datavorous/entrance-exam-dataset huggingface dataset:
A given object takes $n$ times more time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down a perfectly smooth $45^{\circ}$ incline. The coefficient of kinetic friction between the object and the incline is :

Options:
A
$\sqrt{1-\frac{1}{n^2}}$

B
$1-\frac{1}{n^2}$

C
$\frac{1}{2-n^2}$

D
$\sqrt{\frac{1}{1-n^2}}$

The correct answer is:
$1-\frac{1}{n^2}$

Solution:
The coefficients of kinetic friction between the object and the incline
$\mu=\tan \theta\left(1-\frac{1}{\mathrm{n}^2}\right)$
$\Rightarrow \mu=1-\frac{1}{\mathrm{n}^2} \quad\left(\because \theta=45^{\circ}\right)$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A given object takes n times more time to slide down a 45o rough inclined plane as it takes to slide down a perfectly smooth 45o incline. The coefficient of kinetic friction between the object and the incline is:

Options:
A 1-1n2
B 1-1n2
C 12-n2
D 11-n2

The correct answer is:
1-1n2

Solution:
Time taken to slide along a smooth surface is t1.
S=12gsin⁡45ot12
t1=22sg
Time taken to slide along a rough surface is t2.
S=12gsin⁡45o-μgsin⁡45o t22
t2=22sg1-μ
t2=nt1
22sg1-μ=n2×22sg⇒1-μ=1n2⇒μ=1-1n2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (15 Apr)
A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is

Options:
A 23P
B P
C 32P
D 2P

The correct answer is:
P

Solution:
In isothermal process
PV = constant
â´ PdV + VdP = 0 or dP dV = - P V
â´ Bulk modulus, B = - dP dV / V = - dP dV V
â´ B = - - P V V = P
â´ B = P
Note: Adiabatic bulk modulus is given by B = Î³P

Topic: Physics, Thermodynamics, JEE Main
A given ray of light suffers minimum deviation in an equilateral prism $P$. Additional prisms $Q$ and $R$ of identical shape and material are now added to $P$ as shown in the figure. The ray will suffer

Options:
A Greater deviation
B Same deviation
C No deviation
D Total internal reflection

The correct answer is:
Same deviation

Solution:
As the prisms $Q$ and $R$ are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.

Topic: Physics, Ray Optics, JEE Main
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of same material as that of $\mathrm{P}$ are now combined as shown in figure. The ray will now suffer

Options:
A greater deviation
B no deviation
C same deviation as before
D total internal reflection

The correct answer is:
same deviation as before

Solution:
When the ray suffers minimum deviation, it becomes parallel to the base of prism P. As prisms $\mathrm{Q}$ and $\mathrm{R}$ are of same material and have identical shape, therefore, the ray continues to be parallel to base of $Q$ and $R$. Hence final deviation of the ray remains the same as before.

Topic: Physics, Ray Optics, VITEEE, VITEEE 2021
A given sample of milk turns sour at room temperature $\left(27^{\circ} \mathrm{C}\right)$ in $5 \mathrm{~h}$. In a refrigerator at $-3^{\circ} \mathrm{C}$, it can be stored 10 times longer. The energy of activation for the souring of milk is

Options:
A $2.303 \times 5 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$
B $2.303 \times 3 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$
C $2.303 \times 2.7 \mathrm{R} \mathrm{kJ}^{-1} \mathrm{~mol}^{-1}$
D $2.303 \times 10 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$

The correct answer is:
$2.303 \times 2.7 \mathrm{R} \mathrm{kJ}^{-1} \mathrm{~mol}^{-1}$

Solution:
Energy of activation is given by
$$
\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]
$$
Here, $\quad \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{1}{10}$
$\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$\mathrm{T}_{2}=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$
$\therefore \quad \log \frac{1}{10}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{270-300}{270 \times 300}\right)$
$-1=-\frac{E_{a}}{2.303 R} \frac{30}{270 \times 300}$
$\mathrm{E}_{\mathrm{a}}=+2.303 \times 2700 \times \mathrm{R} \mathrm{J} \mathrm{mol}^{-1}$
$=+2.303 \times 2.7 \times \mathrm{R} \mathrm{kJ} \mathrm{mol}=1$

Topic: Chemistry, Chemical Kinetics, JEE Main
A given shaped glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity $\omega$ when

Options:
A water levels in both sections $A$ and $B$ go up
B water level in section $A$ goes up and that in $B$ comes down
C water level in section $A$ comes down and that in $B$ it goes up
D water levels remain same in both sections.

The correct answer is:
water levels in both sections $A$ and $B$ go up

Solution:
Due to rotation about the axis $O O^{\prime}$, a force is acting on the liquid column $A B$ outwards, pushing up the liquid column on the right. This centrifugal force acting on the small element $d x$ of area of cross-section $d x$, distant $x$ from the axis is
$$
d m \omega^2 x=A d x \rho \omega^2 x
$$
The total force due to column of length $L(=A B)$ is
$$
\int_0^L A \rho \omega^2 \cdot x d x=A \rho \omega^2 \frac{L^2}{2}
$$
Pressure at $B=$ atmospheric pressure +
$$
h_1 \rho g+\frac{A \rho \omega^2}{A} \frac{L^2}{2}
$$
Pressure at $B$ due to liquid column on the right hand side $=$ atmospheric pressure $+h_2 \rho g$
Net pressure due to the left hand side $=$ pressure at $B$ due to liquid column
$$
\begin{aligned}
& \quad \text { atm. pressure }+h_1 \rho g+\rho \omega^2 \frac{L^2}{2}=\text { atm. pressure }+h_2 \rho g \\
& \therefore \quad h_2-h_1=H_0=\frac{\omega^2 L^2}{2 g} .
\end{aligned}
$$
$$
\text { Here one has }
$$
$$
H_2=\frac{\omega^2 L_2^2}{2 g}, \quad H_1=\frac{\omega^2 L_1^2}{2 g}
$$
As $\omega$ is the same, and $L_2>L_1, H_2>H_1$. But both will go up.

Topic: Physics, Mechanical Properties of Fluids, AIIMS, AIIMS 2005
A given shaped glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity ${ }^\omega$ then

Options:
A Water levels in both sections $A$ and $B$ go up
B Water level in Section $A$ goes up and that in $B$ comes down
C Water level in Section $A$ comes down and that in $B$ it goes up
D Water levels remains same in both sections

The correct answer is:
Water levels in both sections $A$ and $B$ go up

Solution:
Water level in both A and B will go up. The pressure difference thus created will provide the necessary centripetal force for the water body to rotate around the vertical axis.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A glass bulb of the volume 400 cm3 is connected to another bulb of volume 200 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20 Â°C and 1.0 atm, respectively. The larger bulb is immersed in steam at 100 Â°C and the smaller in melting ice at 0 Â°C. Find the final common pressure.

Options:
A 1.13 atm

B 1.23 atm

C 1.43 atm

D 1.53 atm

The correct answer is:
1.13 atm

Solution:
Let n1 and n2 denote the number of moles of gas in the large and small bulbs, in the final configuration, respectively. Denoting the final temperatures by T1 and T2 and the final pressure by Pf, the ideal gas law implies that
p fâ¡ V 1 = n 1 RT 1 ...(i)
and p fâ¡ V 2 = n 2 RT 2 ...(ii)
where p0, V0 = V1 + V2 and T0 are the initial pressure, volume, and temperature, respectively.
p fâ¡ V 1 RT 1 + p fâ¡ V 2 RT 2 = p 0 V 0 RT 0
Solving for Pf, we obtain p fâ¡ = p 0 V 0 T 0 V 1 T 1 + V 2 T 2
Inserting the numerical values; p0 = 1.00 atm, T0 = 293.15 K(t0 = 20oC),
T1 = 373.15 K(t1 = 100oC), and T2 = 237.15 K(t2 = 0oC),
we find p fâ¡ = 1.00 600 293.15 4 0 0 373.15 + 2 0 0 273.15 = 1.13 atm

Topic: Physics, Thermodynamics, JEE Main
A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross-sections of different radii. When dipped in water vertically, the water rises in it to a height h, where the radius of its cross-section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, then the value of h will be (g is the acceleration due to gravity)

Options:
A 2Sbρgcos⁡(θ-α)
B 2Sbρgcos⁡θ+α
C 2Sbρgcos⁡θ-α2
D 2Sbρgcos⁡θ+α2

The correct answer is:
2Sbρgcos⁡θ+α2

Solution:
If R be the meniscus radius,
⇒Rcos⁡θ+α2=b
Excess pressure on concave side of meniscus =2SR
hρg=2SR=2Sbcos⁡θ+α2
⇒h=2Sbρgcos⁡θ+α2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross-sections of different radii. When dipped in water vertically, the water rises in it to a height h, where the radius of its cross-section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, then value of h will be (g is the acceleration due to gravity)

Options:
A 2Sbρgcos⁡(θ-α)
B 2Sbρgcos⁡θ+α
C 2Sbρgcos⁡θ-α2
D 2Sbρgcos⁡θ+α2

The correct answer is:
2Sbρgcos⁡θ+α2

Solution:
If R be the meniscus radius,
⇒Rcos⁡θ+α2=b
Excess pressure on concave side of meniscus =2SR
hρg=2SR=2Sbcos⁡θ+α2
⇒h=2Sbρgcos⁡θ+α2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A glass convex lens $\left(\mu_g=1.5\right)$ has a focal length of $8 \mathrm{~cm}$ when placed in air. What would be the focal length of the lens when it is immersed in water $\left(\mu_w=1.33\right)$

Options:
A 2 cm
B 4 cm
C 16 cm
D 32 cm

The correct answer is:
32 cm

Solution:
$\frac{f_l}{f_a}=\frac{\left({ }_a \mu_g-1\right)}{\left({ }_l \mu_g-1\right)} \Rightarrow \frac{f_w}{f_a}=\frac{(1.5-1)}{\left(\frac{1.5}{1.33}-1\right)} \Rightarrow f_w=32 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
A glass hemisphere Î¼=1.5 has a radius of curvature of 16 cm. A small object O is located on its axis halfway between the plane and spherical surface. The distance between two images, when viewed along the axis from the sides of the hemisphere, is

Options:
A 32 15  cm
B 48 15  cm
C 64 15 âcm
D 80 15  cm

The correct answer is:
64 15  cm

Solution:
When viewed from spherical side,
1v1+32×8=12×16
⇒ 1v1=132-316=-532
Image position is 325 cm inside the hemisphere from its periphery.
And when viewed from plane side then image is at 163 cm inside the plane surface.
∴ Δx=16-163-325=6415 cm

Topic: Physics, Experimental Physics, JEE Main
A glass prism is immeresed in water as shown in the figure. When a beam of light incident is incident normally on the face ab, it is internally reflected at the face ad. If the refractive index of glass is 1.5 and that of water is 43, then

Options:
A sinθ<(2/3)
B sinθ>(8/9)
C (2/3)<sinθ<(8/9)
D None of these

The correct answer is:
sinθ>(8/9)

Solution:
θ>θc

sinθ>sinθc

sinθ>89.

Topic: Physics, Ray Optics, JEE Main
A glass prism of angle A=60â gives a minimum angle of deviation Î¸ = 3 0 â with the maximum error of 1â when a beam of parallel light passed through the prism during an experiment. The permissible error in the measurement of refractive index Î¼ of the material of the prism is :

Options:
A 1 0 0 π 1 8 0 %
B 5 π 1 8 0 %
C 5 0 π 1 8 0 %
D 5 Ï 1 8 %

The correct answer is:
5 π 1 8 %

Solution:
μ=sinδ+A2sinA2, where δ is the minimum deviation
dμdδ=cosδ+A22sinA2

⇒ dμμ=cos δ+A22sin δ+A2dδ
⇒ dμμ=12cotδ+A2dδ

δ+A2=45°

⇒ dμμ×100=12cot45°π180×100

Percentage error = 5 π 1 8 %

Topic: Physics, Experimental Physics, JEE Main
A glass prism of refractive index $1.5$ is immersed in water (refractive index $\frac{4}{3}$ ) as shown in figure. A light beam incident normally on the face $A B$ is
$$
\text { totally reflected to reach the face } B C \text {, if }
$$

Options:
A
$\sin \theta>\frac{5}{9}$

B
$\sin \theta>\frac{2}{3}$

C
$\sin \theta>\frac{8}{9}$

D
$\sin \theta>\frac{1}{3}$

The correct answer is:
$\sin \theta>\frac{8}{9}$

Solution:
For total internal reflection on face $A C$ $\theta>$ critical angle $(C)$ and $\sin \theta \geq \sin C$
$$
\begin{aligned}
& \sin \theta \geq \frac{1}{w_{\mu_g}} \\
& \sin \theta \geq \frac{\mu_w}{\mu_g} \Rightarrow \sin \theta \geq \frac{\frac{4}{3}}{\frac{3}{2}} \\
& \therefore \quad \sin \theta \geq \frac{8}{9} .
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2012 (19 May Online)
A glass rod of rectangular cross-section is bent into the shape shown in the figure below. A parallel beam of light falls perpendicularly on the flat surface A. Determine the minimum value of the ratio Rd for which all light entering the glass through surface A will emerge from the glass through the surface B. The index of refraction of the glass is 1.5.

Options:

The correct answer is:
2

Solution:
Consider the representative rays shown in diagram (b). A ray entering the glass through surface A and passing along the inner side of the rod will be reflected by the outer side with the smallest angle α , at which the reflected ray is tangent to the inner side. We have to consider the conditions under which the ray will undergo total internal reflection before reaching B.
If α > θ c , the critical angle, at which total internal reflection occurs, all the incident beam will emerge through the surface B. Hence we require
sin α > 1 n
The geometry gives sin α = R R + d
Therefore R R + d ≥ 1 n ,
or R d min = 1 n - 1 = 1 1.5 - 1 = 2

Topic: Physics, Ray Optics, JEE Main
A glass slab of thickness $4 \mathrm{~cm}$ contains the same number of waves as in 'x' cm of water column when both are transversed by the same monochromatic light. If the
refractive indices of glass and water for that light are $\frac{5}{3}$ and $\frac{4}{3}$ respectively, the
value of $\mathrm{x}$ will be

Options:
A $\frac{9}{20} \mathrm{~cm}$
B $\frac{5}{4} \mathrm{~cm}$
C $5 \mathrm{~cm}$
D $\frac{20}{9} \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
Correct option is D)
Number of waves in glass slab = number of waves in water
$\mu_{1} \mathrm{~d}_{1}=\mu_{2} \mathrm{~d}_{2}$
$\frac{5}{3} \times 4=\frac{4}{3} \times x$
$\mathrm{x}=5 \mathrm{~cm}$

Topic: Physics, Wave Optics, JEE Main
A glass sphere having refractive index 3/2 is having a small irregularity at its centre. It is placed in a liquid of refractive index 43 such that the surface of the liquid is at a distance r above the sphere, where r=20 cm is radius of the sphere. If the irregularity is viewed from above then what is it's distance (in cm) from the centre where eye will observe the irregularity?

Options:

The correct answer is:
13

Solution:
Consideration refraction at glass-water interface.
Î¼2v-Î¼1u=Î¼2-Î¼1R
â 43v-3-2r=4/3-3/2r
â 43v+32r=-16r
â´ v=r5
Now refraction at water air surface
u=-r+45r=-9r5
Î¼2v-Î¼1u=Î¼2-Î¼1â
â1v+4Ã53Ã9r=1-4/3â
1v=-2027r
v=-27r20
So height above center =2r-27r20
=40r-27r20=1320r=13cm

Topic: Physics, Ray Optics, JEE Main
A glass tube is open at both the ends. A tuning fork of frequency $f$ resonates with the air column inside the tube. Now the tube is placed vertically inside water so that half the length of the tube is filled with water. Now the air column inside the tube is in unison with another fork of frequency $f^{\prime}$. Then

Options:
A $f^{\prime}=f$
B $f^{\prime}=4 f$
C $f^{\prime}=2 f$
D $f^{\prime}=\frac{f}{2}$

The correct answer is:
$f^{\prime}=f$

Solution:
For fundamental frequency of vibration of air column
$$
l=\frac{\lambda}{2} \Rightarrow \lambda=2 l
$$
Frequency, $f=\frac{v}{\lambda}$
$\Rightarrow \quad f=\frac{v}{2 l}$
When tube is placed inside water with half length immersed in water then
Frequency, $f^{\prime}=\frac{v}{\lambda^{\prime}}=\frac{v}{2 l}$
From Eqs. (i) and (ii), we get
$$
f^{\prime}=f
$$

Topic: Physics, Waves and Sound, JEE Main
A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water μ=43 into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled ?

Options:
A 10 cm
B 11.7 cm
C 7.5 cm
D 8.75 cm

The correct answer is:
10 cm

Solution:
Height of water observed by observer
=Hμw=H(4/3)=3H4

Height of air observed by observer =17.5-H

According to question, both height observed by observer is same.
3H4=17.5-H

⇒ H=10 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A goods train accelerating uniformly on a straight railway track, approaches an electric pole standing on the side of track. Its engine passes the pole with velocity $u$ and the guard's room passes with velocity $v$. The middle wagon of the train passes the pole with a velocity.

Options:
A
$\frac{u+v}{2}$

B
$\frac{1}{2} \sqrt{u^2+v^2}$

C
$\sqrt{u v}$

D
$\sqrt{\left(\frac{u^2+v^2}{2}\right)}$

The correct answer is:
$\sqrt{\left(\frac{u^2+v^2}{2}\right)}$

Solution:
Let ' $S$ ' be the distance between two ends ' $a$ ' be the constant acceleration As we know $V^2-u^2=2 a S$
or, $a S=\frac{V^2-u^2}{2}$
Let $V$ be velocity at mid point.
Therefore, $V_c^2-u^2=2 a \frac{S}{2}$
$$
\begin{aligned}
& V_c^2=u^2+a S \\
& V_c^2=u^2+\frac{V^2-u^2}{2} \\
& V_c=\sqrt{\frac{u^2+v^2}{2}}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2012 (19 May Online)
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103 kg m-3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

Options:
A 7.5 kHz
B 5 kHz
C 2.5 kHz
D 10 kHz

The correct answer is:
5 kHz

Solution:
L=⋋2⇒⋋=2L=1.2 m
v=yρ=5.86×103 m s-1
∴ Fundamental frequency=v⋋=4.88 kHz
≈5 kHz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (08 Apr)
A granum is made up of:

Options:
A Stacks of membrane-bounded, flattened discoid sacs called thylakoids containing the molecules of chlorophyll.
B A coin-shaped stack of thylakoids, membrane-like structures in the chloroplasts of plant cells. Grana are separately present to form stromal thylakoids.
C Many thylakoids arranged one beside the other
D None of the above

The correct answer is:
Stacks of membrane-bounded, flattened discoid sacs called thylakoids containing the molecules of chlorophyll.

Solution:
Thylakoids are flattened sacs arranged like the stacks of coins. One stack of thylakoids is called a granum. Grana are stacks of membrane-bounded, flattened discoid sacs called thylakoids containing the molecules of chlorophyll. Granum is the site of the light reaction during photosynthesis. They are the main functional units of chloroplasts.

Topic: Biology, Photosynthesis in Higher Plants, NEET
A graph is plotted between log (x/m) and log P according to the equation xm=kP1n .

Which of the following statements about this graph is not correct?

Options:
A The figure shows Freundlich adsorption isotherm
B The figure shows Langmuir adsorption isotherm
C The adsorption varies with pressure
D The factor 1/n can have values between 0 and 1

The correct answer is:
The figure shows Langmuir adsorption isotherm

Solution:
This is Freundlich isotherm not Langmuir.
It is also written as
log⁡xm=log⁡k+1nlog⁡p
where
x = mass of adsorbate
m = mass of adsorbent
p = Equilibrium pressure of adsorbate
K and n are constants for a given adsorbate and adsorbent at a particular temperature.

Topic: Chemistry, Surface Chemistry, JEE Main
A graph is plotted between pressure and volume at different temperature. On the basis of the graph what changes will you observe in the volume if

(i) the pressure is increased at constant temperature

(ii) the temperature is decreased at constant pressure

Options:
A volume increases in both the cases
B volume decreases in both the cases
C volume increases in (i) and decreases in (ii)
D volume decreases in (i) and increases in (ii)

The correct answer is:
volume decreases in both the cases

Solution:
Volume decrease when pressure is increased at constant temperature. When temperature is decreased at constant pressure, the volume decreases.

Topic: Chemistry, States of Matter, JEE Main
A graph is plotted between the fringe-width (z) and the distance (D) between the slit and eye-piece, keeping other adjustment same. The correct graph is

Options:
A (B)
B (A)
C (C)
D (D)

The correct answer is:
(C)

Solution:
(A)
$z=\frac{\lambda D}{d} \quad \therefore z \propto D .$

Topic: Physics, Wave Optics, JEE Main
A graph of pressure versus volume for an ideal gas for different processes is as shown. In the graph curve $O C$ represents

Options:
A isochoric process
B isothermal process
C isobaric process
D adiabatic process

The correct answer is:
adiabatic process

Solution:
(i) Curve OA represents isobaric process (since pressure is constant)
Since, the slope of adiabatic process is more steeper than the isothermal process.
(ii) Curve OB represents isothermal process.
(iii) Curve OC represents adiabatic process.
(iv) Curve OD represents isochoric process (since volume is constant).

Topic: Physics, Thermodynamics, JEE Main
A graph of vapour pressure and temperature for three different liquids X,Y and Z is shown below:
The following inferences are made:
AX has higher intermolecular interactions compared to Y.
BX has lower intermolecular interactions compared to Y.
CZ has lower intermolecular interactions compared to Y.
The correct inferences is/are:

Options:
A A and C
B A
C B
D C

The correct answer is:
B

Solution:
At a particular temperature as intermolecular force of attraction increases vapour pressure decreases.
order of intermolecular interactions Z>Y>X

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A graph plotted between $\log K$ vs $1 / T$ for calculating activation energy is shown by

Options:
A

B

C

D

The correct answer is:

Solution:
A graph plotted between $\log k \mathrm{Vs} \frac{1}{T}$ for calculating activation energy is shown as from Arrhenius equation
$\log k=\log A-\frac{E_A}{2.303 R T}$

Topic: Chemistry, Chemical Kinetics, JEE Main
A graph plotted between log k versus 1/T for calculating activation energy is shown by

Options:
A
B
C
D

The correct answer is:

Solution:
k=Ae-Ea/RT
loge⁡k=loge⁡A-EaRT
log10⁡k=log10⁡A-Ea2.303R×1T
y=mx+C
Plot of log k against 1/T will give straight line with negative slope.

Topic: Chemistry, Chemical Kinetics, JEE Main
A graph plotted between log⁡t50 vs log concentration is a straight line. What conclusion can you draw from the given graph?

[Given: n = order of reaction]

Options:
A n = 1,  t 1 2 = 1 k.a
B n=2,t12=1a
C n = 1,  t 1 2  =  0.693 k
D None of the above

The correct answer is:
n = 1,  t 1 2  =  0.693 k

Solution:
t12∝a1-n
or t12=Za1-n
or log⁡t12=log⁡Z+(1-n)log⁡a
or (y)=c+mx
Thus, slope = m = 1 - n
or 1 - n = 0
So n = 1
and for I order reaction t12=0.693k.

Topic: Chemistry, Chemical Kinetics, JEE Main
A graph plotted between t 1 2 vs. log concentration is a straight line. What conclusion can you draw from the given graph

Options:
A n=1,t12=1k.a
B n=2,t12=1a
C n=1,t12=0.693k
D none of these

The correct answer is:
n=1,t12=0.693k

Solution:
t12αa1-n
or t12=ka1-n
log⁡t12=log⁡k+1-nlog⁡a
so slope =1-n
or 1-n=0
or n=1

Topic: Chemistry, Chemical Kinetics, JEE Main
A green light is incident from the water to the air - water interface at the critical angle θc. Select the correct statement.

Options:
A The entire spectrum of visible light will come out of the water at an angle of 90o to the normal
B The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
C The spectrum of visible light whose frequency is more than that of green light will come out to the air medium
D The entire spectrum of visible light will come out of the water at various angles to the normal

The correct answer is:
The spectrum of visible light whose frequency is less than that of green light will come out to the air medium

Solution:
sin θ c = 1 μ

For greater wavelength (i.e lesser frequency) μ is less So, θ c would be more. So, they will not suffer reflection and come out at angles less than 90 ∘ .

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (06 Apr)