datavorous/entrance-exam-dataset huggingface dataset:
A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _____ hours.

Options:

The correct answer is:
15

Solution:
Given here, T=150 minutes,

A0=64x, where, x is the safe limit.

Using radioactive decay law relation, A=A0×2-tT, where, t is the minimum time in hours after which it would be possible to work safely with source.

Putting the values, we have

x=64x×2-tT

∴ t=6 T=6×2.5=15 hours

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
A frictionless piston-cylinder based enclosure contains some amount of gas at a pressure of $400 \mathrm{kPa}$. Then heat is transferred fo the gas at constant pressure in a quasi-static process. The piston moves up slowly through a height of $10 \mathrm{cm} .$ If the piston has a cross-sectional area of $0.3 \mathrm{m}^{2},$ the work done by the gas in this process is

Options:
A $6 \mathrm{kJ}$
B $12 \mathrm{kJ}$
C $7.5 \mathrm{kJ}$
D $24 \mathrm{kJ}$

The correct answer is:
$12 \mathrm{kJ}$

Solution:
$W_{\text {constant pressure }}=p \times \Delta V$ $=400 \times 10^{3} \times 0.3 \times 10 \times 10^{-2}$
$=400 \times 10 \times 3$
$=12000=12 \mathrm{kJ}$

Topic: Physics, Thermodynamics, JEE Main
A frictionless track $A B C D E$ ends in a circular loop of radius $R$. A body slides down the track from point $A$ which is at $a$ height $h=5 \mathrm{~cm}$. Maximum value of $R$ for the body to successfully complete the loop is

Options:
A $5 \mathrm{~cm}$
B $\frac{15}{4} \mathrm{~cm}$
C $\frac{10}{3} \mathrm{~cm}$
D $2 \mathrm{~cm}$

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

Solution:
Condition for vertical looping $\quad h=\frac{5}{2} r=5 \mathrm{~cm} \quad \therefore r=2 \mathrm{~cm}$

Topic: Physics, Work Power Energy, JEE Main
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is x nm. The value of x to the nearest integer is ______.

Options:

The correct answer is:
600

Solution:
β=λDd

λ=βdD

λ=6×10-3×10-310

λ=6×10-7 m=600×10-9 m

λ=600 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A fruit developed from hypanthodium inflorescence is called :

Options:
A Caryopsis
B Hesperidium
C Sorosis
D Syconus

The correct answer is:
Syconus

Solution:
No Solution Available

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2009 (Mains)
A fruity smell is produced by the reaction of $\mathrm{C}_2 \mathrm{H}_3 \mathrm{OH}$ with

Options:
A $\mathrm{PCl}_3$
B $\mathrm{CH}_3 \mathrm{COCH}_3$
C $\mathrm{CH}_3 \mathrm{COOH}$
D None of these

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOH}$

Solution:
Alcohols react with acids to form esters which have fruity smell

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A fruity smell is produced by the reaction of $\mathrm{C}_2 \mathrm{H}_3 \mathrm{OH}$ with

Options:
A $\mathrm{PCl}_{\text {, }}$
B $\mathrm{CH}_3 \mathrm{COCH}_3$
C $\mathrm{CH}_3 \mathrm{COOH}$
D None of these

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOH}$

Solution:
Alcohols react with acids to form esters which have fruity smell.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A fuel cell involves combustion of the butane at at 1 atm and 298 K
C4H10g+132O2gâ4CO2g+5H2Ol
ÎGÂ°=-2744 kJ/mole
The value of Ecello
Report your answer by rounding it upto nearest whole number.

Options:

The correct answer is:
1

Solution:
C4H10g+132O2g→4CO2g+5H2Ol
Here total change in oxidation number of C-atoms =+16-(-10)=+26
It means total number of e- involved = 26
As Ecello=-ΔG°nF=-(-2744)×100026×96500
=+1.09V
≈1.0V

Topic: Chemistry, Electrochemistry, JEE Main
A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self inductance $L$ at $t=0$. The time at which the energy is stored equally between the electric and the magnetic field is :

Options:
A
$\frac{\pi}{4} \sqrt{L C}$

B
$2 \pi \sqrt{L C}$

C
$\sqrt{L C}$

D
$\pi \sqrt{L C}$

The correct answer is:
$\frac{\pi}{4} \sqrt{L C}$

Solution:
Charge oscillates simple harmonic motion $q=q_0 \sin \omega t, U=\frac{1}{2} \frac{q^2}{C}$
$$
\begin{aligned}
& q=\frac{q_0}{\sqrt{2}} \Rightarrow \omega t=\frac{\pi}{4} \\
& \Rightarrow t=\frac{T}{8}=\frac{2 \pi}{8} \sqrt{L C}=\frac{\pi}{4} \sqrt{L C}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self inductance $L$ at $t=0$. The time at which the energy is stored equally between the electric and the magnetic field is :

Options:
A
$\frac{\pi}{4} \sqrt{L C}$

B
$2 \pi \sqrt{L C}$

C
$\sqrt{L C}$

D
$\pi \sqrt{L C}$

The correct answer is:
$\frac{\pi}{4} \sqrt{L C}$

Solution:
Charge oscillates simple harmonic motion $q=q_0 \sin \omega t, U=\frac{1}{2} \frac{q^2}{C}$
$$
\begin{aligned}
& q=\frac{q_0}{\sqrt{2}} \Rightarrow \omega t=\frac{\pi}{4} \\
& \Rightarrow t=\frac{T}{8}=\frac{2 \pi}{8} \sqrt{L C}=\frac{\pi}{4} \sqrt{L C}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2011
A fully charged capacitor C with initial charge Q0 is connected to an ideal inductor of self-inductance L at t=0. The time at which energy is stored equally between the electric and the magnetic fields is

Options:
A π4LC
B 2πLC
C LC
D πLC

The correct answer is:
π4LC

Solution:
The charge on the capacitor at any instant of time is

Q=Q0cosωt

⇒i=dQdt=-ωQ0sinωt

12Li2=Q22C

⇒12L ωQ0sinωt2=Q0cosωt22C

ω=1LC

⇒tanωt=1

ωt=tan-11=π4

t=π4ω=π4LC

Topic: Physics, Electromagnetic Induction, JEE Main
A fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire, embedded in a block of specific heat $s$ and mass $m$ under thermally isolated conditions. If the temperature of the block is raised by $\Delta T$, the potential difference $V$ across the capacitor initially is

Options:
A $\left(\frac{2 m s \Delta T}{C}\right)^2$
B $\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$
C $\left(\frac{2 m s \Delta T}{C}\right)$
D $2 m s \Delta T C$

The correct answer is:
$\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$

Solution:
Let $V$ be the potential across the capacitor when it is fully charged, then, energy stored in the capacitor will be
$=\frac{1}{2} C V^2$
(where $C=$ capacitance of capacitor)
When the capacitor is fully discharged, loss of energy is in the form of heat $=\Delta H$ As the system is thermally isolated so,
$\Delta H=\frac{1}{2} C V^2$
Here,
$\Delta H=m s \Delta T$
$\therefore \quad \frac{1}{2} C V^2=m s \Delta T$
or $V=\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$

Topic: Physics, Capacitance, JEE Main
A fully charged capacitor has a capacitance ' $C$ ' it is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ' $s$ ' and mass ' $\mathrm{m}$ '. If the temperature of the block is raised by ' $\Delta \mathrm{T}$ '. The potential difference $V$ across the capacitance is

Options:
A
$\sqrt{\frac{2 m C \Delta T}{s}}$

B
$\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{s}}$

C
$\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{c}}$

D
$\sqrt{\frac{2 m s \Delta T}{C}}$

The correct answer is:
$\sqrt{\frac{2 m s \Delta T}{C}}$

Solution:
Dimensionally only $4^{\text {th }}$ option is correct.

Topic: Physics, Capacitance, JEE Main, JEE Main 2005
A function $\mathrm{f}$ from the set of natural numbers to integers defined by $\mathrm{f}(\mathrm{n})=\left\{\begin{array}{l}\frac{\mathrm{n}-1}{2} \text {, when } n \text { isodd } \\ \frac{\mathrm{n}}{2} \text {, when } \mathrm{n} \text { is even }\end{array}\right.$ is

Options:
A
neither one-one nor onto

B
one-one but not onto

C
onto but not one-one

D
one-one and onto both.

The correct answer is:
one-one and onto both.

Solution:
$f: N \rightarrow 1$
$f(1)=0, f(2)=-1, f(3)=-1, f(4)=-2$
$f(5)=2$, and $f(6)=-3$ so on.
In this type of function every element of set $\mathrm{A}$ has unique image in set $\mathrm{B}$ and there is no element left in set $\mathrm{B}$. Hence $\mathrm{f}$ is one-one and onto function.

Topic: Mathematics, Functions, JEE Main, JEE Main 2003
A function f is defined on [-3,3] as

fx=min|x|,2-x2,-2≤x≤2[|x|] , 2<|x|≤3

where [x] denotes the greatest integer ≤x. The number of points, where f is not differentiable in -3,3 is ___ .

Options:

The correct answer is:
5

Solution:
fx=min|x|,2-x2,-2≤x≤2[|x|] , 2<|x|≤3

⇒x∈[-3,-2)∪(2,3]

Number of points of non-differentiability in -3,3=5

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is such that $f(\mathrm{l})=2$ and $f(x+y)=f(x) \cdot f(y) \forall x, y$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed interms of $f(1), f(2)$ and $f(4)$ is

Options:
A $\frac{f(4)}{f(1)+2 f(2)}$
B $\frac{f(4)}{1+f(2)}$
C $\frac{2 f(4)}{2 f(1)+f(2)}$
D $\frac{f(4)}{2 f(1)+f(2)}$

The correct answer is:
$\frac{f(4)}{1+f(2)}$

Solution:
Given, $f(x+y)=f(x) \cdot f(y)$
$\therefore \quad f(x)=a^x$
$f(\mathrm{1})=a^{\mathrm{l}}=2 \Rightarrow a=2$
$\therefore \quad f(x)=2^x$
Area enclosed by the lines $2|x|+5|y| \leq 4$
$4\left(\frac{1}{2} \times 2 \times \frac{4}{5}\right)=\frac{16}{5}$
$=\frac{16}{1+(2)^2}=\frac{(2)^4}{1+(2)^2}=\frac{f(4)}{1+f(2)}$

Topic: Mathematics, Functions, JEE Main
A function $f: R \rightarrow R$ satisfies the relation $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$ and $f(x) \neq 0$, $\forall x \in R$. If $f$ is differentiable at $x=0$ and $f^{\prime}(0)=4$ and $f(6)=3$, then $f^{\prime}(6)$ is equal to

Options:
A 0
B 12
C 3
D 6

The correct answer is:
12

Solution:
$$
\begin{aligned}
& \text { } f: R \rightarrow R \\
& f(x+y)=f(x) \cdot f(y), \forall x, y \in R \\
& f(x) \neq 0, \forall x \in R \\
& \text { and } f^{\prime}(0)=4 \text { and } f(6)=3 \\
& \because \quad f(x+y)=f(x) \cdot f(y) ...(i)\\
& \text { Put } x=0 \quad y=0 \\
& \quad f(0)=f(0) \cdot f(0) \\
& \Rightarrow f(0)[f(0)-1]=0 \\
& \quad f(0)=0 \text { and } f(0)-1=0
\end{aligned}
$$
not possible and $f(0)=1$
$$
\begin{aligned}
& f^{\prime}(0)=4 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=4 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(f(h)-1)}{h}=4 ...(ii)\\
&
\end{aligned}
$$
Now,
$$
\text { } \begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} f(x) \cdot\left[\frac{f(h)-1}{h}\right] \\
& =f(x) \lim _{h \rightarrow 0} \frac{f(h)-1}{h} \\
& =f(x) \times 4[by Eq. (ii)] \\
\therefore \quad f^{\prime}(x) & =4 f(x) \\
\therefore \quad f^{\prime}(6) & =4 f(6)=4 \times 3=12
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
A function f(x) is given by fx=5x5x+5, then the sum of the series

f120+f220+f320+…+f3920

is equal to:

Options:
A 192
B 492
C 392
D 292

The correct answer is:
392

Solution:
fx=5x5x+5 and f2-x=55x+5

fx+f2-x=1

⇒f120+f220+…+f3920

=f120+f3920+…+f1920+f2120+f2020

=19+12=392

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A function f: ZâZ is defined as fn=n+1nâodd integern2nâeven integer. If kâodd integer and fffk=33, then the sum of the digits of k is

Options:
A 7
B 5
C 9
D 8

The correct answer is:
5

Solution:
∵ k∈odd integer
fk=k+1 which is even
∴ffk=k+12
If k+12 is odd ⇒33=k+12+1⇒k=63 not possible (because k+12 is even)
If k+12 is even ⇒33=k+122⇒33=k+14⇒k=131
Sum of digits =5

Topic: Mathematics, Functions, JEE Main
A function is $f(x)= \begin{cases}1+x, & x \leq 2 \\ 5-x, & x\gt2\end{cases}$is

Options:
A Not continuous at $x=2$
B Differentiable at $x=2$
C Continuous but not differentiable at $x=2$
D None of these

The correct answer is:
Continuous but not differentiable at $x=2$

Solution:
$\lim _{h \rightarrow 0^{-}} 1+(2-h)=3, \lim _{h \rightarrow 0^{+}} 5-(2+h)=3, f(2)=3$
Hence, $f$ is continuous at $x=2$
Now $R f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{5-(2+h)-3}{h}=-1$
$L f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{1+(2-h)-3}{-h}=1$
$\Rightarrow R f^{\prime}(x) \neq L f^{\prime}(x) ; \therefore f$ is not differentiable at $x=2$

Topic: Mathematics, Continuity and Differentiability, JEE Main
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
$Interval \rightarrow Function$

Options:
A
$(-\infty, \infty) \rightarrow x^3-3 x^2+3 x+3$

B
$[2, \infty) \rightarrow 2x^3-3 x^2-12 x+6$

C
$\left(-\infty, \frac{1}{3}\right] \rightarrow \quad 3 x^2-2 x+1$

D
$(-\infty, -4] \rightarrow x^3+6 x^2+6$

The correct answer is:
$\left(-\infty, \frac{1}{3}\right] \rightarrow \quad 3 x^2-2 x+1$

Solution:
Clearly function $f(x)=3 x^2-2 x+1$ is increasing when $f^{\prime}(x)=6 x-2 \geq 0 \Rightarrow x \in[1 / 3, \infty)$
Hence $(C)$ is incorrect.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
A function $y=f(x)$ has a second order derivative $f^{\prime \prime}(x)=6(x-1)$. If its graph passes through the point $(2,1)$ and at that point the tangent to the graph is $y=3 x-5$, then the function is

Options:
A
$(x-1)^2$

B
$(x-1)^3$

C
$(x+1)^3$

D
$(x+1)^2$

The correct answer is:
$(x-1)^3$

Solution:
$f^{\prime \prime}(x)=6(x-1) \Rightarrow f^{\prime}(x)=3(x-1)^2+c$
and $f^{\prime}(2)=3 \Rightarrow c=0$
$\Rightarrow f(x)=(x-1)^3+k$ and $f(2)=1 \Rightarrow k=0$
$\Rightarrow f(x)=(x-1)^3$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2004
A(g)⇌2 B( g)+C(g)

For the given reaction, if the initial pressure is 450 mmHg and the pressure at time t is720 mmHg at a constant temperatureT and constant volume V. The fraction of A(g) decomposed under these conditions is x×10-1. The value of x is (nearest integer)

Options:

The correct answer is:
3

Solution:
Ag⇌2Bg+Cg intial 450 0 0at time=t 450-p 2p p

(450 – p) + (2p) + (p) = 720

450 + 2p = 720

2p= 270

p= 135

Fraction of A(g) decomposed=Amount of A decomposedInitial amount =135450=0.3

The fraction Ag decomposed = 3×10-1

⇒x=3

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (10 Apr Shift 2)
AgBr(s)+2S2O32-(aq)⇌Ag(S2O3)23-(aq)+Br-(aq)
Given Ksp(AgBr)=5×10-13;KfAg(S2O3)23-=5×1013
What is the molar solubility of AgBr in 0.1 M Na2S2O3 ?

Options:
A 0.5 M
B 0.45 M
C 0.045 M
D None of these

The correct answer is:
0.045 M

Solution:
AgBr(s)⇌Ag+(aq)+Br-(aq)Ksp
Ag+(aq)+2S2O32-(aq)⇌Ag(S2O3)23-(aq)Kf
AgBrs+2S2O32-(aq)0.1-2x⇌Ag(S2O3)23-(aq)x+Br-xaq
Here, K=Ksp×Kf=25
K=x2(0.1-2x)2
25=x2(0.1-2x)2
On solving x=0.045 M.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A Ge specimen is doped with Al. The concentration of acceptor atoms is ~1021atoms m-3. Given that the intrinsic concentration of electron-hole pair at this temperature is ~1019 m-3, the concentration of electrons in the specimen is:

Options:
A 1017 m-3
B 1015 m-3
C 104 m-3
D 102 m-3

The correct answer is:
1017 m-3

Solution:
ne=ni2NA
=101921021
ne=1017 m-3

Topic: Physics, Semiconductors, JEE Main
a g KHC2O4 is used to neutralize 100 mL of 0.02M KMnO4 in acid medium, where as b g KHC2O4 is used to neutralize 100 mL of 0.02M Ca(OH)2 then,

Options:
A a=b
B 4a=5b
C 5a=4b
D 2a=b

The correct answer is:
4a=5b

Solution:
milli eq. of  KHC 2 O 4  = milli eq. of  KMnO 4
(C3+)2→2C4++2e
Mn7++5e→Mn2+
aM2×1000=100×0.02×5
a=5×10–3M
milli eq. of  KHC 2 O 4  = milli eq. of  Ca(OH) 2
KHC2O4 acts as acidic salt
bM1×1000=100×0.02×2
b=4×10–3M
ab=54

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A galaxy is moving away from the earth at a speed of 286 km s-1. The shift in the wavelength of a red line at 630 nm is x×10-10 m. The value of x, to the nearest integer, is ___ .

[Take the value of the speed of the light c, as 3×108 m s-1]

Options:

The correct answer is:
6

Solution:
⇒Δλλc=v,

⇒Δλ=vc×λ=2863×105×630×10-9=6×10-10 m

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A galvanic cell has electrical potential of $1.1 \mathrm{~V}$. If an opposing potential of $1.1 \mathrm{~V}$ is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

Options:

None

Solution:
Alternating current is used in electrolysis so that concentration of ions in the solution remains constant and exact value of resistance can be measured.

Topic: Chemistry, Electrochemistry
A galvanometer acting as a voltmeter should have

Options:
A Low resistance in series with its coil
B Low resistance in parallel with its coil
C High resistance in series with its coil
D High resistance in parallel with its coil

The correct answer is:
High resistance in series with its coil

Solution:
Voltmeter has high resistance and is always connected in parallel with the circuit. So, to convert a galvanometer into voltmeter, a high resistance must be connected in series with it so that it draws negligible current from the circuit.

Topic: Physics, Current Electricity, JEE Main
A galvanometer acting as a voltmeter will have:

Options:
A a high resistance in parallel with its coil
B a high resistance in series with its coil
C a low resistance in parallel with its coil
D a coil resistance in series with its coil

The correct answer is:
a high resistance in series with its coil

Solution:
A galvanometer acting as a voltmeter will have a high resistance in series with its coil.

Topic: Physics, Current Electricity, JEE Main
A galvanometer can be used as a voltmeter by connecting a

Options:
A High resistance in series
B Low resistance in series
C High resistance in parallel
D Low resistance in parallel

The correct answer is:
High resistance in series

Solution:
To convert a galvanometer into a voltmeter, a high value resistance is to be connected in series with it.

Topic: Physics, Current Electricity, JEE Main
A galvanometer coil has 500 turns and each turn has an average area of 3×10-4m2. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5A is flowing through it, the strength of the field (in T) is _________ .

Options:

The correct answer is:
20

Solution:
τ=M→×B→=NiAB sin90°

=NiAB=Ni S B

⇒ B=τNiS=1.5500×0.5×3×10-4=20

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A galvanometer coil has a resistance 90 Î© and full scale deflection current 10 mA. A 910 Î© resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is 0.1 V, the number of divisions on its scale is

Options:

The correct answer is:
100

Solution:
V max = I max R g + R s
= 1 0 Ã 1 0 - 3 1 0 0 0
= 1 0 V
L.C. = 0.1 V
N L.C. = V max
â N = 1 0 0.1 = 1 0 0

Topic: Physics, Experimental Physics, JEE Main
A galvanometer coil has a resistance 90 Î© and full-scale deflection current 10 mA. A 910 Î© resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is 0.1 V, the number of divisions on its scale is

Options:

The correct answer is:
100

Solution:
V max = I max R g + R s
= 1 0 Ã 1 0 - 3 1 0 0 0
= 1 0 V
L.C. = 0.1 V
N L.C. = V max
â N = 1 0 0.1 = 1 0 0

Topic: Physics, Experimental Physics, JEE Main
A galvanometer G of 2Ω resistance is connected in the given circuit. The ratio of charge stored in C1 and C2 is:

Options:
A 23
B 32
C 1
D 12

The correct answer is:
12

Solution:
In steady state:

Req=12 Ω

Current through the branches with resistance, I=612=0.5 A

Now, potential drop across C1=3 V

and potential drop C2=4 V

Charge on the first capacitor, q1=C1 V1=12 μC

Charge on the second capacitor, q2=C2 V2=24 μC

Therefore, q1q2=12

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R.B.=2400 Ω. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 â¦. Then we can conclude :
Note: This question is awarded as the bonus. Now the question is corrected.

Options:
A Current sensitivity of galvanometer is 20 Î¼A/division
B Resistance of galvanometer is 200Â Î©.
C Resistance required on R.B. for a deflection of 10 divisions is 9800Â Î©.
D Full scale deflection current is 2 mA.

The correct answer is:
Current sensitivity of galvanometer is 20 Î¼A/division

Solution:
Current in the circuit,
I=V2400+RG and I2=V4900+RG
âRG=100Î©
I=22500=8Ã10-4A
Current Sensitivity =8Ã10-440
20Î¼A per division

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (10 Apr Online)
A galvanometer has a current range of 15 mA and a voltage range of 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is

Options:
A 0.8 Î©
B 0.93 Î©
C 0.03 Î©
D 2.0 Î©

The correct answer is:
0.03 Î©

Solution:
V=750×10-3 V
Ig=15×10-3A
and I=25 A
Using the relation, R=VIg, resistance of galvanometer is
=750×10-315×10-3
=50 Ω
Ig=SS+R×I
15×10-3=SS+50×25
S=0.03 Ω

Topic: Physics, Current Electricity, JEE Main
A galvanometer has a resistance of 50 Ω and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance.

Options:
A 5975 Ω
B 20050 Ω
C 19950 Ω
D 19500 Ω

The correct answer is:
19950 Ω

Solution:
If we connect a high resistance(R) in series with the galvanometer, it can be successfully converted into a voltmeter. Therefore,

Voltage drop across voltmeter:

V=IgRg+R⇒100=Ig50+R

⇒R=1005×10-3-50

=19950 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter given full scale deflection for a potential difference of 10V?

Options:
A 10kΩ
B 8.9 kΩ
C 7.9 kΩ
D 9.9 kΩ

The correct answer is:
9.9 kΩ

Solution:
Vg=igRg=0.1V
V=10V
R=RgVVg-1
=100×99=9.9KΩ

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A galvanometer having a coil resistance of $60 \Omega$ shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto $5.0 \mathrm{amp}$ by :

Options:
A Putting in parallel a resistance of $15 \Omega$
B Putting in parallel a resistance of $240 \Omega$
C Putting in series a resistance of $15 \Omega$
D Putting in series a resistance of $240 \Omega$

The correct answer is:
Putting in parallel a resistance of $15 \Omega$

Solution:
$$
\begin{aligned}
\mathrm{Ig} & =1.0 \mathrm{~A}, \mathrm{G}=60 \mathrm{q}, \mathrm{I}=5.0 \mathrm{~A} \\
\mathrm{~S} & =\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}} \mathrm{G} \\
\mathrm{G} & =\frac{1.0}{5.0-1.0}=60 \\
& =15 \Omega \text { in parallel }
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2009 (Mains)
A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is:

Options:
A 0.1Â Î©
B 3Â Î©
C 0.01 Î©
D 2Â Î©

The correct answer is:
0.01 Î©

Solution:
As shown in figure a shunt resistance S is connected parallel to the galvanometer to convert it into an ammeter. Let G is the resistance of the galvanometer.

Where G = 100 Ω and full-scale deflection current is Ig = 10 mA = 10×10-3 A

For galvanometer Vg = Ig G = 10×10-3 ×10 = 0.1 volt

From the figure, the shunt is parallel to galvanometer

∴VS = Vg ⇒IS S = Vg ...1

For ammeter full-scale deflection current is

IS =I-Ig ≈ 10 A

∴ from equation 1

10 S = 0·1 ⇒S = 0·01 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (03 Apr)