datavorous/entrance-exam-dataset huggingface dataset:
A rubber cord has a cross-sectional area 10-6 m2 and total unstretched length 0.1 m. It is stretched to 0.125 m and then released to project a particle of mass 5.0 g. The velocity of projection is [Given, Young's modulus of rubber, Y=5×108 N m-2]

Options:
A 45 m s-1
B 30 m s-1
C 25 m s-1
D 15 m s-1

The correct answer is:
25 m s-1

Solution:
Force constant K of rubber is given by

K=YAl=5×108×10-60.1=5×103 N m-1

Now, from conservation of energy,

elastic potential energy of cord = kinetic energy of particle

i.e. 12K∆l2=12mv2

⇒v=Km·∆l

=5×1035×10-30.125-0.1

=103×0.025=25 m s-1

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2019
A running track of $440 \mathrm{ft}$ is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then the lengths of its side are

Options:
A
$70 \mathrm{ft}$ and $110 \mathrm{ft}$

B
$80 \mathrm{ft}$ and $120 \mathrm{ft}$

C
$35 \mathrm{ft}$ and $110 \mathrm{ft}$

D
$35 \mathrm{ft}$ and $120 \mathrm{ft}$

The correct answer is:

$35 \mathrm{ft}$ and $110 \mathrm{ft}$

Solution:
Perimeter $=440 \mathrm{ft}$
$\Rightarrow \quad 2 x+\pi r+\pi r=440$
$2 x+2 \pi r=440$
$$
\begin{aligned}
A & =\text { Area of the rectangular portion }=x 2 r \\
A & =x \frac{(440-2 x)}{\pi}=\frac{1}{\pi}\left(440 x-2 x^2\right)
\end{aligned}
$$
Let $\frac{d A}{d x}=\frac{1}{\pi}(440-4 x)=0$
$$
\begin{aligned}
& \Rightarrow x=110 \text { for which } \frac{d^2 A}{d x^2} < 0 \\
& \Rightarrow A \text { is maximum when } x=110 \\
& \Rightarrow 2 r=\frac{440-2 x}{\pi}=\frac{440-220}{22 / 7}=70 \\
& \therefore \quad r=35 \mathrm{ft} \text { and } x=110 \mathrm{ft}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2022
A sample of gas expands from volume $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$. The amount of work done by the gas is greatest when the expansion is

Options:
A isothermal
B isobaric
C adiabatic
D equal in all cases

The correct answer is:
isothermal

Solution:
No Solution available

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2009
A sample of ideal monoatomic gas is taken round the cycle ABCA as shown in the figure. The work done during the cycle is

Options:
A $3 \mathrm{pV}$
B zero
C $9 \mathrm{pV}$
D $6 \mathrm{pV}$

The correct answer is:
$3 \mathrm{pV}$

Solution:
$\Delta \mathrm{W}=$ area under the $\mathrm{p}-\mathrm{V}$ curve
$=\frac{1}{2} \times 3 \mathrm{p} \times 2 \mathrm{~V}=3 \mathrm{pV}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2012
A semicircular disc of mass M and radius R is free to rotate about its diameter. The moment of inertia of semicircular disc about a line perpendicular to its plane through centre is

Options:
A 34MR2
B MR22
C MR23
D MR24

The correct answer is:
MR22

Solution:
The moment of inertia of disc of radius R about its axis is

I=M'R22

where, M'= mass of complete disc =2M

I=2MR22

I=MR2 ...i

If I1 be the moment of inertia of semicircular disc about its. axis, then disc may be assumed as combination of two semi-circular parts,

Thus, I=I1+I1⇒I1=I2=MR22

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2019
A semicircular wire of radius r rotates in uniform magnetic field B about its diameter with angular velocity ω. If the total resistance of the circuit is R, then the mean power generated per period of rotation is

Options:
A πBr2ω28 R
B πBrω28R
C πBrω28 R
D 0

The correct answer is:
πBr2ω28 R

Solution:
The flux linked with coil (made of wire of radius r) of area A and magnetic field B, is given by

ϕ=BA cos θ=B πr22 cos ωt

ϕ=πBr22 cos ωt [∵ A=πr22 and θ=ωt]

By Faraday's law of electromagnetic induction, induced emf, e=-dϕdt

Hence, after differentiating,

e=-ddtπBr22 cos ωt=πBr2ω2 sin ωt (where, R= resistance in the circuit)

∴ Power P=e2R=π2B2r4ω2 sin2ωt4 R

∴ Mean power P¯=π2B2r4ω24 R·sin2ωt¯

=π2B2r4ω24 R·12 ∵ sin2ωt¯=12

=πBr2ω28 R

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2019
A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid?

Options:
A Three
B Four
C One
D Two

The correct answer is:
Three

Solution:
The sequence of bases in mRNA are read in a serial order in groups of three at a time. Each triplet of nucleotides (having a specific sequence of bases) in known as codon. Each codon specifies one amino acid. Further since, there are four bases. therefore, $4^{3}=64$ triplets or codons are possible.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2015
A set $A$ has 3 elements and another set B has 6 elements. Then

Options:
A $3 \leq \mathrm{n}(\mathrm{A} \cup \mathrm{B}) \leq 6$
B $3 \leq \mathrm{n}(\mathrm{A} \cup \mathrm{B}) \leq 9$
C $6 \leq \mathrm{n}(\mathrm{A} \cup \mathrm{B}) \leq 9$
D $0 \leq n(A \cup B) \leq 9$

The correct answer is:
$6 \leq \mathrm{n}(\mathrm{A} \cup \mathrm{B}) \leq 9$

Solution:
We have
$\min n(A \cup B)=\max \{n(A), n(B)\}=\max \{3,6\}=6$
$\operatorname{Max} n(A \cup B)=n(A)+n(B)=9$
$\therefore 6 \leq n(A \cup B) \leq 9$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A shopkeeper wants to purchase two articles A and $\mathrm{B}$ of cost price 4 and 3 respectively. He thought that he mayearn 30 paise byselling article A and 10 paise by selling article $B$. He has not to purchase total articles worth more than ' 24 . If he purchases the number of articles of $\mathrm{A}$ and $\mathrm{B}, \mathrm{x}$ and y respectively, then linear constraints are

Options:
A $x \geq 0, y \geq 0,4 x+3 y \leq 24$
B $x \geq 0, y \geq 0,30 x+10 y \leq 24$
C $x \geq 0, y \geq 0,4 x+3 y \geq 24$
D $x \geq 0, y \geq 0,30 x+40 y \geq 24$

The correct answer is:
$x \geq 0, y \geq 0,4 x+3 y \leq 24$

Solution:
$x, y \geq 0$ and $4 x+3 y \leq 24$.

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2013
A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration $3 \mathrm{~ms}^{-2}$ is

Options:
A $t \sqrt{\frac{9.8}{12.8}}$
B $t \sqrt{\frac{12.8}{9.8}}$
C $\mathrm{t} \sqrt{\frac{9.8}{6.8}}$
D $\mathrm{t} \sqrt{\frac{6.8}{9.8}}$

The correct answer is:
$t \sqrt{\frac{9.8}{12.8}}$

Solution:
$\mathrm{t} \propto \frac{1}{\sqrt{9.8}}, \mathrm{t}^{\prime} \propto \frac{1}{\sqrt{12.8}}$
$\left(\because \mathrm{g}^{\prime}=9.8+3=12.8\right)$
$\therefore \frac{\mathrm{t}^{\prime}}{\mathrm{t}}=\sqrt{\frac{9.8}{12.8}} \Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{9.8}{12.8}} \mathrm{t}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2010
A single slit of a width $a$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The value of ' $a$ ' for which first minimum appears at $\theta$ $=30^{\circ}$ on the screen will be :

Options:
A $0.6 \mu \mathrm{m}$
B $1.2 \mu \mathrm{m}$
C $1.8 \mu \mathrm{m}$
D $3 \mu \mathrm{m}$

The correct answer is:
$1.2 \mu \mathrm{m}$

Solution:
Wavelength of monochromatic light,
$\lambda=600 \times 10^{-9} \mathrm{~m}$
As fir first minima a $\sin \theta=\lambda$
$\begin{aligned}
& \Rightarrow \mathrm{a} \sin 30^{\circ}=600 \times 10^{-9} \\
& \Rightarrow \mathrm{a}=1200 \times 10^{-9} \mathrm{~m}=1.2 \mu \mathrm{m}
\end{aligned}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A six faced die is a biased one. It is thrice more likely to show an odd numbers than show an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even, is

Options:
A
$\frac{5}{9}$

B
$\frac{5}{8}$

C
$\frac{1}{2}$

D
None of these

The correct answer is:

$\frac{5}{8}$

Solution:
Let $P$ be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is $3 P$. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.
$$
\therefore \quad P+3 P=1 \Rightarrow P=1 / 4
$$
Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$.
If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd.
$$
\therefore \text { Required probability }=\frac{3}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{4}=\frac{10}{16}=\frac{5}{8}
$$

Topic: Mathematics, Probability, BITSAT, BITSAT 2022
A small block of mass $\mathrm{m}$ is kept on a rough inclined surface of inclination $\theta$ fixed in a elevator. The elevator goes up with a uniform velocity $\mathrm{v}$ and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be:

Options:
A zero
B $\mathrm{mgvt} \cos ^{2} \theta$
C $\mathrm{mgvt} \sin ^{2} \theta$
D $\mathrm{mgvt} \sin 2 \theta$

The correct answer is:
$\mathrm{mgvt} \sin ^{2} \theta$

Solution:
\(\mathrm{W}_{\mathrm{f}}=\mathrm{f} \sin \theta \mathrm{S}\)

\(=(\mathrm{mg} \sin \theta)(\sin \theta)(\mathrm{vt})\)

\(=\mathrm{mg} \sin ^{2} \theta \mathrm{vt}\)

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2014
A solenoid is $2 \mathrm{~m}$ long and $3 \mathrm{~cm}$ in diameter. It has 5 layers of winding of 1000 turns each and carries a current of $5 \mathrm{~A}$. What is the magnetic field at its centre ? Use the standard value of $\mu_{0}$.

Options:
A $1.57 \times 10^{-4} \mathrm{~T}$
B $2.57 \times 10^{-3} \mathrm{~T}$
C $1.57 \times 10^{-2} \mathrm{~T}$
D $3.57 \times 10^{-6} \mathrm{~T}$

The correct answer is:
$1.57 \times 10^{-2} \mathrm{~T}$

Solution:
Length of solenoid, $l=2 \mathrm{~m}$

Total number of turns, $\mathrm{N}=5 \times 1000=5000$

Number of turns per unit length,

$\mathrm{n}=\frac{\mathrm{N}}{l}=\frac{5000}{2}=2500 \mathrm{~m}^{-1}$

Current, I = 5A

Now, B $=\mu_{0} \mathrm{nI}=4 \pi \times 10^{-7} \times 2500 \times 5 \mathrm{~T}$

$=1.57 \times 10^{-2} \mathrm{~T}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2021
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube $2 \mathrm{~m}$ long and $0.2 \mathrm{~m}$ in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:

Options:
A $2.4 \times 10^3 \mathrm{Am}^{-1}$
B $1.2 \times 10^3 \mathrm{Am}^{-1}$
C $1 \mathrm{Am}^{-1}$
D $2.4 \times 10^{-3} \mathrm{~A} \mathrm{~m}^{-1}$

The correct answer is:
$1.2 \times 10^3 \mathrm{Am}^{-1}$

Solution:
Given that number of turns, $n=1200$
Current, $\mathrm{I}=2 \mathrm{~A}$
Magnetic field at centre inside the solenoid is given by,
$\mathrm{B}=\mu_0 \mathrm{nI}$
So magnetic intensity at centre of the solenoid,
$\begin{aligned}
& \mathrm{H}=\frac{\mathrm{B}}{\mu_0}=\mathrm{nI}=\left(\frac{1200}{2}\right)(2) \quad\left(\because \mathrm{B}=\mu_0 \mathrm{H}\right) \\
& \mathrm{H}=1.2 \times 10^3 \mathrm{Am}^{-1}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A solid body of constant heat capacity $1 \mathrm{~J} /{ }^{\circ} \mathrm{C}$ is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature $100^{\circ} \mathrm{C}$ to final temperature $200^{\circ} \mathrm{C}$. Entropy change of the body in the two cases respectively is:

Options:
A $\ln 2,2 \ln 2$
B $2 \ln 2,8 \ln 2$
C $\ln 2,4 \ln 2$
D $\ln 2, \ln 2$

The correct answer is:
$\ln 2, \ln 2$

Solution:
The entropy change of the body in the two cases is same as entropy is a state function.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2012
A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

Options:
A Solid cylinder
B Both together
C One with higher density
D Hollow cylinder

The correct answer is:
Solid cylinder

Solution:
Solid cylinder reaches the bottom first because for solid cylinder $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2}$ and for hollow cylinder, $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1$.

Acceleration down the inclined plane $\propto \frac{1}{1+K^{2} / R^{2}} .$ Solid cylinder has greater acceleration. It reaches the bottom first.

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2020
A solid sphere of $80 \mathrm{~kg}$ and radius $15 \mathrm{~m}$ moving in a space becomes a circular disc of radius $20 \mathrm{~m}$ in $1 \mathrm{~h}$. The rate of change of moment of Inertia in this process is

Options:
A
$\frac{30}{9} \mathrm{~kg}^2 \mathrm{~m}^2 \mathrm{~s}^{-1}$

B
$\frac{25}{9} \mathrm{~kg}^{-\mathrm{m}^2 \mathrm{~s}^{-1}}$

C
$\frac{10}{9} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$

D
$\frac{22}{9} \mathrm{~kg}^{-\mathrm{m}^2 \mathrm{~s}^{-1}}$

The correct answer is:

$\frac{22}{9} \mathrm{~kg}^{-\mathrm{m}^2 \mathrm{~s}^{-1}}$

Solution:
Given, mass of solid sphere $=80 \mathrm{~kg}$ radius of solid sphere, $R_s=15 \mathrm{~m}$
radius of circular disc, $R_c=20 \mathrm{~m}$ and time $=1$ hour $=60$ minute $=60 \times 60 \mathrm{sec}$
$\therefore$ Moment of inertia of solid sphere, $I_s=\frac{2}{5} M R^2$
$$
\begin{aligned}
& =\frac{2}{5} \times 80 \times(15)^2 \\
& =7200 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$$
Similarly,
$$
\text { moment of inertia of the disc, } \begin{aligned}
I_c & =\frac{1}{2} M R_c^2 \\
& =\frac{1}{2} \times 80 \times(20)^2 \\
& =16000 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$$
$$
\begin{aligned}
\text { Rate of change of moment of Inertia } & =\frac{I_c-I_s}{t} \\
& =\frac{16000-7200}{60 \times 60} \\
& =\frac{22}{9} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2022
A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho$ $=k r^{a}$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=$ $\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R,$ the value of $a$ is

Options:
A 3
B 5
C 2
D 7

The correct answer is:
2

Solution:
Using Gauss's law, we have

$$

\oint \vec{E} \cdot d \vec{A}=\frac{1}{\epsilon_{0}} \int(\rho d v)=\frac{1}{\epsilon_{0}} \int_{0}^{R} k r^{a} \times 4 \pi r^{2} d r

$$

or $E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\epsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}$

$$

\therefore \quad E_{1}=\frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}

$$

For $r=\frac{R}{2} \cdot E_{2}=\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}$

Given, $E_{2}=\frac{E_{1}}{8}$

or $\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}$

$$

\begin{array}{ll}

\therefore & \quad \frac{1}{a+1}=\frac{1}{8} \\

\text { or } & a=2 .

\end{array}

$$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2015
A solution of $\mathrm{NH}_{4} \mathrm{Cl}$ and $\mathrm{NH}_{3}$ has $\mathrm{pH}=8.0 .$ Which of the following hydroxides may be precipitated when this solution is mixed with equal volume of 0.2 M of metal ion.

Options:
A $\mathrm{Ba}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=1.1 \times 10^{-4}\right)$
B $\mathrm{Mg}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=3.5 \times 10^{-4}\right)$
C $\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=8.1 \times 10^{-16}\right)$
D $\mathrm{Ca}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=2.1 \times 10^{-5}\right)$ ).

The correct answer is:
$\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=8.1 \times 10^{-16}\right)$

Solution:
$\mathrm{pH}=8, \mathrm{pOH}=6 ;\left[\mathrm{OH}^{-}\right]=10^{-6} \mathrm{M}$

Ionic product of $\mathrm{Fe}(\mathrm{OH})_{2}=0.2 \times\left(1 \times 10^{-6}\right)^{2}$

$=2 \times 10^{-13}>\mathrm{K}_{\mathrm{sp}}\left(=8.1 \times 10^{-16}\right)$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2020
A sound souorce is moving towards stationary listener with $\frac{1}{10}$ th of the speed of sound. The ratio of apparent to read frequency is

Options:
A $\left(\frac{9}{10}\right)^{2}$
B $\left(\frac{10}{9}\right)$
C $\left(\frac{11}{10}\right)$
D $\left(\frac{11}{10}\right)^{2}$

The correct answer is:
$\left(\frac{10}{9}\right)$

Solution:
$v_{\mathrm{s}}=\frac{\mathrm{v}}{10} \quad \mathrm{n}^{\prime}=\mathrm{n} \frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}$
$\frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\frac{\mathrm{v}}{\left(\mathrm{v}-\frac{\mathrm{v}}{10}\right)}=\frac{10}{9}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2012
A sound source, emitting sound of constant frequency, moves with a constant speed and crosses a stationary observer. The frequency (n) of sound heard by the observer is plotted against time (t). Which of the following graphs represents the correct variation?

Options:
A
B
C
D

The correct answer is:

Solution:
No Solution

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2014
A source of sound produces waves of wavelength $60 \mathrm{~cm}$ when it is stationary. If the speed of sound in air is $320 \mathrm{~m} \mathrm{~s}^{-1}$ and source moves with speed $20 \mathrm{~m} \mathrm{~s}^{-1},$ the wavelength of sound in the forward direction will be nearest to

Options:
A $56 \mathrm{~cm}$
B $60 \mathrm{~cm}$
C $64 \mathrm{~cm}$
D $68 \mathrm{~cm}$

The correct answer is:
$56 \mathrm{~cm}$

Solution:
No Solution available

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2010
A source of sound S emitting waves of frequency $100 \mathrm{~Hz}$ and an observor $\mathrm{O}$ are located at some distance from each other. The source is moving with a speed of $19.4 \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air $330 \mathrm{~ms}^{-1}$ )

Options:
A $103 \mathrm{~Hz}$
B $106 \mathrm{~Hz}$
C $97 \mathrm{~Hz}$
D $100 \mathrm{~Hz}$

The correct answer is:
$103 \mathrm{~Hz}$

Solution:
Here, original frequency of sound, $\mathrm{f}_{0}=100 \mathrm{~Hz}$ Speed of source $V_{s}=19.4 \cos 60^{\circ}=9.7$

From Doppler's formula

$\mathrm{f}^{\mathrm{l}}=\mathrm{f}_{0}\left(\frac{\mathrm{V}-\mathrm{V}_{0}}{\mathrm{~V}-\mathrm{V}_{\mathrm{s}}}\right)$

$\mathrm{f}^{\mathrm{l}}=100\left(\frac{\mathrm{V}-0}{\mathrm{~V}-(+9.7)}\right)$

$\mathrm{f}^{\mathrm{l}}=100 \frac{\mathrm{V}}{\mathrm{V}\left(1-\frac{9.7}{\mathrm{~V}}\right)}$

$\mathrm{f}^{1}=100\left(1+\frac{9.7}{330}\right)=103 \mathrm{~Hz}$

Apparent frequency $\mathrm{f}^{1}=103 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2016
A spherical ball is dropped in a long column of a viscous liquid. The speed (v) of the ball as a function of time (t) may be best represented by

Options:
A
B
C
D

The correct answer is:

Solution:
No Solution available

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2010
A spherical balloon is filled with $4500 ~\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 ~\pi$ cubic meters per minute then the rate (in meters per minute) at which the radius of the balloon decreases $49 \mathrm{~min}$ after the leakage began is

Options:
A
$\frac{9}{7}$

B
$\frac{7}{9}$

C
$\frac{2}{9}$

D
$9$

The correct answer is:

$\frac{2}{9}$

Solution:
$V=\frac{4}{3} \pi r^3 \Rightarrow 4500 \pi=\frac{4 \pi r^3}{3}$
$\Rightarrow \quad r=15 \mathrm{~m}$
After $49 \min =(4500-49.72) \pi$
$=972 \pi \mathrm{m}^3$
$\begin{aligned} \Rightarrow & & 972 \pi & =\frac{4}{3} \pi r^3 \\ \Rightarrow & & r^3 & =3 \times 243=3 \times 3^5\end{aligned}$
$\begin{aligned} r & =9 \\ \frac{d v}{d t} & =4 \pi r^2\left(\frac{d r}{d t}\right) \\ 72 \pi & =4 \pi \times 9 \times 9\left(\frac{d r}{d t}\right) \\ \frac{d r}{d t} & =\left(\frac{2}{9}\right)\end{aligned}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2022
A spherical drop of radius $R$ is divided into eight equal droplets. If surface tension is $\mathrm{T}$, then the work done in this process is

Options:
A $2 \pi \mathrm{R}^{2} \mathrm{~T}$
B $3 \pi R^{2} T$
C $4 \pi R^{2} T$
D $2 \pi \mathrm{RT}^{2}$

The correct answer is:
$4 \pi R^{2} T$

Solution:
Volume of 8 small droplets $=$ Volume of 1 big drop $\Rightarrow \frac{4}{3} \pi r^{3} \times 8=\frac{4}{3} \pi \mathrm{R}^{3} \Rightarrow r=\frac{\mathrm{R}}{(8)^{1 / 3}}$

Work done $=($ Change in area $) \times$ surface tension $=\left(4 \pi r^{2} n-4 \pi \mathrm{R}^{2}\right) \mathrm{T}$

$\left\{4 \pi\left(\frac{\mathrm{R}^{2}}{(8)^{2 / 3}}\right) \times 8-4 \pi \mathrm{R}^{2}\right\} \mathrm{T}=4 \pi \mathrm{R}^{2} \mathrm{~T}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2021
A spherically symmetric gravitational system of particles has a mass density $\rho=\left\{\begin{array}{ll}\rho_{0} & \text { for } \mathrm{r} \leq \mathrm{R} \\ 0 & \text { for } \mathrm{r}>\mathrm{R}\end{array}\right.$

where $\mathrm{r}_{0}$ is a constant. A test mass can undergo

circular motion under the influence of the gravitational field of particles. Its speed $\mathrm{V}$ as a function of distance $r(0

PhysicsElectrostaticsJEE Main

Options:

A
B
C
D

Solution:

1262 Upvotes

Verified Answer

The correct answer is:

Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\)

Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).

\(\begin{aligned}

&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\

&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }

\end{aligned}\)

Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE &
NEET with chapter-wise PYQs, revision notes, formula sheets,
custom tests & much more.

Use on iOS or Desktop Download from Google Play

Options:
A
B
C
D

The correct answer is:

Solution:
Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\)

Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).

\(\begin{aligned}

&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\

&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }

\end{aligned}\)

Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.

Topic: Physics, Electrostatics, JEE Main