datavorous/entrance-exam-dataset huggingface dataset:
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is,
[Given, the Young's Modulus for steel and brass are, respectively, 120×109 N/m2 and 60×109 N/m2 ]

Options:
A 8.0×106 N/m2
B 1.2×106 N/m2
C 0.2×106 N/m2
D 1.8×106 N/m2

The correct answer is:
8.0×106 N/m2

Solution:
Stress will be same in both rods
Stress =Y1Δl1l=Y2Δl2l … (1)
And
Δl1+Δl2=0.2×10-3 m … (2)
From equation (1) and (2)
Stress ×lY1+lY2=0.2×10-3 m
Stress ×1120×109+160×109=0.2×10-3 m
Stress =8×106 N/m2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2019 (10 Apr Shift 2)
In an experiment, electrons are accelerated, from rest, by applying a voltage of $500 \mathrm{~V}$. Calculate the radius of the path if a magnetic field $100 \mathrm{mT}$ is then applied. [Charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$

Mass of the electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right]$

Options:
A $7.5 \times 10^{-3} \mathrm{~m}$
B $7.5 \times 10^{-2} \mathrm{~m}$
C $7.5 \mathrm{~m}$
D $7.5 \times 10^{-4} \mathrm{~m}$

The correct answer is:
$7.5 \times 10^{-4} \mathrm{~m}$

Solution:
Radius of the path (r) is given by $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$

\begin{aligned}

\mathrm{r} &=\frac{\sqrt{2 \mathrm{mk}}}{\mathrm{eB}} \quad(\because \mathrm{p}=\mathrm{mv}=\sqrt{2 \mathrm{mk}}) \\

&=\frac{\sqrt{2 \mathrm{meV}}}{\mathrm{eB}} \quad(\because \mathrm{k}=\mathrm{eV}) \\

\mathrm{r} &=\frac{\sqrt{\frac{2 \mathrm{~m}}{\mathrm{e}} \mathrm{V}}}{\mathrm{B}}=\frac{\sqrt{\frac{2 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}}(500)}}{100 \times 10^{-3}}

\end{aligned}

$r=\frac{\sqrt{\frac{9.1}{0.16} \times 10^{-10}}}{10^{-1}}=\frac{3}{4} \times 10^{-4}$

$=7.5 \times 10^{-4}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In an experiment for determination of refractive index of glass of a prism by i v/s δ plot, it was found that a ray incident at angle 35o , suffers a deviation of 40o and that it emerges at angle 79o . In that case which of the following is closest to the maximum possible value of the refractive index?

Options:
A 1.7
B 1.8
C 1.5
D 1.6

The correct answer is:
1.5

Solution:
i=35o, δ=40o, e=79o
δ=i+e-A
40o=35o+79o-A
A=74o
And r1+r2=A=74o
Since δmin<40o
μ<sin⁡74+402sin⁡37
μmax=1.44

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (03 Apr)
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 40 cm from the pole of the mirror is formed at distance 120 cm from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in 1 cm. The value of error in measurement of focal length of the mirror is 1K cm. The value of K is ______.

Options:

The correct answer is:
32

Solution:
Given here, u=-40 cm, v=-120 cm

Using mirror formula, 1v+1u=1f

-1120-140=1f⇒f=-30 cm

Now, differentiating mirror equation, we get

-1v2dv-1u2du=-1f2df

Also, the least count of scale, dv=du=120 cm

Putting in above equation,

∴ 120(120)2+120(40)2=df(30)2

On solving, we get

df=132 cm

∴ K=32.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
In an experiment for the determination of the speed of sound in air using the resonance method, the length of the air column that resonates in the fundamental mode with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. The end correction is

Options:
A 0.012 m
B 0.025 m
C 0.05 m
D 0.024 m

The correct answer is:
0.025 m

Solution:
λ4=e+l1

3λ4=e+l2

e+l2=3e+3l1

e⁡=l⁡2-3l⁡12=0.35-0.32=0.052=0.025 m

Topic: Physics, Experimental Physics, JEE Main
In an experiment of CE configuration of n-p-n transistor, the transfer characteristics are observed as given in figure.

If the input resistance is 200Ω and output resistance is 60Ω, the voltage gain in this experiment will be _____ .

Options:

The correct answer is:
15

Solution:
Voltage gain, AV=?
Input Resistance =200 Ω
Output Resistance =60 Ω

β=ΔICΔIB=5-0×10-3100-0×10-6=50

AV=βRCRB=50×60200=15

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (25 Jun Shift 2)
In an experiment of determine the Young's modulus of wire of a length exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of ±0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ±0.01 mm. The error in the measurement of Young's modulus ΔY is found to be x×1010 N m-2. The value of x is _____ .

Options:

The correct answer is:
2

Solution:
Given here,

Length, L=1 m and extension in length, ΔL=0.4×10-3 m

Mass of load, m=1 kg and diameter of wire, d=0.4×10-3 m

Young's modulus is given by

Y=FLAΔL=mg×1πd240.4×10-3

⇒Y=10×4π0.4×10-32×0.4×10-3

⇒Y=40×722×64×10-3×10-9

⇒Y=0.199×10-12 N m-2

Error in measurement of Young's modulus,

ΔYY=ΔFF+ΔLL+ΔAA+ΔΔLΔL

=0.020.4+2Δdd=0.24+2×0.010.4

=0.12+0.12=0.1

=0.199×1011≈2×1010

Hence, the value of x=2.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell$ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be

Options:
A
$\ell$

B
$2 \ell$

C
$\ell / 2$

D
$\ell / 4$

The correct answer is:
$\ell / 2$

Solution:
Balancing length $l$ will give emf of cell $\therefore E=K l$
Here $\mathrm{K}$ is potential gradient.
If the cell is short circuited by resistance 'R'
Let balancing length obtained be $l^{\prime}$ then
$$
\begin{aligned}
& V=k l^{\prime} \\
& r=\left(\frac{E-V}{V}\right) R \\
\Rightarrow & V=E-V \\
\Rightarrow & 2 V=E \\
\text { or, } & 2 K l^{\prime}=K l \\
\therefore \quad & l^{\prime}=\frac{l}{2}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (26 May Online)
In an experiment of simple pendulum, the errors in the measurement of length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$, respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is

Options:
A $\mathbf{5} \%$
B $7 \%$
C $8 \%$
D $1 \%$

The correct answer is:
$7 \%$

Solution:
Maximum percentage in the value of $\frac{L}{T^3}$
$\begin{aligned} & =\frac{\Delta L}{L} \times 100 \%+2 \frac{\Delta T}{T} \times 100 \% \\ & =3+2 \times 2=7 \%\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is $6600 Å$, then wavelength of first maximum will be:

Options:
A
$3300 Å$

B
$4400 Å$

C
$5500 Å$

D
$6600 Å$

The correct answer is:
$4400 Å$

Solution:
In a single slit experiment, For diffraction maxima, $a \sin \theta=\left(2 n+1 \frac{Å}{2}\right.$ and for diffraction minima, $\mathrm{a} \sin \theta=\mathrm{n} \lambda$ According to question,
$$
\begin{aligned}
&\left(2 \times 1+1 \frac{\lambda}{2}=1 \times 6600\right. \\
&\left(\because \lambda_{\mathrm{R}}=6600 Å\right) \\
&\lambda=\frac{6600 \times 2}{3} \\
&\lambda=4400 Å
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (12 Apr Online)
In an experiment on a circuit as shown in the figure, the voltmeter shows \(8 \mathrm{~V}\) reading. The resistance of the voltmeter is,

Options:
A \(20 \Omega\)
B \(320 \Omega\)
C \(160 \Omega\)
D \(1.44 \mathrm{k} \Omega\)

The correct answer is:
\(160 \Omega\)

Solution:
Hint: Voltage across \(20 \Omega=2 \mathrm{~V}\)
Main current \(=\frac{2}{20}=0.1 \mathrm{~A}\)
Current through \(160 \Omega=\frac{8}{160}=\frac{1}{20}=0.05 \mathrm{~A}\)
Also, \(8 \mathrm{~V}=0.05 \times \mathrm{R}\)
\(R=\frac{8}{0.05}=160 \Omega\)

Topic: Physics, Current Electricity, WBJEE, WBJEE 2023
In an experiment on measurements of emf of a cell by a potentiometer, the balancing length for a cell of emf E and internal resistance r is found to be l. Now if another cell of emf E and internal resistance 2r is connected in parallel to the first cell and balancing length determined, then the balancing length will be-

Options:
A l
B 2 l
C l2
D 2l3

The correct answer is:
l

Solution:
Since potentiometer doesn't draw any current to measure the emf of the cell, the balancing length will remain the same.
balancing length =l

Topic: Physics, Experimental Physics, JEE Main
In an experiment on photoelectric effect, a student plots stopping potential $\mathrm{V}_0$ against reciprocal of the wavelength $\lambda$ of the incident light for two different metals $\mathrm{A}$ and $\mathrm{B}$. These are shown in the figure.
Looking at the graphs, you can most appropriately say that:

Options:
A
Work function of metal $\mathrm{B}$ is greater than that of metal $\mathrm{A}$

B
For light of certain wavelength falling on both metal, maximum kinetic energy of electrons emitted from A will be greater than those emitted from $\mathrm{B}$.

C
Work function of metal $\mathrm{A}$ is greater than that of metal $\mathrm{B}$

D
Students data is not correct

The correct answer is:
Students data is not correct

Solution:
$\begin{aligned}
& \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_0 \\
& \mathrm{v}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi}{\mathrm{e}}
\end{aligned}$
As the value of $\frac{1}{\lambda}$ (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function $(\phi)$.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2013 (25 Apr Online)
In an experiment on photoelectric effect photons of wavelength $300 \mathrm{nm}$ eject electrons from a metal of work function $2.25 \mathrm{eV}$. A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom:

Options:
A $n=2$ to $n=1$ state
B $n=3$ to $n=1$ state
C $n=3$ to $n=2$ state
D $n=4$ to $n=3$ state

The correct answer is:
$n=3$ to $n=2$ state

Solution:
Given $\phi=2.25 \mathrm{eV}$
$$
\lambda=300 \times 100^{-3} \mathrm{~m}
$$
Energy of the given photon,
$$
\mathrm{E}=\mathrm{hv}
$$
$$
=\frac{\mathrm{hc}}{\lambda}=\frac{663 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}}=6.63 \times 10^{-19} \mathrm{~J}=4.13 \mathrm{eV}
$$
Now,
$$
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi=4.137-2.25=1.88 \mathrm{eV} \rightarrow(1)
$$
So, $1.88 \mathrm{eV}$ energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen
$$
\begin{array}{rrrr}
\Rightarrow \mathrm{n}=1 & 2 & 3 & 4 \\
\mathrm{E}=-13.6 \mathrm{ev} & -3.4 & -1.51 & -0.85
\end{array}
$$
And from statement (1),
$$
1.18=E_i-E_f
$$
$\left\{E_i \& E_f \quad\right.$ are energy of initial and final orbit $\}$ And also from the above table, we can observe that,
$$
E_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}}=(-1.51)-(-3.4)=1.89 \mathrm{eV}
$$
$$
=\mathrm{E}_2-\mathrm{E}_3
$$
That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 3oonm wavelength hydrogen electron can jump from 3 to 2 .

Topic: Physics, Dual Nature of Matter, JEE Main
In an experiment on photoelectric emission from a metallic surface, the wavelength of incident light is 2Ã10-7 m and stopping potential is 2.5 V. The threshold frequency of the metal (in Hz ) approximately is (charge on electron e=1.6Ã10-19 C, Planck's constant h=6.6Ã10-34 J s)

Options:
A 12×1015
B 9×1015
C 9×1014
D 12×1013

The correct answer is:
9×1014

Solution:
eV0=hν-hν0
∴ Threshold frequency,
ν0=v-eV0h
= cλ-eV0h
∴ ν0=3×1082×10-7-1.6×10-19×2.56.6×10-34
= 9.0×1014 Hz

Topic: Physics, Dual Nature of Matter, JEE Main
In an experiment on the photoelectric effect, the following graphs were obtained between the photoelectric current (I) and the anode potential \( (V) \). Name the characteristic of the incident radiation that was kept constant in this experiment.

Options:
A Intensity
B Photon flux
C Stopping Potential
D Resistance

The correct answer is:
Stopping Potential

Solution:
Stopping potential is same for the curves as all the curves start from same point when current is zero.

Topic: Physics, Dual Nature of Matter, JEE Main
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. The specific heat of the metal is -

Options:
A 0.1 cal gâ1 Â°Câ1
B 0.2 cal g–1 °C–1
C 0.3 cal g–1 °C–1
D 0.4 cal g–1 °C–1

The correct answer is:
0.1 cal g–1 °C–1

Solution:
Heat lost by the metal = Heat gain by water + calorimeter

0.20×103×c×150-40=175×13

c=175×13110×200≈0.1 cal g-1 °C-1

Topic: Physics, Experimental Physics, JEE Main
In an experiment, setup A consists of two parallel wires which carry currents in opposite directions as shown in the figure. A second setup B is identical to setup A, except that there is a metal plate between the wires
Let $\mathrm{F}_{\mathrm{A}}$ and $\mathrm{F}_{\mathrm{B}}$ be the magnitude of the force between the two wires in setup A and setup $\mathrm{B}$, respectively.

Options:
A $F_{A}>F_{B} \neq 0$
B $\mathrm{F}_{\mathrm{A}} < \mathrm{F}_{\mathrm{B}}$
C $\mathrm{F}_{\mathrm{A}}=\mathrm{F}_{\mathrm{B}} \neq 0$
D $\mathrm{F}_{\mathrm{A}}>\mathrm{F}_{\mathrm{B}}=0$

The correct answer is:
$\mathrm{F}_{\mathrm{A}}=\mathrm{F}_{\mathrm{B}} \neq 0$

Solution:
In setup B, A metal is placed, due to which metal may get magnetized and it may also exert force on current carrying wire but force between two wire remain same however net force on wire may get charge due to magnetic field produced by magnetized metal.
Ans. (C)

Topic: Physics, Magnetic Effects of Current, KVPY, KVPY 2016 (SA)
In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $\left(=0.5^{\circ}\right)$, then the least count of the instrument is

Options:
A One degree
B Half degree
C One minute
D Half minute

The correct answer is:
One minute

Solution:
30 divisions of the vernier scale coincide with 29 divisions of the main scale.
Therefore, I V.S.D $=(29 / 30)$ M.S.D
Least Count $=1 \mathrm{MSD}-1 \mathrm{VSD}$
$=1$ MSD $-(29 / 30)$ MSD
$=(1 / 30)$ MSD
$=(1 /30)\left(0.5^0\right)$
$=(1 / 30) \times 30 \mathrm{~min}=1 \mathrm{~min}$
Answer: (3) One minute

Topic: Physics, Mathematics in Physics, JEE Main
In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $\left(=0.5^{\circ}\right)$, then the least count of the instrument is

Options:
A
one minute

B
half minute

C
one degree

D
half degree

The correct answer is:
one minute

Solution:
$$
\text { Least count }=\frac{\text { value of main scale division }}{\text { No of divisions on vernier scale }}=\frac{1}{30} \mathrm{MSD}=\frac{1}{30} \times \frac{1}{2}^{\circ}=\frac{1}{60}^{\circ}=1 \text { minute }
$$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2009
In an experiment the angles are required to be measured using an instrument in which 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half a degree $\left(=0.5^{\circ}\right)$, then the least count of the instrument is

Options:
A half minute
B one degree
C half degree
D one minute

The correct answer is:
one minute

Solution:
According to the question, given that
29 main scale division $($ MSD $)=30$ vernier scale division (VSD)
$\begin{aligned}
& \Rightarrow 29 \mathrm{MSD}=30 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{29}{30} \times 1 \mathrm{MSD} \\
& \Rightarrow \quad 1 \mathrm{VSD}=\frac{29}{30}\left(0.5^{\circ}\right) \quad\left[\because 1 \mathrm{MSD}=0.5^{\circ}\right] \\
& \Rightarrow \quad 1 \mathrm{VSD}=\left(\frac{29}{60}\right)^{\circ}
\end{aligned}$
$\therefore$ Least count of vernier calliperse is given as
$\begin{aligned}
\text { LC } & =1 \text { MSD }-1 \text { VSD } \\
& =0.5^{\circ}-\left(\frac{29}{60}\right)^{\circ}=\left(\frac{30-29}{60}\right)^{\circ}=\left(\frac{1}{60}\right)^{\circ} \\
& =1 \min \quad\left[\because 1 \min =\frac{1}{60} \text { degree }\right]
\end{aligned}$

Topic: Physics, Experimental Physics, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
In an experiment to determine the focal length f of a concave mirror by the u-Ï method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then

Options:
A x<f
B f<x<2f
C x=2f
D x>2f

The correct answer is:
f<x<2f

Solution:
Since object and image move in opposite directions, the positioning should be as shown in the figure. Object lies between focus and centre of curvature f<x<2f.

â´ Correct option is (b)

Topic: Physics, Ray Optics, JEE Main
In an experiment to determine the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure. What is the value of g at the place?

Options:
A 10.0 m s-2
B 9.87 m s-2
C 9.91 m s-2
D 9.81 m s-2

The correct answer is:
9.87 m s-2

Solution:
T=2πLg

⇒ g=4π2LT2

T2L=4π2g=tanθ=slope of graph

⇒ g=4π2tanθ

∵ tanθ=8−22−0.5=61.5=4

⇒ g=4π24=(3.14)2=π2=3.142=9.87 m s-2

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2014 (19 Apr Online)
In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii r1 and r2 . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be 5×10-4 s, the difference in radii, r1-r2 is best-given by

Options:
A 0.01 cm
B 0.05 cm
C 0.5 cm
D 1 cm

The correct answer is:
0.05 cm

Solution:
T=2π lg
T∝l
l=1
ΔTT=12 Δll
Δl=r1-r2⇒ ∆l=2∆TT×l
=2×5×10-42×1 m
= 5×10-2 cm
=0.05 cm

Topic: Physics, Oscillations, JEE Main, JEE Main 2017 (09 Apr Online)
In an experiment to determine the resistance of a galvanometer by half deflection method, the circuit shown is used. In one set of readings, if R=10 Î© and S =4 Î©, then the resistance of the galvanometer is

Options:
A 20 3 Î©
B 40 3 Î©
C 50 3 Ω
D 70 3 Ω

The correct answer is:
20 3 Î©

Solution:
In 1st case,

I1=ER+G

In 2nd case,

I2=ER+SGS+G
for half deflection,
I2G=12I1

12ER+G=ER+SGS+GSS+G

âG=RSR-S

G=10Ã410-4=203 Î©

Topic: Physics, Experimental Physics, JEE Main
In an experiment to find acceleration due to gravity g using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x%. The value of x is

Options:

The correct answer is:
5

Solution:
Time period of oscillation is given by T=2πlg , here, l is length and g is acceleration due to gravity.

From above relation, we have g=14π2T2l.

Fractional error in g is stated as

Δgg=2ΔTT+Δll

⇒Δgg×100=2×1100×0.5×100+1 mm10 cm×100

⇒Δgg×100=5100×100=5%

Hence, value of x=5.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
In an Experiment to find the loss of energy with respect to time in case of swinging simple pendulum, the graph between square of amplitude and time is best represented by

Options:
A
B
C
D

The correct answer is:

Solution:
Energy = 1 2 K A 2
For a damped harmonic oscillation
A= A o e − bt 2m sin( w't+δ )
amplitude can be written as A= A 0 e − bt 2m
So curve between A 2  vs t
will be an exponential decay

Topic: Physics, Oscillations, JEE Main
In an experiment to measure focal length $(f)$ of convex lens, the least counts of the measuring scales for the position of object $(\mathrm{u})$ and for the position of image $(\mathrm{v})$ are $\Delta \mathrm{u}$ and $\Delta \mathrm{v}$, respectively. The error in the measurement of the focal length of the convex lens will be:

Options:
A $2 f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$
B $\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}$
C $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$
D $f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$

The correct answer is:
$f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$

Solution:
$\begin{aligned} & \mathrm{f}^{-1}=\mathrm{v}^{-1}-\mathrm{u}^{-1} \\ & -\mathrm{f}^{-2} \mathrm{df}=-\mathrm{v}^{-2} \mathrm{dv}-\mathrm{u}^{-2} \mathrm{du} \\ & \frac{\mathrm{df}}{\mathrm{f}^2}=\frac{\mathrm{dv}}{\mathrm{v}^2}+\frac{\mathrm{du}}{\mathrm{u}^2} \\ & \mathrm{df}=\mathrm{f}^2\left[\frac{\mathrm{dv}}{\mathrm{v}^2}+\frac{\mathrm{du}}{\mathrm{u}^2}\right]\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2âm when the cell is shunted by a 4 Î© resistance and at 3âm when cell is shunted by a 8 Î© resistance. The internal resistance of cell is -

Options:
A 12Â Î©
B 8 Î©
C 16Â Î©
D 1Â Î©

The correct answer is:
8 Î©

Solution:
VA-VB=ÏÃl
E-Err+4=ÏÃ2 ....(i).
E4r+4=ÏÃ2 .... (i)
Similarly, E8r+8=ÏÃ3 ....(ii)
eq1eq2âr+82r+4=23â3r+24=4r+16âr=8 Î©

Topic: Physics, Current Electricity, JEE Main
In an experiment to verify Stokes law, a small spherical ball of radius r and density ρ falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to: (ignore viscosity of air)

Options:
A r4
B r
C r3
D r2

The correct answer is:
r4

Solution:
The velocity of ball just before striking the surface of water is given by using kinematical equation of motion under the gravity

v2=u2-2gy

⇒v2=-2g(-h)

⇒v=2gh ...(1)

Now, the terminal velocity of spherical ball inside the water tank is given by,

vT=29r2(ρ-σ)gη ...(2)

After falling through h the velocity of ball should be equal to its terminal velocity.

From equation (1) and (2) we get,

⇒2gh=29r2(ρ-σ)gη

Squaring the above equation, we get

⇒h=281r4g(ρ-σ)2η2

⇒h∝r4

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (05 Sep Shift 2)
In an experiment with 9 observations on x, the following results are available ∑x2=360 and ∑x=34. One observation that was 8, was found to be wrong and was replaced by the correct value 10, then the corrected variance is

Options:
A 2509
B 28
C 2409
D 28

The correct answer is:
28

Solution:
∑x2=360,∑x=34
corrected∑x'=34–8+10=36
corrected ∑x'2=360+100–64=396
The corrected variance =∑x'29–∑x92=3969–3692
=44–16=28

Topic: Mathematics, Statistics, JEE Main
In an experiment with 10 observations on x the following results are available ∑x2=354 and ∑x=58. If one observation that was 8 was found to be wrong and was replaced by the corrected value 10, then the corrected variance is

Options:
A 5
B 3
C 4
D 6

The correct answer is:
3

Solution:
âx2=354,âx=58
∑ x ′ =58−8+10=60
∑ x ′ 2 =354+100−64=390
Variance = ∑ x ′ 2 10 − ∑ x ′ 10 2 = 390 10 − 60 10 2
=39-36
=3

Topic: Mathematics, Statistics, JEE Main