datavorous/entrance-exam-dataset huggingface dataset:
Two mercury drops (each of radius $\mathrm{r}$ ) merge to form a bigger drop. The surface energy of the bigger drop, if $\mathrm{T}$ is the surface tension, is

Options:
A $2^{5 / 3} \pi r^{2} T$
B $4 \pi \mathrm{r}^{2} \mathrm{~T}$
C $2 \pi^{2} T$
D $2^{8 / 3} \pi \mathrm{r}^{2} \mathrm{~T}$

The correct answer is:
$2^{8 / 3} \pi \mathrm{r}^{2} \mathrm{~T}$

Solution:
$\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3} \Rightarrow \mathrm{R}=2^{1 / 3} \mathrm{r}$
Surface energy of bigger drop,
$\mathrm{E}=4 \pi R^{2} T=4 \times 2^{2 / 3} \pi r^{2} T=2^{8 / 3} \pi r^{2} T$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2010
Two metal plates having a potential difference of 800 V are 2 cm apart. It is found that a particle of mass 1.96×10-15 kg remains suspended in the region between the plates. The charge on the particle must be (e= elementary charge)

Options:
A 2 e
B 3 e
C 6 e
D 8 e

The correct answer is:
3 e

Solution:
V=800 V

E=Vd=8002×10-2 V m-1

=4×104 V m-1

At equilibrium electrostatic force must be equal to force due to weight.

qE=mg

⇒ q=mgE

=1.96×10-15×9.84×104 coulomb

=1.96×10-15×9.84×104×1.6×10-19 elementary charges

=3 e

Topic: Physics, Electrostatics, BITSAT, BITSAT 2017
Two metallic spheres of radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are given charges of $-1 \times 10^{-2} \mathrm{C}$ and $5 \times 10^{-2} \mathrm{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Options:
A $2 \times 10^{-2} \mathrm{C}$
B $3 \times 10^{-2} C$
C $4 \times 10^{-2} C$
D $1 \times 10^{-2} C$

The correct answer is:
$3 \times 10^{-2} C$

Solution:
$\quad \frac{k x}{1 \mathrm{~cm}}=\frac{k(Q-x)}{3 \mathrm{~cm}}$
$3 x=Q-x \Rightarrow 4 x=Q$
$x=\frac{Q}{4}=\frac{4 \times 10^{-2}}{4} C=1 \times 10^{-2}$
$Q^{\prime}=\mathrm{Q}-x=3 \times 10^{-2} \mathrm{C}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2010
Two monochromatic light waves are travelling with same frequency and constant phase difference. If both waves interfere, then

Options:
A there is a gain of energy
B there is a loss of energy
C the energy is redistributed and the distribution remains constant in time
D the energy is redistributed and the distribution changes with time

The correct answer is:
the energy is redistributed and the distribution remains constant in time

Solution:
When two waves with same frequency and constant phase difference interfere to each other, then after interference, intensity of resultant wave is distributed such that it is maximum at some points [constructive interference or bright fringes] and minimum at another points. [destructive interference or dark fringes]. Hence, energy is distributed and remains constant [in the form of bright and dark fringes]

Topic: Physics, Wave Optics, BITSAT, BITSAT 2019
Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of $10.0 \mathrm{~A}$. Point $\mathrm{C}$ is a distance

$\frac{\mathrm{d}}{2}$ to the right of the $10.0 \mathrm{~A}$ current. If $\mathrm{d}=18 \mathrm{~cm}$ and I is adjusted so that the magnetic field at $\mathrm{Cis}$ zero, the value of the current $I$ is

Options:
A $10.0 \mathrm{~A}$
B $30.0 \mathrm{~A}$
C $8.0 \mathrm{~A}$
D $18.0 \mathrm{~A}$

#

The correct answer is:
$30.0 \mathrm{~A}$

Solution:
The magnetic field at $\mathrm{C}$ due to first conductor

is $\mathrm{B}_{1}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{3 \mathrm{~d} / 2}$ (since, point $\mathrm{C}$ is separated

by $\mathrm{d}+\frac{\mathrm{d}}{2}=\frac{3 \mathrm{~d}}{2}$ from 1 st conductor). The direction of field is perpendicular to the plane of paper and directed outwards. The magnetic field at $\mathrm{C}$ due to second conductor is $\mathrm{B}_{2}=\frac{\mu_{0}}{2 \pi} \frac{10}{\mathrm{~d} / 2}$ (since, point $\mathrm{C}$

is separated by $\frac{\mathrm{d}}{2}$ from 2 nd conductor) The direction of field is perpendicular to the plane of paper and directed inwards. Since, direction of $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ at point $\mathrm{C}$ is in opposite direction and the magnetic field at C is zero, therefore,

$$

\begin{array}{c}

\mathrm{B}_{1}=\mathrm{B}_{2} \\

\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{3 \mathrm{~d} / 2}=\frac{\mu_{0}}{2 \pi} \frac{10}{\mathrm{~d} / 2}

\end{array}

$$

On solving $\mathrm{I}=30.0 \mathrm{~A}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2013
Two particles $\mathrm{P}$ and $\mathrm{Q}$ describe S.H.M. of same amplitude a, same frequency $\mathrm{f}$ along the same straight line. The maximum distance between the two particles is $\mathrm{a} \sqrt{2}$. The initial phase difference between the particle is $-$

Options:
A zero
B $\pi / 2$
C $\pi / 6$
D $\pi \beta$

The correct answer is:
$\pi / 2$

Solution:
$\mathrm{x}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{1}\right), \mathrm{x}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{2}\right)$

$\Rightarrow \quad\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|=2 \mathrm{a} \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$

Tomaximize $\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|: \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right)=1$

$\Rightarrow \mathrm{a} \sqrt{2}=2 \mathrm{a} \times 1 \times \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$

$\Rightarrow \frac{1}{\sqrt{2}}=\cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right) \Rightarrow \frac{\pi}{4}=\frac{\phi_{1}-\phi_{2}}{2}$

$\Rightarrow \phi_{1}-\phi_{2}=\frac{\pi}{2}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2014
Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$, respectively. The ratio of masses of $\mathrm{X}$ and $\mathrm{Y}$ is

Options:
A $\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{1 / 2}$
B $\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)$
C $\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}$
D $\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)$

The correct answer is:
$\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}$

Solution:
When a charge particle is allowed to move in a uniform magnetic field, then it describes spiral or circular path

$$

\begin{array}{l}

\text { Centripetal force, } \frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB} \quad \therefore \\

\mathrm{v}=\left(\frac{\mathrm{qB}}{\mathrm{m}} \mathrm{R}\right) \\

\text { Hence, } \sqrt{\frac{2 \mathrm{q} \mathrm{V}}{\mathrm{m}}}=\left(\frac{\mathrm{qB}}{\mathrm{m}}\right) \mathrm{R}\left[\because \mathrm{V}=\sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}}\right] \\

\Rightarrow \mathrm{R}=\left(\frac{2 \mathrm{mV}}{\mathrm{q}}\right)^{1 / 2} \times \frac{1}{\mathrm{~B}} \\

\text { or, } \mathrm{m} \propto \mathrm{R}^{2}

\end{array}

$$

$[\because \mathrm{V}, \mathrm{q}$ and $\mathrm{B}$ are constant $]$

$$

\text { or, } \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}

$$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2015
Two planets $\mathrm{A}$ and $\mathrm{B}$ have the same material density. If the radius of $\mathrm{A}$ is twice that of $\mathrm{B}$, then the ratio of the escape velocity $\mathrm{v}_{\mathrm{A}} / \mathrm{v}_{\mathrm{B}}$ is

Options:
A 2
B $\sqrt{2}$
C $1 / \sqrt{2}$
D $1 / 2$

The correct answer is:
2

Solution:
$v_{e}=\sqrt{2 g R}=\sqrt{\frac{2 G M}{R^{2}}} R=\sqrt{\frac{2 G d \frac{4}{3} \pi R^{3}}{R^{2}}} R$
$=\mathrm{R} \sqrt{2 \mathrm{Gd} \frac{4}{3} \pi}$
as $v_{e} \propto R$ for same density, $\frac{v_{A}}{v_{B}}=2$

Topic: Physics, Gravitation, BITSAT, BITSAT 2010
Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to A is :

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

The correct answer is:
$3: 4$

Solution:
Acceleration due to gravity,
$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{4}{3} \pi \mathrm{G} \rho \mathrm{R}$
$\therefore \frac{g_2}{g_1}=\frac{\rho_2}{\rho_1} \times \frac{R_2}{R_1}=\frac{1}{2} \times 1.5=\frac{3}{4}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Two point dipoles $\mathrm{p} \hat{\mathrm{k}}$ and $\frac{\mathrm{p}}{2} \hat{\mathrm{k}}$ are located at (0,0,0) and $(1 \mathrm{~m}, 0,2 \mathrm{~m})$ respectively. The resultant electric field due to the two dipoles at the point $(1 \mathrm{~m}, 0,0)$ is

Options:
A $\frac{9 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}$
B $\frac{-7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}$
C $\frac{7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}$
D None of these

The correct answer is:
$\frac{-7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}$

Solution:
The given point is at axis of $\frac{\overrightarrow{\mathrm{P}}}{2}$ dipole and

at equatorial line of $\overrightarrow{\mathrm{P}}$ dipole so that field at given point.

Topic: Physics, Electrostatics, BITSAT, BITSAT 2014
Two projectiles $\mathrm{A}$ and $\mathrm{B}$ are thrown with the same speed but angles are $40^{\circ}$ and $50^{\circ}$ with the horizontal. Then

Options:
A A will fall earlier
B $\mathrm{B}$ will fall earlier
C both will fall at the same time
D None of these

The correct answer is:
A will fall earlier

Solution:
$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}},$ lesser is the value of $\theta$, lesser is $\sin \theta$ and hence lesser will be the time taken. Hence A will fall earlier.

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2010
Two radioactive elements $A$ and $B$ initially have same number of atoms. The half life of $A$ is same as the average life of $B$. If $\lambda_A$ and $\lambda_B$ are decay constants of $A$ and $B$ respectively, then choose the correct relation from the given options.

Options:
A $\lambda_A=\lambda_B$
B $\lambda_A=2 \lambda_B$
C $\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$
D $\lambda_{\mathrm{A}} \ln 2=\lambda_{\mathrm{B}}$

The correct answer is:
$\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$

Solution:
It is given that
$\begin{aligned}
& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\
& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\
& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2
\end{aligned}$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Two resistances are given as $R_1=(10 \pm$ $0.5) \Omega$ and $R_2=(15 \pm 0.5) \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Options:
A
6.33
B
2.33
C
4.33
D
5.33

The correct answer is:
4.33

Solution:
Equivalent resistance in parallel combination is given by
$\begin{aligned}
& \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{10 \times 15}{10+15}=6
\end{aligned}$
Differentiating both sides, we get
$\begin{aligned} & \frac{\Delta R_{\text {eq }}}{\mathrm{R}_{\mathrm{eq}}^2}=\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2} \\ & \Rightarrow \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}}=\left(\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2}\right) \mathrm{R}_{\mathrm{eq}} \\ & =\left(\frac{0.5}{100}+\frac{0.5}{225}\right) \cdot 6=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300} \\ & \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}} \times 100=\frac{13}{3}=4.33 \%\end{aligned}$

Topic: Physics, Mathematics in Physics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Two resistances at $0^{\circ} \mathrm{C}$ with temperature coefficient of resistance $\alpha_{1}$ and $\alpha_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

Options:
A $\alpha_{1}+\alpha_{2}$
B $\frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$
C $\frac{\alpha_{1}-\alpha_{2}}{2}$
D $\frac{\alpha_{1}+\alpha_{2}}{2}$

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

Solution:
$\begin{aligned} R_{1} &=R_{0}\left(1+\alpha_{1} t\right)+R_{0}\left(1+\alpha_{2} t\right) \\ &=2 R_{0}\left(1+\frac{\alpha_{1}+\alpha_{2}}{2} t\right) \end{aligned}$

Comparing with $R=R_{0}(1+\alpha t)$

$\alpha=\frac{\alpha_{1}+\alpha_{2}}{2}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2016
Two rings of radius $\mathrm{R}$ and $\mathrm{nR}$ made of same material have the ratio of moment of inertia about an axis passing through centre is 1: 8 . The value of $\mathrm{n}$ is

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

The correct answer is:
2

Solution:
Ratio of moment of inertia of the rings

$$

\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right)\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}

$$

$\begin{aligned}=&\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}=\left(\frac{2 \pi \mathrm{R}}{2 \pi \mathrm{n} \mathrm{R}}\right)\left(\frac{\mathrm{R}}{\mathrm{n} \mathrm{R}}\right)^{2} \\ &(\lambda=\text { linear density of wire }=\mathrm{constant}) \\ \Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{\mathrm{n}^{3}}=\frac{1}{8} \quad \text { (given) } \\ \therefore \quad \mathrm{n}^{3}=8 \Rightarrow \mathrm{n}=2 \end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
Two rings of radius $\mathrm{R}$ and $\mathrm{nR}$ made of same material have the ratio of moment of inertia about an axis passing through centre is 1: 8 . The value of $\mathrm{n}$ is

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

The correct answer is:
2

Solution:
Ratio of moment of inertia of the rings

$$

\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right)\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}

$$

$\begin{aligned}=&\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}=\left(\frac{2 \pi \mathrm{R}}{2 \pi \mathrm{n} \mathrm{R}}\right)\left(\frac{\mathrm{R}}{\mathrm{n} \mathrm{R}}\right)^{2} \\ &(\lambda=\text { linear density of wire }=\mathrm{constant}) \\ \Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{\mathrm{n}^{3}}=\frac{1}{8} \quad \text { (given) } \\ \therefore \quad \mathrm{n}^{3}=8 \Rightarrow \mathrm{n}=2 \end{aligned}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2013
Two separate soap bubbles of radii 3×10-3 m and 2×10-3 m respectively, formed of same liquid (surface tension 65×10-2 N m-1) come together to form a double bubble. The radius of interface of double bubble is

Options:
A 6×10-3 m
B 4×10-3 m
C 15×10-3 m
D 0.66×10-3 m

The correct answer is:
6×10-3 m

Solution:
Excess pressure inside first bubble of radius r1

p1=p0+4Tr1

Excess pressure inside second bubble of radius r2
p2=p0+4Tr2

Excess pressure inside double bubble,

p=p2-p1

p=4Tr1-r2r1r2

If R be the radius of double bubble, then

p=4TR

or 4Tr1-r2r1r2=4TR

⇒R=r1r2r1-r2=3×10-3×2×10-33-2×10-3=6×10-3 m

R=6×10-3 m

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2019
Two simple pendulums of lengths 5 m and 20 m respectively are given small displacement in one direction at the same time they will again be in the same sense when the pendulum of shorter length has completed n oscillations. Then, n is

Options:
A 5
B 1
C 2
D 3

The correct answer is:
2

Solution:
According to question

n1T1=n2T2

⇒ n·2π5g=n-12π20g

nn-1=2⇒n=2

Topic: Physics, Oscillations, BITSAT, BITSAT 2017
Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1: 2 . The ratio of self-inductance of two solenoids is

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

The correct answer is:
1: 2

Solution:
From $L=\frac{\mu_{0} N^{2} A}{1} \alpha \frac{N^{2}}{1}$
we get, $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{(1 / 2)^{2}}{1 / 2}=\frac{1}{2}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2010
Two sources of sound A and B are moving towards and away from a stationary observer with the same speed respectively. If the frequency of the sound produced by both the sources is equal to 400 Hz, then the speed of sources (approximately) if the observer detects 4 beats per second, is [Given, speed of sound =340 m s-1]

Options:
A 1.7 m s-1
B 3.4 m s-1
C 2.4 m s-1
D 1 m s-1

The correct answer is:
1.7 m s-1

Solution:
Given, frequency of sound produced by sources A and B, fA=fB=400 Hz

Speed of sound in air, v=340 m s-1

Apparent frequency heard by the observer when source A is moving towards him,

fA'=vv-vS×fA=340340-vS×400

Apparent frequency heard by the observer when source B is moving away from him,

fB'=vv+vs×fB=340340+vs×400

Given, Beats =fA'-fB'=4

340340-vs×400-340340+vs×400=4

⇒34000×2vS3402-vS2=1

34000×2vS3402=1 ∵ vS2<<3402

⇒vs=1.7 m s-1

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2019
Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is 25 minute and for smaller one is 16 minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

Options:
A 4: 5
B 5: 4
C 25: 8
D 8: 25

The correct answer is:
8: 25

Solution:
Radius of small sphere $=\mathrm{r}$ Thickness of small sphere $=\mathrm{t}$ Radius of bigger sphere $=2 \mathrm{r}$ Thickness of bigger sphere $=\mathrm{t} / 4$ Mass of ice melted $=($ volume of sphere $) \times$ (density of ice) Let $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ be the thermal conductivities of larger and smaller sphere.

For bigger sphere, $\frac{K_{1} 4 \pi(2 r)^{2} \times 100}{t / 4}=\frac{\frac{4}{3} \pi(2 r)^{3} \rho L}{25 \times 60}$

For smaller sphere,

$\frac{\mathrm{K}_{2} \times 4 \pi \mathrm{r}^{2} \times 100}{\mathrm{t}}=$ $\frac{\frac{4}{3} \pi(2 \mathrm{r})^{3} \rho \mathrm{L}}{25 \times 60}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2016
Two substances $\mathrm{R}$ and $\mathrm{S}$ decompose in solution independently, both following first order kinetics. The rate constant of $\mathrm{R}$ is twice that of $\mathrm{S}$. In an experiment, the solution initially contained 0.5 millimoles of $\mathrm{R}$ and 0.25 of $\mathrm{S}$. The molarities of $\mathrm{R}$ and $\mathrm{S}$ will be equal just at the end of time equal to

Options:
A twice the half life of $\mathrm{R}$
B twice the half life of $S$
C the half life of $\mathrm{S}$
D the half life of $\mathrm{R}$

The correct answer is:
twice the half life of $\mathrm{R}$

Solution:
Substance $\mathrm{R}$ Substance $\mathrm{S}$ $2 \mathrm{k} \quad \mathrm{k}$ rate constant $\mathrm{t}_{1 / 2}=\mathrm{n} \times \mathrm{t}_{1 / 2}$

where $\mathrm{n}=$ number of half life period

Amount of $\mathrm{R}$ left $=\frac{0.5}{(2)^{\mathrm{T} / \mathrm{t}_{/ 2}}}$

Amount of $\mathrm{S}$ left $=\frac{0.25}{(2)^{\mathrm{T} / 2 \mathrm{t} / 2}}$

Equating both $\frac{0.5}{0.25}=\frac{(2)^{\mathrm{T} / \mathrm{t}_{1 / 2}}}{(2)^{\mathrm{T} / 2 \mathrm{t}_{1 / 2}}}$

or $2=(2)^{\mathrm{T} / \mathrm{t}_{1 / 2}}$

$\therefore \mathrm{T}=2 \mathrm{t}_{1 / 2} \cdot 2 \mathrm{t}_{1 / 2}$ is half life of $\mathrm{S}$ and

twice the half -life of $\mathrm{R}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2014
Two tangents \(P Q\) and \(P R\) drawn to the circle \(x^2+y^2-2 x-4 y-20=0\) from point \(P(16,7)\). Ifthe centre of the circle is \(C\) then the area of quadrilateral \(\mathrm{PQCR}\) is

Options:
A 75 sq. unit
B 73 sq. unit
C 72 sq. unit
D 74 sq. unit

The correct answer is:
75 sq. unit

Solution:
Area \(\mathrm{PQCR}=2~ \triangle ~\mathrm{PQC}=2 \times \frac{1}{2} \mathrm{~L} \times r\)
where \(\mathrm{L}=\) length of tangent and \(r=\) radius of circle.
\(\mathrm{L}=\sqrt{\mathrm{S}_1}=15 \text { and } \mathrm{r}=\sqrt{1+4+20}=5\)
Hence the required area \(=75\) sq. units.

Topic: Mathematics, Limits, BITSAT, BITSAT 2009
Two tuning forks $A$ and $B$ sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are $24 \mathrm{~cm}$ and 25 $\mathrm{cm}$ respectively. Calculate the frequencies of forks.

Options:
A $120 \mathrm{~Hz}, 124 \mathrm{~Hz}$
B $110 \mathrm{~Hz}, 114 \mathrm{~Hz}$
C $150 \mathrm{~Hz}, 144 \mathrm{~Hz}$
D $170 \mathrm{~Hz}, 118 \mathrm{~Hz}$

The correct answer is:
$150 \mathrm{~Hz}, 144 \mathrm{~Hz}$

Solution:
Let the frequency of the first fork be $\mathrm{f}_{1}$ and that of second be $\mathrm{f}_{2}$
We then have, $\mathrm{f}_{1}=\frac{\mathrm{v}}{4 \times 24}$ and $\mathrm{f}_{2}=\frac{\mathrm{v}}{4 \times 25}$
We also see that $\mathrm{f}_{1}>\mathrm{f}_{2}$
$$
\therefore \mathrm{f}_{1}-\mathrm{f}_{2}=6
$$
and $\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{24}{25}$
Solving (i) and (ii), we get $\mathrm{f}_{1}=150 \mathrm{~Hz}$ and $\quad \mathrm{f}_{2}=144 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2009
Two tuning forks with natural frequencies 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats offrequency $3 \mathrm{~Hz}$. Find the speed of the tuning forks.

Options:
A $1.5 \mathrm{~m} / \mathrm{s}$
B $2 \mathrm{~m} / \mathrm{s}$
C $1 \mathrm{~m} / \mathrm{s}$
D $2.5 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1.5 \mathrm{~m} / \mathrm{s}$

Solution:
Let $v=$ speed of sound and $v_{\mathrm{S}}=$ speed of tuning forks. Apparent frequency of fork moving towards the observer is

$$

\mathrm{n}_{1}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}

$$

Apparent frequency of the fork moving away from the observer is

$$

\mathrm{n}_{2}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}

$$

If $f$ is the number of beats heard per second. then $\mathrm{f}=\mathrm{n}_{1}-\mathrm{n}_{2}$

$$

\begin{array}{l}

\Rightarrow \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}-\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n} \\

\Rightarrow \mathrm{f}=\frac{\mathrm{v}\left(\mathrm{v}+\mathrm{v}_{\mathrm{s}}\right)-\mathrm{v}\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}\right)}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}(\mathrm{n}) \\

\Rightarrow \frac{2 \mathrm{vv}_{\mathrm{s}} \mathrm{n}}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}=\mathrm{f} \Rightarrow 2\left(\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}}\right)=\mathrm{n}=\mathrm{f}\left\{\right.\text { if } \mathrm{v}_{\mathrm{s}}<\mathrm{K}v

\end{array}

$$

$$

\Rightarrow v_{\mathrm{s}}=\frac{\mathrm{fv}}{2 \mathrm{n}}

$$

$$

\begin{array}{l}

\text { putting } \mathrm{v}=340 \mathrm{~m} / \mathrm{s}, \mathrm{f}=3, \mathrm{n}=340 \mathrm{~Hz} \text { we get, } \\

\mathrm{v}_{\mathrm{s}}=\frac{340 \times 3}{3 \times 340}=1.5 \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2015