datavorous/entrance-exam-dataset huggingface dataset:
In amplitude modulation

Options:
A Amplitude remains constant but frequency change
B Both amplitude and frequency do not change
C Both amplitude and frequency change
D Amplitude of the carrier wave changes according to information signal

The correct answer is:
Amplitude of the carrier wave changes according to information signal

Solution:
In amplitude modulation amplitude of the carrier wave changes according to information signal.

Topic: Physics, Semiconductors, JEE Main
In an $A C$-circuit containing only capacitance, the current

Options:
A leads the voltage by $180^{\circ}$
B remains in phase with the voltage
C leads the voltage by $90^{\circ}$
D lags the voltage by $90^{\circ}$

The correct answer is:
leads the voltage by $90^{\circ}$

Solution:
In an AC-circuit containing only capacitance, the current always leads the voltage by $90^{\circ}$ phase angle.

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
In an $A C$ circuit, voltage $V=V_0 \sin \omega t$ and inductor $L$ is connected across the circuit. Then the instantaneous power will be

Options:
A $\frac{V_0^2}{2 \omega L} \sin \omega t$
B $\frac{-V_0^2}{2 \omega L} \sin \omega t$
C $\frac{-V_0^2}{2 \omega L} \sin 2 \omega t$
D $\frac{V_0^2}{\omega L} \sin 2 \omega t$

The correct answer is:
$\frac{-V_0^2}{2 \omega L} \sin 2 \omega t$

Solution:
If $\frac{d I}{d t}$ is the rate of change of current through inductor $L$ of any instant, then induced emf in the inductor of the same instant is
$=\left(-\frac{L d I}{d t}\right)$
To maintain the flow of current in the circuit applied voltage must be equal and opposite to the induced voltage.
$\therefore \quad V=-\left(-L \frac{d I}{d t}\right) \Subset$
Given, $V=V_0 \sin \omega t$
$\therefore \quad V_0 \sin \omega t=\frac{L d I}{d t}$
$\frac{d I}{d t}=\frac{V_0}{L} \sin \omega t$
or $I=\frac{-V_0}{L} \frac{\cos \omega t}{\omega}$
Instantaneous power,
$\because \quad P=V I$
$=\left(V_0 \sin \omega t\right)\left(-\frac{V_0 \cos \omega t}{L \omega}\right) \quad(\mathrm{Using}$ (i) \& (ii))
$=\frac{-V_0^2 \sin 2 \omega t}{2 L \omega}$

Topic: Physics, Alternating Current, AIIMS, AIIMS 2012
In an $\mathrm{AC}$ circuit an alternating voltage $e=200 \sqrt{2} \sin 100 t$ volt is connected to a capacitor of capacity $1 \mu \mathrm{F}$. The rms value of the current in the circuit is

Options:
A $100 \mathrm{~mA}$
B $200 \mathrm{~mA}$
C $20 \mathrm{~mA}$
D $10 \mathrm{~mA}$

The correct answer is:
$20 \mathrm{~mA}$

Solution:
Given, $e=200 \sqrt{2} \sin 100 t$
and
$$
\begin{aligned}
C & =1 \mu \mathrm{F} \\
E_{\mathrm{rms}} & =200 \mathrm{~V} \\
X_C & =\frac{1}{\omega C}=\frac{1}{1 \times 10^{-6} \times 100}=10^4 \Omega \\
i_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{X_C} \\
i_{\mathrm{rms}} & =\frac{200}{10^4}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
In an ac circuit, an alternating voltage \( e=200 \sqrt{2} \sin 100 t \) volts is connected to a capacitor of capacity \( 1 \mu \mathrm{F} \). The rms value of the current in the circuit is

Options:
A \( 10 \mathrm{~mA} \)
B \( 100 \mathrm{~mA} \)
C \( 200 \mathrm{~mA} \)
D \( 20 \mathrm{~mA} \)

The correct answer is:
\( 20 \mathrm{~mA} \)

Solution:
The given equation of alternating voltage is,
e=2002 sin 100t ...(i)
The standard equation of alternating voltage is,
e=e0sinωt ...(ii)
Comparing equations (i) and (ii), we get,
e0=2002 V, ω=100 rad s-1
The capacitive reactance is,
XC=1ωC=1100×1×10-6 Ω
The rms value of current in the circuit is,
irms=vrmsXC=e0/21/ωC=2002/21/100×10-6
=200×100×10-6 A=2×10-2 A=20 mA

Topic: Physics, Alternating Current, JEE Main
In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL=R=XC. Impedance of this circuit is :

Options:
A Zero
B R
C R2
D 2R2

The correct answer is:
R

Solution:
Recall the formula of impedance is in terms of resistance, Z=XL-XC2+R2, here, the condition of resonance, XL=XC, it means,
Z=XL-XL2+R2=R

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (31 Aug Shift 1)
In an ac circuit $I=100 \sin 200 \pi t$. The time required for the current to achieve its peak value will be

Options:
A $\frac{1}{100} \mathrm{sec}$
B $\frac{1}{200} \mathrm{sec}$
C $\frac{1}{300} \mathrm{sec}$
D $\frac{1}{400} \mathrm{sec}$

The correct answer is:
$\frac{1}{400} \mathrm{sec}$

Solution:
The current takes $\frac{T}{4} \sec$ to reach the peak value.
In the given question $\frac{2 \pi}{T}=200 \pi \Rightarrow T=\frac{1}{100} \mathrm{sec}$
$\therefore$ Time to reach the peak value $=\frac{1}{400} \mathrm{sec}$

Topic: Physics, Alternating Current, JEE Main
In an A.C. circuit the current

Options:
A Always leads the voltage
B Always lags behind the voltage
C Is always in phase with the voltage
D May lead or lag behind or be in phase with the voltage

The correct answer is:
May lead or lag behind or be in phase with the voltage

Solution:
In an A.C. circuit, the current may depend on any circumstances. It may lag or lead or in phase of voltage.

Topic: Physics, Alternating Current, JEE Main
In an a.c. circuit the e.m.f. (e) and the current (i ) at any instant are given respectively by
$$
\begin{aligned}
& e=E_0 \sin \omega t \\
& i=I_0 \sin (\omega t-\phi)
\end{aligned}
$$
The average power in the circuit over one cycle of a.c. is

Options:
A $\mathrm{E}_0 \mathrm{I}_0$
B $\frac{E_0 I_0}{2}$
C $\frac{E_0 I_0}{2} \sin \phi$
D $\frac{E_0 l_0}{2} \cos \phi$

The correct answer is:
$\frac{E_0 l_0}{2} \cos \phi$

Solution:
Since phase difference between current and e.m.f is $\phi$
$$
\therefore P_{a V}=\frac{E_0 l_0}{2} \cos \phi
$$

Topic: Physics, Alternating Current, NEET, NEET 2008 (Mains)
In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to :
A. pure inductor.
B. pure capacitor.
C. pure resistor.
D. combination of an inductor and capacitor.
Choose the correct answer from the options given below :

Options:
A A, B and C only
B A and B only
C B, C and D only
D A, B and D only

The correct answer is:
A, B and D only

Solution:
This is possible when phase difference is $\frac{\pi}{2}$ between current and voltage so correct answer will be (4)

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (04 Apr Shift 1)
In an A.C. circuit, the instantaneous e.m.f. and current are given by, E=100sin30t,I=20sin30t-π4. In one cycle of A.C., the average power consumed by the circuit (in watt) and the watt-less current (in ampere) are, respectively:

Options:
A 50, 0
B 50, 10
C 10002, 10
D 502, 0

The correct answer is:
10002, 10

Solution:
Instantaneous emf and current is given by, E=100sin30t, I=20sin30t-π4.

Phase difference between emf and current is

ϕ=30t-30t-π4=π4

Average power is given by,

Pav=ErmsIrmscosϕ
=1002×202×12=10002

Watt-less current is given by,

IWL=Irmssinϕ=Io√2sin45°⇒IWL=202.12=10 A.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2018 (08 Apr)
In an AC circuit, the instantaneous values of emf and current are Î=200sin300t V and i=2sinâ¡300t+Ï3 A respectively. The average power consumed (in watt) is

Options:
A 200
B 100
C 50
D 400

The correct answer is:
100

Solution:
The average power consumed in an AC circuit is given by
P=V0I02cos⁡ϕ
Where ϕ is phase angle and V0 and I0 are the peak values of voltage and current respectively.
Given, V0=200 V, I0=2 A, ϕ=π3.
P=200×22cos⁡π3
=200×22×12=100 W

Topic: Physics, Alternating Current, JEE Main
In an AC circuit, the instantaneous values of emf and current are $e=200 \sin 314 t$ volt and $I=\sin \left(314 t+\frac{\pi}{3}\right)$ amp. The average power
consumed in watt is

Options:
A 200
B 100
C 50
D 25

The correct answer is:
50

Solution:
$V_{\mathrm{rms}}=\frac{200}{\sqrt{2}}, I_{\mathrm{rms}}=\frac{1}{\sqrt{2}}$
$\therefore \quad P=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi$
$$
=\frac{200}{\sqrt{2}} \frac{1}{\sqrt{2}} \cos \frac{\pi}{3}=50 \mathrm{~W}
$$

Topic: Physics, Alternating Current, JEE Main
In an ac circuit, the power factor

Options:
A Is zero when the circuit contains an ideal resistance only
B Is unity when the circuit contains an ideal resistance only
C Is unity when the circuit contains an ideal inductance only
D Is unity when the circuit contains an ideal inductance only

The correct answers are:
Is unity when the circuit contains an ideal resistance only

Solution:
Power factor, $\cos \phi=\frac{R}{Z}$
When circuit contains only resistance, then
$\begin{aligned}
& \mathrm{Z}=\mathrm{R} \\
& \therefore \cos \phi=1
\end{aligned}$
When circuit contains only inductance, then
$\begin{aligned}
& \mathrm{R}=\mathrm{o} \\
& \therefore \cos \phi=\mathrm{o}
\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
In an a.c. circuit the voltage applied is $\mathrm{E}=\mathrm{E}_0 \sin \omega \mathrm{t}$. The resulting current in the circuit is $\mathrm{I}=\mathrm{I}_0 \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$. The power consumption in the circuit is given by

Options:
A
$\mathrm{P}=\frac{\mathrm{E}_0 \mathrm{I}_0}{\sqrt{2}}$

B
$\mathrm{P}=$ zero

C
$\mathrm{P}=\frac{\mathrm{E}_0 \mathrm{I}_0}{2}$

D
$\mathrm{P} \sqrt{2} \mathrm{E}_0 \mathrm{I}_0$

The correct answer is:
$\mathrm{P}=$ zero

Solution:
$\cos \phi=0$
So power $=0$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2007
In an ac circuit, $V$ and I are given by $V=100 \sin (100 t)$ volts, $I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$

Options:
A $10^4$ watt
B $10 watt$
C $2.5 watt$
D $5 watt$

The correct answer is:
$2.5 watt$

Solution:
$\begin{aligned} & P=V_{r m, s .} \times i_{r m, s .} \times \cos \phi=\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \frac{\pi}{3} \\ & =\frac{10^4 \times 10^{-3}}{2} \times \frac{1}{2}=\frac{10}{4}=2.5 \mathrm{watt}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
In an a.c. circuit, voltage and current are given by: V=100sin(100t) V and I=100sin100t+π3 mA respectively. The average power dissipated in one cycle is :

Options:
A 5 W
B 10 W
C 2.5 W
D 25 W

The correct answer is:
2.5 W

Solution:
Standard expression of voltage is given by, V=V0sinωt and for current is given by, I=I0sinωt+ϕ. Clearly, ϕ is the phase difference here. From the given expression, ϕ=π3.

Average power dissipated in one cycle is given by,

Pavg=VrmsIrmscos(ϕ)

=1002×100×10-32×cosπ3

=1042×10-3×12

=104=2.5 W

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (29 Jan Shift 2)
In an $\mathrm{AC}$ generator, a coil with $\mathrm{N}$ turns, all of the same area $\mathrm{A}$ and total resistance $\mathrm{R}$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is "

Options:
A
N.A.B. $\omega$

B
N.A.B.R. $\omega$

C
N.A.B

D
N.A.B.R

The correct answer is:
N.A.B. $\omega$

Solution:
NBA $\omega$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2006
In an AC generator, a coil with \( N \) turns, all of the same area \( A \) and total resistance \( R \), rotates with frequency \( \omega \) in a magnetic field \( B \). The maximum value of emf generated in the coil is

Options:
A \( N A B R \omega \)
B \( N A B \)
C \( N A B R \)
D \( N A B \omega \)

The correct answer is:
\( N A B \omega \)

Solution:
The emf generated would be maximum when flux (cutting) would be maximum ie, angle between area vector of coil and magnetic field is 0°. The emf generated is given by [as a function of time]
e=NBAωcos⁡ωt
⇒ emax=NABω

Topic: Physics, Electromagnetic Induction, JEE Main
In an ac generator, a rectangular coil of 100 turns each having area 14×10–2 m2 is rotated at 360 rev min-1 about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T. The maximum value of the emf produced will be ______ V.

[Take π=227]

Options:

The correct answer is:
1584

Solution:
Maximum value of induced EMF can be written as,

ξmax=NABω

=100×14×10-2×3×360×2π60

=1584 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (30 Jan Shift 2)
In an acid-base titration, 0.1 M HCl solution was added to the NaOH solution of unknown strength. Which of the following correctly shown the change of pH of the titration mixture in this experiment?

Options:
A A
B B
C D
D C

The correct answer is:
A

Solution:
For titration of strong base with strong acid pH changes sharply near equivalence point and at equivalent point pH sharply decreases because strong base & strong acid dissociate completely.
HClSA + NaOHSB→NaCl+ H2Ol
at equivalent point pH=7 and at endpoint solution will be neutral.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (09 Apr Shift 2)
In an adiabatic expansion of an ideal gas the product of pressure and volume

Options:
A decreases
B increases
C remains constant
D at first increases and then decreases

The correct answer is:
decreases

Solution:
In an adiabatic expansion, internal energy decreases and hence temperature also decreases. Hence, from ideal gas equation,
$p V=n R T$
Since, $T$ decreases.
Hence, $p V$ also decreases.

Topic: Physics, Thermodynamics, KCET, KCET 2020
In an adiabatic expansion of an ideal gas

Options:
A $W=-\Delta E$
B $W=\Delta E$
C $\Delta E=0$
D $W=0$

The correct answer is:
$W=\Delta E$

Solution:
$q=\Delta E-W$ if $q=0$ for adiabatic process, than $\Delta E=W$.

Topic: Chemistry, Thermodynamics (C), JEE Main
In an adiabatic expansion, the volume of air is increased by 5%, then the approximate percentage decrease in its pressure will be

Options:
A 5%
B 6%
C 7%
D 8%

The correct answer is:
7%

Solution:
ΔPΔV=-γPV →Air→diatomicγ= 1.4
∴ΔPP=-γΔVV=-1.4×5=7%

Topic: Physics, Thermodynamics, JEE Main
In an adiabatic process, the density of a diatomic gas becomes 32n times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is:

Options:
A 32
B 326
C 128
D 132

The correct answer is:
128

Solution:
Given,

Final density of diatomic gas is ρf=32nρi

Final pressure is Pf=nPi

In adiabatic process, equation of state is given by,

PVγ=constant ...1

Where, V is the volume, P is the pressure and γ=75 is the ratio of specific heat of diatomic gas

Pmργ=constant ...2, As V=mρ

and mass is constant, then the pressure is given by

∴P∝ργ

⇒PfPi=ρfρiγ

⇒nPiPi=32ρiρi7/5=(32)7/5

∴ n=27=128

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (05 Sep Shift 2)
In an adiabatic process, what remains constant during the change of state?

Options:
A Pressure
B Volume
C Internal energy
D Temperature

The correct answer is:
Internal energy

Solution:
In an adiabatic process, the internal energy remains constant during the change of state.

Topic: Chemistry, Thermodynamics (C), NEET
In an adiabatic process, when work is done on the system, what happens to its internal energy?

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

The correct answer is:
It increases

Solution:
In an adiabatic process where work is done on the system, its internal energy increases.

Topic: Chemistry, Thermodynamics (C), NEET
In an adiabatic process,

Options:
A temperature remains constant
B pressure remains constant
C volume remains constant
D there is no transfer of heat

The correct answer is:
there is no transfer of heat

Solution:
In an adiabatic process, there is no transfer of heat takes place between a thermodynamic system and its surroundings.

Topic: Physics, Thermodynamics, COMEDK, COMEDK 2016
In an admission test, there are 15 multiple choice questions. Each question is followed by 4 alternatives to choose. Out of these there may be one or more than one correct answers. If a student attempts all the 15 questions and marks the answers randomly, then number of different ways he can answer the question paper is

Options:
A ${ }^{4 \times 15} \mathrm{C}_4$
B $15^{15}$
C $4^{15}$
D $4 ! \cdot 15 !$

The correct answer is:
$15^{15}$

Solution:
These are 15 multiple type questions. Students attempts all 15 questions.
The number of way student can answer the questions is $=m^n$
$$
\begin{aligned}
m & =15 \\
n & =15 \Rightarrow(15)^{15}
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be:

Options:
A 1499G
B 499500G
C 1500G
D 500499G

The correct answer is:
1500G

Solution:
As shown in figure a shunt S is connected in parallel with the galvanometer to convert it into the ammeter, let G is the resistant of the galvanometer. Where I is the current in the circuit as shown, according to question 0·2% current passing through the galvanometer, i.e.,

IG = 0·2% of I = 0·2100I= I500;

so, the current through the shunt is I=IG + IS ⇒ IS = I-IG⇒IS = I-1500 =499500I
As G and the shunt S are in parallel combination ,

VG = VS

igRg=isRs⇒i500G=499500iS
⇒S=G499
Equivalent resistance of the ammeter
1Req=1G+1G499⇒Req=G500

Topic: Physics, Current Electricity, NEET, NEET 2014
In an ammeter, 5% of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be:

Options:
A G20
B G199
C 199 G
D 200 G

The correct answer is:
G20

Solution:
As potential drop across both branches would be the same, we get ISS=IgG

95100IS=5I100G

⇒S=G19

Now, resistance of ammeter will be

RA=SGS+G=G21920G19

⇒RA=G20

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 2)
In an ammeter $10 \%$ of main current is passing through the galvanometer. If the resistance of the galvanometer is $G$, then the shunt resistance (in $\Omega$ ) is

Options:
A $9 G$
B $\frac{G}{9}$
C $90 G$
D $\frac{G}{90}$

The correct answer is:
$\frac{G}{9}$

Solution:
If $I$ be the main current, then current passing through galvanometer,
$I_{g}=10 \% \text { of } I=\frac{I}{10}$
Resistance of galvanometer, $R_{g}=G$
Shunt resistance, $R_{S}=\frac{I_{g} R_{g}}{I-I_{g}}=\frac{\frac{I}{10} \times G}{I-\frac{I}{10}}=\frac{G}{9}$

Topic: Physics, Current Electricity, COMEDK, COMEDK 2017
In an amplifier the load resistance $R_L$ is equal to the plate resistance $\left(r_p\right)$. The voltage amplification is equal to

Options:
A $\mu$
B $2 \mu$
C $\mu / 2$
D $\mu / 4$

The correct answer is:
$\mu / 2$

Solution:
Voltage gain $A_v=\frac{\mu}{1+\frac{r_p}{R_L}}$, for $r_p=R_L \Rightarrow A_v=\frac{\mu}{2}$

Topic: Physics, Semiconductors, JEE Main
In an amplitude modulation, a modulating signal having amplitude of X V is superimposed with a carrier signal of amplitude Y V in first case. Then, in second case, the same modulating signal is superimposed with different carrier signal of amplitude 2Y V. The ratio of modulation index in the two case respectively wil be :

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

The correct answer is:
2 : 1

Solution:
Modulating index is stated as μ=AmAc, where, Am=amplitude of modulating signal and Ac=amplitude of carrier signal.

Now, for first case, μ1=XY

For second case, μ2=X2Y

Thus, the ratio of μ1μ2=XYX2Y=21⇒μ1:μ2=2:1

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (01 Feb Shift 2)
In an A.P., the sixth term a6=2. If the a1a4a5 is the greatest, then the common difference of the A.P., is equal to

Options:
A 32
B 85
C 23
D 58

The correct answer is:
85

Solution:
Given: a6=2. Also, a1a4a5 is greatest.

⇒a+5d=2 ...i

Let, y=a1a4a5

⇒y=a(a+3d)(a+4d)

⇒y=(2-5d)(2-2d)(2-d)

⇒y=4-4d-10d+10d22-d

⇒y=4-14d+10d22-d

⇒y=8-4d-28d+14d2+20d2-10d3

⇒y=8-32d+34d2-10d3

⇒y'=-32+68d-30d2

⇒y'=-215d2-34d+16

⇒y'=-2(5d-8)(3d-2)

So, the function attains maxima at d=85

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (29 Jan Shift 1)
In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Options:
A Specific conductivity decreases because the number of ions per unit volume decreases.
B Specific conductivity increases with the addition of water.
C Specific conductivity remains constant with the addition of water.
D Specific conductivity depends on the type of electrolyte.

The correct answer is:
Specific conductivity decreases because the number of ions per unit volume decreases.

Solution:
Correct Option is : (A)
Specific conductivity decreases because the number of ions per unit volume decreases.

Topic: Chemistry, Electrochemistry, NEET
In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Options:

None

Solution:
pH of solution depends upon concentration of $\mathrm{H}^{+}$present in solution. $\mathrm{pH}$ of the solution will not be affected as $\left[\mathrm{H}^{+}\right]$ remains constant.
At anode : $2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-}$
At cathode : $4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2$

Topic: Chemistry, Electrochemistry
In an aqueous solution, the ionization constants for carbonic acid are

K 1 = 4.2 Ã 1 0 - 7 and K 2 = 4.8 Ã 1 0 - 11

Select the correct statement for a saturated 0.034 M solution of the carbonic acid:

Options:
A The concentration of H + is double that of CO 3 2 - .
B The concentration of CO 3 2 - is 0.034 M .
C The concentration of CO 3 2 - is greater than that of HCO 3 - .
D The concentration of H + and HCO 3 - are approximately equal.

The correct answer is:
The concentration of H + and HCO 3 - are approximately equal.

Solution:
H 2 CO 3 â H + + HCO 3 - ; K 1 = 4.2 Ã 1 0 - 7 HCO 3 - â H + + CO 3 2 - ; K 2 = 4.8 Ã 1 0 - 1 1
âµ K 1 â« K 2 , so H 2 CO 3 ionises more than HCO 3 - and hence, contribution of H + is mostly due to ionization of carbonic acid. Thus, the concentrations of H + and HCO 3 - are approximately equal.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In an arithmetic progression containing 99 terms, the sum of all the odd numbered terms is 2550. If the sum of all the 99 terms of the arithmetic progression is k, then k100 is equal to

Options:

The correct answer is:
50.49

Solution:
Let the terms of AP be a,a+d,a+2d,…..,a+98d
For sum of all odd numbered terms A=a,D=2d,N=50.
i.e. 5022a+492d=2550
⇒a+49d=255050=51
⇒2a+98d=102
Sum of all 99 terms is,
9922a+98d=992×102=99×51
⇒k=5049⇒k100=50.49

Topic: Mathematics, Sequences and Series, JEE Main
In an arithmetic progression the p+1th term is twice the q+1th term. If its 3p+1th term is Î» times the p+q+1th term, then Î» is equal to

Options:
A 2
B 12
C 3
D 13

The correct answer is:
2

Solution:
For the A.P. let a be the first term and d be the common difference

Given Tp+1=2Tq+1
⇒a+pd=2a+qd⇒a=p-2qd
Also,
T3p+1=λTp+q+1
a+3pd=λa+p+qd
aλ-1=d3p-λp-λq
So, p-2qd=λp+λq-3pd1-λ
⇒p-pλ-2q+2qλ=λp+λq-3p
⇒λp+q+p-2q=p-2q+3p
⇒λ=4p-2q2p-q=22p-q2p-q=2

Topic: Mathematics, Sequences and Series, JEE Main
In an astronomical telescope, the focal lengths of two lenses are $180 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively. In normal adjustment, the magnifying power will be

Options:
A 1080
B 200
C 30
D 186

The correct answer is:
30

Solution:
$|m|=\frac{f_o}{f_e}=\frac{180}{6}=30$

Topic: Physics, Ray Optics, JEE Main
In an atom, an electron is moving with a speed of $600 \mathrm{~m} / \mathrm{s}$ with an accuracy of $0.005 \%$. Certainity with which the position of the electron can be located is $\left(h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right.$, mass of electron, $e_m=9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A
$1.52 \times 10^{-4} \mathrm{~m}$

B
$5.10 \times 10^{-3} \mathrm{~m}$

C
$1.92 \times 10^{-3} \mathrm{~m}$

D
$3.84 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$1.92 \times 10^{-3} \mathrm{~m}$

Solution:
$$
\begin{aligned}
& \Delta \mathrm{x} . \mathrm{m} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi} \\
& \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}} \\
& \Delta \mathrm{v}=600 \times \frac{0.005}{100}=0.03 \\
& \Rightarrow \Delta \mathrm{x}=\frac{6.625 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03}=1.92 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2009
In an atom how many orbital(s) will have the quantum numbers; $n=3, l=2$ and $m_l=+2$ ?

Options:
A
$5$

B
$3$

C
$1$

D
$7$

The correct answer is:
$1$

Solution:
$n=3, l=2$ means $3 d$ orbital
i.e. in an atom only one orbital can have the value $m_l=+2$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2013 (09 Apr Online)