datavorous/entrance-exam-dataset huggingface dataset:
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Options:
A 64
B 129
C 84
D None of these

The correct answer is:
64

Solution:
Atleast one black ball can be drawn in the following ways

(1) one black and two other colour balls

$$

={ }^{3} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2}=3 \times 15=45

$$

(ii) two black and one other colour balls

$$

={ }^{3} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{1}=3 \times 6=18

$$

(iii) All the three are black $={ }^{3} \mathrm{C}_{3} \times{ }^{6} \mathrm{C}_{0}=1$

$\therefore$ Req. no. of ways $=45+18+1=64$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2013
A boy is throwing stones at a target. The probability of hitting the target at any trial is $\frac{1}{2}$ The probability of hitting the target 5 th time at the 10 th throw is:

Options:
A $\frac{5}{2^{10}}$
B $\frac{63}{2^{9}}$
C $\frac{10}{2^{10}}$
D None

The correct answer is:
$\frac{63}{2^{9}}$

Solution:
The probability of hitting the target 5 th time at the 10 th throw $=\mathrm{P}($ the probability of hitting the target 4 times in the first 9 throws) $\mathrm{P}($ the probability of hitting the target at the 10 th throw $)=$
$\left[{ }^{9} \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{5}\right]\left(\frac{1}{2}\right)=\frac{9 !}{4 ! 5 !} \times\left(\frac{1}{2}\right)^{10}=\frac{63}{2^{9}}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2012
A boy pushes a toy box $2.0 \mathrm{~m}$ along the floor by means of a force of $10 \mathrm{~N}$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

Options:
A $6 \mathrm{~J}$
B $8 \mathrm{~J}$
C $10 \mathrm{~J}$
D $12 \mathrm{~J}$

The correct answer is:
$10 \mathrm{~J}$

Solution:
$\mathrm{W}=\mathrm{F} \mathrm{s} \cos \theta=10 \times 2 \cos 60^{\circ}=10 \mathrm{~J}$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2020
A capacitor of capacitance $C$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is:

Options:
A $\frac{C V}{2 \varepsilon_0}$
B $\frac{2 \mathrm{CV}}{\varepsilon_0}$
C $\frac{\mathrm{CV}}{\varepsilon_0}$
D
Zero

The correct answer is:
$\frac{\mathrm{CV}}{\varepsilon_0}$

Solution:
From the Gauss's law
$\begin{aligned}
& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{Q}}{\epsilon_0} \quad \because \mathrm{Q}=\mathrm{CV} \\
& \therefore \phi=\frac{\mathrm{CV}}{\epsilon_0}
\end{aligned}$

Topic: Physics, Capacitance, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A capillary tube is attached horizontally to a constant heat arrangement. If the radius of the capillary tube is increased by $25 \%$, then the rate of flow of liquid will change nearly by

Options:
A
$100 \%$

B
$112 \%$

C
$124 \%$

D
$144 \%$

The correct answer is:

$144 \%$

Solution:
Liquid volume coming out of the tube per second
$$
\begin{aligned}
& \quad V=\frac{p \pi r^4}{8 \eta l} \\
& \Rightarrow \\
& \Rightarrow \frac{V_2}{V_1}=\left[\frac{r_2}{r_1}\right]^4 \Rightarrow V_2=V_1\left[\frac{125}{100}\right]^4=V_1\left(\frac{5}{4}\right)^4 \\
& \Rightarrow \\
& \text { So, now } \frac{\Delta V}{V}=\frac{V_2-V_1}{V_1}=\frac{(244) V_1-V_1}{V_1}=144 \%
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2022
A capillary tube of radius $\mathrm{r}$ is immersed vertically in a liquid such that liquid rises in it to height $\mathrm{h}$ (less than the length of the tube). Mass of liquid in the capillary tube is $\mathrm{m}$. If radius of the capillary tube is increased by $50 \%$, then mass of liquid that will rise in the tube, is

Options:
A $\frac{2}{3} \mathrm{~m}$
B $\mathrm{m}$
C $\frac{3}{2} \mathrm{~m}$
D $\frac{9}{4} \mathrm{~m}$

The correct answer is:
$\frac{3}{2} \mathrm{~m}$

Solution:
$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}} \Rightarrow \frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{2}{3}$

$$

\begin{aligned}

\left(\because \mathrm{r}_{1}=\mathrm{r}, \quad \mathrm{r}_{2}\right.&=\mathrm{r}+50 \% \text { of } \left.\mathrm{r}=\frac{3}{2} \mathrm{r}\right) \\

\text { New mass } \mathrm{m}_{2}=\pi \mathrm{r}_{2}^{2} \mathrm{~h}_{2} \rho &=\pi\left(\frac{3}{2} \mathrm{r}_{1}\right)^{2}\left(\frac{2}{3} \mathrm{~h}_{1}\right) \rho \\

&=\frac{3}{2}\left(\pi \mathrm{r}_{1}^{2} \mathrm{~h}_{1}\right) \rho=\frac{3}{2} \mathrm{~m}

\end{aligned}

$$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2016
A carbon resistor of 56±5.6 kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be

Options:
A green, blue, orange, silver
B blue, yellow, orange, silver
C green, red, orange, silver
D red, green, blue, gold

The correct answer is:
green, blue, orange, silver

Solution:
Given, R=56±5.6 k Ω

=56 k Ω±10 % of 56 k Ω
=56×103±10 % Ω

As per the colour code for carbon resistors, the colour assigned to numbers

5→ Green
6→ Blue
3→ Orange

For 10 % accuracy or tolerance, the colour is silver. Hence, the bands of colours of carbon resistor in sequence are green, blue, orange and silver.

Topic: Physics, Current Electricity, BITSAT, BITSAT 2019
A Carnot's heat engine works between the temperatures $427^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. What amount of heat should it consume per second to deliver mechanical work at the rate of $1.0 \mathrm{~kW} ?$

Options:
A $0.417 \mathrm{kcal} / \mathrm{s}$
B $4.17 \mathrm{kcal} / \mathrm{s}$
C $41.7 \mathrm{kcal} / \mathrm{s}$
D $0.212 \mathrm{kcal} / \mathrm{s}$

The correct answer is:
$0.417 \mathrm{kcal} / \mathrm{s}$

Solution:
The efficiency of the heat engine is

$$

\eta=1-\frac{T_{2}}{T_{1}}=1-\left(\frac{273+27 K}{273+427 K}\right)=\frac{4}{7}

$$

But $\quad \eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$

$$

\therefore \mathrm{Q}_{1}=\frac{\mathrm{W}}{\eta}=\frac{1.0 \mathrm{~kW}}{4 / 7}=1.75 \mathrm{~kW}=0.417 \mathrm{kcal} / \mathrm{s}

$$

Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2013
A certain metal sulphide, $M S_2$, is used extensively as a high temperature lubricant. If $\mathrm{MS}_2$ is $40.06 \%$ by mass sulphur, metal $M$ has atomic mass.

Options:
A
$160 \mathrm{u}$

B
$64 \mathrm{u}$

C
$40 \mathrm{u}$

D
$96 \mathrm{u}$

The correct answer is:

$96 \mathrm{u}$

Solution:
$M S_2=M+32 \times 2=M+64$
$\%$ of sulphur $=\left(\frac{64}{M+64}\right) \times 100=40.06$
$M+64=\frac{6400}{40.06}$
$M+64=160$
$M=160-64=96$ u

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2022
A chain of mass $\mathrm{M}$ is placed on a smooth table with $1 / \mathrm{n}$ of its length $\mathrm{L}$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is

Options:
A $\mathrm{MgL} / \mathrm{n}$
B $\mathrm{MgL} / 2 \mathrm{n}$
C $\mathrm{MgL} / \mathrm{n}^{2}$
D $\mathrm{MgL} / 2 \mathrm{n}^{2}$

The correct answer is:
$\mathrm{MgL} / 2 \mathrm{n}^{2}$

Solution:
$\quad \mathrm{W}=$ change in $\mathrm{PE}$ of $\mathrm{COM}$ of hanging part
$$
=\frac{\mathrm{M}}{\mathrm{n}} \mathrm{g} \frac{\mathrm{L}}{2 \mathrm{n}}=\frac{\mathrm{MgL}}{2 \mathrm{n}^{2}}
$$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2009
A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) \mathrm{T}$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}^2$, then the value of $\alpha$ will be :

Options:
A 3
B 6
C 12
D 2

The correct answer is:
6

Solution:
Given that uniform magnetic field, $\vec{B}=(2 \hat{i}+3 \hat{j}) T$
Acceleration $\vec{a}=(\alpha \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}^2$ We know that
$F=q(\vec{v} \times \vec{B}) \Rightarrow m \vec{a}=q(\vec{v} \times \vec{B})$
Here, $\vec{a} \perp \vec{B}$, so, $\vec{a} \cdot \vec{B}=0$
$(\alpha \hat{i}-4 \hat{j})(2 \hat{i}+3 \hat{j})=0 \Rightarrow 2 \alpha-12=0 \Rightarrow \alpha=6$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A charge $\mathrm{Q}_{1}$ exerts some force on a second charge $\mathrm{Q}_{2}$. If a 3 rd charge $\mathrm{Q}_{3}$ is brought near, then the force of $\mathrm{Q}_{1}$ exerted on $\mathrm{Q}_{2}-$

Options:
A will increase
B will decrease
C will remain unchanged
D will increase if $\mathrm{Q}_{3}$ is of the same sign as $\mathrm{Q}_{1}$ and will decrease if $\mathrm{Q}_{3}$ is of opposite sign.

The correct answer is:
will remain unchanged

Solution:
No Solution available

Topic: Physics, Electrostatics, BITSAT, BITSAT 2011
A charge $+q$ is at a distance $L / 2$ above a square of side $\mathrm{L}$. Then what is the flux linked with the surface?

Options:
A $\frac{\mathrm{q}}{4 \varepsilon_{0}}$
B $\frac{2 q}{3 \varepsilon_{0}}$
C $\frac{\mathrm{q}}{6 \varepsilon_{0}}$
D $\frac{6 \mathrm{q}}{\varepsilon_{0}}$

The correct answer is:
$\frac{\mathrm{q}}{6 \varepsilon_{0}}$

Solution:
m{L}

The given square of side $L$ may be considered as one of the faces of a cube with edge $L$. Then given charge q will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by

$\phi=\mathrm{q} / \varepsilon_{0}$

Hence, electric flux through one face of the cube for the given square will be

$\phi^{\prime}=\frac{1}{6} \phi=\frac{\mathrm{q}}{6 \varepsilon_{0}}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2020
A charge $+q$ is at a distance $L / 2$ above a square of side $\mathrm{L}$. Then what is the flux linked with the surface?

Options:
A $\frac{\mathrm{q}}{4 \varepsilon_{0}}$
B $\frac{2 q}{3 \varepsilon_{0}}$
C $\frac{\mathrm{q}}{6 \varepsilon_{0}}$
D $\frac{6 \mathrm{q}}{\varepsilon_{0}}$

The correct answer is:
$\frac{\mathrm{q}}{6 \varepsilon_{0}}$

Solution:
The given square of side $\mathrm{L}$ may be considered as one of the faces of a cube with edge $\mathrm{L}$. Then given charge $\mathrm{q}$ will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by
$$
\phi=\mathrm{q} / \varepsilon_{0}
$$
Hence, electric flux through one face of the cube for the given square will be
$\phi^{\prime}=\frac{1}{6} \phi=\frac{\mathrm{q}}{6 \varepsilon_{0}}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2010
A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $\frac{Q}{q}$ equals:

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

The correct answer is:
$-2 \sqrt{2}$

Solution:
It is given that $\vec{F}_{A}+\vec{F}_{B}+\vec{F}_{D}=0$

Where $\vec{F}_{A}, \vec{F}_{B}$ and $\vec{F}_{D}$ are the forces applied by charges placed at $A, B$ and $D$ on the charge placed at $C$.

$\begin{array}{l}

\Rightarrow \overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=-\overrightarrow{\mathrm{F}}_{\mathrm{A}} \\

\left|\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}\right|=\sqrt{2} \frac{\mathrm{KqQ}}{\mathrm{a}^{2}} \\

\Rightarrow\left|\overrightarrow{\mathrm{F}}_{\mathrm{A}}\right|=\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\

\therefore \sqrt{2} \frac{\mathrm{KqQ}}{a^{2}}=-\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\

\Rightarrow \frac{\mathrm{Q}}{q}=-2 \sqrt{2}

\end{array}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2021
A charged capacitor of capacitance C is discharging through a resistor of resistance R. At what time the charge on the capacitor is equal to one half of its initial value?

Options:
A RC2
B RCloge2
C RCloge 2
D 1RCloge2

The correct answer is:
RCloge 2

Solution:
The charge on the capacitor at time t in a discharging circuit is given by q=q0e-t/τ

where, τ=RC= time constant
∴ q=q0e-t/RC

when, q=q02

then q02=q0e-t/RC

12=e-t/RC

Taking natural log on both side loge12=log0e-t/RC

-loge2=-tRClogee

⇒t=RCloge2

Topic: Physics, Capacitance, BITSAT, BITSAT 2019
A charged particle moving in a uniform magnetic field and losses $4 \%$ of its kinetic energy. The radius of curvature of its path changes by

Options:
A $2 \%$
B $4 \%$
C $10 \%$
D $12 \%$

The correct answer is:
$2 \%$

Solution:
As we know $\mathrm{F}=\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$

$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}$

And $\mathrm{KE}=\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$

$\therefore \mathrm{mv}=\sqrt{2 \mathrm{~km}}$

$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~km}}}{\mathrm{qB}}$

$\Rightarrow r \propto \sqrt{k}$ or $r=c^{1 / 2}(c$ is a constant $)$ $\frac{\mathrm{dr}}{\mathrm{dr}}=\mathrm{c} \frac{\mathrm{dk}^{1 / 2}}{\mathrm{dr}}$ or $\frac{\mathrm{c} \Delta \mathrm{k}}{\Delta \mathrm{r}}=2 \sqrt{\mathrm{k}}$

$\frac{\Delta r}{r}=\frac{\mathrm{c} \Delta \mathrm{k}}{2 \sqrt{\mathrm{k}} \mathrm{c} \sqrt{\mathrm{k}}}=\frac{\Delta \mathrm{k}}{2 \mathrm{k}}$

Therefore percentage changes in radius of path,

$\frac{\Delta r}{r} \times 100=\frac{\Delta k}{2 k} \times 100=2 \%$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2015
A circle touches both the $y$-axis and the line $x+$ $y=0$. Then the locus of its center is

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

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

Solution:
Let $(h, k)$ is centre of circle
$\left|\frac{h-k}{\sqrt{2}}\right|=|h|, k^2-h^2+2 h k=0$

Topic: Mathematics, Circle, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A circular disc of radius $\mathrm{R}$ and thickness $\frac{R}{6}$ has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is

Options:
A $I$
B $\frac{2 I}{8}$
C $\frac{I}{5}$
D $\frac{I}{10}$

The correct answer is:
$\frac{I}{5}$

Solution:
According to problem disc is melted and recasted into a solid sphere so their volume

$$

\begin{array}{c}

\text { will be same. } \\

V_{\text {Disc }}=V_{\text {Sphere }} \Rightarrow \pi R_{\text {Disc }}^{2} t=\frac{4}{3} \pi R_{\text {Sphere }}^{3} \\

\Rightarrow \pi R_{\text {Disc }}^{2}\left(\frac{R_{\text {Disc }}}{6}\right)=\frac{4}{3} \pi R_{\text {Sphere }}^{3}\left[t=\frac{R_{\text {Disc }}}{6}, \text { given }\right] \\

\Rightarrow R_{\text {Disc }}^{3}=8 R_{\text {Sphere }}^{3} \Rightarrow R_{\text {Sphere }}=\frac{R_{\text {Disc }}}{2}

\end{array}

$$

Moment of inertia of disc

$$

\begin{array}{l}

I_{\text {Disc }}=\frac{1}{2} M R_{\text {Disc }}^{2}=I \text { (given) } \\

\therefore \text { M }\left(R_{\text {Disc }}\right)^{2}=2 I

\end{array}

$$

Moment of inertia of sphere

$$

\begin{array}{l}

I_{\text {Sphere }}=\frac{1}{2} M R_{\text {Sphere }}^{2} \\

=\frac{2}{5} M\left(\frac{R_{\text {Disc }}}{2}\right)^{2}=\frac{M}{10}\left(R_{\text {Disc }}\right)^{2}=\frac{2 I}{10}=\frac{I}{5}

\end{array}

$$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2016
A class has 175 students. The following data shows the number of students obtaining one or more subjects. Mathematics 100 , Physics 70 , Chemistry 40; Mathematics and Physics 30 , Mathematics and Chemistry 28 , Physics and Chemistry 23; Mathematics, Physics and Chemistry $18 .$ How many students have offered Mathematics alone?

Options:
A 35
B 48
C 60
D 22

The correct answer is:
60

Solution:
$\begin{aligned} & n(\mathrm{M} \text { alone }) \\ &=n(\mathrm{M})-n(\mathrm{M} \cap \mathrm{C})-n(\mathrm{M} \cap \mathrm{P})+n(\mathrm{M} \cap \mathrm{C} \cap \mathrm{P}) \\ &=100-28-30+18=60 \end{aligned}$

Topic: Mathematics, Sets and Relations, JEE Main
A class has 175 students. The following data shows the number of students obtaining one or more subjects. Mathematics 100 , Physics 70 , Chemistry 40; Mathematics and Physics 30 , Mathematics and Chemistry 28 , Physics and Chemistry 23; Mathematics, Physics and Chemistry $18 .$ How many students have offered Mathematics alone?

Options:
A 35
B 48
C 60
D 22

The correct answer is:
60

Solution:
$\begin{aligned} & n(\mathrm{M} \text { alone }) \\ &=n(\mathrm{M})-n(\mathrm{M} \cap \mathrm{C})-n(\mathrm{M} \cap \mathrm{P})+n(\mathrm{M} \cap \mathrm{C} \cap \mathrm{P}) \\ &=100-28-30+18=60 \end{aligned}$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2013
A coil 10 turns and a resistance of $20 \Omega$ is connected in series with B.G of resistance $30 \Omega$. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction $10^{-2} \mathrm{~T}$. If it is now turned through an angle of $60^{\circ}$ about an axis in its plane. Find the charge induced in the coil. (Area of a coil $=10^{-2} \mathrm{~m}^{2}$ )

Options:
A $2 \times 10^{-5} \mathrm{C}$
B $3.2 \times 10^{-5} \mathrm{C}$
C $1 \times 10^{-5} \mathrm{C}$
D $5.5 \times 10^{-5} \mathrm{C}$

The correct answer is:
$1 \times 10^{-5} \mathrm{C}$

Solution:
Given: $\mathrm{n}=10$ turns, $\mathrm{R}_{\text {coil }}=20 \Omega, \mathrm{R}_{\mathrm{G}}=30 \Omega$, Total resistance in the circuit $=20+30$

$=50 \Omega$

$\mathrm{A}=10^{-2} \mathrm{~m}^{2}, \mathrm{~B}=10^{-2} \mathrm{~T}, \phi_{1}=0^{\circ}, \phi_{2}=60^{\circ}$

$\mathrm{q}=\frac{\phi_{1}-\phi_{2}}{\mathrm{R}}=\frac{\mathrm{Bn} \mathrm{A} \cos \theta_{1}-\mathrm{Bn} \mathrm{A} \cos \theta_{2}}{\mathrm{R}}$

$=\frac{\operatorname{Bn} \mathrm{A}(\cos 0-\cos 60)}{\mathrm{R}}=\frac{\mathrm{BnA}(1-0.5)}{\mathrm{R}}$

$=\frac{0.5 \times 10^{-2} \times 10 \times 10^{-2}}{50}=\frac{50 \times 10^{-5}}{50}$

$=1 \times 10^{-5} \mathrm{C} \quad$ (Charge induced in a coil)

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2014
A coil has resistance 30 ohm and inductive reactance 20 ohm at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz},$ is connected across the coil, the current in the coil will be

Options:
A $4.0 \mathrm{~A}$
B $8.0 \mathrm{~A}$
C $7.2 \mathrm{~A}$
D $2.0 \mathrm{~A}$

The correct answer is:
$4.0 \mathrm{~A}$

Solution:
If $\omega=50 \times 2 \pi$ then $\omega \mathrm{L}=20 \Omega$ If $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} \mathrm{L}=40 \Omega$
Current flowing in the coil is
$$
\mathrm{I}=\frac{200}{\mathrm{Z}}=\frac{200}{\sqrt{\mathrm{R}^{2}+\left(\omega^{\prime} \mathrm{L}\right)^{2}}}=4 \mathrm{~A}
$$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2009
A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B is in plane of the coil. If due to a current i in the triangle, a torque τ rests on it, the side l of the triangle is

Options:
A 2τ3Bi12
B 23τBi
C 2τBi12
D 13τBi

The correct answer is:
2τ3Bi12

Solution:
Torque acting on equilateral triangle in a magnetic field B is

τ=i AB sin θ

Area of triangle LMN,

A=32l2 and θ=90°

Substituting the given values in expression for torque, we get

τ=i×34 l2 B sin 90°

τ=34il2B ∵ sin 90°=1

l2=4τ3 Bi

Here, l=2 τ3 Bi12

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2017
A coin is tossed 7 times. Each time a man calls head. Find the probability that he wins the toss on more occasions.

Options:
A $\frac{2}{3}$
B $\frac{1}{2}$
C $\frac{3}{4}$
D $\frac{1}{3}$

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

Solution:
The man has to win at least 4 times.

$$

\begin{array}{r}

\therefore \text { Reqd. probability }= \\

\qquad \begin{array}{r}

7 \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4} \cdot\left(\frac{1}{2}\right)^{3}+{ }^{7} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5} \cdot\left(\frac{1}{2}\right)^{2} \\

+{ }^{7} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6} \cdot \frac{1}{2}+{ }^{7} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{7} \\

=\left({ }^{7} \mathrm{C}_{4}+{ }^{7} \mathrm{C}_{5}+{ }^{7} \mathrm{C}_{6}+{ }^{7} \mathrm{C}_{7}\right) \cdot \frac{1}{2^{7}}=\frac{64}{2^{7}}=\frac{1}{2}

\end{array}

\end{array}

$$

Topic: Mathematics, Probability, BITSAT, BITSAT 2014
A coin is tossed 7 times. Each time a man calls head. The probability that he wins the toss at least four times is

Options:
A 14
B 58
C 12
D 16

The correct answer is:
12

Solution:
The man has to win atleast 4 times.

∴ Required probability

=C47124123+C57125122+C6712612+C77127

=C47+C57+C67+C77×127=6427=12

Topic: Mathematics, Probability, BITSAT, BITSAT 2017
A coin is tossed twice. Then, the probability that atleast one tail occurs is

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{1}{3}$
D $\frac{3}{4}$

The correct answer is:
$\frac{3}{4}$

Solution:
The sample space is $\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$
Let $\mathrm{E}$ be the event of getting atleast one tail
$\begin{aligned}
& \therefore \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}, \mathrm{TT}\} \\
& \therefore \text { Required probability } \mathrm{p} \\
& =\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{3}{4}
\end{aligned}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is

Options:
A 500
B 1000
C 1250
D 50

The correct answer is:
1250

Solution:
Voltage gain =β× impedance gain

⇒ 50=β×200100

⇒ β=25

Also power gain =β2× impedance gain

=252×200100=1250

Topic: Physics, Semiconductors, BITSAT, BITSAT 2017
A compound of Xe and $\mathrm{F}$ is found to have $53.5 \%$ of Xe. What is oxidation number of Xe in this compound?

Options:
A -4
B 0
C 4
D 6

The correct answer is:
6

Solution:
$\mathrm{Xe}=53.5 \% \therefore \mathrm{F}=46.5 \%$

Relative number of atoms Xe

$$

=\frac{53.5}{131.2}=0.4 \text { and } \mathrm{F}=\frac{46.5}{19}=2.4

$$

Simple ratio $\mathrm{Xe}=1$ and $\mathrm{F}=6 ;$ Molecular formula is $\mathrm{XeF}_{6}$

O. $\mathrm{N}$.of $\mathrm{Xe}$ is +6

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2016
A conducting circular loop is placed in a uniform magnetic field, B=0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s-1. The induced emf when the radius is 2 cm, is

Options:
A 2π μV
B π μV
C π2μV
D 2 μV

The correct answer is:
π μV

Solution:
Here, magnetic field,B=0.025 T

∴ Magnetic flux ϕ=B·A=B·πr2

Induced emf e=dϕdt

=Bπ2r·drdt

=0.025×π×2×2×10-2×1×10-3

=π×10-6 V

=π μV

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2017
A conducting circular loop is placed in a uniform magnetic field of $0.04 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 \mathrm{~mm} / \mathrm{s}$. The induced emf in the loop when the radius is $2 \mathrm{~cm}$ is

Options:
A $4.8 \pi \mu \mathrm{V}$
B $0.8 \pi \mu \mathrm{V}$
C $1.6 \pi \mu \mathrm{V}$
D $3.2 \pi \mu \mathrm{V}$

The correct answer is:
$3.2 \pi \mu \mathrm{V}$

Solution:
Induced emf in the loop is given by

$\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}$ where $\mathrm{A}$ is the area of the loop.

$$

\begin{array}{l}

\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(\pi \mathrm{r}^{2}\right)=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\

\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \\

\mathrm{dr}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\

\mathrm{dt}=1 \mathrm{~s} \\

\mathrm{e}=-0.04 \times 3.14 \times 2 \times 2 \times 10^{-2} \times \frac{2 \times 10^{-3}}{1} \mathrm{~V} \\

=0.32 \pi \times 10^{-5} \mathrm{~V} \\

=3.2 \pi \times 10^{-6} \mathrm{~V} \\

=3.2 \pi \mu \mathrm{V}

\end{array}

$$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2016