datavorous/entrance-exam-dataset huggingface dataset:
Water is flowing at a speed of 1.5 m s-1 through a horizontal tube of cross-sectional area 10-2 m2 and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water = 103 kg m-3)

Options:
A 33.7 N
B 45 N
C 15 N
D 22.5 N

The correct answer is:
22.5 N

Solution:
υ = 1.5 m/s A = 1 0 - 2 m 2
Now, F = d dt P
= d dt m υ = υ dm dt
= A υ · ρ · υ
= A ρ υ 2
= 1 0 - 2 × 1 0 0 0 × 1.5 2
= 1 0 - 2 × 1 0 0 0 × 2.25
= 22.5 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (09 Apr Online)
Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 \times 10^{-1} \mathrm{~m}$ below the lap is close to :

Options:
A
$7.5 \times 10^{-3} \mathrm{~m}$

B
$9.6 \times 10^{-3} \mathrm{~m}$

C
$3.6 \times 10^{-3} \mathrm{~m}$

D
$5.0 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$3.6 \times 10^{-3} \mathrm{~m}$

Solution:
Equation of continuity
$$
\begin{aligned}
& \Rightarrow(\mathrm{a} \times \mathrm{v}) \text { top }=(\mathrm{a} \times \mathrm{v}) \text { bottom } \\
& \mathrm{v}_{\mathrm{b}}^2-(0.4)^2=2 \times 9.8 \times 0.2\left[\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh} \text { is used }\right] \\
& \mathrm{v}_{\mathrm{b}}=2 \mathrm{~m} / \mathrm{s} \text { (nearly) } \\
& \pi\left[8 \times 10^{-3}\right] \times 0.4=\pi \mathrm{d}^2 \times 4 \\
& \mathrm{~d} \approx 3.6 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2011
Water is flowing through a channel that is 12 m wide with a speed of 0.75 m s-1. The water then flows into four identical channels that have a width of 4.0 m each. The depth of the water does not change as it flows into the four channels. The speed of the water in one of the smaller channels is

Options:
A 0.56 m s-1
B 2.3 m s-1
C 0.25 m s-1
D 0.75 m s-1

The correct answer is:
0.56 m s-1

Solution:
Av = A1v1 + A2v2 + A3v3 + A4v4
Av = 4(A1v1)
12 × 0.75 = 4 [4 × v]
v= 34× 0.75
= 0.75 × 0.75
v= 0.56 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water is flowing through a horizontal pipe in stream line flow. At the narrowest part of the pipe

Options:
A velocity is maximum and pressure is minimum
B pressure is maximum and velocity is minimum
C both pressure and velocity are minimum
D both pressure and velocity are maximum

The correct answer is:
velocity is maximum and pressure is minimum

Solution:
Using equation of continuity,
$\mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2$
In streamlined flow, as the product of $\mathrm{Av}$ is a constant, the velocity at the narrowest part will be maximum.
By using Bernoulli's principle,
$\mathrm{P}+\frac{1}{2} \rho v^2+\rho g h=\text { Constant }$
We see that $\rho g h$ is constant because the pipe is horizontal.
As velocity increases, for the equation to remain constant, $\mathrm{P}$ has to decrease.
$\therefore \quad$ Velocity will be maximum and pressure is minimum

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2023 (09 May Shift 1)
Water is flowing through a horizontal tube having cross-sectional areas of its two ends being $A$ and $A^{\prime}$ such that the ratio $A / A^{\prime}$ is 5 . If the pressure difference of water between the two ends is $3 \times$ $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, the velocity of water with which it enters the tube will be (neglect gravity effects)

Options:
A
$5 \mathrm{~m} \mathrm{~s}^{-1}$

B
$10 \mathrm{~m} \mathrm{~s}^{-1}$

C
$25 \mathrm{~m} \mathrm{~s}^{-1}$

D
$50 \sqrt{10} \mathrm{~m} \mathrm{~s}^{-1}$

The correct answer is:
$5 \mathrm{~m} \mathrm{~s}^{-1}$

Solution:
According to Bernoulli's theorem
$$
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \ldots(\mathrm{i})
$$
From question,
$$
P_1-P_2=3 \times 10^5, \frac{A_1}{A_2}=5
$$
According to equation of continuity
$$
\begin{aligned}
& A_1 v_1=A_2 v_2 \\
& \text { or, } \frac{A_1}{A_2}=\frac{v_2}{v_1}=5 \\
& \Rightarrow \quad v_2=5 v_1
\end{aligned}
$$
From equation (i)
$$
\begin{aligned}
& P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\
& \text { or } 3 \times 10^5=\frac{1}{2} \times 1000\left(5 v_1^2-v_1^2\right. \\
& \Rightarrow \quad 600=6 v_1 \times 4 v_1 \\
& \Rightarrow \quad v_1^2=25 \\
& \therefore \quad v_1=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (12 May Online)
Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as

Options:
A relative velocity
B terminal velocity
C critical velocity
D particle velocity

The correct answer is:
critical velocity

Solution:
The velocity of water below which the flow remains streamline flow is known as critical velocity.

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2012
Water is indispensable for life. What properties of water make it useful for all biological process on the earth?

Options:

None

Solution:
Following are the properties of water that make it useful for all biological processes.
(i) Water is the major solvent through which mineral nutrients enter a Plant from the soil solution.
(ii) It is an ideal solvent with neutral $\mathrm{pH}$.
(iii) Water is the major constituent of protoplasm. it constitutes approximately $90 \%$ of the protoplasm.
(iv) Water acts as a medium for translocation of nutritive substances. Mineral nutrients are absorbed by the roots. Carbohydrates that are formed during photosynthesis are transported by water from cell to cell, tissue to tissue and organ to organ.
(v) Water is involved in photosynthesis in plants, as it incorporates hydrogen atom into carbohydrate and releases oxygen atoms as $\mathrm{O}_2$.
(vi) Water acts as an agent for temperature control. The specific heat of water helps plant in maintaining a relatively stable internal temperature.
(vii) Water is necessary for pollination in some plants in bryophytes and pteridophytes, water are essentially requires for the fertilisation process.

Topic: Biology, Transport in Plants
Water level is maintained in a cylindrical vessel upto a fixed height $H$. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel, so that the water stream coming out of the hole strikes the horizontal plane of the greatest distance from the vessel?

Options:
A $h=\frac{H}{2}$
B $h=\frac{3 H}{2}$
C $h=\frac{2 H}{3}$
D $h=\frac{3}{4} H$

The correct answer is:
$h=\frac{H}{2}$

Solution:
The velocity with which water will came out
$v=\sqrt{2 g(H-h)}$
The time of flight of water ejected from the hole.
$\begin{aligned} t & =\sqrt{\frac{2 h}{g}}=\text { time of flight } \\ \therefore \quad x & =v \times t \\ & =\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}=\sqrt{4\left(H h-h^2\right)}\end{aligned}$
For $x$ to be maximum
$\begin{aligned} & \frac{d x}{d h}=0 \\ \Rightarrow & d \frac{\sqrt{4\left(H h-h^2\right)}}{d h}=0 \\ \Rightarrow & 0=H-2 h \Rightarrow h=\frac{H}{2}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JIPMER, JIPMER 2015
Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27oC to 77oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Options:
A 8 min 20 s
B 7 min
C 14 min
D 6 min 2 s

The correct answer is:
8 min 20 s

Solution:
P = 1 0 0 0 watt
∴ Pin=1000-160 J/s=840 J/s
Now, d Q dt = ms dT dt
∴ 840=msdTdt
∴ dt = ms 8 4 0 dT
Integrating both sides we have;
t = ms 8 4 0 ∫ dT
= 2 × 1 0 0 0 × 1 0 - 6 × 1 0 0 0 × 4.2 × 1 0 0 0 8 4 0 × 5 0
= 5 0 0 sec
= 8 min 2 0 sec

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2014 (09 Apr Online)
Water (prevention and control of pollution) Act was passed in the year

Options:
A 1974
B 1987
C 1995
D 1986

The correct answer is:
1974

Solution:
Realizing the importance of maintaining the cleanliness of the water bodies, the Government of India has passed the Water (prevention and control of pollution) Act was passed in the year 1974 to safeguard our water resources from pollution.

Topic: Biology, Environmental Issues, NEET
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{-4} \mathrm{~N}$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{-2} \mathrm{Nm}^{-1}$, the inner circumference of the capillary must be

Options:
A $\quad 1.25 \times 10^{-2} \mathrm{~m}$
B $0.50 \times 10^{-2} \mathrm{~m}$
C $6.5 \times 10^{-2} \mathrm{~m}$
D $12.5 \times 10^{-2} \mathrm{~m}$

The correct answer is:
$12.5 \times 10^{-2} \mathrm{~m}$

Solution:
$6 \times 10^{-2} \times$ Circumference $=$ Force
$\therefore$ Circumference $=\frac{75 \times 10^{-4}}{6 \times 10^{-2}}=12.5 \times 10^{-2} \mathrm{~m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water rises in a capillary tube to a height of 1 cm. In another capillary where the radius is one-third of it, how high will the water rise (in cm)?

Options:

The correct answer is:
3

Solution:
h=2Tcos⁡θrlg

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water rises to a height 'h' in capillary tube. If the length of capillary tube above the surface of water is made less than 'h', then:

Options:
A water rises upto the top of capillary tube and stays there without overflowing.
B water rises upto a point a little below the top and stays there.
C water does not rise at all.
D water rises upto the tip of capillary tube and then starts overflowing like a fountain.

The correct answer is:
water rises upto the top of capillary tube and stays there without overflowing.

Solution:
It won't overflow. That should be obvious since doing so would create a constant flow, constantly using energy, but without any energy input. The same force that pulls the water along the inside of the capillary tube also holds it there when it reaches the end. This force doesn't just pull upward, it pulls the water along the glass. At a certain height, the weight of the water column balances this pull. In that case the pull is upward since there is water below but not above.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water sample is called cleanest on the basis of which one of the BOD values given below :

Options:
A 11 ppm
B 15 ppm
C 3 ppm
D 21 ppm

The correct answer is:
3 ppm

Solution:
Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Water sample is reported to be highly polluted if BOD (Biological Oxygen Demand) value of sample becomes

Options:
A
more than $17 \mathrm{ppm}$

B
equal to $10 \mathrm{ppm}$

C
equal to $5 \mathrm{ppm}$

D
less than $5 \mathrm{ppm}$

The correct answer is:
more than $17 \mathrm{ppm}$

Solution:
Water sample is reported to be highly polluted if BOD value of sample is more than 17 ppm.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2012 (12 May Online)
Water samples with BOD values of 4 ppm and 18 ppm, respectively, are:

Options:
A Highly polluted and Highly polluted
B Clean and Clean
C Highly polluted and Clean
D Clean and Highly polluted

The correct answer is:
Clean and Highly polluted

Solution:
Clean water has a BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17ppm so more polluted.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Watt-hour meter measures

Options:
A Electric energy
B Current
C Voltage
D Power

The correct answer is:
Electric energy

Solution:
Watt-hour meter measures electric energy.

Topic: Physics, Current Electricity, JEE Main
Wave pulse can travel along a tense string like a violin spring. Aseries of experiments showed that the wave velocity $\mathrm{V}$ of a pulse depends on the following quantities, the tension T of the string, the cross-section area A of the string and then as per unit volume $\rho$ of the string. Obtain an expression for $V$ in terms of the $\mathrm{T}, \mathrm{A}$ and $\rho$ using dimensional analysis.

Options:
A $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$
B $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A}}}$
C $V=k \sqrt{\frac{A \rho}{T}}$
D None of these

The correct answer is:
$\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$

Solution:
Let $\mathrm{V}=\mathrm{kT}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \rho^{\mathrm{c}}$,

$\mathrm{k}=$ dimensional constant

Writing dimension on both we side

$$

\begin{aligned}

\left[\mathrm{LT}^{-1}\right] &=\left[\mathrm{MLT}^{-2}\right]^{a}\left[\mathrm{~L}^{2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}} \\

&=\left[\mathrm{M}^{\mathrm{a}+\mathrm{c}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}-3 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}}\right]

\end{aligned}

$$

Comparing power on both sides we have

$$

\begin{array}{l}

a+c=0, a+2 b-3 c=1, \quad-2 a=-1 \\

\therefore \quad a=\frac{1}{2}, c=-\frac{1}{2} \Rightarrow b=-\frac{1}{2} \therefore V=k \sqrt{\frac{T}{A \rho}}

\end{array}

$$

Topic: Physics, Units and Dimensions, JEE Main
Wavefront means

Options:
A All particles in it have same phase
B All particles have opposite phase of vibrations
C Few particles are in same phase, rest are in opposite phase
D None of these

The correct answer is:
All particles in it have same phase

Solution:
Wavefront is the locus of all the particles which vibrates in the same phase

Topic: Physics, Wave Optics, JEE Main
Wavefront of a wave has direction with wave motion

Options:
A Parallel
B Perpendicular
C Opposite
D At an angle of $\theta$

The correct answer is:
Perpendicular

Solution:
Direction of wave is perpendicular to the wavefront.

Topic: Physics, Wave Optics, JEE Main
Wavelength of first line in Lyman series is λ. The wavelength of first line in Balmer series is

Options:
A 527λ
B 365λ
C 275λ
D 536λ

The correct answer is:
275λ

Solution:
According to Bohr, the wavelength emitted when an electron jumps from n1th to n2th orbit is
E = hcλ = E2 –E1
1λ = R 1n12-1n22
For first-line in Lyman series
1λL =R112-122 = 3 R4 ...(i)
For first-line in Balmer series,
1λB =R122-132 = 5 R36 ...(ii)
From Eqs. (i) and (ii)
∴ λBλL =3 R4×365R = 275
∴ λB=275λ (∵ λL=λ)

Topic: Physics, Atomic Physics, JEE Main
Wavelength of light used in an optical instrument are $\lambda_1=4000 Å$ and $\lambda_2=5000 Å$, then ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$ ) is

Options:
A
16 : 25

B
9 : 1

C
4 : 5

D
5 : 4

The correct answer is:
5 : 4

Solution:
Resolving power $\alpha(1 / \lambda)$. Hence, $\frac{(\text { R.P })_1}{(\text { R.P })_2}=\frac{\lambda_2}{\lambda_1}=\frac{5}{4}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2002
Wavelength of spectral line emitted is inversely proportional to

Options:
A Radius
B Energy
C Velocity
D Quantum number

The correct answer is:
Energy

Solution:
Wavelength of spectral line emitted is inversely proportional to energy $\lambda \propto \frac{1}{E}$

Topic: Chemistry, Structure of Atom, JEE Main
Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons. (Take hc=1242 eV nm.)

Options:

The correct answer is:
0.55

Solution:
Wavelengths corresponding to minimum wavelength (Î»min ) or maximum energy will emit photoeiectrons having kinetic energy which is maximum. (Î»min ) belonging to Balmer series and Lying in the given range (450 nm to 750 nm) corresponds to transition from (n = 4 to n = 2). Here,
Eâ¡ 4 = - 1 3 Â· 6 4 2 = - 0 Â· 8 5 eV
and Eâ¡ 2 = - 1 3 Â· 6 2 2 = - 3 Â· 4 eV
â´ Î Eâ¡ = Eâ¡ 4 - Eâ¡ 2 = 2 Â· 5 5 eV
Kâ¡ max = Energy of photon - work function = 2 Â· 5 5 - 2 Â· 0 = 0 Â· 5 5 eV

Topic: Physics, Atomic Physics, JEE Main
Wavenumber for a radiation having \(5800 Ã…\) wavelength is \(x \times 10 \mathrm{~cm}^{-1}\). The value of \(x\) is ________ (Integer answer)

Options:

The correct answer is:
1724

Solution:
$\bar{v}($ wave no. $)=\frac{1}{\lambda}=\frac{1}{5800 \times 10^{-8} \mathrm{~cm}}=17241$
OR
$1724 \times 10 \mathrm{~cm}^{-1} \Rightarrow \mathrm{x}=1724$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Wax is coated on the inner wall of a capillary tube and the tube is then dipped in water. Then, compared to the unwaxed capillary, the angle of contact $\theta$ and the height $h$ upto which water rises change. These changes are :

Options:
A
$\theta$ increases and $h$ also increases

B
$\theta$ decreases and $h$ also decreases

C
$\theta$ increases and $h$ decreases

D
$\theta$ decreases and $h$ increases

The correct answer is:
$\theta$ increases and $h$ decreases

Solution:
Angle of $\operatorname{contact} \theta$
$$
\cos \theta=\frac{\mathrm{T}_{\mathrm{SA}}-\mathrm{T}_{\mathrm{SL}}}{\mathrm{T}_{\mathrm{LA}}}
$$
when water is on a waxy or oily surface
$\mathrm{T}_{\mathrm{SA}} < \mathrm{T}_{\mathrm{SL}} \cos \theta$ is negative i.e.,
$$
90^{\circ} < \theta < 180^{\circ}
$$
i.e., angle of contact $\theta$ increases
And for $\theta>90^{\circ}$ liquid level in capillary
tube fall. i.e., h decreases

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2013 (23 Apr Online)
Waxes are esters of

Options:
A glycerol
B long chain alcohols
C glycerol and fatty acid
D long chain alcohols and long chain fatty acids

The correct answer is:
long chain alcohols and long chain fatty acids

Solution:
Wax is an ester of fatty acid and an alcohol.
They are formed by combining one fatty acid with one fatty alcohol molecule.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
We do not transmit low frequency signal to long distances because
(a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength.
(b) Effective power radiated by a long wavelength baseband signal would be high.
(c) We want to avoid mixing up signals transmitted by different transmitter simultaneously.
(d) Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well.

Therefore, the most suitable option will be :

Options:
A All statements are true
B (a), (b) and (c) are true only
C (a), (c) and (d) are true only
D (b), (c) and (d) are true only

The correct answer is:
(a), (c) and (d) are true only

Solution:
For longer wavelength, size of antenna would increase. The size of the antennas should be atleast λ4 for high efficiency of signal radiation.

The effective power radiated by an antenna is inversely proportional to the square of the wavelength, so, for the same antenna length, the power radiated increases with decreasing wavelength or increasing frequency.Hence, the effective power radiated by a long-wavelength baseband signal would be small.

Also, mixing of signals needs to be avoided, as all the signals get mixed together and a receiver can not separate them from each other. Also, we can use modulation to send low frequency signal by superimposing them with high frequency signals. Electrical audio signals are superimposed on a high frequency wave called the carrier wave. The carrier wave is generated from radio-frequency oscillators and is undamped in nature.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (27 Jun Shift 2)
We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M and 0.001 M, respectively. The value of van t’ Haft factor i for these solutions will be in the order.

Options:
A iA<iB<iC
B iA<iC<iB
C

iA=iB=iC

D iA>iB>iC

The correct answer is:
iA<iB<iC

Solution:
It is because the extent or degree of dissociation increases with increase in dilution of a solution. 0.001 M NaCl solution 'C' is most diluted as compared to the other two NaCl solutions. The vant Hoff factor (i) depends on extent of dilution.This confirms answer at (i) iA<iB<iC

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (01 Feb Shift 1)
We have three beakers $A, B$ and $C$ containing glycerine, water and kerosene respectively. They are stirred vigorously and placed on a table. The liquid which comes to rest at the earliest is

Options:
A Glycerine
B Water
C Kerosene
D All of them at the same time

The correct answer is:
Glycerine

Solution:
As the coefficient of viscosity of glycerin is greater than that of water and kerosene, therefore the relative motion between adjacent layers of glycerin is more strongly opposed compared to water and kerosene. Hence, glycerin comes to rest earlier than water and kerosene.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
We wish to make a microscope with the help of two positive lenses both with a focal length of 20 $\mathrm{mm}$ each and the object is positioned $25 \mathrm{~mm}$ from the objective lens. How far apart the lenses should be so that the final image is formed at infinity?

Options:
A
$20 \mathrm{~mm}$

B
$100 \mathrm{~mm}$

C
$120 \mathrm{~mm}$

D
$80 \mathrm{~mm}$

The correct answer is:
$120 \mathrm{~mm}$

Solution:
To obtain final image at infinity, object which is the image formed by objective should be at focal distance of eye-piece. By lens formula (for objective)
$$
\begin{aligned}
& \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\
\text { or, } \quad & \frac{1}{v_0}-\frac{1}{-25}=\frac{1}{20} \\
\Rightarrow \quad & \frac{1}{v_0}=\frac{1}{20}-\frac{1}{25}=\frac{5-4}{100}=\frac{1}{100} \mathrm{~mm} \\
\therefore \quad & v_0=100 \mathrm{~mm}
\end{aligned}
$$
Therefore the distance between the lenses
$$
=v_0+f_e=100 \mathrm{~mm}+20 \mathrm{~mm}=120 \mathrm{~mm}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2012 (12 May Online)
We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If αiron=1.2× 10-5/K and αiron=1.8× 10-5/K , what should we take as length of each strip?

Options:
A 10cm
B 20cm
C 30cm
D 40cm

The correct answer is:
20cm

Solution:
Correct Option is : (B)
20cm

Topic: Chemistry, Thermal Properties of Matter, NEET
We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say $10 \mathrm{~cm}$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their length $B$ remain constant. If $\alpha_{\text {iron }}=1.2 \times 10^{-5} / \mathrm{K}$ and $\alpha_{\text {brass }}=1.8$ $\times 10^{-5} / \mathrm{K}$, what should we take as length of each strip?

Options:

None

Solution:
According to question
$l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm}=$ constant at all temperatures
Let $l_0$ be length at temperature $0^{\circ} \mathrm{C}$ and $l$ be the length after change in temperature of $\Delta t$.
Now, we can write,
As per question,
$$
\begin{aligned}
&l_{\text {iron }}\left(1+\alpha_{\text {iron }} \Delta t\right)-l_{\text {brass }}\left(1+\alpha_{\text {brass }} \Delta t\right)=10 \mathrm{~cm} \\
&l_{\text {iron }}-l_{\text {brass }}-\alpha_{\text {iron }} l_{\text {iron }} \Delta t-\alpha_{\text {bras }} l_{\text {brass }} \Delta t=0 \\
&\left(\because l_{\text {iron }}-l_{\text {brass }}=10\right) \\
&l_{\text {iron }} \alpha_{\text {iron }}-\alpha_{\text {brass }} l_{\text {brass }}=0 \\
&l_{\text {iron }} \alpha_{\text {iron }}=l_{\text {braass }} \alpha_{\text {brass }} \\
&\frac{l_{\text {iron }}}{l_{\text {brass }}}=\frac{1.8}{1.2}=\frac{3}{2} \Rightarrow \frac{1}{2} l_{\text {brass }}=10 \mathrm{~cm} \\
&l_{\text {brass }}=20 \mathrm{~cm} \text { and } l_{\text {iron }}=30 \mathrm{~cm} \text {. } \\
&
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter