datavorous/entrance-exam-dataset huggingface dataset:
Three rods of identical cross-sectional area and made from the same metal, form the sides of an isosceles triangle ABC right angled at B as shown in figure. The point A and B are maintained at temperature T and 2T respectively, in the steady state. Now, assuming that only heat conduction takes place. The temperature of point C will be

Options:
A T2+1
B T2-1
C 3T2+1
D 3T(2+1)

The correct answer is:
3T2+1

Solution:
Let T0 be the temperature of point C and x be the length of rod AB or BC.

Then, CA=x2+x2=2x

At steady state, the rate of heat flowing from B to C= rate of heat flowing from C to A.

So, KA2T-T0x=KAT0-T2x

22T-T0=T0-T

By solving, T0=3T(2+1)

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2018
Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. HCl solution and one formed a black precipitate with $\mathrm{H}_{2} \mathrm{~S}$. The salt could be

Options:
A $\mathrm{AgNO}_{3}$
B $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$
C $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$
D $\mathrm{MnSO}_{4}$

The correct answer is:
$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$

Solution:
$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NH}_{4} \mathrm{OH}$
$$
\begin{array}{r}
\rightarrow \mathrm{Pb}(\mathrm{OH})_{2} \downarrow+2 \mathrm{NH}_{4} \mathrm{NO}_{3}(\text { white ppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{HCl} \rightarrow \mathrm{PbCl}_{2} \downarrow+2 \mathrm{HNO}_{3}(\text { whiteppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{PbS} \downarrow+2 \mathrm{HNO}_{3} \text { (black) }
\end{array}
$$

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2010
Three specimens A, B, C of same radioactive element has activities 1 microcurie, 1 rutherford and 1 becquerel respectively. Which specimen has maximum mass?

Options:
A A
B B
C C
D all have equal masses

The correct answer is:
A

Solution:
1 becquerel $=1 \mathrm{dis} / \mathrm{s}$

1 rutherford $=10^{6} \mathrm{dis} / \mathrm{s}$

1 curie $=3.7 \times 10^{10} \mathrm{dis} / \mathrm{s}$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2021
Three voltmeters $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ having resistances $\mathrm{R}, 1.5 \mathrm{R}$ and $3 \mathrm{R},$ respectively, are connected as shown. When some potential difference is applied between $\mathrm{X}$ and $\mathrm{Y},$ the voltmeter readings are $\mathrm{V}_{\mathrm{A}}, \mathrm{V}_{\mathrm{B}}$ and $\mathrm{V}_{\mathrm{C}}$ respectively. Then -

Options:
A $V_{A} \neq V_{B}=V_{C}$
B $V_{\mathrm{A}}=V_{\mathrm{B}} \neq V_{\mathrm{C}}$
C $\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{C}}$
D $V_{A}=V_{B}=V_{C}$

The correct answer is:
$V_{A}=V_{B}=V_{C}$

Solution:
$V_{A}=I R$
$V_{B}=\left(\frac{2 I}{3}\right) 1.5 R=I R \quad V_{C}=\left(\frac{1}{3}\right) 3 R=I R$
$\therefore \quad V_{A}=V_{B}=V_{C}$

Topic: Physics, Current Electricity, JEE Main
Three voltmeters $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ having resistances $\mathrm{R}, 1.5 \mathrm{R}$ and $3 \mathrm{R},$ respectively, are connected as shown. When some potential difference is applied between $\mathrm{X}$ and $\mathrm{Y},$ the voltmeter readings are $\mathrm{V}_{\mathrm{A}}, \mathrm{V}_{\mathrm{B}}$ and $\mathrm{V}_{\mathrm{C}}$ respectively. Then -

Options:
A $V_{A} \neq V_{B}=V_{C}$
B $V_{\mathrm{A}}=V_{\mathrm{B}} \neq V_{\mathrm{C}}$
C $\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{C}}$
D $V_{A}=V_{B}=V_{C}$

The correct answer is:
$V_{A}=V_{B}=V_{C}$

Solution:
$V_{A}=I R$
$V_{B}=\left(\frac{2 I}{3}\right) 1.5 R=I R \quad V_{C}=\left(\frac{1}{3}\right) 3 R=I R$
$\therefore \quad V_{A}=V_{B}=V_{C}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2012
Through the vertex $\mathrm{O}$ of a parabola $\mathrm{y}^{2}=4 \mathrm{x}$, chords OP and $O Q$ are drawn at right angles to one another. The locus of the middle point of $\mathrm{PQ}$ is

Options:
A $y^{2}=2 x+8$
B $\quad y^{2}=x+8$
C $y^{2}=2 x-8$
D $\quad y^{2}=x-8$

The correct answer is:
$y^{2}=2 x-8$

Solution:
Given parabola is $\mathrm{y}^{2}=4 ...(1)

\mathrm{x}$ Let $\mathrm{P} \equiv\left(\mathrm{t}_{1}^{2}, 2 \mathrm{t}_{1}\right)$ and $\mathrm{Q} \equiv\left(\mathrm{t}_{2}^{2}, 2 \mathrm{t}_{2}\right)$

Slope of $\mathrm{OP}=\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}$ and slope of $\mathrm{OQ}=\frac{2}{\mathrm{t}_{2}}$

Since $\mathrm{OP} \perp \mathrm{OQ}, \therefore \frac{4}{\mathrm{t}_{1} \mathrm{t}_{2}}=-1$ or $\mathrm{t}_{1} \mathrm{t}_{2}=-4 \ldots(2)$

Let $\mathrm{R}(\mathrm{h}, \mathrm{k})$ be the middle point of $\mathrm{PQ},$ then $\mathrm{h}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}}{2} \quad \ldots(3) \quad$ and $\quad \mathrm{k}=\mathrm{t}_{1}+\mathrm{t}_{2} \ldots(4)$

From $(4), \mathrm{k}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}=2 \mathrm{~h}-8 \quad[\mathrm{From}$

(2) and (3)] Hence locus of $\mathrm{R}(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}^{2}=2 \mathrm{x}-8$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2014
To increase the current sensitivity of a moving coil galvanometer by 50 %, its resistance is increased, so that new resistance becomes twice its initial resistance, by what factor does its voltage sensitivity change?

Options:
A Increases by 15 %
B Decreases by 15 %
C Increases by 25 %
D Decreases by 25 %

The correct answer is:
Decreases by 25 %

Solution:
New, current sensitivity, Is'=Is+50100Is

=150100Is=32Is

New resistance, R'=2 R

Initial voltage sensitivity, Vs=IsR

Now, new voltage sensitivity.

Vs=is'R'=32 Ic2 R=34Vs

% decrease in voltage sensitivity

=Vs-Vs'Vs×100

=1-VsVs×100

=1-34×100=25 %

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2018
Total number of ways of selecting 2 white squares on a normal chess board if the squares are not from the same row or column is equal to

Options:
A 480
B 496
C 412
D 400

The correct answer is:
400

Solution:
2 white squares can be selected in ${ }^{32} \mathrm{C}_{2}$ ways If they belong to some row or same column then $2\left(8 \cdot{ }^{4} \mathrm{C}_{2}\right)$ ways.

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2020
Transfer characteristics [output voltage $\left(V_{0}\right)$ vs input voltage $\left.\left(V_{1}\right)\right]$ for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used

Options:
A in region (III)
B both in region (I) and (II)
C in region (II)
D in region (I)

The correct answer is:
both in region (I) and (II)

Solution:
$\mathrm{I} \rightarrow \mathrm{ON}$

$\mathrm{II} \rightarrow \mathrm{OFF}$

In $I^{\text {nd }}$ state it is used as a amplifier it is active region.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2013
Transverse waves of the same frequency are generated in two steel wires $A$ and $B$. The diameter of $A$ is twice that of $B$ and the tension in $A$ is half that in $B$. The ratio of the velocities of the waves in $A$ and $B$ is

Options:
A
$1: 2$

B
$1: \sqrt{2}$

C
$1: 2 \sqrt{2}$

D
$3: 2 \sqrt{2}$

The correct answer is:

$1: 2 \sqrt{2}$

Solution:
The velocity of transverse waves is given by $v=\sqrt{\frac{T}{m}}$ where, $T=$ Tension and $m=$ mass per unit length of the wire. If $r$ is the radius of the wire and $\rho$ its density then,
$m=\pi r^2 \rho$
$\therefore \quad v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\pi r^2 \rho}}$
$\therefore \quad v_A=\frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}}$
and $\quad v_B=\frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}}$
Now, $\quad \frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}$
$\because \quad r_A=2 r_B$ and $T_A=\frac{1}{2} T_B$
Hence, $\quad \frac{v_A}{v_B}=\frac{1}{2 \sqrt{2}}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2022
Two balloons are filled one with pure he gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is

Options:
A more in He gas filled balloon
B same in both balloons
C more in air filled balloon
D in the ratio 1: 4

The correct answer is:
same in both balloons

Solution:
Assuming the balloons have the same volume, as $\mathrm{pV}=\mathrm{nRT}$. If $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are the same, $\mathrm{n}$ the number of moles present will be the same, whether it is He or air. Hence, number of molecules per unit volume will be same in both the balloons.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2014
Two blocks A and B are placed one over the other on a smooth horizontal surface. The maximum horizontal force that can be applied on lower block A, so that A and B move without separation is 49 N. The coefficient of friction between A and B is

Options:
A 0.2
B 0.3
C 0.5
D 0.8

The correct answer is:
0.5

Solution:
Equation of motion of lower block is

F=μMA+MB·g

μ=FMA+MBg

=4910×9.8=12=0.5

Topic: Physics, Laws of Motion, JEE Main
Two blocks A and B are placed one over the other on a smooth horizontal surface. The maximum horizontal force that can be applied on lower block A, so that A and B move without separation is 49 N. The coefficient of friction between A and B is

Options:
A 0.2
B 0.3
C 0.5
D 0.8

The correct answer is:
0.5

Solution:
Equation of motion of lower block is

F=μMA+MB·g

μ=FMA+MBg

=4910×9.8=12=0.5

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2017
Two blocks are connected over a massless pulley as shown in fig. The mass of block $A$ is $10 \mathrm{~kg}$ and the coefficient of kinetic
friction is 0.2 . Block A slides down the incline at constant speed. The mass of block $\mathrm{B}$ in $\mathrm{kg}$ is:

Options:
A 3.5
B 3.3
C 3
D 2.5

The correct answer is:
3.3

Solution:
Considering the equilibrium of A, we get
\begin{array}{l}
10 \mathrm{a}=10 \mathrm{~g} \sin 30^{\circ}-\mathrm{T}-\mu \mathrm{N} \\
\text { where } \mathrm{N}=10 \mathrm{~g} \cos 30^{\circ}
\end{array}
$\therefore 10 \mathrm{a}=\frac{10}{2} \mathrm{~g}-\mathrm{T}-\mu \times 10 \mathrm{~g} \cos 30^{\circ}$
but $\mathrm{a}=0, \mathrm{~T}=\mathrm{m}_{\mathrm{B}} \mathrm{g}$
$0=5 \mathrm{~g}-\mathrm{m}_{\mathrm{B}} \mathrm{g}-\frac{0.2 \sqrt{3}}{2} \times 10 \times \mathrm{g}$
$\Rightarrow \mathrm{m}_{\mathrm{B}}=3.268 \approx 3.3 \mathrm{~kg}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2010
Two bodies are having kinetic energies in the ratio $16: 9$. If they have same linear momentum, the ratio of their masses respectively is:

Options:
A
4 : 3
B
3 : 4
C
16 : 9
D
9 : 16

The correct answer is:
9 : 16

Solution:
$\begin{aligned} & \text { K.E. }=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \\ & \therefore \frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\mathrm{p}_1^2}{2 \mathrm{~m}_1} \times \frac{2 \mathrm{~m}_2}{\mathrm{p}_2^2}=\frac{\mathrm{m}_2}{\mathrm{~m}_1}=\frac{16}{9} \\ & \therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{9}{16}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. It is found that by connecting them together the potential on each one can be made zero. Then

Options:
A $5 C_{1}=3 C_{2}$
B $3 C_{1}=5 C_{2}$
C $3 C_{1}+5 C_{2}=0$
D $9 C_{1}=4 C_{2}$

The correct answer is:
$3 C_{1}=5 C_{2}$

Solution:
For potential to be made zero, after connection
$120 \mathrm{C}_{1}=200 \mathrm{C}_{2} \quad\left[\because C=\frac{q}{v}\right]$
$\Rightarrow 3 C_{1}=5 C_{2}$

Topic: Physics, Capacitance, BITSAT, BITSAT 2012
Two circular plates of radius $5 \mathrm{~cm}$ each, have a 0.01 mm thick water film between them. Then what will be the force required to separate these plate (S.T. of water $=73$ dyne/cm)?

Options:
A $125 \mathrm{~N}$
B $95 \mathrm{~N}$
C $115 \mathrm{~N}$
D $105 \mathrm{~N}$

The correct answer is:
$115 \mathrm{~N}$

Solution:
$\begin{aligned} \mathrm{F}=& \frac{2 \mathrm{AT}}{\mathrm{d}}=\frac{2 \times \pi \times(0.05)^{2} \times 73 \times 10^{-3}}{0.01 \times 10^{-3}} \\=& 36.5 \pi \approx 115 \text { newton } \end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2010
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $\mathrm{A}=10 \mathrm{~cm}^{2}$ and length $=20 \mathrm{~cm} .$ If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right)$

Options:
A $2.4 \pi \times 10^{-5} \mathrm{H}$
B $4.8 \pi \times 10^{-4} \mathrm{H}$
C $4.8 \pi \times 10^{-5} \mathrm{H}$
D $2.4 \pi \times 10^{-4} \mathrm{H}$

The correct answer is:
$2.4 \pi \times 10^{-4} \mathrm{H}$

Solution:
$\begin{aligned} M &=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{~N}_{2} \mathrm{~A}}{\ell} \\ &=\frac{4 \pi \times 10^{-7} \times 300 \times 400 \times 100 \times 10^{-4}}{0.2} \\ &=2.4 \pi \times 10^{-4} \mathrm{H} \end{aligned}$

#

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2013
Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is -

Options:
A $\frac{1}{3}$
B $\frac{25}{36}$
C $\frac{25}{216}$
D None of these

The correct answer is:
$\frac{25}{216}$

Solution:
(c) The probability of showing same number by both dice $p=\frac{6}{36}=\frac{1}{6}$
In binomial distribution here $n=4, r=2, p=$
$\begin{array}{l}
\frac{1}{6}, \mathrm{q}=\frac{5}{6} \\
\text { req. probability }={ }^{n} C_{r} q^{n-r} p^{r}={ }^{4} C_{2}\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)^{2} \\
=6\left(\frac{25}{36}\right)\left(\frac{1}{36}\right)=\frac{25}{216}
\end{array}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2012
Two electrons are moving with equal speed of 5×105 m s-1 parallel to each other. The ratio of the electrostatic to the magnetic force between them is

Options:
A 5×105
B 2.5×104
C 3.6×105
D 4.4×103

The correct answer is:
3.6×105

Solution:
Given, speed of electrons, v=5×105 m s-1

Electrostatic force between the electrons,

Fe=k·e2r2 ...i

Magnetic field produced due to moving electron with velocity v,

B=μ04π·evr2 ...ii

Magnetic force between the electrons,

Fm=Bev=μ04π·evr2·ev (from Eq. (ii))

Fm=μ04π·e2v2r2

∴ From Eqs. (i) and (iii), we have

FeFm=ke2/r2μ0e2v24π·r2=k·4πμ0v2

=9×109×4×3.144π×10-75×1052=3.6×105

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2019
Two elements A and B have electronegativities 1.2 and 3.0 respectively. The nature of bond between A and B would be

Options:
A ionic
B covalent
C co-ordinate
D metallic

The correct answer is:
ionic

Solution:
∵ Difference of electronegativity between (A) and (B) =3.0-1.2=1.8 and, a bond having electronegativity difference greater than 1.65 is of ionic nature.

Hence, bond between (A) and (B) would be of ionic nature.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2017
Two elements $A$ and $B$ have similar chemical properties. They don't form solid hydrogencarbonates, but react with nitrogen to form nitrides. A and B, respectively, are :

Options:
A $Na$ and $Rb$
B $Na$ and $Ca$
C $Cs$ and $Ba$
D $Li$ and $Mg$

The correct answer is:
$Li$ and $Mg$

Solution:
Li and $\mathrm{Mg}$ do not form solid bicarbonate, but react with $\mathrm{N}_{2}$ to give nitrides.

$\begin{array}{l}

6 \mathrm{Li}+\mathrm{N}_{2} \stackrel{\Delta}{\rightarrow} 2 \mathrm{Li}_{3} \mathrm{~N} \\

3 \mathrm{Mg}+\mathrm{N}_{2} \stackrel{\Delta}{\rightarrow} \mathrm{Mg}_{3} \mathrm{~N}_{2}

\end{array}$

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2021
Two equal sides of an isosceles triangle are 7x-y+3=0 and x+y-3=0 and its third side passes through the point (1,-10). The equation of the third side is

Options:
A x-3y=-31
B x-3y=31
C x+3y=31
D x+3y=-31

The correct answer is:
x-3y=31

Solution:
The third side is parallel to a bisector of the angle between equal sides.

The bisectors are 7x-y+372+-12=±(x+y-3)12+12

⇒7x-y+352=±(x+y-3)2

⇒7x-y+3=±5(x+y-3)

⇒2x-6y+18=0 or 12x+4y-12=0

⇒x-3y+9=0 or 3x+y-3=0

Let the third side be x-3y=k or 3x+y=L

It passes through (1,-10).

⇒k=31, L=-7

Hence, required lines are x-3y=31 or 3x+y=-7.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2017
Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is 112 more than that of the second set. The values of $m$ and $n$ respectively are,

Options:
A 4,7
B 7,4
C 4,4
D 7,7

The correct answer is:
7,4

Solution:
$2^{m}-2^{n}=112 \Rightarrow 2^{n}\left(2^{m-n}-1\right)=16.7$

$\quad \therefore \quad 2^{n}\left(2^{m-n}-1\right)=2^{4}\left(2^{3}-1\right)$

Comparing we get $n=4$ and $m-n=3$ $\Rightarrow n=4$ and $m=7$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2016
Two gases occupy two containers $A$ and $B$ the gas in $\mathrm{A},$ of volume $0.10 \mathrm{~m}^{3},$ exerts a pressure of $1.40 \mathrm{MPa}$ and that in $\mathrm{B}$ of volume $0.15 \mathrm{~m}^{3}$ exerts a pressure $0.7 \mathrm{MPa}$. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Options:
A 0.7
B 0.98
C 1.4
D 210

The correct answer is:
0.98

Solution:
We know that $\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}=\mathrm{n}_{\mathrm{A}} \mathrm{RT}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{n}_{\mathrm{B}} \mathrm{RT}$

$\begin{array}{ll} & \text { and } P_{f}\left(V_{A}+V_{B}\right)=\left(n_{A}+n_{B}\right) R T \\ & P_{f}\left(V_{A}+V_{B}\right)=P_{A} V_{A}+P_{B} V_{B} \\ \therefore & P_{f}=\left(\frac{P_{A} V_{A}+P_{B} V_{B}}{V_{A}+V_{B}}\right) \\ & =\frac{1.4 \times 0.1+0.7 \times 0.15}{0.1+0.15} \mathrm{MPa}=0.98 \mathrm{MPa}\end{array}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2013
Two identical capacitors, have the same capacitance $\mathrm{C}$. One of them is charged to potential $\mathrm{V}_{1}$ and the other to $\mathrm{V}_{2}$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is -

Options:
A $\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)$
B $\frac{1}{4} C\left(V_{1}^{2}+V_{2}^{2}\right)$
C $\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}$
D $\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$

The correct answer is:
$\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}$

Solution:
Initial energy of combined system

$$

\mathrm{U}_{1}=\frac{1}{2} \mathrm{CV}_{1}^{2}+\frac{1}{2} \mathrm{CV}_{2}^{2}

$$

Final common potential, $\mathrm{V}=\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2},$

Final energy of system,

$$

\mathrm{U}_{2}=2 \times \frac{1}{2} \mathrm{C}\left(\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2}\right)^{2}

$$

Hence loss of energy $=\mathrm{U}_{1}-\mathrm{U}_{2}$

$$

=\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}

$$

Topic: Physics, Capacitance, BITSAT, BITSAT 2013
Two identical charged spheres suspended from a common point by two massless strings of lengths I, are initially at a distance d $(\mathrm{d}<<1)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v$. Then $v$ varies as a function of the distance $\mathrm{x}$ between the spheres, as:

Options:
A $\quad v \propto x^{\frac{1}{2}}$
B $v \propto x$
C $v \propto x^{-\frac{1}{2}}$
D $\quad v \propto x^{-1}$

The correct answer is:
$v \propto x^{-\frac{1}{2}}$

Solution:
From figure $\tan \theta=\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{mg}} \simeq \theta$

$$

\frac{\mathrm{kq}^{2}}{\mathrm{x}^{2} \mathrm{mg}}=\frac{\mathrm{x}}{2 \ell}

$$

or $x^{3} \propto q^{2} \ldots$ (1)

or $x^{3 / 2} \propto q \ldots(2)$

Differentiating eq. (1) w.r.t. time

$3 \mathrm{x}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} \propto 2 \mathrm{q} \frac{\mathrm{dq}}{\mathrm{dt}}$ but $\frac{\mathrm{dq}}{\mathrm{dt}}$ is constant

$$

\text { so } \mathrm{x}^{2}(\mathrm{v}) \propto \mathrm{q} \text { Replace } \mathrm{q} \text { from eq. }(2)

$$

$\mathrm{x}^{2}(\mathrm{v}) \propto \mathrm{x}^{3 / 2} \quad$ or $\quad \mathrm{v} \propto \mathrm{x}^{-1 / 2}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2016
Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Options:
A $\sqrt{\frac{G m}{2 a^3}}$
B $\sqrt{\frac{\mathrm{Gm}}{8 \mathrm{a}^3}}$
C $\sqrt{\frac{G m}{4 a^3}}$
D $\sqrt{\frac{G m}{a^3}}$

The correct answer is:
$\sqrt{\frac{G m}{4 a^3}}$

Solution:
Radius of circle, $\mathrm{r}=\mathrm{a}$
Centrifugal force, $F=m \omega^2 r$
$\begin{aligned}
& \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\
& \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{~d}}\right)
\end{aligned}$
Here $M_1=m ; M_2=m$ and $d=2 a$
$\Rightarrow$ angular speed, $\omega=\sqrt{\frac{G m}{4 a^3}}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Two inductors L1 and L2 are connected in parallel and a time varying current flows as shown in figure. The ratio ot current i1i2 at any ime t Is

Options:
A L2L1
B L1L2
C L22L1+L22
D L12L1+L22

The correct answer is:
L2L1

Solution:
As the inductors are in parallel, induced emf across the two inductors is the same, i.e.

e1=e2

L1di1dt=L2di2dt

On integrating both sides, we get
L1∫di1dt=L2∫di2dt

L1i1=L2i2

⇒i1i2=L2L1

Topic: Physics, Electromagnetic Induction, JEE Main
Two inductors L1 and L2 are connected in parallel and a time varying current flows as shown in figure. The ratio ot current i1i2 at any ime t Is

Options:
A L2L1
B L1L2
C L22L1+L22
D L12L1+L22

The correct answer is:
L2L1

Solution:
As the inductors are in parallel, induced emf across the two inductors is the same, i.e.

e1=e2

L1di1dt=L2di2dt

On integrating both sides, we get
L1∫di1dt=L2∫di2dt

L1i1=L2i2

⇒i1i2=L2L1

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2018
Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$

$\mathrm{L}_{2}: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$ are coplanar. Then, $\alpha \mathrm{can}$ take value (s)

Options:
A 1,4,5
B 1,2,5
C 3,4,5
D 2,4,5

The correct answer is:
1,4,5

Solution:
The equations of given lines can be written

as

$$

\begin{array}{l}

\mathrm{L}_{1}: \mathrm{x}-5=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2} \\

\mathrm{~L}_{2}: \mathrm{x}-\alpha=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}

\end{array}

$$

Since, these lines are coplanar.

$$

\begin{array}{l}

\text { Therefore, }\left|\begin{array}{ccc}

5-\alpha & 0-0 & 0-0 \\

0 & 3-\alpha & -2 \\

0 & -1 & 2-\alpha

\end{array}\right|=0 \\

\Rightarrow \quad(5-\alpha)(3-\alpha)(2-\alpha)-2=0 \\

\Rightarrow \quad(5-\alpha)\left(6-3 \alpha-2 \alpha+\alpha^{2}-2\right]=0 \\

\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right]=0 \\

\Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\

\Rightarrow \quad \alpha=1,4,5

\end{array}

$$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2015
Two long coaxial and conducting cylinders of radius a and $\mathrm{b}$ are separated by a material of conductivity $\sigma$ and a constant potential difference $\mathrm{V}$ is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is -

Options:
A $\frac{4 \pi \sigma}{\ln (b / a)} V$
B $\frac{4 \pi \sigma}{(b+a)} V$
C $\frac{2 \pi \sigma}{\ln (b / a)} V$
D $\frac{2 \pi \sigma}{(b+a)} V$

The correct answer is:
$\frac{2 \pi \sigma}{\ln (b / a)} V$

Solution:
$\mathrm{V}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{E} \cdot \mathrm{d} \ell=\frac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}} \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$

Now, $I=\sigma \int \vec{E} \cdot \overrightarrow{d A}=\sigma \int \frac{\lambda}{2 \pi \varepsilon_{0} r} \cdot 2 \pi d r=\frac{\sigma \lambda}{\varepsilon_{0}}$

From (1)$: I=\frac{2 \sigma \pi \varepsilon_{0}}{\varepsilon_{0} \ln (b / a)}=\frac{2 \pi \sigma}{\ln (b / a)} v$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2014