datavorous/entrance-exam-dataset huggingface dataset:
In an uniformly charged sphere of total charge Q and radius $\mathrm{R},$ the electric field $\mathrm{E}$ is plotted as function of distance from the centre. The graph which would correspond to the above will be

Options:
A
B
C
D

The correct answer is:

Solution:
Electric field inside the uniformly charged sphere varies linearly, $E=\frac{k Q}{R^{3}} r,(r \leq R)$ while outside the sphere, it varies as inverse
square of distance, $E=\frac{k Q}{r^{2}} ;(r \geq R)$ which is correctly represented in option (c).

Topic: Physics, Electrostatics, JEE Main
In an uniformly charged sphere of total charge Q and radius $\mathrm{R},$ the electric field $\mathrm{E}$ is plotted as function of distance from the centre. The graph which would correspond to the above will be

Options:
A
B
C
D

The correct answer is:

Solution:
Electric field inside the uniformly charged sphere varies linearly, $E=\frac{k Q}{R^{3}} r,(r \leq R)$ while outside the sphere, it varies as inverse
square of distance, $E=\frac{k Q}{r^{2}} ;(r \geq R)$ which is correctly represented in option (c).

Topic: Physics, Electrostatics, BITSAT, BITSAT 2011
In case of a forced vibration, the resonance wave becomes very sharp when the

Options:
A restoring force is small
B applied periodic force is small
C quality factor is small
D damping force is small

The correct answer is:
damping force is small

Solution:
No Solution available

Topic: Physics, Oscillations, BITSAT, BITSAT 2009
In CE transistor amplifier, the audio signal voltage across the collector resistance of $2 \mathrm{k} \Omega$ is $2 \mathrm{~V}$. If the base resistance is $1 \mathrm{k} \Omega$ and the current amplification of the transistor is $100,$ the input signal voltage is

Options:
A $2 \mathrm{mV}$
B $3 \mathrm{mV}$
C $10 \mathrm{mV}$
D $0.1 \mathrm{mV}$

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

Solution:
Given : Voltage across the collector $\mathrm{V}_{0}=2$ $\mathrm{V} ;$ collector resistance, $\mathrm{R}_{\mathrm{c}}=2 \times 10^{3} \Omega ;$ Base resistance $\mathrm{R}_{\mathrm{B}}=1 \times 10^{3} \Omega ;$ Input signal voltage, $\mathrm{V}_{\mathrm{i}}=?$

$$

\begin{array}{l}

\mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\

\Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{~A} \\

\text { Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\

\Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\

\quad \mathrm{~V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times 10^{3} \times 10^{-5} \\

\Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{~V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV}

\end{array}

$$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2015
In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block $A$ weighs $400 \mathrm{~N}$, block $B$ weighs $300 \mathrm{~N}$, and the strut $A B$ weigh $200 \mathrm{~N}$. If $\mu=0.25$ under $B,$ determine the minimum coefficient of friction under $A$ to prevent motion.

Options:
A 0.4
B 0.2
C 0.8
D 0.1

The correct answer is:
0.4

Solution:
Consider FBD of structure.

Applying equilibrium equations, $A v+B v=200 \mathrm{~N}$

$A_{H}=B_{H} \ldots$ (ii)

From FBD of block $B$,

$B_{H}+F_{B} \cos 60^{\circ}-N_{B} \sin 60^{\circ}=0$

$N_{B} \cos 60^{\circ}-B_{V}-300+F_{B} \sin 60^{\circ}=0$

$F_{B}=0.25 N_{B}$

$B_{H}-0.74 N_{B}=0$

(iii) $-B_{V}+0.71 N_{B}=300$

FBD of block $A$

$F_{A}-A_{H}=0$

$N_{A}-A_{V}=400 \quad \ldots$

$F_{A}=\mu_{A} N_{A}$

$\therefore \mu_{A} N_{A}-A_{H}=0 \quad \ldots$

on solving the above equation

$N_{A}=650 \mathrm{~N}, F_{A}=260 \mathrm{~N}, F_{A}=\mu_{A} N_{A}$

$\therefore \mu_{A}=\frac{260}{250}=0.4$

Topic: Physics, Laws of Motion, JEE Main
In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block $A$ weighs $400 \mathrm{~N}$, block $B$ weighs $300 \mathrm{~N}$, and the strut $A B$ weigh $200 \mathrm{~N}$. If $\mu=0.25$ under $B,$ determine the minimum coefficient of friction under $A$ to prevent motion.

Options:
A 0.4
B 0.2
C 0.8
D 0.1

The correct answer is:
0.4

Solution:
Consider FBD of structure.

Applying equilibrium equations, $A v+B v=200 \mathrm{~N}$

$A_{H}=B_{H} \ldots$ (ii)

From FBD of block $B$,

$B_{H}+F_{B} \cos 60^{\circ}-N_{B} \sin 60^{\circ}=0$

$N_{B} \cos 60^{\circ}-B_{V}-300+F_{B} \sin 60^{\circ}=0$

$F_{B}=0.25 N_{B}$

$B_{H}-0.74 N_{B}=0$

(iii) $-B_{V}+0.71 N_{B}=300$

FBD of block $A$

$F_{A}-A_{H}=0$

$N_{A}-A_{V}=400 \quad \ldots$

$F_{A}=\mu_{A} N_{A}$

$\therefore \mu_{A} N_{A}-A_{H}=0 \quad \ldots$

on solving the above equation

$N_{A}=650 \mathrm{~N}, F_{A}=260 \mathrm{~N}, F_{A}=\mu_{A} N_{A}$

$\therefore \mu_{A}=\frac{260}{250}=0.4$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2015
In Freundlich adsorption isotherm, the value of 1/n is

Options:
A between 0 and 1 in all cases
B between 2 and 4 in all cases
C always 1 in case of physical adsorption
D always 1 in use of chemical adsorption

The correct answer is:
between 0 and 1 in all cases

Solution:
Freundlich isotherm is related with the pressure of gas adsorbed, as follows xm=k·p1/n

where, xm= amount of gas x adsorbed over the surface of mass m.

p= pressure of gas

k= freundlich constant

Thus, value of 'n' is always greater than 1 and therefore value of 1/n lies between 0 and 1.

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2019
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together?

Options:
A $4 !$
B $5 !$
C $4 !+5 !$
D $4 ! \times 5 !$

The correct answer is:
$4 ! \times 5 !$

Solution:
Leaving one seat vacant between two boys, 5 boys may be seated in $4 !$ ways. Then at remaining 5 seats, 5 girls any sit in $5 !$ ways. Hence the required number $=4 ! \times 5 !$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2013
In how many ways can 5 prizes be distributed among 4 boys when every boy can take one or more prizes?

Options:
A 1024
B 625
C 120
D 600

The correct answer is:
1024

Solution:
First prize may be given to any one of the 4 boys, hence first prize can be distributed in 4 ways. similarly every one of second, third, fourth and fifth prizes can also be given in 4 ways. the number of ways of their distribution $=4 \times 4 \times 4 \times 4 \times 4=4^{5}=1024$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2012
In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?

Options:
A 246
B 222
C 186
D None of these

The correct answer is:
246

Solution:
A committee of 5 out of $6+4=10$ can be made in ${ }^{10} \mathrm{C}_{5}=252$ ways. If no woman is to be included, thennumber of ways $={ }^{5} \mathrm{C}_{5}=6$

$\therefore$ the required number $=252-6=246$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2014
In non-reducing disaccharide, the reducing group of monosaccharides i.e. aldehydic or ketonic group are bounded.

Which of the following disaccharide is a non-reducing sugar?

Options:
A
B
C
D

The correct answer is:

Solution:
This structure represents sucrose in which α-D-glucose and β-D-fructose is attached to each other by C1-C2 glycosidic linkage. Since reducing groups of glucose and fructose are involved in glycosidic bond formation, this is considered as non-reducing sugar.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2018
In paper chromatography:

Options:
A Mobile phase is liquid and stationary phase is solid.
B Mobile phase is solid and stationary phase is liquid.
C Both phases are liquids.
D Both phases are solids.

The correct answer is:
Both phases are liquids.

Solution:
Paper chromatography is a special case of partition chromatography where the special quality paper containing water trapped in it acts as a stationary phase and solvent as a mobile phase. Thus, both phases are liquids.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2012
In $\mathrm{PO}_{4}^{3-},$ the formal charge on each oxygen atom and the $\mathrm{P}-\mathrm{O}$ bond order respectively are

Options:
A -0.75,0.6
B -0.75,1.0
C -0.75,1.25
D -3,1.25

The correct answer is:
-0.75,1.25

Solution:
Bond order

$=\frac{\text { Number of bonds }}{\text { Number of Resonating structures }}$

$=\frac{5}{4}=1.25$

Three unit negative charge is being shared by four $\mathrm{O}$ atoms. Formal charge $=-3 / 4$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In $\mathrm{PO}_{4}^{3-},$ the formal charge on each oxygen atom and the $\mathrm{P}-\mathrm{O}$ bond order respectively are

Options:
A -0.75,0.6
B -0.75,1.0
C -0.75,1.25
D -3,1.25

The correct answer is:
-0.75,1.25

Solution:
Bond order

$=\frac{\text { Number of bonds }}{\text { Number of Resonating structures }}$

$=\frac{5}{4}=1.25$

Three unit negative charge is being shared by four $\mathrm{O}$ atoms. Formal charge $=-3 / 4$

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2016
In reaction $\mathrm{BF}_3+3 \mathrm{LiBH}_4 \rightarrow 3 \mathrm{LiF}+X ; X$ is

Options:
A $\mathrm{B}_4 \mathrm{H}_{10}$
B $\mathrm{B}_2 \mathrm{H}_6$
C $\mathrm{BH}_3$
D $\mathrm{B}_3 \mathrm{H}_8$

The correct answer is:
$\mathrm{B}_2 \mathrm{H}_6$

Solution:
$2 \mathrm{BF}_3+3 \mathrm{LiBH}_4 \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiF}$

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2023 (Memory Based Paper 2)
In redox reaction $1 \mathrm{~g}$ -eq of reducing agent requires $\mathrm{Pgm}-\mathrm{eq} .$ of oxidising agent. The value of $\mathrm{P}$ is

Options:
A 1
B 2
C 3
D Depends on reaction

The correct answer is:
1

Solution:
In redox reaction, $g$ equivalent of reducing agent $=\mathrm{g}$. equivalent of oxidising agent Hence lg equ. of reducing agent $=\mathrm{Pg}$ equ. of oxidising agent.

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2010
In series combination of $R, L$ and $C$ with an A.C. source at resonance, if $\mathrm{R}=20 \mathrm{ohm},$ then impedence $Z$ of the combination is

Options:
A $20 \mathrm{ohm}$
B zero
C 10 ohm
D $400 \mathrm{ohm}$

The correct answer is:
$20 \mathrm{ohm}$

Solution:
No Solution

Topic: Physics, Alternating Current, BITSAT, BITSAT 2020
In the anion $\mathrm{HCOO}^{-}$ the two carbon-oxygen bonds are found to be of equal length. What is the reason for it?

Options:
A The $\mathrm{C}=\mathrm{O}$ bond is weaker than the $\mathrm{C}-\mathrm{O}$ bond.
B The anion $\mathrm{HCOO}^{-}$ has two resonating structures.
C The anion is obtained byremoval of a proton from the acid molecule.
D Electronic orbitals of carbon atom are hybridised.

The correct answer is:
The anion $\mathrm{HCOO}^{-}$ has two resonating structures.

Solution:

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2011
In the arrangement shown in figure, when the switch $S_2$ is open, the galvanometer, shows no deflection for $l=50 \mathrm{~cm}$ when the switch $S_2$ is closed, the galvanometer shows no deflection for $l=0.416 \mathrm{~m}$. The internal resistance $(r)$ of $6 \mathrm{~V}$ cell is

Options:
A
$2 \Omega$

B
$3 \Omega$

C
$5 \Omega$

D
$9 \Omega$

The correct answer is:

$2 \Omega$

Solution:
When $S_2$ open.
Assume resistance of $X Y=R$.
Resistance of wire per unit length, $x=\frac{R}{L}=R \Omega \mathrm{m}^{-1}$
$\because \quad I=E_0 / R$
Now, the potential drop across $50 \mathrm{~cm}$ length is $6 \mathrm{~V}$, So
$$
\begin{array}{cc}
& \frac{E_0}{R} \times R \times \frac{50}{100}=6 \\
\Rightarrow & E_0=12 \mathrm{~V}
\end{array}
$$
When $S_2$ closed, potential drop across $0.416 \mathrm{~cm}$ length,
$$
V_1=\frac{E_0}{R} \times R \times 0.416=12 \times 0.416 \approx 5 \mathrm{~V}
$$
Hence, $\quad E-I r=5 \mathrm{~V}$
$$
\begin{aligned}
\Rightarrow & & 6-I r & =5 \\
& \because & I & =\frac{5}{10} \\
\therefore & & 6-5 & =\frac{5}{10} r \\
\Rightarrow & & r & =2 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2022
In the binomial \(\left(2^{1 / 3}+3^{-1 / 3}\right)^{\mathrm{n}}\), if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is \(1 / 6\), then \(n\) equal to

Options:
A 6
B 9
C 12
D 15

The correct answer is:
9

Solution:
\(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} a^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{b}^{\mathrm{r}}\) where \(\mathrm{a}=2^{1 / 3}\) and \(\mathrm{b}=3^{-1 / 3}\)
\(\mathrm{T}_7\) from beginning \(={ }^n \mathrm{C}_6 a^{\mathrm{n}-6} \mathrm{~b}^6\) and \(\mathrm{T}_7\) from end \(={ }^{\mathrm{n}} \mathrm{C}_6 \quad \mathrm{~b}^{\mathrm{n}-6} \quad \mathrm{a}^6\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{a}^{\mathrm{n}-12}}{\mathrm{~b}^{\mathrm{n}-12}}=\frac{1}{6} \Rightarrow 2^{\frac{\mathrm{n}-12}{3}} \cdot 3^{\frac{\mathrm{n}-12}{3}}=6^{-1} \\
& \Rightarrow \mathrm{n}-12=-3 \Rightarrow \mathrm{n}=9
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2011
In the cell reaction

$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$

$\mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V}$. By doubling the concentration of $\mathrm{Cu}^{2+}, \mathrm{E}_{\text {cell }}^{0}$ is

Options:
A doubled
B halved
C increases but less than double
D decreases by a small fraction

The correct answer is:
decreases by a small fraction

Solution:
No Solution

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2014
In the circuit given below, the charge in $\mu \mathrm{C},$ on the capacitor having $5 \mu \mathrm{F}$ is

Options:
A 4.5
B 9
C 7
D 15

The correct answer is:
9

Solution:
Potential difference across the branch de is

$6 \mathrm{~V}$. Net capacitance of de branch is $2.1 \mu \mathrm{F}$ So, $q=C V$ $\Rightarrow \mathrm{q}=2.1 \times 6 \mu \mathrm{C} \quad \Rightarrow \mathrm{q}=12.6 \mu \mathrm{C}$

Potential across $3 \mu \mathrm{F}$ capacitance is $\mathrm{V}=\frac{12.6}{3}=4.2 \mathrm{volt}$

Potential across 2 and 5 combination in parallel is $6-4.2=1.8 \mathrm{~V}$

So, $q^{\prime}=(1.8)(5)=9 \mu C$

Topic: Physics, Capacitance, JEE Main
In the circuit given below, the charge in $\mu \mathrm{C},$ on the capacitor having $5 \mu \mathrm{F}$ is

Options:
A 4.5
B 9
C 7
D 15

The correct answer is:
9

Solution:
Potential difference across the branch de is

$6 \mathrm{~V}$. Net capacitance of de branch is $2.1 \mu \mathrm{F}$ So, $q=C V$ $\Rightarrow \mathrm{q}=2.1 \times 6 \mu \mathrm{C} \quad \Rightarrow \mathrm{q}=12.6 \mu \mathrm{C}$

Potential across $3 \mu \mathrm{F}$ capacitance is $\mathrm{V}=\frac{12.6}{3}=4.2 \mathrm{volt}$

Potential across 2 and 5 combination in parallel is $6-4.2=1.8 \mathrm{~V}$

So, $q^{\prime}=(1.8)(5)=9 \mu C$

Topic: Physics, Capacitance, BITSAT, BITSAT 2020
In the circuit given below, the value of resistance X, when the potential difference between the points B and D is zero, will be

Options:
A 9 Ω
B 8 Ω
C 4 Ω
D 6 Ω

The correct answer is:
8 Ω

Solution:
P=15+6=21 Ω, Q=8X8+X+3

R=15+6×66+6=18 Ω, S=4+4×44+4=6 Ω

Since potential difference across BD is zero hence it satisfies Wheatstone's Bridge condition,

∴PQ=RS, so Q=PSR=21×618=7 Ω

3+8x8+x=7⇒x=8 Ω

Topic: Physics, Current Electricity, BITSAT, BITSAT 2018
In the circuit shown assume the diode to be ideal. When $V_i$ increases from $-2 \mathrm{~V}$ to $6 \mathrm{~V}$, the change in current is (in $\mathrm{mA})$

Options:
A
Zero

B
$20$

C
$\frac{25}{8}$

D
$32$

The correct answer is:

$20$

Solution:
$$
l_{\text {initial }}=0 \text { for } V_i \leq+1 \mathrm{~V}
$$
This diode conduct only beyond $V_i=+1 \mathrm{~V}$
$$
I_{\text {final }}=\frac{5}{250}=0.02 \mathrm{~A}
$$
So, change in $I=0.02 \mathrm{~A}=20 \mathrm{~mA}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2022
In the circuit shown below, the ac source has voltage $V=20 \cos (\omega t)$ volt with $\omega=2000 \mathrm{rad} / \mathrm{s}$

The amplitude of the current will be nearest to

Options:
A $2 \mathrm{~A}$
B $3.3 \mathrm{~A}$
C $2 / \sqrt{5} \mathrm{~A}$
D $\sqrt{5} \mathrm{~A}$

The correct answer is:
$2 \mathrm{~A}$

Solution:
$$

\begin{aligned}

Z=& \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\

&=\sqrt{10^{2}+\left(2000 \times 5 \times 10^{3}-\frac{1}{2000 \times 50 \times 10^{-6}}\right)^{2}} \\

=& 10 \Omega \\

\quad i=\frac{V_{0}}{Z}=\frac{20}{10}=2 A .

\end{aligned}

$$

Topic: Physics, Alternating Current, JEE Main
In the circuit shown below, the ac source has voltage $V=20 \cos (\omega t)$ volt with $\omega=2000 \mathrm{rad} / \mathrm{s}$

The amplitude of the current will be nearest to

Options:
A $2 \mathrm{~A}$
B $3.3 \mathrm{~A}$
C $2 / \sqrt{5} \mathrm{~A}$
D $\sqrt{5} \mathrm{~A}$

The correct answer is:
$2 \mathrm{~A}$

Solution:
$$

\begin{aligned}

Z=& \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\

&=\sqrt{10^{2}+\left(2000 \times 5 \times 10^{3}-\frac{1}{2000 \times 50 \times 10^{-6}}\right)^{2}} \\

=& 10 \Omega \\

\quad i=\frac{V_{0}}{Z}=\frac{20}{10}=2 A .

\end{aligned}

$$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2015
In the circuit shown in figure potential difference between points $A$ and $B$ is $16 \mathrm{~V}$. the current passing through $2 \Omega$ resistance will be

Options:
A $2.5 \mathrm{~A}$
B $3.5 \mathrm{~A}$
C $4.0 \mathrm{~A}$
D zero

The correct answer is:
$3.5 \mathrm{~A}$

Solution:
$\therefore 4 \mathrm{i}_{1}+2\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)-3+4 \mathrm{i}_{1}=16 \mathrm{~V}$

Using Kirchhoff's second law in the closed loop we have

$$

9-i_{2}-2\left(i_{1}+i_{2}\right)=0

$$

Solving equations (i) and (ii), we get $\mathrm{i}_{1}=1.5 \mathrm{~A}$ and $\mathrm{i}_{2}=2 \mathrm{~A}$

$\therefore$ current through $2 \mathrm{~W}$ resistor $=2+1.5=3.5 \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in figure potential difference between points $A$ and $B$ is $16 \mathrm{~V}$. the current passing through $2 \Omega$ resistance will be

Options:
A $2.5 \mathrm{~A}$
B $3.5 \mathrm{~A}$
C $4.0 \mathrm{~A}$
D zero

The correct answer is:
$3.5 \mathrm{~A}$

Solution:
$\therefore 4 \mathrm{i}_{1}+2\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)-3+4 \mathrm{i}_{1}=16 \mathrm{~V}$

Using Kirchhoff's second law in the closed loop we have

$$

9-i_{2}-2\left(i_{1}+i_{2}\right)=0

$$

Solving equations (i) and (ii), we get $\mathrm{i}_{1}=1.5 \mathrm{~A}$ and $\mathrm{i}_{2}=2 \mathrm{~A}$

$\therefore$ current through $2 \mathrm{~W}$ resistor $=2+1.5=3.5 \mathrm{~A}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2013
In the circuit shown in the figure, find the current in $45 \Omega$.

Options:
A $4 \mathrm{~A}$
B $2.5 \mathrm{~A}$
C $2 \mathrm{~A}$
D None of these

The correct answer is:
$2 \mathrm{~A}$

Solution:
$i=\frac{180}{90}=2 A$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure, find the current in $45 \Omega$.

Options:
A $4 \mathrm{~A}$
B $2.5 \mathrm{~A}$
C $2 \mathrm{~A}$
D None of these

The correct answer is:
$2 \mathrm{~A}$

Solution:
$i=\frac{180}{90}=2 A$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2016
In the circuit shown in the figure, the $\mathrm{AC}$ source gives a voltage $V=20 \cos (2000 t)$ neglecting source resistance, the voltmeter and ammeter reading will be

Options:
A
$0 \mathrm{~V}, 0.47 \mathrm{~A}$

B
$282 \mathrm{~V}, 1.41 \mathrm{~A}$

C
$1.41 \mathrm{~V}, 0.47 \mathrm{~A}$

D
$15 \mathrm{~V}, 837 \mathrm{~A}$

The correct answer is:

$282 \mathrm{~V}, 1.41 \mathrm{~A}$

Solution:
Given, $R_1=8 \Omega$
$\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}$
and $\quad V_0=20 \cos (2000 t)$
Impedance, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
As, here, $X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega$
Similarly, $X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega$ $\because \quad X_L=X_C$, Hence, $i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{~A}$
Hence, $\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{~A}$ and
$$
\begin{aligned}
V & =R_2 i_{\mathrm{rms}}=1.41 \times 2 \\
& =282 \mathrm{~V}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
In the circuit shown in the figure, the $\mathrm{AC}$ source gives a voltage $V=20 \cos (2000 t)$ neglecting source resistance, the voltmeter and ammeter reading will be

Options:
A
$0 \mathrm{~V}, 0.47 \mathrm{~A}$

B
$282 \mathrm{~V}, 1.41 \mathrm{~A}$

C
$1.41 \mathrm{~V}, 0.47 \mathrm{~A}$

D
$15 \mathrm{~V}, 837 \mathrm{~A}$

The correct answer is:

$282 \mathrm{~V}, 1.41 \mathrm{~A}$

Solution:
Given, $R_1=8 \Omega$
$\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}$
and $\quad V_0=20 \cos (2000 t)$
Impedance, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
As, here, $X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega$
Similarly, $X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega$ $\because \quad X_L=X_C$, Hence, $i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{~A}$
Hence, $\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{~A}$ and
$$
\begin{aligned}
V & =R_2 i_{\mathrm{rms}}=1.41 \times 2 \\
& =282 \mathrm{~V}
\end{aligned}
$$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2022
In the circuit shown, the heat produced in 5 Ω resistor is 10 cal s-1. The heat produced per sec in 4 Ω resistor will be

Options:
A 1 cal
B 2 cal
C 3 cal
D 4 cal

The correct answer is:
2 cal

Solution:
H=I2Rt

∴ H1H2=l1I22×R1R2

or 10H2=112×54

H2=2 cal

Topic: Physics, Current Electricity, BITSAT, BITSAT 2017
In the compound,

CH2(I)=CH(II)-C≡CH(III)

The most acidic hydrogen atom is

Options:
A Only I
B Only II
C Only III
D All are equally acidic

The correct answer is:
Only III

Solution:
The hydrogen attached with most electronegative carbon will be the most acidic.

The hydrogen, attached with sp-hybrid C-atom is most acidic because of highest electronegativity of sp hybrid carbon atom, H atom becomes relatively more acidic.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2017
In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is incorrect.

Options:
A CO and CO2 are produced in the process.
B Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity.
C Al3+ is reduced at the cathode to form Al.
D Na3AlF6 serves as the electrolyte.

The correct answer is:
Na3AlF6 serves as the electrolyte.

Solution:
(i) In Hall-Heroult process for extraction of Al, carbon anode is oxidised to CO and CO2.

(ii) When Al2O3 is mixed with CaF2, it lowers the melting point of the mixture and brings conductivity.

(iii) Al3+ is reduced at cathode to form Al.

(iv) Al2O3 serves as the electrolyte, undergoing the redox process. Na3AlF6 although is an electrolyte but serves as a solvent, not electrolyte.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2018
In the diagram shown below, both the strings $A B$ and $C D$ are made of same material and have same cross-section. The pulleys are light and fictionless. If the speed of wave in string $A B$ is $v_1$ and in $C D$ is $v_2$, then $\frac{v_1}{v_2}$ is

Options:
A
1

B
$\sqrt{2}$

C
2

D
$1 / \sqrt{2}$

The correct answer is:

$1 / \sqrt{2}$

Solution:
Let the speed of wave in string be $v$ and tension in string be $T$.
$$
\begin{gathered}
v \propto \sqrt{T} \\
\frac{v_{A B}}{v_{C D}}=\sqrt{\frac{T}{2 T}}=\frac{1}{\sqrt{2}}
\end{gathered}
$$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2022
In the equation $\mathrm{PV}^{\gamma}=$ constant, the value of $\gamma$ is unity. Then the process is

Options:
A isothermal
B adiabatic
C isobaric
D irreversible

The correct answer is:
isothermal

Solution:
$\mathrm{PV}=$ constant represents isothermal process.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2010
In the expansion of 1+x+x3+x410, the coefficient of x4 is

Options:
A 235
B 310
C 285
D 325

The correct answer is:
310

Solution:
1+x+x3+x410=1+x1+x310

=1+x101+x310

=1+C110x+C210x2+C310x3+C410x4…×1+C110x3+C210x6+….

∴ Coefficient of x4=C110C110+C410

=100+210=310

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2019