datavorous/entrance-exam-dataset huggingface dataset:
In a Carnot's engine, as the gas absorbs heat energy from the source, then the temperature of the source

Options:
A decreases
B increases
C remains constant
D becomes zero

The correct answer is:
remains constant

Solution:
In a carnot's engine, as the gas absorbs heat energy from the source, then the temperature of the source remain constant.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2023 (17 May Shift 1)
In a Carnot's engine, as the gas gives heat energy to the sink, then the temperature of the sink

Options:
A increases
B decreases
C remains constant
D becomes zero

The correct answer is:
remains constant

Solution:
In carnot engine, sink is thermal energy reservoir hence as heat is rejected or extracted, temperature does not change.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2023 (16 May Shift 2)
In a Carnot's engine if the temperatures of the source and the sink are decreased by $100 \mathrm{~K}$ each then the efficiency of the engine

Options:
A increases
B decreases
C remains constant
D becomes one

The correct answer is:
increases

Solution:
Initial efficiency is given by
$\eta=1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$
Decreased by $100 \mathrm{~K}$ each the efficiency
$\begin{aligned} & \eta=1-\frac{T_2-100}{T_1-100} \\ & =\frac{T_1-100-T_2+100}{T_1-100}=\frac{T_1-T_2}{T_1-100}\end{aligned}$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2023 (17 May Shift 2)
In a catalytic reaction \( \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \) (all in gaseous state)
rate \( =\frac{\Delta\left[\mathrm{NH}_{3}\right]}{\Delta \mathrm{t}}=4.0 \times 10^{-4} \) mole \( \mathrm{L}^{-1} \mathrm{~S}^{-1} \)
The rate of reaction when expressed in terms of change in the concentration of (i) \( \mathrm{N}_{2} \) (ii) \( \mathrm{H}_{2} \) are, respectively
(Assume that no side reactions have taken place).

Options:
A \( 2 \times 10^{-4} ; 6 \times 10^{-4} \)
B \( 1 \times 10^{-4} ; 3 \times 10^{-4} \)
C \( 2 \times 10^{-4} ; 3 \times 10^{-4} \)
D \( 4 \times 10^{-4} ; 3 \times 10^{-4} \)

The correct answer is:
\( 2 \times 10^{-4} ; 6 \times 10^{-4} \)

Solution:
N2+3H2→2NH3Rate = ∆[NH3]∆t = 4.0 ×10-4Change in the concentration of N2:- -∆[N2]∆t = 12 ∆[NH3]∆t ∆[N2]∆t = 10.5×4.0 ×10-4=2.0 ×10-4Change in the concentration of H2:--13∆[H2]∆t = 12 ∆[NH3]∆t ∆[H2]∆t = 32×4.0 ×10-4=6.0 × 10-4

Topic: Chemistry, Chemical Kinetics, JEE Main
In a cathode ray tube if $\mathrm{V}$ is the potential difference between the cathode and anode, the speed of the electrons, when they reach the anode is proportional to (Assume initial velocity $=0$ )

Options:
A $\frac{1}{V}$
B $\sqrt{\mathrm{V}}$
C $\left(\frac{v^2}{2 e m}\right)$
D $V$

The correct answer is:
$\sqrt{\mathrm{V}}$

Solution:
Here when an electron is accelerated by the potential difference towards the anode then the kinetic energy of electrons is equal to work done by potential on electrons
$\begin{aligned}& \frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \\& \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\& \mathrm{v}=\sqrt{\mathrm{V}}\end{aligned}$
where $\mathrm{e} ~\&~ \mathrm{~m}$ are constant
Answer is $\mathrm{C}$.

Topic: Physics, Dual Nature of Matter, JEE Main
In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kÎ© is 2 V, if the base resistance is 1 kÎ© and the current amplification of the transistor is 100 then the input signal voltage is

Options:
A 10 mV
B 1 mV
C 0.1 V
D 1 V

The correct answer is:
10 mV

Solution:
Current amplification factor,

β=ΔICΔIB

Collector current,

ΔIC=2V2×103Ω=1×10-3 A

Base current,

ΔIB=VBRB=VB1×103=VB×10-3

Given, β=100

Now, 100=10-3VB×10-3

⇒VB=1100 V=10 mV

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (12 Apr Shift 2)
In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kÎ© is 2 V, if the base resistance is 1 kÎ© and the current amplification of the transistor is 100 then the input signal voltage is

Options:
A 10 mV
B 1 mV
C 0.1 V
D 1 V

The correct answer is:
10 mV

Solution:
We know that voltage amplification is given by,
V0Vin=Î²RLRi; 2Vm=1002Ã1031Ã103

Topic: Physics, Semiconductors, JEE Main
In a cell that utilises the reaction
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})$
addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ to cathode compartment, will

Options:
A increase the $\mathrm{E}$ and shift equilibrium to the right.
B lower the $E$ and shift equilibrium to the right.
C lower the $E$ and shift equlibrium to the left.
D increase the $E$ and shift equilibrium to the left.

The correct answer is:
increase the $\mathrm{E}$ and shift equilibrium to the right.

Solution:
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}+(\mathrm{aq}) \rightleftharpoons$
$$
\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})
$$
$$
\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\circ} \mathrm{cell}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{H}^{+}\right]^{2}}
$$
Addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will increase $\left[\mathrm{H}^{+}\right]$ and $\mathrm{E}_{\text {cell }}$ will also increase and the equilibrium will shift towards RHS.

Topic: Chemistry, Electrochemistry, JEE Main
In a cell that utilises the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g})$ addition of $\mathrm{H}_2 \mathrm{SO}_4$ to cathode compartment, will

Options:
A
lower the $E$ and shift equilibrium to the left

B
increases the $E$ and shift equilibrium to the left

C
increase the $E$ and shift equilibrium to the right

D
Lower the $E$ and shift equilibrium to the right

The correct answer is:
increase the $E$ and shift equilibrium to the right

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
In a cell that utilizes the reaction
Zn(s) +  2H + (aq)→ Zn 2+ (aq) +  H 2 (g)
addition of H2SO4 to cathode compartment,

Options:
A Lowers the E and shifts equilibrium to the left
B Lowers the E and shifts the equilibrium to the right
C Increases the E and shifts the equilibrium to the right
D Increases the E and shifts the equilibrium to the left

The correct answer is:
Increases the E and shifts the equilibrium to the right

Solution:
Zn(s) +  2H + (aq)→ Zn 2+ (aq) +  H 2 (g)
E cell  =  E cell o − 0.059 2 log  Zn 2+  ×  p H 2 H + 2
On adding H 2 SO 4 , the H + will increase therefore E cell will also increase and the equilibrium will shift towards the right.

Topic: Chemistry, Electrochemistry, JEE Main
In a cell, the number of chromosomes is the same, but the amount of DNA content has changed from C to 2C. Also, the amount of histones in the cell becomes double. In which stage of the cell cycle, this process could have happened?

Options:
A G1 phase
B S phase
C G2 phase
D G0 phase

The correct answer is:
S phase

Solution:
G1 phase corresponds to the interval between mitosis and initiation of DNA replication. During G1 phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA and histones per cell doubles. If the initial amount of DNA is denoted as C then it increases to 2C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G1, even after the S phase the number of chromosomes remains the same, i.e., 2n.

Topic: Biology, Cell Cycle and Cell Division, NEET
In a certain culture of bacteria, the rate of increase is proportional to the number
present. It is found that the number doubles in 4 hours. Then the number of times
the bacteria are increased in 12 hours is

Options:
A 6
B 8
C 12
D 4

The correct answer is:
8

Solution:
Let $x$ be the number of bacteria in certain culture at time $t$. Then rate of increase is $\frac{d x}{d t}$ which is proportional to $x$.
$\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{Kx} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{Kdt} \Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{K} \int \mathrm{dt} \\
\therefore \log \mathrm{x}=\mathrm{Kt}+\mathrm{c} ...(1)\\
\text { Initially when } \mathrm{t}=0, \operatorname{let} \mathrm{x}=\mathrm{x}_{0} \\
\therefore \log \mathrm{x}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{x}_{0} \\
\therefore \text { from }(1) \log \mathrm{x}=\mathrm{kt}+\log \mathrm{x}_{0} \\
\therefore \log \left(\frac{\mathrm{x}}{\mathrm{x}_{0}}\right)=\mathrm{Kt}...(2)
\end{array}$
Given number doubles in 4 hrs i.e. when $t=4, x=2 x_{0}$ $\therefore \log \left(\frac{2 \mathrm{x}_{0}}{\mathrm{x}_{0}}\right)=4 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{1}{4} \log 2$
$\therefore$ from (2) $\log \left(\frac{x}{x_{0}}\right)=\frac{t}{4} \log 2$
When $\mathrm{t}=12$
$\log \left(\frac{x}{x_{0}}\right)=\frac{12}{4} \log 2=3 \log 2=\log 2^{3}=\log 8$
$\therefore \frac{x}{x_{0}}=8 \Rightarrow x=8 x_{0}$
This problem can be alternatively solved as follows :
Let the intial number of bacteria be $\mathrm{x}$.
$\therefore$ Number of bacteria after $4 \mathrm{hrs}=2 \mathrm{x}$
Number of bacteria after $8 \mathrm{hrs}=2(2 \mathrm{x})=4 \mathrm{x}$
Number of bacteria after $12 \mathrm{hrs}=2(4 \mathrm{x})=8 \mathrm{x}$

Topic: Mathematics, Differential Equations, JEE Main
In a certain region static electric and magnetic fields exist. The magnetic field is given by B→=B0i^+2j^-4k^. If a test charge moving with a velocity v→=v03i^-j^+2k^ experiences no force in that region, then the electric field in the region, in SI units, is:

Options:
A E→= -v0 B0i^+j^+7k^
B E→= -υ0 B03i^-2j^-4k^
C E→= v0 B014j^+7k^
D E→=-v0 B014j^+7k^

The correct answer is:
E→=-v0 B014j^+7k^

Solution:
The total force experienced by test charge is Fe→+ F→m=0.
Fe=-Fm
Fe=−q0v0B0V→×B→
Fe=-q0 v0B03i^- j^+2k^×i^+2j^-4k^
F→e= -qv0B0 (14j^ +7k^)
The electric field produced by the test charge is ∵ E→= f→eq= -v0B0 ( 14j^+7k^)

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2017 (08 Apr Online)
In a certain test, there are n questions. In this test $2^{n-i}$ students gave wrong answers to at least i questions, where i $=1,2, \ldots \ldots \mathrm{n}$. If the total number of wrong answers given is 2047 , then $\mathrm{n}$ is equal to

Options:
A 10
B 11
C 12
D 13

The correct answers are:
11

Solution:
Hint:
Total students was gave wrong answer to exactly i - questions $=2^{n-i}-2^{n-(i+1)}$
Total wrong answer given \(\sum_{i=0}^{n-1} i \times \left(2^{n-i}-2^{n-(i+1)}\right)+n \times 1\)
$\Rightarrow 2^{n-1}+\ldots \ldots 1=2047$
$\Rightarrow 2^{n}=2048$
$\Rightarrow \mathrm{n}=11$

Topic: Mathematics, Permutation Combination, JEE Main
In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 100 °C to 300 °C will be xnR where n denotes number of moles of a gas find x ;

Options:

The correct answer is:
50

Solution:
P=kV3

Ti=100 °C & Tf=300 °C

ΔT=300-100

ΔT=200 °C

P=kV3

now PV=nRT

∴ kV4=nRT

now 4kV3dV=nRdT

∴ PdV=nRdT/4

∴ Work =∫PdV=∫nRdT4=nR4ΔT

=2004×nR=50nR

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
In a certain town 25 % families own a cell phone, 15 % families own a scooter and 65 % families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is

Options:
A 10000
B 20000
C 30000
D 40000

The correct answer is:
30000

Solution:
Let the total population of town be x.
â´ 25x100+15x100-1500+65x100=x
â 105x100-x=1500
â 5x100=1500
â x=30000

Topic: Mathematics, Sets and Relations, JEE Main
In a certain town, 25% of the families own a phone and 15% own a car; 65% families own neither a phone nor a car and 2000 families own both a car and a phone. Consider the following three statements:
i 5% families own both a car and a phone.
ii 35% families own either a car or a phone.
iii 40000 families live in the town.
Then,

Options:
A Only ii and iii are correct
B Only i and ii are correct
C All i, ii and iii are correct
D Only i and iii are correct

The correct answer is:
All i, ii and iii are correct

Solution:
Let set A contains families which own a phone and set B contain families which own a car.
If 65% families own neither a phone nor a car, then 35% will own either a phone or a car
∴nA∪B=35%
Also we know that
nA∪B=nA+nB-n(A∩B)
35=25+15-n(A∩B)
nA∩B=5%
5% families own both phone and car and it is given to be 2000 .
∴5% of x=2000
5x100 =2000
x=40000
Hence, i, ii and iii are correct.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2015 (10 Apr Online)
In a chamber, a uniform magnetic field of $6.5 \mathrm{G}(1 \mathrm{G}$ $\left.=10^{-4} \mathrm{~T}\right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $\left(e=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$

Options:

None

Solution:
Given, $\mathrm{B}=6.5 \mathrm{G}=6.5 \times 10^{-4} \mathrm{~T}$
$\mathrm{v}=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}, \mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
$\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$.
$\because \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \Rightarrow \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$
$=\frac{91 \times 48 \times 10^{-27}}{16 \times 65 \times 10^{-25}}=4.2 \times 10^{-2} \mathrm{~m}=4.2 \mathrm{~cm}$

Topic: Physics, Magnetic Effects of Current
In a charged capacitor, the energy resides

Options:
A The positive charges
B Both the positive and negative charges
C The field between the plates
D Around the edge of the capacitor plates

The correct answer is:
The positive charges

Solution:
As the electric field outside the capacitor plates is zero and field only exists in the between the plates, the energy will reside in field between the plates.

Topic: Physics, Capacitance, JEE Main
In a chemical reaction, A+2B⇌K2C+D , the initial concentration of B was 1.5 times of the concentration of A , but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant K for the chemical reaction is:

Options:
A 1
B 4
C 14
D 16

The correct answer is:
4

Solution:
A + 2B â 2C + D
Initial conc. of [A] = a , [B] = 1.5a , [C] = 0 and [D] = 0
At equilibrium [A] = (a - x) , [B] = 1.5(a - 2x) , [C] = 2x and [D] = x
At equilibrium a-x=1.5a-2x
0.5âa=x
a=2x
kC=2x2xa-x1.5a-2x2=4x2.xxx2=4

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2019 (12 Jan Shift 1)
In a chemical reaction $A$ is converted into $B$. The rates of reaction, starting with initial concentrations of $A$ as $2 \times 10^{-3} \mathrm{M}$ and $1 \times 10^{-3}$ $\mathrm{M}$, are equal to $2.40 \times 10^{-4} \mathrm{Ms}^{-1}$ and $0.60 \times 10^{-4} \mathrm{Ms}^{-1}$ respectively. The order of reaction with respect to reactant $A$ will be

Options:
A
0

B
$1.5$

C
1

D
2

The correct answer is:
2

Solution:
$A \longrightarrow B$
Initial concentration Rate of reaction
$$
\begin{array}{ll}
2 \times 10^{-3} \mathrm{M} & 2.40 \times 10^{-4} \mathrm{Ms}^{-1} \\
1 \times 10^{-3} \mathrm{M} & 0.60 \times 10^{-4} \mathrm{Ms}^{-1}
\end{array}
$$
rate of reaction
$$
r=k[A]^x
$$
where $x=$ order of reaction hence
$$
\begin{aligned}
& 2.40 \times 10^{-4}=k\left[2 \times 10^{-3}\right]^x \\
& 0.60 \times 10^{-4}=k\left[1 \times 10^{-3}\right]^x
\end{aligned}
$$
on dividing eqn.(i) from eqn. (ii) we get $4=(2)^x$
$\therefore \quad x=2$
i.e. order of reaction $=2$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (12 May Online)
In a chemical reaction i.e., $A+2 B \longrightarrow$ products, when concentration of $A$ is doubled, rate of the reaction increases 4 times and when concentration of $B$ alone is doubled rate continues to be the same. The order of the reaction is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Let the order of reaction w.r.t. $A$ is $x$ and w.r.t. $B$ is $y$.
$r_{1}=k[A]^{x}[B]^{y} \quad \text{...(i)}$
$r_{2}=k[2 A]^{x}[B]^{y} \quad \text{...(ii)}$
$r_{3}=k[A]^{x}[2 B]^{y} \quad \text{...(iii)}$
$\begin{aligned} \frac{r_{1}}{r_{2}} &=\frac{k[A]^{x}[B]^{y}}{k[2 A]^{x}[B]^{y}} \Rightarrow \frac{1}{4}=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad\left(\frac{1}{2}\right)^{2} &=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad x &=2 \end{aligned}$
Similarly,
$$
\begin{aligned}
& \frac{r_{1}}{r_{3}}=\frac{k[A]^{x}[B]^{y}}{k[A]^{x}[2 B]^{y}} \Rightarrow 1=\left(\frac{1}{2}\right)^{y} \\
\Rightarrow \quad\left(\frac{1}{2}\right)^{0} &=\left(\frac{1}{2}\right)^{y} \\
y &=0
\end{aligned}
$$
Hence, the rate of equation is
$$
\text { Rate }=k[A]^{2}[B]^{0}
$$
Order of reaction $=2$

Topic: Chemistry, Chemical Kinetics, JEE Main
In a circle with centre $\mathrm{O}$, suppose A, P, B are three points on its circumference such that $\mathrm{P}$ is the mid-point of minor arc $\mathrm{AB}$. Suppose when $\angle \mathrm{AOB}=\theta$,
$\frac{\operatorname{area}(\Delta \mathrm{AOB})}{\text { area }(\Delta \mathrm{APB})}=\sqrt{5}+2$
If $\angle \mathrm{AOB}$ is doubled to $2 \theta$, then the ratio $\frac{\text { area }(\Delta \mathrm{AOB})}{\text { area }(\triangle \mathrm{APB})}$ is -

Options:
A $\frac{1}{\sqrt{5}}$
B $\sqrt{5}-2$
C $2 \sqrt{3}+3$
D $\frac{\sqrt{5}-1}{2}$

The correct answer is:
$\frac{1}{\sqrt{5}}$

Solution:
$\frac{\Delta(\mathrm{AOB})}{\Delta \mathrm{APB}}=2+\sqrt{5}$
$\frac{\cos \frac{\theta}{2}}{1-\cos \frac{\theta}{2}}=2+\sqrt{5} \quad \Rightarrow \cos \frac{\theta}{2}=\frac{1+\sqrt{5}}{4}$
So $\cos \theta=\frac{\sqrt{5}-1}{4}$
If $\theta \rightarrow 2 \theta$
$\frac{\Delta \mathrm{AOB}}{\Delta \mathrm{APB}}=\frac{\cos \theta}{1-\cos \theta}=\frac{1}{\sqrt{5}}$

Topic: Mathematics, Straight Lines, JEE Main
In a circuit consisting of a capacitance and a generator with alternating emf, \( E_{g}=E_{g o} \sin \omega t, V_{C} \) and \( I_{C} \) are the voltage and current. Correct phasor diagram for such circuit is

Options:
A
B
C
D

The correct answer is:

Solution:
In capacitor, current lead voltage by π2.

Topic: Physics, Alternating Current, JEE Main
In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA division-1. In the absence of shunt resistance, the galvanometer produces a deflection of θ=9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ2, is closest to:

Options:
A 550Â Î©
B 220Â Î©
C 55Â Î©
D 110 Î©

The correct answer is:
110 Î©

Solution:
We are given a galvanometer connected in the first circuit in the above figure, where E=6 V, R = 11 kΩ and the deflection θ = 9 divisions. The galvanometer has a resistance of G. The current in the circuit,

I=ER+G ...(1)

Also, since the figure of merit is 60 μA/division

I = 60×9 μA = 540 μA ...(2)

Substituting equation (2) in (1),

540×10-6 = 611,000+G⇒G = 6540×10-6-11,000⇒G = 111 Ω ...(3)

In the second circuit, it is given that the deflection is half that of the previous circuit. So, current flowing through the galvanometer half the first circuit. Hence,
I2= ER+GSG+S ×SS+G ⇒ I2=ESRS+G+GS
⇒S= RG ×I2E-(R+G)I2

Substituting the values from equations (1), (2) and (3),
S= 11 ×103 ×111 ×5402 ×10-66-62=110 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (16 Apr Online)
In a circuit $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^{\circ}$. The value of $\mathrm{C}$ is:

Options:
A $\frac{1}{\pi f(2 \pi f(L-R)}$
B $\frac{1}{2 \pi f(2 \pi f L-R)}$
C $\frac{1}{\pi f(2 \pi f L+R)}$
D $\frac{1}{2 \pi f(2 \pi f L+R)}$

The correct answer is:
$\frac{1}{2 \pi f(2 \pi f L+R)}$

Solution:
$$
\begin{aligned}
\tan \phi & =\frac{X_C-X_L}{R} \\
\tan \left(\frac{\pi}{4}\right) & =\frac{\frac{1}{\omega C}-\omega L}{R} \\
\Rightarrow \quad R & =\frac{1}{\omega C}-\omega L \\
(R+2 \pi f L) & =\frac{2}{2 \pi f C} \\
C & =\frac{1}{2 \pi f(R+2 \pi f L)}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
In a circular motion of a particle, the tangential acceleration of the particle is given by at=9 m s-2. The radius of the circle is 4 m. The particle was initially at rest. Time after which total acceleration of the particle makes an angle of 45Â° with the radial acceleration is

Options:
A 13 s
B 23 s
C 53 s
D 43 s

The correct answer is:
23 s

Solution:
At 45°, at=ar=9
v 2 r = v 2 4 = 9
v=6 m s-1
6=0+9t
t=23 s.

Topic: Physics, Laws of Motion, JEE Main
In a circus, a performer throws an apple towards a hoop held at 45 m height by another performer standing on a high platform (see figure below). The thrower aim s for the hoop and throws the apple with a speed of 24 m s-1. At the exact moment that the thrower released the apple, the other performer drops the hoop. The hoop falls straight down. At what height above the ground does the apple go through the hoop?

Options:
A 21 m
B 22 m
C 23 m
D 24 m

The correct answer is:
22 m

Solution:
Velocity of projection =24 m s-1

Distance between point of projection and hoop =252+452

∴ Time taken by ball to reach the hoop

=252+45224

(Note :- We are analysing the motion wrt hoop)

∴ Distance by which hoop will fall

=12at2=12×10×252+452242

∴ Height above the ground where apple go through the hoop is given by

45-12×10×252+452242=22 m

Topic: Physics, Motion In Two Dimensions, JEE Main
In a circus, a stuntman rides a motor bike in a vertical circular track of radius \(r\). Find the minimum speed, he must maintain at highest point of track.

Options:
A \(\sqrt{2 g r}\)
B \(2 \sqrt{g r}\)
C \(\sqrt{g r}\)
D \(\sqrt{5 g r}\)

The correct answer is:
\(\sqrt{g r}\)

Solution:
For motion in vertical circle, to complete the revolution of moving body, the minimum speed of body at lowest point and highest point should be \(\sqrt{5 g r}\) and \(\sqrt{g r}\) respectively, where \(r\) is the radius of circular path.
Hence, in circus, for completing vertical circular track, minimum speed of stuntman at highest point of track should be \(\sqrt{g r}\).

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 2)
In a city, the total income of all people with salary below $Rs. 10000$ per annum is less than the total income of all people with salary above $Rs. 10000$ per annum. If the salaries of people in the first group increases by $5 \%$ and the salaries of people in the second group decreases by $5 \%$ then the average income of all people

Options:
A increases
B decreases
C remains the same
D cannot be determined from the data

The correct answer is:
decreases

Solution:
Let total number of people with salary below $\mathrm{Rs. 10000}$ per annum is $x$ and salary is $A$. Let total number of people with salary above $\mathrm{Rs. 10000}$ per annum is $y$ and salary is $B$ then.
$\mathrm{xA}-\mathrm{yB} < 0$
$\frac{\text { average after }}{\text { average before }}=\frac{x\left(\frac{105}{100} A\right)+y\left(\frac{95}{100} B\right)}{x A+y B}$
$=1+\frac{5}{100}\left(\frac{x A-y B}{x A+y B}\right)$

Topic: Mathematics, Statistics, KVPY, KVPY 2013 (SA)
In a class of \( 60 \) students, \( 25 \) students play cricket and \( 20 \) students play tennis and \( 10 \) students play both the games, then the number of students who play neither is

Options:
A \( 45 \)
B \( 0 \)
C \( 25 \)
D \( 35 \)

The correct answer is:
\( 25 \)

Solution:
Let student play cricket =C
Student play tennis =T
and total number of students =S
∴ nS=60, nC=25, nT=20
and nC∩T=10
Now, nC∪T=nC+nT-nC∩T
=25+20-10=35
∴ The number of students who play neither game
=nC∩T′=nS-nC∪T
=60-35=25

Topic: Mathematics, Sets and Relations, JEE Main