datavorous/entrance-exam-dataset huggingface dataset:
The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is

Options:
A $4$
B $8$
C $6$
D $10$

The correct answer is:
$6$

Solution:
Relative lowering of vapour pressure,
$\begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}} & =x_A \\ & =\frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A}+\frac{w_B}{m_B}}\end{aligned}$
(where, $w_A$ and $m_A$ are the mass and molar mass of solute and $w_B$ and $m_B$ are the mass and molar mass of water.)
$\therefore \quad \begin{aligned} 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+\frac{90}{18}} \\ 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+5} \\ \frac{1}{0.02} & =\frac{\frac{x}{60}+5}{\frac{x}{60}} \\ 50 & =1+\frac{5 \times 60}{x} \\ 49 & =\frac{300}{x} \\ x & =\frac{300}{49}=6 \mathrm{~g}\end{aligned}$

Topic: Chemistry, Solutions, JEE Main
The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth Re=6400 km)

Options:
A 9.8 N
B 4.9 N
C 19.6 N
D 8 N

The correct answer is:
8 N

Solution:
As height of the object is considerable compared to the radius of the earth, therefore approximation formula can not be used.

So weight at given height,

F=mg'=GMmR+R22=49GMmR2

⇒mg'=49(mg)

=49×18

=8 N

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The weight of a body on the earth is 400 N. Then weight of the body when taken to a depth half of the radius of the earth will be:

Options:
A 200 N
B Zero
C 100 N
D 300 N

The correct answer is:
200 N

Solution:
The acceleration due to gravity changes with depth from the surface, which is given by

g'=g1-dR ...(i)

The given data is

W=400 Nd=R2

Multiplying equation (i) with the mass we get the weight.

mg'=mg1-R2R=mg12⇒W'=400 N2=200 N

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The weight of a body on the surface of the earth is 63 N. What is the gravitational force on it due to the earth at a height equal to half the radius of earth?

Options:
A 28 N

B 23 N

C 12 N

D 67 N

The correct answer is:
28 N

Solution:
Given h = R⁡ e⁡ 2
Acceleration due to gravity at altitude h is given by
g ′ = g 1 + h⁡ R⁡ e⁡ 2
= g 1 + R⁡ e⁡ / 2 R⁡ e⁡ 2
= g 1 + 1 2 2
= g 3 / 2 2 =g' = 4 9 g ...(i)
Weight of the body at earth's surface
w=mg=63 N ...(ii)
Weight of the body at altitude h=Re2
w⁡ ′ = m g⁡ ′ = 4 9 mg ...(iii)
Using Eq. (ii), we get
w⁡ ′ = 4 9 × 6 3
w'=28 N

Topic: Physics, Gravitation, JEE Main
The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

Options:
A 64 N
B 25 N
C 50 N
D 100 N

The correct answer is:
64 N

Solution:
The acceleration due to gravity changes as one goes to a height from the surface of the earth. Let the new acceleration due to gravity be g'. The formula for acceleration due to gravity at a height is given by g'=gR2R+h2

The weight of the body on surface is W=mg=100 N

The weight at the given height is W=mg'.

⇒W'=100×R2R+R42⇒W'=100×R225R216=64 N

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The weight of an object in the coal mine, sea level, at the top of the mountain are $W_1, W_2$ and $W_3$ respectively, then

Options:
A $W_1 \lt W_2 \gt W_3$
B $W_1=W_2=W_3$
C $W_1 \lt W_2 \lt W_3$
D $W_1 \gt W_2 \gt W_3$

The correct answer is:
$W_1 \lt W_2 \gt W_3$

Solution:
$W_1 \lt W_2 \gt W_3$ Because value of $g$ decreases when we move either in coal mine or at the top of mountain.

Topic: Physics, Gravitation, JEE Main
The weight of glucose dissolved in 100 gm of water produce same lowering of vapor pressure as 1 gm urea dissolved in 50gm of water at same temperature is

Options:

The correct answer is:
6

Solution:
since same lowering in vapour pressure.
X solute of urea = X solute of glucose
1 60 1 60 + 50 18 = w 180 w 180 + 100 18 â 1 60 50 18 = w 180 100 18 (For dilute solution)
ââw=6âofâglucose

Topic: Chemistry, Solutions, JEE Main
The weight $W$ of a certain stock of fish is given by $W=n w$, where $n$ is the size of stock and $w$ is the average weight of a fish. If $n$ and $w$ change with time $t$ as $n=2 t^2+3$ and $w=t^2-t+2$, then the rate of change of $W$ with respect to $t$ at $t=1$ is

Options:
A
1

B
8

C
13

D
5

The correct answer is:
13

Solution:
Let $\mathrm{W}=n w$
$$
\Rightarrow \frac{d W}{d t}=n \frac{d w}{d t}+w \cdot \frac{d n}{d t}
$$
Given : $w=t^2-t+2$ and $n=2 t^2+3$ $\Rightarrow \frac{d w}{d t}=2 t-1$ and $\frac{d n}{d t}=4 t$
$\therefore$ Equation (1)
$$
\Rightarrow \frac{d w}{d t}=\left(2 t^2+3\right)(2 t-1)+\left(t^2-t+2\right)(4 t)
$$
Thus, $\left.\frac{d W}{d t}\right|_{t=1}=(2+3)(2-1)+(2)(4)$
$$
=5(1)+8=13
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (19 May Online)
The wettability of a surface by a liquid depends primarily on

Options:
A viscosity
B surface tension
C density
D angle of contact between the surface and the liquid

The correct answer is:
angle of contact between the surface and the liquid

Solution:
The value of angle of contact determines whether a liquid will spread on the surface.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as R1 has the colour code (orange, red, brown). The resistors R2 and R4 are 80 Ω and 40 Ω, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be

Options:
A brown, blue, black.
B brown, blue, brown.
C grey, black, brown.
D red, green, brown.

The correct answer is:
brown, blue, brown.

Solution:
Colour band A and B indicate the first two significant figures of resistance in 'ohm', while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicated value or in other word, it represents the percentage accuracy of the indicated value.

For R1, the given colour of the rings of a carbon resistor are orange, red, brown then, the value of resistance is

R1=32×101

Condition for balance in the Wheat stone's bridge is

R1R3=R2R4
So, 320R3=8040
∴R3=160 Ω, so colour coding of resistor R3 is

Brown→1

Blue→6

Brown→1
16×101⇒160 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The width of fringe is 2 mm on the screen in a double slit experiment for the light of wavelength of 400 nm. The width of the fringe for the light of wavelength 600 nm will be:

Options:
A 4 mm
B 2 mm
C 1.33 mm
D 3 mm

The correct answer is:
3 mm

Solution:
For a light of wavelength λ, the fringe width in double slit experiment is given by

β=λDd ...(1)

For a light of wavelength λ', the fringe width is given by

β'=λ'Dd ...2

Divide equation (2) by equation (1) and simplify to obtain the fringe width for the later case.

β'β=λ'DdλDd= λ'λ⇒β'= λ'λβ ...3

Substitute the values of the known parameters into equation (3) to calculate the required fringe width.

β'=600 nm400 nm×2 mm= 3 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

Options:
A $1: 1$
B $4: 1$
C $9: 1$
D $16: 1$

The correct answer is:
$9: 1$

Solution:
Since, Intensity $\propto$ width of slit $(\omega)$
$\begin{aligned}
& \text { so, } I_1=I, I_2=4 \mathrm{I} \\
& I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I \\
& I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I} \\
& \frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}
\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x: 4 where x is _______.

Options:

The correct answer is:
1

Solution:
Given amplitude $\propto$ slit width
Also intensity $\propto(\text { Amplitude })^2 \propto(\text { Slit width })^2$
$\frac{I_1}{I_2}=\left(\frac{3}{1}\right)^2=9 \Rightarrow I_1=9 I_2$
$\frac{I_{\min }}{I_{\max }}=\left(\frac{\sqrt{I_1}-\sqrt{I_2}}{\sqrt{I_1}+\sqrt{I_2}}\right)^2=\left(\frac{3-1}{3+1}\right)^2=\frac{1}{4}=\frac{x}{4}$
$\Rightarrow x=1$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The Wien's displacement law expresses the relation between:

Options:
A wavelength corresponding to maximum energy and temperature
B relation energy and wavelength
C temperature and wavelength
D colour of light and temperature

The correct answer is:
wavelength corresponding to maximum energy and temperature

Solution:
Using Wein's displacement law : the wavelength corresponding to maximum energy is inversely proportional to the abolute temperature of the body. i.e.
$\lambda_m \propto \frac{d}{T} \Rightarrow \lambda_m T=\text { constant }$

Topic: Physics, Thermal Properties of Matter, JEE Main
The work done by a gas molecule in an isolated system is given by, W=αβ2e-x2αk T, where x is the displacement, k is the Boltzmann constant and T is the temperature. α and β are constants. Then the dimensions of β will be:

Options:
A M2 L T2
B ML2 T-2
C MLT-2
D M0 L T0

The correct answer is:
MLT-2

Solution:
x2αkT→ dimensionless

⇒α=x2kT=L2ML2 T-2=M-1 T2

Now W=αβ2

β=ML2 T-2M-1 T2=M1 L1 T-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The work done by the force Fâ=6i^+2j^ N in displacing an object from r1â=3i^+8j^ m to r2â=5i^â4j^ m, is

Options:
A 12 J
B -36 J
C 36 J
D -12 J

The correct answer is:
-12 J

Solution:
Fâ=6i^+2j^

râ=râ2-râ1=5i^-4j^-3i^-8j^

=2i^-12j^
W=FâÂ·Sâ=(6i^+2j^)Â·(2i^-12j^)
=-12J

Topic: Physics, Work Power Energy, JEE Main
The work done during the expansion of a gas from a volume of $4 \mathrm{dm}^3$ te $6 \mathrm{dm}^3$ against a constant external pressure of $3 \mathrm{~atm}$ is $(1 \mathrm{Latm}=101.32 \mathrm{~J})$

Options:
A $+304 \mathrm{~J}$
B $-304 J$
C $-6 J$
D $-608 \mathrm{~J}$

The correct answer is:
$-608 \mathrm{~J}$

Solution:
$W=-p \Delta V ; W=-3 \times(6-4)$
$W=-6 \times 101.32(\therefore 1 \mathrm{Latm}=101.32 \mathrm{~J})$
$W=-608 \mathrm{~J}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The work done in an open vessel at 300 K, when 112 g iron reacts with dil. HCl is ( R = 2 cal mol-1 K-1 )

Options:
A 1200 cal
B 600 cal
C 300 cal
D 200 cal

The correct answer is:
1200 cal

Solution:
Fe + 2 HCl â FeCl 2 + H 2 1 mol 1 mol 1 1 2 5 6 = 2 mol 2 mol
Now, V 2 = V H 2 and V 1 = 0 (For solid + aqueous state)
Work done = - p ext Î V
=âÎ(PV)=â n H 2 RT
= - 2 Ã 2 Ã 3 0 0
= - 1200 cals (Work of expansion)

Topic: Chemistry, Thermodynamics (C), JEE Main
The work done in blowing a soap bubble of radius 'R' is ' $\mathrm{W}$ '. The work done in blowing a bubble of radius ' $2 \mathrm{R}^{\prime}$ of the same soap solution is

Options:
A $\frac{\mathrm{w}}{4}$
B $\frac{\mathrm{w}}{2}$
C 4w
D 2w

The correct answer is:
4w

Solution:
(B)
$\mathrm{W}_{1}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ where $\mathrm{T}$ is surface tension.
If radius is doubled and surface tension remains same then $\mathrm{W}_{2}=8 \pi(2 \mathrm{R})^{2} \mathrm{~T}=8 \pi \mathrm{R}^{2} \mathrm{~T}=4 \mathrm{~W}_{1}$
However, the solution is heated, the value of surface tension decreases. Hence, $\mathrm{W}_{2} < 4 \mathrm{~W}_{1}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2020 (16 Oct Shift 2)
The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$ ) into 64 equal drops is

Options:
A $4 \pi R^2 T$
B $\frac{\pi R^2 T}{64}$
C $\frac{12 \pi T}{R^2}$
D $12 \pi R^2 T$

The correct answer is:
$12 \pi R^2 T$

Solution:
Radius of big drop $=R$
Number of small drops, $n=64$
Let radius of small drop $=r$
In the process of breaking the drop, the volume of liquid drops remains constant.
$$
V_i=V_f
$$
$$
\begin{aligned}
\Rightarrow \quad \frac{4}{3} \pi R^3 & =n\left(\frac{4}{3} \pi r^3\right) \\
R^3 & =64 r^3 \Rightarrow R=4 r \\
r & =\frac{R}{4}
\end{aligned}
$$
Now, work done in breaking a big drop in 64 small drops.
$$
\begin{aligned}
W & =\text { final energy }- \text { initial energy } \\
W & =T\left(n 4 \pi r^2\right)-T\left(4 \pi R^2\right)=4 \pi T\left[64\left(\frac{R}{4}\right)^2-R^2\right] \\
& =4 \pi T\left[4 R^2-R^2\right]=12 \pi R^2 T
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an initial temperature of 300 K is: [R=2cal/mol−K]

Options:

The correct answer is:
720

Solution:
To calculate the work done, we neet to know the final temperature.

T1=300 K,â P1=1âatm,âân=2âmole,ââP2=2ââatm,ââV1=nRT1P1

w=ÎY=nCvT2âT1=âP2V2âV1

For a monatomic gas, Cv=32R

â ââ32nRT2âT1=âP2V2âV1

32nRT2âT1=âP2nRT2P2ânRT1P1

32T2âT1=âT2+P2P1âT2=23P2P1+32T1

T2=0.42+1.5300=420 K

Work done, W=nCvÎÎ¤=2Ã32RÃ420â300

âââW=2Ã32Ã2Ã120=720âcal

Topic: Chemistry, Thermodynamics (C), JEE Main
The work done in carrying a charge $\mathrm{Q}$ once around a circle of radius $\mathrm{r}$ about a charge $\mathrm{q}$ at the centre is

Options:
A $\frac{q Q}{4 \pi \varepsilon_0 r}$
B $\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{\pi r}$
C $\frac{q Q}{4 \pi \varepsilon_0}\left(\frac{1}{2 \pi r}\right)$
D $0$

The correct answer is:
$0$

Solution:
Work done by conservative force in a round trip is zero

Topic: Physics, Electrostatics, JEE Main
The work done in increasing the size of a soap film from 10 cmÃ6 cm to 10 cmÃ11 cm is 3Ã10-1 J. The surface tension of the film is

Options:
A 1.5Ã10â2 N m-1
B 3.0Ã10â2 N m-1
C 6.0Ã10â2 N m-1
D 11.0Ã10â2 N m-1

The correct answer is:
3.0Ã10â2 N m-1

Solution:
W=TÃÎA
â´ T= W ÎA T= 3Ã 10 â4 2Ã(110â60)Ã 10 â4 (Soap film has two surfaces) = 3Ã10â2 N m-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The work done in moving a point charge of \( 10 \mu \mathrm{C} \) through a distance of \( 3 \mathrm{~cm} \) along the equatorial axis of an electric dipole
is

Options:
A zero
B \( 30 \times 10^{-6} \mathrm{~J} \)
C \( 20 \times 10^{-6} \mathrm{~J} \)
D \( 5 \times 10^{-6} \mathrm{~J} \)

The correct answer is:
zero

Solution:
The electric potential difference along the equatorial axis of an electric dipole is zero.

The change of potential energy stored in charge 10 μC is equal to the work done on 10 μC to bring it from A to B.

∴ W=q∆V=0

Topic: Physics, Electrostatics, JEE Main
The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity 100 microfarad is

Options:
A $3.1 \times 10^{-26}$ joule
B $4 \times 10^{-10}$ joule
C $32 \times 10^{-32}$ joule
D $16 \times 10^{-32}$ joule

The correct answer is:
$32 \times 10^{-32}$ joule

Solution:
Work done $=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\left(8 \times 10^{-18}\right)^2}{2 \times 100 \times 10^{-6}}$
$=32 \times 10^{-32} \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main
The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity 100 micro-farad is

Options:
A
$16 \times 10^{-32}$ joule

B
$3.1 \times 10^{-26}$ joule

C
$4 \times 10^{-10}$ joule

D
$32 \times 10^{-32}$ joule

The correct answer is:
$32 \times 10^{-32}$ joule

Solution:
Work done $=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{c}}=\frac{\left(8 \times 10^{-18}\right)^2}{2 \times 100 \times 10^{-8}}=32 \times 10^{-32} \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2003
The work done in placing the dielectric slab inside one of the capacitors as shown in the diagram.

Options:
A CV22K – 1K+1
B CV24K – 1K+1
C CV24K+1K – 1
D CV22K+1K – 1

The correct answer is:
CV24K – 1K+1

Solution:
Ui=12 C/2V2=CV24

Uf=12C × KCC+KCV2=KCV22 1+K
W=ΔU=Uf – Ui
=KCV22 1+K-CV24=CV22K1+K – 12
=CV222K – 1 – K2 1+K
=CV24K – 1K+1

Topic: Physics, Electrostatics, JEE Main
The work done on a body by a force is minimum when the force and the displacement of the body are in
direction.

Options:
A Same
B Opposite
C Different
D Some

The correct answer is:
Opposite

Solution:
We know that the work done by the force is defined to be the product of a component of the force in the direction of the displacement and the magnitude of the displacement.

Thus the work done is a scalar product of force and displacement.

⇒W=F·d=(Fcos°)×d=F×d×cosθ.

When the force and displacement are in opposite direction, θ=180°.

So, W=F×d×cos180°=F×d×(-1)=-(F×d).

This is the minimum value of work done which is equal to the negative of product of force and displacement.

Thus, the work done on a body by a force is minimum when the force and the displacement of the body are in opposite direction.

Topic: Physics, Work Power Energy, JEE Main
The work done on the gas in taking it from $D$ to $A$ is

Options:
A
$-414 \mathrm{R}$

B
$+414 \mathrm{R}$

C
$-690 \mathrm{R}$

D
$+690 R$

The correct answer is:
$-414 \mathrm{R}$

Solution:
At constant temperature (isothermal process)
$$
\begin{aligned}
W_{D A}=n R T \ln \left(\frac{P_1}{P_2}\right)= & 2.303 \times 2 \mathrm{R} \times 300 \log \left(\frac{10^5}{2 \times 10^5}\right) \\
& =2.303 \times 600 \mathrm{R} \log \left(\frac{1}{2}\right) \\
& =0.693 \times 600 \mathrm{R}=-414 \mathrm{R} .
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
The work done on the gas in taking it from $D$ to $A$ is

Options:
A
$-414 \mathrm{R}$

B
$+414 \mathrm{R}$

C
$-690 \mathrm{R}$

D
$+690 R$

The correct answer is:
$-414 \mathrm{R}$

Solution:
At constant temperature (isothermal process)
$$
\begin{aligned}
W_{D A}=n R T \ln \left(\frac{P_1}{P_2}\right)= & 2.303 \times 2 \mathrm{R} \times 300 \log \left(\frac{10^5}{2 \times 10^5}\right) \\
& =2.303 \times 600 \mathrm{R} \log \left(\frac{1}{2}\right) \\
& =0.693 \times 600 \mathrm{R}=-414 \mathrm{R} .
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2009
The work function of a certain metal is $3.31 \times 10^{-19}$ J. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 Å$ is (Given, $\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J}$ $\left.\mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, e=1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A $2.48 \mathrm{eV}$
B $0.41 \mathrm{eV}$
C $2.07 \mathrm{eV}$
D $0.82 \mathrm{eV}$

The correct answer is:
$0.41 \mathrm{eV}$

Solution:
Work function $W_{0}=3.31 \times 10^{-19} \mathrm{~J}$ Wavelength of incident radiation $\lambda=5000 \times 10^{-10} \mathrm{~m}$
According to Einstein's photoelectric equation $E=W_{0}+K E$
$\begin{aligned} \Rightarrow \frac{h c}{\lambda} &=3.31 \times 10^{-19}+\mathrm{KE} \\ \mathrm{KE} &=-3.31 \times 10^{-19}+\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}} \\ &=-3.31 \times 10^{-19}+\frac{6.62 \times 3}{5} \times 10^{-19} \\ &=(-3.31 \times 1.324 \times 3) \times 10^{-19} \end{aligned}$
$\begin{aligned}
&=(3.972-3.31) \times 10^{-19} \\
&=0.662 \times 10^{-19} \mathrm{~J} \\
\text { or } E &=\frac{0.662 \times 10^{-19}}{1.6 \times 10^{-19}}=0.41 \mathrm{eV}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
The work function of a metal is in the range of 2 âeVââtoââ5â eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Consider, Planck's constant =4Ã 10 â15 eV-s , velocity of light =3Ã108âm s-1)

Options:
A 510 nm
B 650 nm
C 400 nm
D 570 nm

The correct answer is:
650 nm

Solution:
Given that, 2 eV â¤Ïâ¤5 eV
Wavelength corresponding to minimum and maximum values of work function are
Î»max=hcE=4Ã10-15Ã3Ã1082
= 6Ã10-7m=600 nm
And Î»min=hcE=4Ã10-15Ã3Ã1085
= 2.4Ã10-7m=240 nm
Clearly wavelength range would be 240â¤Î»â¤600
Thus, wavelength 650 nm is not suitable for photoelectric effect.

Topic: Physics, Dual Nature of Matter, JEE Main
The work function of a metal surface is $6.2 \mathrm{eV}$. Find the threshold wavelength of the metal ( $\mathrm{hc}=1240 \mathrm{eV} . \mathrm{nm})$.

Options:
A $400 \mathrm{~nm}$
B $360 \mathrm{~nm}$
C $200 \mathrm{~nm}$
D $100 \mathrm{~nm}$

The correct answer is:
$200 \mathrm{~nm}$

Solution:
Given that: $h c=1240 \mathrm{eV} \mathrm{nm}$
Work function $(\boldsymbol{\phi})=6.2 \mathrm{eV}$
Work function $(\phi)=h c / \lambda_{\text {th }}$
Threshold wavelength $\left(\lambda_{\text {th }}\right)=\mathrm{hc} / \phi=1240 / 6.2=200 \mathrm{~nm}$

Topic: Physics, Dual Nature of Matter, JEE Main
The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

Options:
A 215 nm
B 414 nm
C 400 nm
D 200 nm

The correct answer is:
414 nm

Solution:
For longest wavelength, the emitted electron will have nearly zero kinetic energy.

Therefore,

λ=hcWe

⇒λ=1240 nm-eV3 eV

⇒λ=413.33 nm

⇒λmax≈414 nm

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The work function of a substance is $4.0 \mathrm{eV}$. Then longest wavelength of light that can cause photoelectron emission from this substance approximately

Options:
A
$540 \mathrm{~nm}$

B
$400 \mathrm{~nm}$

C
$310 \mathrm{~nm}$

D
$220 \mathrm{~nm}$

The correct answer is:
$310 \mathrm{~nm}$

Solution:
$\frac{\mathrm{hc}}{\lambda}=\mathrm{W}$
$\lambda_{\text {longest }}=\frac{\mathrm{hc}}{\mathrm{W}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 1.6 \times 10^{-19}}$
$\Rightarrow \lambda_{\text {longest }} \approx 310 \mathrm{~nm}$.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2004
The work function of a surface of a photosensitive material is $6.2 \mathrm{eV}$. The wavelength of the incident radiation for which the stopping potential is $5 \mathrm{~V}$ lies in the

Options:
A X-ray region
B Ultraviolet region
C Visible region
D Infrared region

The correct answer is:
Ultraviolet region

Solution:
$$
\begin{aligned}
& \mathrm{eV}_0=E-\phi \\
& E=\mathrm{eV}_0+\phi \\
& =5 \mathrm{eV}+6.2 \mathrm{eV} \\
& =11.2 \mathrm{eV} \\
& \therefore \quad \lambda=\left(\frac{12400}{11.2}\right) Å=1000 Å \\
& \Rightarrow \text { hence lies in ultraviolet region. } \\
&
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2008 (Mains)
The work function of sodium metal is 4.41×10-19J. If photons of wavelength 300 nm are incident on the metal, the kinetics energy of the ejected electrons will be h=6.63×10-34J s; c=3×108 m s-1__________×10-21J

Options:

The correct answer is:
222

Solution:
E=E0+(kE)max
hCλ=4.41×10-19+kE
6.63×10-34×3×108300×10-9=4.41×10-19+kE
So, KEmax=6.63×10-19-4.41×10-19

=2.22×10-19

=222×10-21J

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The work function ϕ of some metal is listed below. The number of metals which will show photoelectric effect when light of 300nm wavelength fall on the metal is

Options:

The correct answer is:
4

Solution:
energy of light =1240300=4.13eV , so the metal having work function lower than 4.13eV will emit photoelectron, so the following 4 metals will show photoelectric effect

Topic: Chemistry, Structure of Atom, JEE Main
The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

Options:
A 1.24
B 2
C 1
D 1.5

The correct answer is:
1

Solution:
The data given is

ϕAl=4.1 eV

ϕAu=5.1 eV

The energy is given by

hν=ϕ+KE⇒hν=ϕ+eV0

where V0 is the stopping potential and K=eV0

Thus, the above equation can be written as

eV0=hν-ϕ

⇒V0=hνe-ϕe

The equation is of the form of y=mx+c. The slope is

m=he.

Since the Planck constant h and charge of an electron is a constant e, the slope does not depend on the work function. Hence the ratio of the slope of the plot of stopping potential and frequency for both the given metals is 1.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (06 Apr Shift 2)