10) !}{24}$
C $\frac{(9) !}{24}$
D None of these

The correct answer is:
$\frac{(10) !}{24}$

Solution:
Selection of 6 guests $=10 \mathrm{C} 6$

Permutation of 6 on round table $=5 !$

Permutation of 4 on round table $=3 !$

Then, total number of arrangements $=10 \mathrm{C} 6.5 ! .3 !$

$=\frac{(10) !}{6 ! 4 !} \cdot 5 ! .3 !=\frac{(10) !}{24}$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2020
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is \(60^{\circ}\). when he retreats 20 feet from the bank, he finds the angle to be \(30^{\circ}\). The breadth of the river in feet is :

Options:
A 15
B \(15 \sqrt{3}\)
C \(10 \sqrt{3}\)
D 10

The correct answer is:
10

Solution:
et \(h\) be the height of tree PQ and breadth of river PS be \(x \mathrm{~ft}\). Angle of elevation subtended by a tree is \(60^{\circ}\). Also, when he retreats 20 feet, the angle becomes \(30^{\circ}\).
Also, in \(\triangle \mathrm{PQS}\),
\(\begin{aligned}
& \tan 60^{\circ}=\frac{h}{x} \\
& \Rightarrow h=\sqrt{3} x
\end{aligned}\)
and in \(\triangle \mathrm{PQR}\),
\(\begin{aligned}
& \tan 30^{\circ}=\frac{h}{x+20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+20} \\
& \Rightarrow x+20=\sqrt{3} h \\
& \Rightarrow x+20=3 x \Rightarrow 2 x=20 \Rightarrow x=10
\end{aligned}\)
Hence breadth of river is \(10 \mathrm{ft}\).

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2009
A person with his hand in his pocket is skating on ice at the rate of $10 \mathrm{~m} / \mathrm{s}$ and describes a circle of radius $50 \mathrm{~m}$. What is his inclination to vertical:

$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)$

Options:
A $\tan ^{-1}(1 / 2)$
B $\tan ^{-1}(1 / 5)$
C $\tan ^{-1}(3 / 5)$
D $\tan ^{-1}(1 / 10)$

The correct answer is:
$\tan ^{-1}(1 / 5)$

Solution:
Since surface (ice) is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}$ where $\mathrm{v}$ is speed of skating $\&$

i is radius of circle in which he moves.

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2014
A person writes letter to six friends and addresses the corresponding envelopes. Let x be the numbers of ways so that at least two of the letters are in wrong envelopes and y be the numbers of ways so that all the letters are in wrong envelopes. Then x-y=

Options:
A 719
B 265
C 454
D 720

The correct answer is:
454

Solution:
Since only one letters can not be in wrong envelope so that at least two letters are in wrong envelope means all the letters are not in right envelopes.

∴ x=6!-1=720-1=719

Y= number of ways, so that all the letters are in wrong envelops

=6!1-11!+12!-13!+14!-15!+16!

=360-120+30-6+1=265

∴ x-y=719-265=454

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2019
A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in time $t$ then force exerted on the surface is ( $c=$ speed of light)

Options:
A $\frac{2 \mathrm{E}}{\mathrm{Atc}}$
B $\frac{\mathrm{E}}{2 \mathrm{c}}$
C $\frac{2 \mathrm{E}}{\mathrm{ct}}$
D zero

The correct answer is:
$\frac{2 \mathrm{E}}{\mathrm{ct}}$

Solution:
Incident momentum, $p=\frac{E}{c}$

For perfectly reflecting surface with normal incidence

$\Delta p=2 p=\frac{2 \mathrm{E}}{c} ; \mathrm{F}=\frac{\Delta p}{\Delta t}=\frac{2 \mathrm{E}}{c t}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2021
A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in timet then average pressure exerted on the surface is $(\mathrm{c}=$ speed of light $)$

Options:
A zero
B $\mathrm{E} / \mathrm{Atc}$
C $2 \mathrm{E} / \mathrm{Atc}$
D $\mathrm{E} / \mathrm{c}$

The correct answer is:
$2 \mathrm{E} / \mathrm{Atc}$

Solution:
Incident momentum, $\mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}}$ For perfectly reflecting surface with normal incidence

$\Delta \mathrm{p}=2 \mathrm{p}=\frac{2 \mathrm{E}}{\mathrm{c}}$

$\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{2 \mathrm{E}}{\mathrm{ct}}$

$\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{2 \mathrm{E}}{\mathrm{ctA}}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2020
A planoconvex lens fits exactly into a planoconcave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices $\mu_{1} \& \mu_{2}$ and $R$ is the radius of curvature of the curved surface of the lenses, then focal length of combination is

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

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

Solution:
If F be the equivalent focal length, then

$$

\begin{aligned}

\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \quad \frac{1}{F}=\left(\mu_{1}-1\right)\left(\frac{1}{\infty}+\frac{1}{R}\right) \\

+\left(\mu_{2}-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)=\frac{\mu_{1}-\mu_{2}}{R} \\

F=\frac{R}{\mu_{1}-\mu_{2}}

\end{aligned}

$$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2015
A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ1 and μ2,R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is

Options:
A R2μ1+μ2
B R2μ1-μ2
C Rμ1-μ2
D 2Rμ1-μ2

The correct answer is:
Rμ1-μ2

Solution:
Focal length of combination

1f=1f1+1f2 …i

Using lens maker's formula
1f2=μ2-11-R-1∞=-μ2-1R

1f1=μ1-11∞-1-R=μ1-1R

or 1f1=μ1-1R

1f2=-μ2-1R

Putting these values in Equation (i), we get

⇒ 1f=μ1-1R-μ2-1R

⇒ 1f=μ1-1-μ2+1R

⇒ 1f=μ1-μ2R

f=Rμ1-μ2

Topic: Physics, Ray Optics, BITSAT, BITSAT 2017
A planoconvex lens has thickness of 4 cm. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be 258 cm. The focal length of the lens is (assume thickness of lens to be negligible)

Options:
A 85 cm
B 59 cm
C 75 cm
D 7.5 cm

The correct answer is:
75 cm

Solution:
In first case refraction of the rays, takes place from a plane surface, so we can use dapp=dactual μ

3=4μ⇒μ=43

where, dapp and dactual = apparent and real depth, respectively.

Now, in second case, as from Fig. (b), refraction takes place from a spherical surface, so

μ2v-μ1u=μ2-μ1Rm

or 1(-25/8)-4/3(-4)=1-4/3-R

or 13R=13-825=175

R=25 cm

Now, using Lens' Makers formula, to calculate focal length

1f=μ-11R1-1R2

1f=43-11∞-1(-25)=175

f=75 cm

Topic: Physics, Ray Optics, BITSAT, BITSAT 2018