datavorous/entrance-exam-dataset huggingface dataset:
A conducting circular loop of radius $r$ carries a constant current $i .$ It is placed in a uniform magnetic field $\overrightarrow{\mathrm{B}}_{0}$ such that $\overrightarrow{\mathrm{B}}_{0}$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Options:
A $i r \mathrm{~B}_{0}$
B $2 \pi i r \mathrm{~B}_{0}$
C zero
D $\pi$ ir $\mathrm{B}_{0}$

The correct answer is:
zero

Solution:
The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element $\mathrm{AB}$ the direction of force will be Leftwards and the magnitude will be
$$
\mathrm{dF}=\operatorname{Idl} \mathrm{B} \sin 90^{\circ}=\mathrm{IdIB}
$$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2009
A conducting loop of radius $\frac{10}{\sqrt{\pi}} \mathrm{cm}$ is placed perpendicular to a uniform magnetic field of $0.5 T$. The magnetic field is decreased to zero in $0.5 s$ at a steady rate. The induced emf in the circular loop at $0.25 s$ is:

Options:
A $\mathrm{emf}=1 \mathrm{mV}$
B $\mathrm{emf}=10 \mathrm{mV}$
C $\mathrm{emf}=100 \mathrm{mV}$
D $\mathrm{emf}=5 \mathrm{mV}$

The correct answer is:
$\mathrm{emf}=10 \mathrm{mV}$

Solution:
As $\left.\varepsilon\right|_{t=0.5 \mathrm{sec}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$
$=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \quad\left[\because \theta=0^{\circ} \Rightarrow \cos \theta=1\right]$
$=-\pi \times\left(\frac{10}{\sqrt{\pi}}\right)^2 \times 10^{-4} \times \frac{0-0.5}{0.5}=10^{-2} \mathrm{~V}=10 \mathrm{mV}$
As $\frac{d B}{d t}=$ constant $\Rightarrow$ Induced emf will not change with time. So, $\left.\mathrm{e}\right|_{0,5 \mathrm{sec}}=\left.\mathrm{e}\right|_{0.25 \mathrm{sec}}=10 \mathrm{mV}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A conducting square loop is placed in a magnetic field $B$ with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate $\alpha$. The induced emf in the loop at an instant when its side is ' $a$ ' is

Options:
A $2 a \alpha B$
B $a^{2} \alpha B$
C $2 a^{2} \alpha B$
D $a \alpha B$

The correct answer is:
$2 a \alpha B$

Solution:
At any time $t$, the side of the square $a=\left(a_{0}\right.$ $-\alpha t),$ where $a_{0}=$ side at $t=0$

At this instant, flux through the square :

$$

\begin{array}{l}

\phi=B A \cos 0^{\circ}=B\left(a_{0}-\alpha t\right)^{2} \\

\therefore \text { emf induced } E=-\frac{d \phi}{d t} \\

\Rightarrow E=-B .2\left(a_{0}-\alpha t\right)(0-\alpha)=+2 \alpha a B

\end{array}

$$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2015
A conducting square loop of side $\mathrm{L}$ and resistance $\mathrm{R}$ moves in its plane with a uniform velocity v perpendicular to one of its side. A magnetic induction $\mathrm{B}$ constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere.

The current induced in the loop is

Options:
A $\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$ clockwise
B $\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$ anticlockwise
C $\frac{2 \mathrm{~B} \ell v}{\mathrm{R}}$ anticlockwise
D zero

The correct answer is:
zero

Solution:
Since the magnetic field is uniform the flux $f$ through the square loop at any time $t$ is constant, because

$\begin{array}{l}

\mathrm{f}=\mathrm{B} \times \mathrm{A}=\mathrm{B} \times \mathrm{L}^{2}=\text { constant } \\

\therefore \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\text { zero }

\end{array}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2020
A constant torque of $31.4 \mathrm{~N}-\mathrm{m}$ is exerted on a pivoted wheel. If angular acceleration of wheel is $4 \pi \mathrm{rad} / \mathrm{s}^{2},$ then the moment of inertia of the wheel is

Options:
A $2.5 \mathrm{~kg} \mathrm{~m}^{2}$
B $3.5 \mathrm{~kg} \mathrm{~m}^{2}$
C $4.5 \mathrm{~kg} \mathrm{~m}^{2}$
D $5.5 \mathrm{~kg} \mathrm{~m}^{2}$

The correct answer is:
$2.5 \mathrm{~kg} \mathrm{~m}^{2}$

Solution:
$I=\frac{\tau}{\alpha}=\frac{31.4}{4 \pi}=\frac{31.4}{4 \times 3.14}=2.5 \mathrm{~kg} \mathrm{~m}^{2}$.

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2011
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

Options:
A both the length and the radius of the wire are halved.
B both the length and the radius of the wire are doubled.
C the radius of the wire is doubled.
D the length of the wire is doubled.

The correct answer is:
both the length and the radius of the wire are doubled.

Solution:
The heat produced is given by

$$

H=\frac{V^{2}}{R} \text { and } R=\frac{\rho \ell}{\pi r^{2}}

$$

$\therefore H=V^{2}\left(\frac{\pi r^{2}}{\rho \ell}\right)$

or $\quad H=\left(\frac{\pi V^{2}}{\rho}\right) \frac{r^{2}}{\ell}$

Thus heat $(H)$ is doubled if both length $(\ell)$ and radius $(r)$ are doubled.

Topic: Physics, Current Electricity, BITSAT, BITSAT 2015
A convex lens of focal length 25 cm produces images of the same magnification 2, when an object is kept at two positions x1 and x2 x1>x2 from the lens. The ratio of x2 and x1 is

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

The correct answer is:
1:3

Solution:
For the given convex lens, focal length, f=25 cm

Magnification of lens in terms of u and f is given by,

m=ff+u=2525+u ...i

When the image is real, then m=-2 and u=-x1

∴ From Eq. (i), -2=2525-x1⇒x1=752

Again, when the image is virtual, then m=2 and u=-x2

∴ From Eq. (i), 2=2525-x2⇒x2=252

Hence, the ratio of x2 and x1 is

∴ x2x1=25/275/2=13

Topic: Physics, Ray Optics, BITSAT, BITSAT 2019
A convex lens shown in the figure is made up of two types of transparent materials. A point sources of light is placed on its principal axis. If reflections from the boundaries between layers are ignored, the lens will form

Options:
A only one image
B two images
C infinite images
D no image at all

The correct answer is:
two images

Solution:
Since, the lens is made up of two kinds of transparent material, it has two refractive indices for the incident beam of light. Hence, there will be two focal lengths of the lens and therefore two images will be observed.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2018
A copper rod of mass m slides under gravity on two smooth parallel rails l distance apart and set an angle θ to the horizontal. At the bottom, the rails are joined by a resistance R in figure. There is a uniform magnetic field B perpendicular to the plane of the rails. The terminal velocity of rod is

Options:
A

mgRtanθB2l2

B

mgRcotθB2l2

C mgRsinθB2l2
D mgRcosθB2l2

The correct answer is:
mgRsinθB2l2

Solution:
Terminal velocity of the rod is attained when magnetic force on the rod (Bil) balances the component of weight of the rod(mgsinθ), as in figure.

So, Bil=mgsinθ

BeRl=mgsinθ

BleR=mgsinθ

BlB/VTR=mgsinθ

vT=mgsinθRB2l2

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2018
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surroundings.

Options:
A $18.01^{\circ} \mathrm{C}$
B $26^{\circ} \mathrm{C}$
C $10.6^{\circ} \mathrm{C}$
D $20^{\circ} \mathrm{C}$

The correct answer is:
$26^{\circ} \mathrm{C}$

Solution:
By Newton's law of cooling, $\frac{\theta_{1}-\theta_{2}}{t}=-k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ ....(1)
A sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. $\frac{62-50}{10}=-\mathrm{k}\left[\frac{62+50}{2}-\theta_{0}\right]$ ....(2)
Now, sphere cools from $50^{\circ} \mathrm{C}$ to $42^{\circ} \mathrm{C}$ in next $10 \mathrm{~min} .$
$$
\frac{50-42}{10}=-\mathrm{k}\left[\frac{50+42}{2}-\theta_{0}\right] ....(3)
$$
Dividing eq $^{\text {n }}$. (2) by (3) we get, $\frac{56-\theta_{0}}{46-\theta_{0}}=\frac{1.2}{0.8}$ or $0.4 \theta_{0}=10.4$
Hence $\theta_{0}=26^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2011
A crystal has a coefficient of expansion $13 \times 10^{-7}$ in one direction and $231 \times 10^{-7}$ in every direction at right angles to it. Then the cubical coefficient of expansion is

Options:
A $462 \times 10^{-7}$
B $244 \times 10^{-7}$
C $475 \times 10^{-7}$
D $257 \times 10^{-7}$

The correct answer is:
$475 \times 10^{-7}$

Solution:
$\begin{aligned} \gamma &=\alpha_{1}+\alpha_{2}+\alpha_{3} \\ &=13 \times 10^{-7}+231 \times 10^{-7}+231 \times 10^{-7} \\ &=475 \times 10^{-7} \end{aligned}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2020
A cube is subjected to a uniform volume compression. If the side of the cube decreases by $2 \%$ the bulk strain is

Options:
A 0.02
B 0.03
C 0.04
D 0.06

The correct answer is:
0.06

Solution:
\(\begin{aligned} \frac{\Delta a}{d} &=2 \% \\ v &=a^{3} \\ \frac{\Delta v}{v} &=3 \frac{\Delta a}{d} \\ &=6 \% \\ &=0.06 \end{aligned}\)

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2013
A current i flows through a loop as shown in figure. The magnetic field at the Centre O is

Options:
A 7μoi12r acting downwards
B 5μoi12r acting upwards
C 7μ0i12r acting upwards
D 5μ0i12r acting downwards

The correct answer is:
5μ0i12r acting downwards

Solution:
The angle, subtended by arc DE at O is π2 and FC at O=3π2 the effective magnetic field at O is

B=BDE+BFC

=μ0i4π3r×π2+μ0i4πr×3π2

=μoi8r13+3

=5μoi12r acting downwards

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2018
A current of 2.0 A passed for 5 hours through a molten metal salt deposits $22.2 \mathrm{~g}$ of metal (At wt. $=177$ ). The oxidation state of the metal in the metal salt is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
$\mathrm{m}=\frac{\mathrm{E} \cdot \mathrm{wt} \times \mathrm{Q}}{96500}$

$\begin{aligned} \therefore \mathrm{E} . \mathrm{wt} &=\frac{\mathrm{m} \times 96500}{\mathrm{Q}} \\=\frac{22.2 \times 96500}{2 \times 5 \times 60 \times 60}=60.3 \end{aligned}$

Oxidation state $=\frac{\text { At wt. }}{\text { Eq. wt. }}=\frac{177}{60.3}=3$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2013
A cyclic process is shown in the p-T diagram. Which of the curves show the same process on a $\mathrm{P}$ - $\mathrm{V}$ diagram?

Options:
A
B
C
D

The correct answer is:

Solution:
No Solution available

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2009
A cylinder of mass 2 kg is released from rest from the top of an inclined plane of inclination 30° and length 1 m. If the cylinder rolls without slipping, then its speed when it reaches the bottom is [g=10 m s-2]

Options:
A 203 m s-1
B 203 m s-1
C 103 m s-1
D 103 m s-1

The correct answer is:
203 m s-1

Solution:
Potential energy of cylinder at top position

U=mgAB

=2×101.sin30° ∵sin30°=ABBC∴AB=1sin30°

=10 J

If v is the linear speed of cylinder when it reaches at bottom, then at bottom, total potential energy of cylinder is converted into kinetic energy,

i.e. K=U

12Iω2+12mv2=10

12·mr22·v2r2+12mv2=10∵ω=vr

⇒ 34mv2=10

v=403m=403×2=203 m s-1

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2019
A d.c. battery of $V$ volt is connected to a series combination of a resistor $R$ and an ideal diode $D$ as shown in the figure below. The potential difference across $R$ will be

Options:
A $2 V$ when diode is forward biased
B Zero when diode is forward biased
C $V$ when diode is reverse biased
D $V$ when diode is forward biased

The correct answer is:
Zero when diode is forward biased

Solution:
In forward biasing, the diode conducts. For ideal junction diode, the forward resistance is zero; therefore, entire applied voltage occurs across external resistance R i.e., there occurs no potential drop, so potential across $\mathrm{R}$ is $\mathrm{V}$ in forward biased.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2010
A deuteron of kinetic energy $50 \mathrm{keV}$ is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field $\mathrm{B}$. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $\mathrm{B}$ is

Options:
A $25 \mathrm{keV}$
B $50 \mathrm{keV}$
C $200 \mathrm{keV}$
D $100 \mathrm{keV}$

The correct answer is:
$100 \mathrm{keV}$

Solution:
No Solution

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2016
A die is loaded such that the probability of throwing the number \(i\) is proportional to its reciprocal. The probability that 3 appears in a single throw is-

Options:
A \(3 / 22\)
B \(3 / 11\)
C \(9 / 22\)
D None of these

The correct answer is:
None of these

Solution:
\(\begin{aligned}
& \mathrm{P}(\mathrm{i})=\frac{\mathrm{k}}{\mathrm{i}} \Rightarrow \mathrm{1}=\sum_{\mathrm{i}=1}^6 \mathrm{P}(\mathrm{i})=\mathrm{k} \sum_{\mathrm{i}=1}^6 \frac{1}{\mathrm{i}}=\mathrm{k} \frac{49}{20} \\
& \Rightarrow \mathrm{k}=\frac{20}{49} . \mathrm{P}(3)=\frac{20}{147}
\end{aligned}\)

Topic: Mathematics, Probability, BITSAT, BITSAT 2011
A die is thrown 7 times. What is the probability that an even number turns up at least 4 times?

Options:
A 12
B 3164
C 51128
D 35128

The correct answer is:
12

Solution:
For atleast 4 times an even number, probability is

=C47124123+C54125122+C67126121+C77127

=12

Topic: Mathematics, Probability, BITSAT, BITSAT 2019
A direct current of $5 \mathrm{~A}$ is superposed on an alternating current $I=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be

Options:
A $(15 / 2) \mathrm{A}$
B $5 \sqrt{3} \mathrm{~A}$
C $5 \sqrt{5} \mathrm{~A}$
D $15 \mathrm{~A}$

The correct answer is:
$5 \sqrt{3} \mathrm{~A}$

Solution:
Total carrent, $1=(5+10 \sin \omega \mathrm{t})$

$$

\begin{aligned}

\Rightarrow I_{\text {eff }} &=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2} \\

&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2} \\

&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2} \\

\text { But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0

\end{aligned}

$$

and

$$

\begin{array}{l}

\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2} \\

\text { So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times 100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A}

\end{array}

$$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2015
A direct current of $6 \mathrm{~A}$ is superimposed on an alternating current $I=10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be

Options:
A
$5 \sqrt{2}$

B
$5 \sqrt{3}$

C
$9.27$

D
$8.37$

The correct answer is:

$9.27$

Solution:
Given, $I=6+10 \sin \omega t$
$$
\begin{aligned}
I_{\mathrm{eft}} & =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omega t)^2 d t\right]^{1 / 2}\right. \\
& =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omega t+100 \sin ^2 \omega t\right) d t\right]^{1 / 2}
\end{aligned}
$$
But as, $\frac{1}{T} \int_0^T \sin \omega t d t=0$ and $\frac{1}{T} \int_0^T \sin ^2 \omega t=\frac{1}{2}$
$$
\Rightarrow \quad I_{\mathrm{eff}}=\left[36+\frac{1}{2} \times 100\right]^{1 / 2}=9.27
$$
Thus, $\quad I_{\text {eff }}=9.27 \mathrm{~A}$.

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2022
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc $-$

Options:
A velocity is v, acceleration is zero
B velocity is zero, acceleration is

zero
C velocity is v, acceleration is $\mathrm{v}^{2} / \mathrm{R}$
D velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$

The correct answer is:
velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$

Solution:
As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure $\mathrm{v}_{\text {net }}$ (for lowest point) $=\mathrm{v}-\mathrm{R} \omega=\mathrm{v}-\mathrm{v}=0$

and acceleration

$=\frac{\mathrm{v}^{2}}{\mathrm{R}}+0=\frac{\mathrm{v}^{2}}{\mathrm{R}}$

(since linear speed is constant)

Topic: Physics, Rotational Motion, JEE Main
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc $-$

Options:
A velocity is v, acceleration is zero
B velocity is zero, acceleration is

zero
C velocity is v, acceleration is $\mathrm{v}^{2} / \mathrm{R}$
D velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$

The correct answer is:
velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$

Solution:
As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure $\mathrm{v}_{\text {net }}$ (for lowest point) $=\mathrm{v}-\mathrm{R} \omega=\mathrm{v}-\mathrm{v}=0$

and acceleration

$=\frac{\mathrm{v}^{2}}{\mathrm{R}}+0=\frac{\mathrm{v}^{2}}{\mathrm{R}}$

(since linear speed is constant)

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2013
A disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t=0$, the top most point on the disc is $\mathrm{A}$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is

Options:
A $\mathrm{R} \sqrt{\left(\pi^2+4\right)}$
B $\mathrm{R} \sqrt{\left(\pi^2+1\right)}$
C $2 \mathrm{R}$
D $2 \mathrm{R} \sqrt{\left(1+4 \pi^2\right)}$

The correct answer is:
$\mathrm{R} \sqrt{\left(\pi^2+4\right)}$

Solution:
(a) From figure,
Displacement,
$\mathrm{BA}=\sqrt{(2 \mathrm{R})^2+(\pi \mathrm{R})^2}=\mathrm{R} \sqrt{4+\pi^2}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A flask contains hydrogen and oxygen in the ratio of $2: 1$ by mass at temperature $27^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :

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

The correct answer is:
1 : 1

Solution:
Average kinetic energy for diatomic gases
$\begin{aligned}
& \mathrm{K}_{\mathrm{av}}=\frac{5}{2} \mathrm{kT} \\
& \therefore \frac{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{H}}}{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{O}}}=\frac{(27+273)}{(27+273)}=1
\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A frictionless wire $\mathrm{AB}$ is fixed on a sphere of radius $\mathrm{R}$. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to $\mathrm{B}$ is

Options:
A $\frac{\sqrt{2 g R}}{g \cos \theta}$
B $2 \sqrt{g R} \cdot \frac{\cos \theta}{g}$
C $2 \sqrt{\frac{R}{g}}$
D $\frac{g R}{\sqrt{g \cos \theta}}$

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

Solution:
Acceleration of body along $\mathrm{AB}$ is g $\cos \theta$ Distance travelled in time $\mathrm{t}$ sec $=$

$A B=\frac{1}{2}(g \cos \theta) t^{2}$

From $\Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{R} \cos \theta$

Thus, $2 \mathrm{R} \cos \theta=\frac{1}{2} g \cos \theta t^{2}$

$\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2016
A gas has molar heat capacity C=37.55 J mol-1 K-1, in this process pT= constant. The number of degree of freedom of the molecule of gas is

Options:
A 2
B 3
C 5
D 7

The correct answer is:
5

Solution:
Given, C=37.55 J mol-1 K-1

Also, PT= constant K

According to ideal gas equation, PV=RT

⇒ p=RTV .....i

Putting the value of p in Eq. (i), we get

RTV×T=K ⇒ V=RT2K .....ii

On differentiating above equation both sides, we get

dVdT=2RTK....iii
But TK=1p fromEq. i
Hence, Eq. iiibecomes

dVdT=2 Rp
So, C=CV+pdVdT
or C=CV+p×2 Rp=CV+2 R
or CV=C-2 R ...iv

As CV=nR2, where n= number of degrees of freedom.
Putting the value of CV in Eq. (iv), we get

nR2=C-2 R

n=2 C-2 RR=2 37.55-2×8.38.3=5.048≈5

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2018
A gas $X$ at 1 atm is bubbled through a solution containing a mixture of $1 \mathrm{M} \mathrm{Y}^{-}$ and $1 \mathrm{M} \mathrm{Z}^{-}$ at $25^{\circ} \mathrm{C}$. If the reduction potential is $Z>Y>X$, then

Options:
A $\mathrm{Y}$ will oxidise $\mathrm{X}$ and $\mathrm{not} \mathrm{Z}$
B $\mathrm{Y}$ will oxidise $Z$ and $\mathrm{not} \mathrm{X}$
C $\mathrm{Y}$ will oxidise both $\mathrm{X}$ and $\mathrm{Z}$
D $\mathrm{Y}$ will reduce both $\mathrm{X}$ and $\mathrm{Z}$

The correct answer is:
$\mathrm{Y}$ will oxidise $\mathrm{X}$ and $\mathrm{not} \mathrm{Z}$

Solution:
The more the reduction potential, the more the oxidising power.

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2011
A gaseous reaction $\mathrm{X}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Y}+\frac{1}{2} \mathrm{Z}(\mathrm{g})$

There is increase in pressure from $100 \mathrm{~mm}$ to 120 $\mathrm{mm}$ in 5 minutes. The rate of disappearance of $\mathrm{X}_{2}$ is

Options:
A $8 \mathrm{~mm} \mathrm{~min}^{-1}$
B $2 \mathrm{~mm} \mathrm{~min}^{-1}$
C $16 \mathrm{~mm} \mathrm{~min}^{-1}$
D $4 \mathrm{~mm} \min ^{-1}$

The correct answer is:
$8 \mathrm{~mm} \mathrm{~min}^{-1}$

Solution:
The increase in pressure shows the increase in conc. of $Z$. Rate of appearance of $Z=$

$$

\frac{120-100}{5}=4 \mathrm{~mm} \mathrm{~min}^{-1}

$$

Rate of disappearance of $\mathrm{X}_{2}=2 \times$ rate of appearance of $Z$ $=2 \times 4 \mathrm{~mm} \mathrm{~min}^{-1}=8 \mathrm{~mm} \mathrm{~min}^{-1}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2014
A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a sky satellite orbiting a few 100 km above the earth's surface R=64000 km will approximately be

Options:
A 12 h
B 1 h
C 2 h
D 4 h

The correct answer is:
2 h

Solution:
By Kepler's law of planetary motion,

T2∝r3, hence T12∝r13

Similarly T22∝r23

∴ T12∝360003

T22∝6400+h3

Therefore,

T22=T126400+h360003

>T12640036003

>2428453

∴ T2>24×8×845×45

>1.8 h

So, T2≅2 h

Topic: Physics, Gravitation, BITSAT, BITSAT 2017
A glass capillary tube of internal radius $\mathrm{r}=0.25$ $\mathrm{mm}$ is immersed in water. The top end of the tube projected by $2 \mathrm{~cm}$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $=0.7 \mathrm{~N} / \mathrm{m}$

Options:
A $\theta=90^{\circ}$
B $\quad \theta=70^{\circ}$
C $\theta=45^{\circ}$
D $\theta=35^{\circ}$

The correct answer is:
$\quad \theta=70^{\circ}$

Solution:
Water wets glass and so the angle of contact is zero. For full rise, neglecting the small mass in the meniscus

$$

\begin{aligned}

2 \pi \mathrm{T} &=\pi \mathrm{r}^{2} \mathrm{~h} \rho \mathrm{g} \Rightarrow \mathrm{h}=\frac{2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{g}}\left[\because \text { water wets glass, } \theta=0^{\circ}\right] \\

&=\frac{2 \times 007}{0.25 \times 10^{-3} \times 1000 \times 9.8}

\end{aligned}

$$

As the tube is only $2 \mathrm{~cm}$ above the water and so, water will rise by $2 \mathrm{~cm}$ and meet the tube at an angle such that,

$$

\begin{aligned}

2 \pi r & \operatorname{T} \cos \theta=\pi r^{2} h^{\prime} \rho g \\

\Rightarrow & 2 T \cos \theta=h^{\prime} r \rho g \\

\Rightarrow \cos \theta &=\frac{h^{\prime} r \rho g}{2 T} \\

&=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}

\end{aligned}

$$

The liquid will meet the tube at an angle, $\theta$ $\cong 70^{\circ}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2015