datavorous/entrance-exam-dataset huggingface dataset:
Phosgene, COCl2 , a poisonous gas decomposes according to the equation
COCl2g⇌COg+Cl2g
If Kc=0.083 at 900°C , what is the value of Kc?

Options:
A 0.125
B 8.0
C 6.1
D 0.16

The correct answer is:
8.0

Solution:
Kp=KcRTΔng
where Δng=1+1-1=1,R,=0.082 L atm mol-1K-1 and T=1173 K
So Kp=Kc(RT)1=0.083×0.082×1173
7.98≃8.0

Topic: Chemistry, Chemical Equilibrium, JEE Main
Phosgene is the common name of

Options:
A Carbonyl chloride
B Phosphine
C Phosphorus oxychloride
D Phosphorus trichloride

The correct answer is:
Carbonyl chloride

Solution:
$\mathrm{CHCl}_3+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\substack{\text { Phosgene or carbonyl } \\ \text { chloride }}}{\mathrm{COCl}_2+\mathrm{HCl}}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, $\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$
If total pressure at equilibrium of the reaction mixture is $\mathrm{P}$ and degree of dissociation of $\mathrm{PCl}_5$ is $\mathrm{x}$, the partial pressure of $\mathrm{PCl}_3$ will be

Options:
A
$\left(\frac{x}{x+1}\right) P$

B
$\left(\frac{2 x}{1-x}\right) P$

C
$\left(\frac{x}{x-1}\right) P$

D
$\left(\frac{x}{1-x}\right) P$

The correct answer is:
$\left(\frac{x}{x+1}\right) P$

Solution:

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2006
Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones, peroxy acetyl nitrile (PAN). X is:

Options:
A CH4
B CO2
C O3
D CO

The correct answer is:
O3

Solution:
Photochemical smog is formed on the chemical reaction of sunlight, nitrogen oxides and volatile organic compounds in the atmosphere.
NO2g→hvNOg+Og
Oxygen atom (nascent) is very reactive and combines with O2 in air to produce ozone.
Og+O2g→O3g

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2015 (10 Apr Online)
Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones, peroxyacetyl nitrates (PAN) and so forth X is

Options:
A CH4
B CO
C CO2
D O3

The correct answer is:
O3

Solution:
Chemical pollutants in photochemical smog are nitrogen oxides (NO and NO2 ), volatile organic compounds, ozone O3 , peroxyacetyl nitrates.
In the presence of sunlight of the following reactions take place
NO2→hvNO+O
O+O2→O3
O3+NO→NO2+O2
Hence, it consists of excessive amount of ozone melecules.

Topic: Chemistry, Environmental Chemistry, JEE Main
Photochemical smog does not contain

Options:
A Nitrogen dioxide
B Carbon dioxide
C PAN (peroxy acyl nitrate)
D Ozone

The correct answer is:
Carbon dioxide

Solution:
Photochemical smog differs from chemical fog by occurring in warm, dry and sunny climate. The main components of the photochemical smog are unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog is made by the deposition of dust and carbon particles on the pollutant gases and water vapours. The gases found in photochemical smog are NO2, PAN, O3, etc.

Topic: Biology, Environmental Issues, NEET
Photochemical smog formed in congested metropolitan cities mainly consists of :

Options:
A ozone, peroxyacetyl nitrate and $\mathrm{NO}_x$
B smoke, peroxyacetyl nitrate and $\mathrm{SO}_2$
C hydrocarbons, $\mathrm{SO}_2$ and $\mathrm{CO}_2$
D hydrocarbons, ozone and $\mathrm{SO}_x$.

The correct answer is:
ozone, peroxyacetyl nitrate and $\mathrm{NO}_x$

Solution:
Photochemical smog mainly consists of ozone, peroxyacetyl nitrate and different oxide of nitrogen and hydrocarbons released from different engines. Nitric oxide (NO) is formed by reaction between oxygen and nitrogen. This reacts with air forming nitrogen dioxide $\left(\mathrm{NO}_2\right)$. This gas absorbs light due to which nitric oxide (NO) and a molecule of nascent oxygen [O] are formed.
$$
\mathrm{NO}_2 \longrightarrow \mathrm{NO}+[\mathrm{O}]
$$
This atomic oxygen reacts with unburnt hydrocarbons to give rise to peroxy radicals which after reaction with oxygen form ozone or reacting with $\mathrm{NO}_2$ give rise to peroxyacetyl nitrate (PAN).

Topic: Biology, Mineral Nutrition, AIIMS, AIIMS 2003
Photochemical smog formed in congested metropolitan cities mainly consists of

Options:
A ozone, peroxyacyl nitrate and $\mathrm{NO}_{\mathrm{x}}$
B smoke, peroxyacyl nitrate and $\mathrm{SO}_2$
C hydrocarbons, $\mathrm{SO}_2$ and $\mathrm{CO}_2$
D hydrocarbons, ozone and $\mathrm{SO}_{\mathrm{x}}$.

The correct answer is:
ozone, peroxyacyl nitrate and $\mathrm{NO}_{\mathrm{x}}$

Solution:
Photochemical smog is a grey or yellowish brown opaque smog having oxidizing environment but little smokes. It is formed at high temperature over cities and towns due to still air, emission of $\mathrm{NO}_2$ oxides and hydrocarbons from automobile exhausts. Nitrogen dioxide splits into nitric oxide and nascent oxygen. Nascent oxygen combines with molecular oxygen to form ozone. Ozone reacts with hydrocarbons to form aldehydes and ketones. Nitrogen oxides, oxygen and ketones combine to form peroxy-acyl-nitrates (PAN).

Topic: Biology, Environmental Issues, AIIMS, AIIMS 2014
Photochemical smog formed in congested metropolitan cities mainly consists of

Options:
A Ozone, peroxyacetyl nitrate and NO X
B Hydrocarbons, SO 2  and C O 2
C Smoke, peroxyacetyl nitrate and SO 2
D Hydrocarbons, ozone and SO X

The correct answer is:
Ozone, peroxyacetyl nitrate and NO X

Solution:
Photochemical smog or oxidizing type of pollution is characterized by the presence of large concentration of ozone, oxides of nitrogen and various hydrocarbons. It occurs in Los Angels.

Topic: Biology, Environmental Issues, NEET
Photochemical smog is related to pollution of

Options:
A Air
B Water
C Soil
D All of the above

The correct answer is:
Air

Solution:
It is related to Air Pollution

Topic: Chemistry, Environmental Chemistry, JEE Main
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.

Options:
A
$\mathrm{NO}_2$

B
$\mathrm{O}_3$

C
$\mathrm{SO}_2$

D
Unsaturated hydrocarbon

The correct answer is:
$\mathrm{SO}_2$

Solution:
The smog which is formed in presence of sunlight is called photochemical smog. The main components of the photochemical smog results from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories.

Topic: Chemistry, Environmental Chemistry
Photodiodes are mostly operated in reverse biased condition because

Options:
A Fractional change in minority carriers produce higher forward current
B Fractional change in majority carriers produce higher reverse current
C Fractional change in minority carriers produce higher reverse current
D Fractional change in majority carriers produce higher forward current

The correct answer is:
Fractional change in minority carriers produce higher forward current

Solution:
The fractional change due to incident light on minority charge caries in reverse bias is much more than that over the majority charge carries in forward bias.

Topic: Physics, Semiconductors, JEE Main
Photoelectric effect experiments are performed using three different metal plates \( p, q \) and \( r \) having work function, \( \phi_{p}=2 \cdot 0 \mathrm{eV}, \phi_{q}=2 \cdot 5 \mathrm{eV} \) and \( \phi_{r}=3 \cdot 0 \mathrm{eV} \) respectively. A light beam containing wavelengths of \( 550 \mathrm{~nm}, 450 \mathrm{~nm} \) and \( 350 \mathrm{~nm} \) with equal intensities illuminates each of the plates. The correct \( I-V \) graph for the experiment is (Take \( h C=1250 \mathrm{eV} \mathrm{nm} \) )

Options:
A
B
C
D

The correct answer is:

Solution:
For photoelectric emission to take place, the wavelength of incident light should be less than the threshold wavelength for that metal. If λm is the threshold wavelength, then work function of metal
ϕ = h⁡ c λ m⁡ or λ m⁡ = h⁡ c ϕ
For metal p, λ m⁡ p⁡ = 1 2 4 0 eV nm 2 · 0 eV = 6 2 0 nm
For metal q, λ m⁡ q⁡ = 1 2 4 0 eV nm 2 · 5 eV = 4 9 6 nm
For metal r, λ m⁡ r⁡ = 1 2 4 0 eV nm 3 · 0 eV = 4 1 3 · 3 nm
Wavelengths in the incident beam are 550 nm, 450 nm and 350 nm. Thus the photoelectric emission is possible by light of wavelength 350 nm (being shorter than λm) from metals p, q and r. For light of wavelength 450 nm (< 620 nm and 496 nm) photoelectric emission is possible from metals p and q. For light of wavelength 550 nm (< 620 nm), the photoelectric emission is possible from metal p.
Current (I) ∝ Intensity
where Intensity = Nh υ of photoelectrons

∴ I is maximum for p metal, less for q metal and least for r metal. Therefore, the graph
is correct.

Topic: Physics, Dual Nature of Matter, JEE Main
Photoelectric effect occurs only if

Options:
A Frequency of incident light is lesser than threshold frequency
B Frequency of incident light is greater than threshold frequency
C Frequency of incident light is equal to threshold frequency
D More than one of the above

The correct answer is:
Frequency of incident light is greater than threshold frequency

Solution:
EXPLANATION:
- The photoelectric effect is said to have occurred when an electron is ejected from the metal surface with light incident on it.
- For an electron to be emitted, it must acquire energy large enough to leave the metal surface. For this to happen, the electron should have an energy greater than the threshold energy.
- Since energy is directly proportional to frequency, the electron must acquir $\epsilon$ a frequency greater than the threshold frequency.
- Therefore, the frequency of incident light has to be greater than the threshold frequency for the photoelectric effect to occur.

Topic: Physics, Dual Nature of Matter, JEE Main
Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $\left(v_1>v_2\right)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1: k$, then the threshold frequency of the metallic surface is

Options:
A $\frac{v_2-v_1}{k-1}$
B $\frac{k v_1-v_2}{k-1}$
C $\frac{k v_2-v_1}{k-1}$
D $\frac{v_2-v_1}{k}$

The correct answer is:
$\frac{k v_1-v_2}{k-1}$

Solution:
Maximum kinetic energy of photoelectrons
$(\mathrm{KE})_{\max }=h v-h v_0$
$v=$ frequency of incident radiation
$v_0=$ threshold frequency
Ist Case $\quad\left(\mathrm{KE}_{\max }\right)_1=h v_1-h v_0$
IInd Case $\left(\mathrm{KF}_{\max }\right)_2=h v_2-h v_0$
$\begin{aligned} \frac{\left(\mathrm{KE}_{\max }\right)_1}{\left(\mathrm{KE}_{\max }\right)_2} & =\frac{h\left(v_1-v_0\right)}{h\left(v_2-v_0\right)} \\ \frac{1}{k} & =\frac{v_1-v_0}{v_2-v_0} \\ v_2-v_0 & =k v_1-k v_0 \\ k v_0-v_0 & =k v_1-v_2 \\ v_0 & =\frac{k v_1-v_2}{k-1}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
Photoelectrons are ejected from a metal when light of frequency $v$ falls on it. Pick out the wrong statement from the following.

Options:
A
No electrons are emitted if $v$ is less than $W / h$, where $W$ is the work function of the metal

B
The ejection of the photoelectrons is instantaneous.

C
The maximum energy of the photoelectrons is hu.

D
The maximum energy of the photoelectrons is independent of the intensity of the light.

The correct answer is:
The maximum energy of the photoelectrons is hu.

Solution:
According to photo-electric equation : $\mathrm{K} . \mathrm{E}_{\max }=\mathrm{h} v-\mathrm{h} v_0$ (Work function) Some sort of energy is used in ejecting the photoelectrons.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2012 (26 May Online)
Photoelectrons are emitted when 4000 â« radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region having Î± -particles to form He+ ion, emitting a single photon. In this process, He+ ions thus formed are in their fourth excited state. The energy released during the combination of He+ ions is

Options:
A 5.38 eV

B 3.38 eV

C 2.38 eV

D 1.38 eV

The correct answer is:
3.38 eV

Solution:
The total energy of electron initially = hc Î» âw=1.18
Final total energy =â2.2
The difference of energy when photoelectron recombines with Î± -particle
=1.18--2.2
=3.38 eV

Topic: Physics, Atomic Physics, JEE Main
Photo-electrons are produced by from a metal surface using a radiation of wavelength 6561 A∘. These photo-electrons, when made to enter a uniform magnetic field of intensity 3×10−4 T, move along different circular paths with a maximum radius of 10 mm, then the work function of the metal is close to

Options:
A 3.1 eV
B 0.1 eV
C 2.1 eV
D 1.1 eV

The correct answer is:
1.1 eV

Solution:
KEmax=E-Ï

=12400Î»( in A)-Ï ( in eV )

â´r=2mKEeB

KEmax=r2e2B22m

=r2eB22m (in J)

â´ Ï=124006561-r2eB22m

=1.1 eV

Topic: Physics, Dual Nature of Matter, JEE Main
Photo-electrons are produced from a metal surface using a radiation of wavelength 6561 Aâ. These photo-electrons, when made to enter a uniform magnetic field of intensity 3Ã10â4 T, move along different circular paths with a maximum radius of 10 mm, then the work function of the metal is close to

Options:
A 3.1 eV
B 0.1 eV
C 2.1 eV
D 1.1 eV

The correct answer is:
1.1 eV

Solution:
KEmax=E-ϕ

=12400λ( in A∘)-ϕ in eV

∴r=2mKEeB

KEmax=r2e2B22m

=r2eB22m (in J)

∴ ϕ=124006561-r2eB22m

=1.1 eV

Topic: Physics, Dual Nature of Matter, JEE Main
Photon of frequency $\mathrm{v}$ has a momentum associated with it. If $\mathrm{c}$ is the velocity of light, the momentum is

Options:
A
$\mathrm{v / c}$

B
$\mathrm{hvc}$

C
$\mathrm{h v / c^2}$

D
$\mathrm{hv/c}$

The correct answer is:
$\mathrm{hv/c}$

Solution:
$\mathrm{P}=\frac{\mathrm{h}}{\lambda}=\frac{\mathrm{h} v}{\mathrm{c}}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2007
Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency $v$ is used to convert 1 $\mathrm{kg}$ of ice at $0^{\circ} \mathrm{C}$ to water at $0^{\circ} \mathrm{C}$. Then, the time T taken for the conversion

Options:
A
decreases with increasing $\mathrm{n}$, with $\mathrm{v}$ fixed

B
decreases with $\mathrm{n}$ fixed, $v$ increasing

C
remains constant with $\mathrm{n}$ and $\mathrm{v}$ changing such that $\mathrm{nv}=$ constant

D
increases when the product nv increases

The correct answers are:
decreases with increasing $\mathrm{n}$, with $\mathrm{v}$ fixed
,
decreases with $\mathrm{n}$ fixed, $v$ increasing
,
remains constant with $\mathrm{n}$ and $\mathrm{v}$ changing such that $\mathrm{nv}=$ constant

Solution:
As given that, the heat energy required to convert $1 \mathrm{~kg}$ of ice at $0^{\circ} \mathrm{C}$ to water at $0^{\circ} \mathrm{C}$ is $\mathrm{Q}=\mathrm{ml}$
So, energy spent to convert ice into water
$=$ mass $\times$ latent heat
$=(1000 \mathrm{~g}) \times(80 \mathrm{cal} / \mathrm{g})$
$=80000 \mathrm{cal}$
Let time $\mathrm{T}$ is taken by radiation to melt the ice at $0^{\circ} \mathrm{C}$ then,
Energy of photons used $=\mathrm{nT} \times \mathrm{E}=\mathrm{nT} \times \mathrm{hv}$
$[\because \mathrm{E}=\mathrm{hv}]$
So, $\quad \mathrm{nTh} v=\mathrm{mL} \Rightarrow \mathrm{T}=\frac{\mathrm{mL}}{\mathrm{nh} v}$
$\therefore \quad \mathrm{T} \propto \frac{1}{\mathrm{n}}$, when $v$ is constant.
$\mathrm{T} \propto \frac{1}{v}$, when $\mathrm{n}$ is fixed.
$$
\Rightarrow \quad \mathrm{T} \propto \frac{1}{\mathrm{n} v}
$$
So, $T$ is constant, if $n v$ is constant and if $n v$ increases then, $\mathrm{T}$ decreases.

Topic: Physics, Dual Nature of Matter
Photons of an electromagnetic radiation has an energy $11 \mathrm{keV}$ each. To which region of electromagnetic spectrum does it belong ?

Options:
A
X-ray region

B
Ultra violet region

C
Infrared region

D
Visible region

The correct answer is:
X-ray region

Solution:
$\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \lambda=\frac{\mathrm{hc}}{\mathrm{E}}$
$\Rightarrow \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{11 \times 1000 \times 1.6 \times 10^{-19}}$
$=12.4 Å$
wavelength range of visible region is $4000 Å$ to $7800 Å$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2013 (09 Apr Online)
Photons of energy $2.4 \mathrm{eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is

Options:
A $2.1 \mathrm{eV}$
B $1.7 \mathrm{eV}$
C $2.8 \mathrm{eV}$
D $2.0 \mathrm{eV}$

The correct answer is:
$2.1 \mathrm{eV}$

Solution:
Let $\phi_0=$ work-function of metal.
Then, according to Einstein's photoelectric equation,
$K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0$
When wavelength is reduced by $50 \%$ i.e, $\lambda^{\prime}=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.
From Eqs. (i) and (ii), we have
$\begin{aligned}
& 9\left(\frac{h c}{\lambda}-\phi_0\right)=\frac{2 h c}{\lambda}-\phi_0 \\
& 7 \frac{h c}{\lambda}=8 \phi_0 \Rightarrow \phi_0=\frac{7}{8} \frac{h c}{\lambda}
\end{aligned}$
Here, photon energy is given as $\frac{h c}{\lambda}=2.4 \mathrm{eV}$
So, work-function is $\phi_0=\frac{7}{8} \times 2.4=2.1 \mathrm{eV}$

Topic: Physics, Dual Nature of Matter, JEE Main
Photons of wavelength $\lambda$ emitted by a source of power $P$ incident on a photo cell. If the current produced in the cell is $I$, then the percentage of incident photons which produce current in the photo cell is. (where, $h$ is Planck's constant and $c$ is the speed of light in vacuum)

Options:
A $\frac{100 e P c}{\ln \lambda}$
B $\frac{100 e P \lambda}{I h c}$
C $\frac{100 / h \lambda}{e P_C}$
D $\frac{100 / h c}{e P \lambda}$

The correct answer is:
$\frac{100 / h c}{e P \lambda}$

Solution:
In photoemission, only one electron can produce one photon.
So, if only $x \%$ of incident photons can produce electrons, then
$$
I=\frac{n}{t}=\frac{100 \times I \times h c}{e P \lambda}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
Photoperiodism was first characterised in

Options:
A tobacco
B potato
C tomato
D cotton

The correct answer is:
tobacco

Solution:
They phenomenon of photoperiodism was first discovered by Garner and Allard (1920, 22) in tobacco plant. They observed that Maryland Mammoth, the variety of tobacco could be made to flower only by reducing the light hours with artificial darkening. It could be made to remain vegetative in winter by providing extra light.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2010 (Screening)
'Photosynthesis' is a:

Options:
A Physico-chemical process
B Physical process
C Chemical process
D Catabolic process

The correct answer is:
Physico-chemical process

Solution:
Physico-chemical processes are the normal intracellular and extracellular physical and chemical processes that are crucial for the maintenance of normal homeostasis of a cell. These reactions are vital for the wellbeing of the cell and provide essential energy for its survival. Absorption of light is a physical process and biochemical phase is synthesis of sugar and other chemical products

Topic: Biology, Photosynthesis in Higher Plants, NEET
Phthalic acid reacts with resorcinol in the presence of concentrated $\mathrm{H}_2 \mathrm{SO}_4$ to give:

Options:
A
Phenolphthalein

B
Alizarin

C
Coumarin

D
Fluorescein

The correct answer is:
Fluorescein

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (12 Apr Online)
Phthalimide is made to undergo following sequence of reactions.
Total number of \(\pi\) bonds present in product 'P' is/are ______

Options:

The correct answer is:
8

Solution:
Total number of $\pi$-bonds present in product $\mathrm{P}$ is 8

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (04 Apr Shift 2)
p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives:

Options:
A
B
C
D

The correct answer is:

Solution:
In electrophilic aromatic substitution, a substituent that favors electrophilic attack ortho or para to the substituent. Most ortho/para directors are also activators, except for the halogens, which are deactivators. A carboxylic acid is a meta director because it experiences resonance, a delocalization of electrons.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Phylogenetic system of classification is based on:

Options:
A Floral characters
B Evolutionary relationships
C Morphological features
D Chemical constituents

The correct answer is:
Evolutionary relationships

Solution:
No Solution Available

Topic: Biology, Biological Classification, NEET, NEET 2009 (Mains)
Physical adsorption is essentially quite appreciable

Options:
A At room temperature
B At higher temperature
C At lower temperature
D None of these

The correct answer is:
At lower temperature

Solution:
In physical adsorption, equilibrium is established between the adsorbate and the fluid phase. In solid or gas systems, at not too high pressures, the extent of physical adsorption increases with increase in gas pressure and usually decreases with increasing temperature. Since physical adsorption is an exothermic process, the rate of physical adsorption is inversely proportional to temperature.

Topic: Chemistry, Surface Chemistry, JEE Main
Physical quantity x and y are related as y=4 tanx . If at x=Ï4 radian error in measurement of x is 2% then find % error in measurement of y at x=Ï4 .

Options:
A 2%
B π2%
C π6%
D π %

The correct answer is:
π2%

Solution:
dydx=4sec2x, Îyy=4sec2Îx4tanx
ÎyyÃ100=4sec2x4tanxÃÎxÃ100
=2ÃÎxÃ100
ÎxxÃ100=2
100 Îx=2ÃÏ4=Ï2
ÎyyÃ100=Ï2

Topic: Physics, Units and Dimensions, JEE Main
Options:
A $\begin{array}{llll}3 & 2 & 1 & 4\end{array}$
B $\begin{array}{llll}4 & 1 & 3 & 2\end{array}$
C $\begin{array}{llll}2 & 3 & 4 & 1\end{array}$
D $\begin{array}{llll}4 & 3 & 2 & 1\end{array}$

The correct answer is:
$\begin{array}{llll}4 & 3 & 2 & 1\end{array}$

Solution:
(A) Time-period in spring-block system is given by, $T=2 \pi \sqrt{\frac{m}{k}}$
If mass is doubled, $T^{\prime}=2 \pi \sqrt{\frac{2 m}{k}}$ $\Rightarrow \quad T^{\prime}=\sqrt{2} T$
Hence, time period becomes $\sqrt{2}$ times, if the mass of the block is doubled.
$$
(A \rightarrow 4)
$$
(B) For spring, $\mathrm{PE}=\frac{1}{2} k x^2$
If spring constant is increased 4 times, PE becomes 4 times.
$$
(B \rightarrow 3)
$$
(C) Speed of the block, $v=\omega A$
Hence, velocity of the block becomes twice when amplitude of the oscillation is doubled.
$$
(C \rightarrow 2)
$$
(D) Energy of the oscillation, $E=\frac{1}{2} m \omega^2 a^2$
$$
\begin{aligned}
& & E^{\prime}=\frac{1}{2} m \omega^{\prime 2} a^2 \\
\Rightarrow & E^{\prime} & =\frac{1}{2} m(2 \omega)^2 a^2=\frac{4}{2} m \omega^2 a^2 \quad\left(\because \omega^{\prime}=2 \omega\right) \\
\Rightarrow & E^{\prime} & =4 E
\end{aligned}
$$
Energy of the oscillation becomes 4 times when angular frequency is doubled.
$$
(D \rightarrow 1)
$$

Topic: Physics, Oscillations, JEE Main
pi (π) bond is formed by the overlap of

Options:
A p-p orbitals
B s-s orbitals
C s-p orbitals
D s-d orbitals

The correct answer is:
p-p orbitals

Solution:
In the formation of pi bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise overlapping consist of two saucer type charged clouds above and below the plane of the participating atoms.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Pick out incorrect statement about K2Cr2O7 .

Options:
A It oxidizes acidified solution of H2S to S
B It oxidizes KI to I2
C It oxidizes HCl to Cl2
D It gives oxygen, when treated with cold dil. H2SO4

The correct answer is:
It gives oxygen, when treated with cold dil. H2SO4

Solution:
Here option D is not correct as oxygen is evolved only when K2Cr2O7 is heated with conc. H2SO4 .
2K2Cr2O7+8H2SO4→Δ2K2SO4+2Cr2SO43+8H2O+3O2
A. Cr2O72-+14H++6e-→2Cr3++7H2O
S2-→S+2e-×3
B. 2I-→I2+2e-
C. 2Cl-→Cl2+2e-

Topic: Chemistry, d and f Block Elements, JEE Main