datavorous/entrance-exam-dataset huggingface dataset:
A man of mass m starts falling towards a planet of mass $\mathrm{M}$ and radius $\mathrm{R}$. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass $2 \mathrm{M} / 3$ and a point mass $\mathrm{M} / 3$ at the centre. Change in the force of gravity experienced by the man is

Options:
A $\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
B 0
C $\frac{1}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
D $\frac{4}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$

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

Solution:
$\quad$ Change in force of gravity
$$
=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}-\frac{\mathrm{G} \frac{\mathrm{M}}{3} \mathrm{~m}}{\mathrm{R}^{2}}
$$
(only due to mass $\mathrm{M} / 3$ due to shell gravitational field is zero (inside the shell))
$$
=\frac{2 \mathrm{GMm}}{3 \mathrm{R}^{2}}
$$

Topic: Physics, Gravitation, BITSAT, BITSAT 2011
A man runs at a speed of $4 \mathrm{~m} / \mathrm{s}$ to overtake a standing bus. When he is $6 \mathrm{~m}$ behind the door at $\mathrm{t}=0,$ the bus moves forward and continuous with a constant acceleration of $1.2 \mathrm{~m} / \mathrm{s}^{2}$. The man reaches the door in time $\mathrm{t}$. Then,

Options:
A $4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
B $1.2 \mathrm{t}^{2}=4 \mathrm{t}$
C $4 t^{2}=1.2 t$
D $6+4 \mathrm{t}=0.2 \mathrm{t}^{2}$

The correct answer is:
$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$

Solution:
Let us draw the figure for given situation,

$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$

$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$

Topic: Physics, Motion In One Dimension, JEE Main
A man runs at a speed of $4 \mathrm{~m} / \mathrm{s}$ to overtake a standing bus. When he is $6 \mathrm{~m}$ behind the door at $\mathrm{t}=0,$ the bus moves forward and continuous with a constant acceleration of $1.2 \mathrm{~m} / \mathrm{s}^{2}$. The man reaches the door in time $\mathrm{t}$. Then,

Options:
A $4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
B $1.2 \mathrm{t}^{2}=4 \mathrm{t}$
C $4 t^{2}=1.2 t$
D $6+4 \mathrm{t}=0.2 \mathrm{t}^{2}$

The correct answer is:
$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$

Solution:
Let us draw the figure for given situation,

$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$

$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2013
A man takes a step forward with probability 0.4 and backward with probability 0.6 . The probability that at the end of eleven steps he is one step away from the starting point is

Options:
A $\frac{2^{5} \cdot 3^{5}}{5^{10}}$
B $462 \times\left(\frac{6}{25}\right)^{5}$
C $231 \times \frac{3^{5}}{5^{10}}$
D none of these

The correct answer is:
$462 \times\left(\frac{6}{25}\right)^{5}$

Solution:
As $0.4+0.6=1$, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure. Then, the probability of success in one step

$=p=0.4=\frac{2}{5}$

The probability of failure in one step

$=\mathrm{q}=0.6=\frac{3}{5}$

In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1 . So, the number of successes $=6$

The number of failures $=5$

or the number of successes $=5$, The number of failures $=6$

$\therefore$ the required probability

$={ }^{11} \mathrm{C}_{6} \mathrm{p}^{6} \mathrm{q}^{5}+{ }^{11} \mathrm{C}_{5} \mathrm{p}^{5} \mathrm{q}^{6}$

$={ }^{11} \mathrm{C}_{6}\left(\frac{2}{5}\right)^{6} \cdot\left(\frac{3}{5}\right)^{5}+{ }^{11} \mathrm{C}_{5}\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{6}$

$=\frac{11 !}{6 ! 5 !}\left(\frac{2}{5}\right)^{5} \cdot\left(\frac{3}{5}\right)^{5}\left\{\frac{2}{5}+\frac{3}{5}\right\}$

$=\frac{11.10 .9 .8 .7}{120} \cdot \frac{2^{5} \cdot 3^{5}}{5^{10}}=462 \times\left(\frac{6}{25}\right)^{5}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2016
A mass $\mathrm{m}$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

Options:
A $2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
B $\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k} / 2}}$
C $\pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k} / 2}}$
D $\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.

The correct answer is:
$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.

Solution:
When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another halfoscillation in the upward direction. The total time period is:
$$
\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}
$$

Topic: Physics, Oscillations, JEE Main
A mass $\mathrm{m}$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

Options:
A $2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
B $\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k} / 2}}$
C $\pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k} / 2}}$
D $\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.

The correct answer is:
$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.

Solution:
When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another halfoscillation in the upward direction. The total time period is:
$$
\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}
$$

Topic: Physics, Oscillations, BITSAT, BITSAT 2011
A massless platform is kept on a light elastic spring as shown in fig. When a sand particle of mass $0.1 \mathrm{~kg}$ is dropped on the pan from a height of $0.24 \mathrm{~m},$ the particle strikes the pan and spring is compressed by $0.01 \mathrm{~m}$. From what height should the particle be dropped to cause a compression of $0.04 \mathrm{~m}$.

Options:
A $3.96 \mathrm{~m}$
B $0.396 \mathrm{~m}$
C $4 \mathrm{~m}$
D $0.4 \mathrm{~m}$

The correct answer is:
$0.396 \mathrm{~m}$

Solution:
No Solution available

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2011
A material particle with a rest mass $\mathrm{m}_{0}$ is moving with a velocity of light $\mathrm{c}$. Then the wavelength of the de Broglie wave associated with it is :

Options:
A $\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)$
B zero
C $\infty$
D $\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)$

The correct answer is:
zero

Solution:
No Solution available

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2010
A maximum of 3 N force can be applied on a block, down the incline, that would still not move the block which is placed on a rough inclined plane as shown in the figure. The maximum external force up the inclined plane that does not move the block is 12 N. The coefficient of static friction between the block and the plane is Take g=10 m s-2

Options:
A 35
B 13
C 335
D 533

The correct answer is:
533

Solution:
Block does not move upto a maximum applied force of 3 N down the inclined plane.

Equation of motion is 3+mg sin30°=F

3+mg sin30°=μ mg cos30° ....(i)

Similarly, block also does not move upto a maximum applied force of 12 N up the plane.

Now, equation of motion is

mg sin30°+F=12

or, mg sin30°+μ mg cos30°=12 ...(ii)

Solving Eqs. (i) and (ii), we have

mg sin30°=92 and μ mg cos30°=152

Hence, dividing

∴ 1μtan30°=9/215/2=35

⇒μ=533

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2019
A metal disc of radius $100 \mathrm{~cm}$ is rotated at a constant angular speed of $60 \mathrm{rad} / \mathrm{s}$ in a plane at right angles to an external field of magnetic induction $0.05 \mathrm{~Wb} / \mathrm{m}^{2}$. The emf induced between the centre and a point on the rim will be

Options:
A $3 \mathrm{~V}$
B $1.5 \mathrm{~V}$
C $6 \mathrm{~V}$
D $9 \mathrm{~V}$

The correct answer is:
$1.5 \mathrm{~V}$

Solution:
Induced emf produced between the centre and a point on the disc is given by

$\begin{array}{l}

e=\frac{1}{2} \omega \mathrm{BR}^{2} \\

\omega=60 \mathrm{rad} / \mathrm{s}, \mathrm{B}=0.05 \mathrm{~Wb} / \mathrm{m}^{2} \\

\text { and } \mathrm{R}=100 \mathrm{~cm}=1 \mathrm{~m} \\

\text { We get } e=\frac{1}{2} \times 60 \times 0.05 \times(1)^{2}=1.5 \mathrm{~V}

\end{array}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2021
A metal rod of length $1 \mathrm{~m}$ is rotated about one of its ends in a plane right angles to a field of inductance $2.5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$. If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.

Options:
A $2.471 \mathrm{~V}$
B $3.171 \mathrm{~V}$
C $0.471 \mathrm{~V}$
D $1.771 \mathrm{~V}$

The correct answer is:
$0.471 \mathrm{~V}$

Solution:
Given : $\ell=1 \mathrm{~m}, \mathrm{~B}=5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$
$$
\mathrm{f}=\frac{1800}{60}=30 \text { rotations } / \mathrm{sec}
$$
In one rotation, the moving rod of the metal traces a circle of radius $\mathrm{r}=\ell$
$\therefore$ Area swept in one rotation $=\pi r^{2}$
$$
\begin{array}{l}
\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{B} \pi \mathrm{r}^{2}}{\mathrm{~T}} \\
=\mathrm{B} \mathrm{f} \pi \mathrm{r}^{2}=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \mathrm{~V}
\end{array}
$$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2011
A metallic element has a cubic lattice. Each edge of the unit cell is 2 A. and the density of metal is 25 g cm3. The unit cell in 200 g of metal are

Options:
A 1×1024
B 1×1022
C 1×1020
D 1×1025

The correct answer is:
1×1024

Solution:
Edge length of the unit cell =2A.=2×10-8 cm

Volume of unit cell =2×10-83=8×10-24 cm3

Mass of one unit cell = volume × density

=8×10-24×25

Number of unit cell in 200 g of metal

= mass of the metal mass of one unit cell

=2008×10-24×25=1×1024

Topic: Chemistry, Solid State, BITSAT, BITSAT 2018
A metallic rod breaks when strain produced is $0.2 \% .$ The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} .$ What should be its area of cross-section to support a load of $10^{4} \mathrm{~N}$ ?

Options:
A $7.1 \times 10^{-8} \mathrm{~m}^{2}$
B $7.1 \times 10^{-6} \mathrm{~m}^{2}$
C $7.1 \times 10^{-4} \mathrm{~m}^{2}$
D $7.1 \times 10^{-2} \mathrm{~m}^{2}$

The correct answer is:
$7.1 \times 10^{-4} \mathrm{~m}^{2}$

Solution:
$\quad$ Maximum possible strain $=0.2 / 100$
$$
\begin{array}{l}
\therefore \mathrm{A}=\frac{\mathrm{F}}{\mathrm{Y} \times \text { strain }} \\
=\frac{10^{4} \times 100}{\left(7 \times 10^{9}\right) \times 0.2}=7.1 \times 10^{-4} \mathrm{~m}^{2}
\end{array}
$$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2009
A mica slit of thickness t and refractive index $\mu$ is introduoed in theray from the first source S$_{1}$. By how much distance of fringes pattern will be displaced?

Options:
A $\frac{\mathrm{d}}{\mathrm{D}}(\mu-1) \mathrm{t}$
B $\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}$
C $\frac{\mathrm{d}}{(\mu-1) \mathrm{D}}$
D $\frac{D}{d}(\mu-1)$

The correct answer is:
$\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}$

Solution:
Hint:
The laws of Electromagnetic Induction give us idea about the various ways in which we can produce Electric Current using Electric or Magnetic field.
Correct Option is D.
Explanation for correct answer:
- Galvanometer is a device which detects current in the circuit.
- Voltmeter is a device which measures potential difference between two points in the circuit.
- Electric motor is a device which converts Electrical Energy into Mechanical Energy.
- Electric Generator is an device which produces Electric Current from Mechanical energy, Chemical energy, or by using Magnetic or Electric Fields. The laws if Electromagnetic Induction are used in Electric Generator which produces Electric Current from Magnetic or Electric Fields.

Topic: Physics, Wave Optics, BITSAT, BITSAT 2012
A mild steel wire of length 2 L and cross-sectional area A is stretched, well with in the elastic limit, horizontally between two pillars as shown in figure. A mass m is suspended from the mid-point of the wire strain in the wire is

Options:
A x22 L2
B xL
C x2L
D x22 L

The correct answer is:
x22 L2

Solution:
Increase in length

Δ L=BO+OC-BC

=2BO-2BD

⇒ ΔL=2BO-2 L

=2 L2+x212-2 L

or ΔL=2 L 1+x2L212-2 L [∵ using Bionomial theorem]

⇒ ΔL≈2 L 1+12x2L2-1

=x2L ∵ x<<L

∴ Strain =ΔL2 L

=12 L×∆L=12 L×x2L=x22 L2.

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2017
A mixture of Boron trichloride and hydrogen is subjected to silent electric discharge to form x and HCl. x is mixed with NH3 and heated to 2000C to form y. The formula of y is

Options:
A B2O3
B B3N3H6
C H3BO3
D B2H6

The correct answer is:
B3N3H6

Solution:
When a mixture of BCl3 and H2 is subjected to silent electric discharge, diborane and HCl are formed. Diborane reacts with NH3 at 200°C to give Borazine (inorganic benzene).

2BCl3+6H2→dischargeElectricB2H6x+6HCl

3B2H61+6NH32⟶200°C2B3N3H6Borazine(y)+12H2

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2018
A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with $\mathrm{HCl}$ and hydrogen sulphide gas was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filterate shall give test for.

Options:
A Sodium and iron
B Sodium and aluminium
C Aluminium and iron
D Sodium, iron, cadmium and Al

The correct answer is:
Sodium and aluminium

Solution:
CuS and CdS are precipitated by $\mathrm{H}_{2} \mathrm{~S}$. Hydroxide of Al will pass into the solution in the form of $\mathrm{NaAlO}_{2}$ being amphoteric in nature. Hence filtrate will give test for sodium and aluminium.

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2013
A monoatomic gas of molar mass $m$ is kept in a insulated container. Container is moving with velocity $v$. If the container is suddenly stopped, then the change in the temperature of the gas is

Options:
A
$m v^2 / 4 R$

B
$m v^2 / 2 R$

C
$m v^2 / R$

D
$m v^2 / 3 R$

The correct answer is:

$m v^2 / 3 R$

Solution:
Let the insulated container has $n$ moles of the monoatomic gas. The loss of kinetic energy of the gas,
$$
\Delta E_K=\frac{1}{2}(m n) v^2
$$
If the change in temperature of the gas is $\Delta T$, then heat gained by the gas is,
$$
\Delta Q=\frac{3}{2} n R \Delta T
$$
Now, according to question,
$$
\begin{aligned}
\Delta Q & =\Delta E_K \\
\frac{3}{2} n R \Delta T & =\frac{1}{2} m n v^2 \\
\Rightarrow \quad \Delta T & =\frac{m v^2}{3 R}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2022
A mosquito $O$ is sitting infront of a glass rod having spherical end of radius of curvature $40 \mathrm{~cm}$. The image would be formed at

Options:
A
$40 \mathrm{~cm}$ left

B
infinity

C
$20 \mathrm{~cm}$ to the right

D
$15 \mathrm{~cm}$ to the left

The correct answer is:

$40 \mathrm{~cm}$ left

Solution:
As, we know, $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1.5}{v}-\frac{1}{(-20)}=\frac{1.5-1}{+40} \\
& v=-40 \mathrm{~cm} \\
&
\end{aligned}
$$
Negative sign shows that image is obtained on the same side of object i.e., towards left.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2022
A moving coil galvanometer has resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of

Options:
A $9.90 \Omega$ in series
B $10 \Omega$ in series
C $990 \Omega$ in series
D $0.10 \Omega$

The correct answer is:
$990 \Omega$ in series

Solution:
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{10}{0.01}-10=990 \Omega$ in series.

Topic: Physics, Current Electricity, BITSAT, BITSAT 2021
A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is

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

The correct answer is:
1: 1

Solution:
Since, the radius of circular path of a charged

particle in magnetic field is $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\rho}{\mathrm{qB}}$

Now, the radius of circular path of charged particle of given momentum $\rho$ and magnetic field $\mathrm{B}$ is given by $\mathrm{r} \propto \frac{1}{\mathrm{q}}$

But charge on both charged particles, protons and deuterons, is same. Therefore,

$$

\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{D}}}=\frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{q}_{\rho}}=\frac{1}{\mathrm{l}}

$$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2014
A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is

Options:
A $10.2 \mathrm{eV}$
B $20.4 \mathrm{eV}$
C $12.1 \mathrm{eV}$
D $16.8 \mathrm{eV}$

The correct answer is:
$20.4 \mathrm{eV}$

Solution:
Let speed of neutron before collision $=\mathrm{V}$ Speed of neutron after collision $=\mathrm{V}_{1}$ Speed of proton or hydrogen atom after collision $=\mathrm{V}_{2}$ Energy of excitation $=\Delta \mathrm{E}$ From the law of conservation of linear momentum, $\mathrm{mv}=\mathrm{mv}_{1}+\mathrm{mv}_{2}$

And for law of conservation of energy,

$$

\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{mv}_{2}^{2}

$$

From squaring eq. (i), we get

$$

v^{2}=v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2}

$$

From squaring eq. (ii), we get

$$

v^{2}=v_{1}^{2}+v_{2}^{2}+\frac{2 \Delta E}{m}

$$

From eqn (3) \& (4)

$$

\begin{aligned}

\therefore 2 \mathrm{v}_{1} \mathrm{v}_{2} &=\frac{2 \Delta \mathrm{E}}{\mathrm{m}} \\

\therefore\left(\mathrm{v}_{1}-\mathrm{v}_{2}\right)^{2} &=\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)^{2}-4 \mathrm{v}_{1} \mathrm{v}_{2}=\mathrm{v}^{2}-\frac{4 \Delta \mathrm{E}}{\mathrm{m}}

\end{aligned}

$$

As, $v_{1}-v_{2}$ must be real, $v^{2}-\frac{4 \Delta E}{m} \geq 0$

$$

\Rightarrow \quad \frac{1}{2} \mathrm{mv}^{2} \geq 2 \Delta \mathrm{E}

$$

The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2 \mathrm{eV}$. Therefore, the maximum kinetic energy needed is

$$

\frac{1}{2} m v_{\min }^{2}=2 \times 10.2=20.4 \mathrm{eV}

$$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2015
A normal is drawn at the point $P$ to the parabola $y^2=8 x$, which is inclined at $60^{\circ}$ with the straight line $y=8$. Then the point $P$ lies on the straight line

Options:
A
$2 x+y-12-4 \sqrt{3}=0$

B
$2 x-y-12+4 \sqrt{3}=0$

C
$2 x-y-12-4 \sqrt{3}=0$

D
None of these

The correct answer is:

$2 x-y-12-4 \sqrt{3}=0$

Solution:
For the parabola $y^2=4 a x$, the equation of normal at $P\left(a m^2,-2 a m\right)$ is $y=m x-2 a m-a m^3$.
Here, $m=\tan 60^{\circ}=\sqrt{3}$
$\therefore P \equiv\left(a(\sqrt{3})^2,-2 a(\sqrt{3})\right) \equiv(6,-4 \sqrt{3}) \quad(\because a=2)$
Thus, $P$ satisfies $2 x-y-12-4 \sqrt{3}=0$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2022
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :

Options:
A the helium nucleus has less momentum than the thorium nucleus
B the helium nucleus has more momentum than the thorium nucleus
C the helium nucleus has less kinetic energy than the thorium nucleus
D the helium nucleus has more kinetic energy than the thorium nucleus

The correct answer is:
the helium nucleus has more kinetic energy than the thorium nucleus

Solution:
In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity. $\mathrm{U} \rightarrow \mathrm{Th}+\mathrm{He}$

$$

\mathrm{KE}_{\mathrm{Th}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{Th}}}, \mathrm{KE}_{\mathrm{He}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{He}}}

$$

sinc $\mathrm{m}_{\mathrm{He}}$ is less so $\mathrm{KE}_{\mathrm{He}}$ will be more.

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2014
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1 .$ What will be the ratio of their nuclear radius?

Options:
A $2^{1 / 3}: 1$
B $1: 2^{1 / 3}$
C $3^{1 / 2}: 1$
D $1: 3^{1 / 2}$

The correct answer is:
$1: 2^{1 / 3}$

Solution:
As momentum is conserved, therefore

$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$

$\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}=\left(\frac{1}{2}\right)^{1 / 3}=1: 2^{1 / 3}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2020
A pair of tangents are drawn from the origin to the circle $x^{2}+y^{2}+20(x+y)+20=0,$ then the equation of the pair of tangent are

Options:
A $x^{2}+y^{2}-5 x y=0$
B $x^{2}+y^{2}+2 x+y=0$
C $x^{2}+y^{2}-x y+7=0$
D $2 x^{2}+2 y^{2}+5 x y=0$

The correct answer is:
$2 x^{2}+2 y^{2}+5 x y=0$

Solution:
Equation of pair of tangents is given by $\mathrm{SS}_{1}=\mathrm{T}^{2}$

$$

\begin{array}{l}

\text { or } \quad \mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20, \mathrm{~S}_{1}=20 \\

\mathrm{~T}=10(\mathrm{x}+\mathrm{y})+20=0 \\

\therefore \quad \mathrm{SS}_{1}=\mathrm{T}^{2} \\

\Rightarrow \quad 20\left(\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20\right)=10^{2} \\

\quad(\mathrm{x}+\mathrm{y}+2)^{2} \\

\Rightarrow \quad 4 \mathrm{x}^{2}+4 \mathrm{y}^{2}+10 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}^{2}+2 \mathrm{y}^{2}+5 \mathrm{xy}=0

\end{array}

$$

Topic: Mathematics, Circle, BITSAT, BITSAT 2013
A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K=2, the level of liquid isd3,initially.
Suppose, the liquid level decreases at a constant speed v, the time constant as a function of time is

Options:
A 6ε0R5d+3vt
B (15d+9vt)ε0R2d3-3dvt-gv2t2
C 6ε0R5d-3vt
D (15d-9vt)ε0R2d3+3dvt-9v2t2

The correct answer is:
6ε0R5d+3vt

Solution:
We have, the time constant, τ=RC' …i

Now, C'=C1C2C1+C2

=Aε0d-xKAε0xAε0d-x+KAε0x

∴ C'=KAε0x+K(d-x)

Putting the value of C' in Equation. (i), we get

τ=RKAε0d3-vt+kd-d3+vt∵x=d3-vt

Given, A=1 and K=2

∴ τ=3×2ε0×Rd-3vt+6d-2d+6vt=6Rε05d+3vt

Topic: Physics, Capacitance, BITSAT, BITSAT 2018
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, JEE Main