COMEDK Mock Test - 1 - JEE MCQ

The magnification is given by

⇒  f_e = 22 mm

So displacement of B = ΔL₁

= 3.92 × 1⁰⁻⁶ m

displacement of C = ΔL₁ + ΔL₂

= 8.82 × 10⁻⁶m

and displacement of D = ΔL₁ + ΔL₂ + ΔL₃

= 23.5 × 10⁻⁶ m

t = 34.65 s.

∴ Possible frequencies = nf (n is odd) = 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz below 1250 Hz.

⇒ 100 - 10 =   = 90 J

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

Magnetic force 
⇒ Wb = 0
⇒ ΔKE = 0 and speed remains constant.

According to curies law, magnetic susceptibility is inversely proportional to temperature for a fixed value of external magnetic field i.e. x = C/T.
The same is applicable for ferromagnet & the relation is given as  (TC is curie temperature) Diamagnetism is due to non-cooperative behaviour of orbiting electrons when exposed to external magnetic field. 

Reduction of nitrobenzene in alkaline medium results in different products depending upon the nature of the reducing agent.

And the reduction in strongly acidic medium with metals results in the formation of aniline, while electrolytic reduction gives the different product in strong and weak acidic medium.

• The greenhouse effect arises due to the presence of gases like freons, methane, carbon dioxide, ozone and water vapours in the atmosphere.
• These gases can trap the heat of the sun, hence, increase the temperature of the atmosphere.
• And also, maintains the average temperature of the earth.

The shape of a molecule depends upon the hybridisation and number of lone pair(s) on the central atom of the molecule. The hybridisation in a molecule can be determined by the total steric number around the central atom, which is equal to the sum of sigma bonds and lone pair(s) of electrons on the central atom. 
The number of sigma bonds and lone pairs can be determined from the Lewis structure of the molecule.
OSF₂
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 3
Number of lone pair = 1
Total steric number = 3 + 1 = 4
The hybridistion of a molecule with a steric effect of 4 is sp³. A hybridisation of sp³ is expected to have a tetrahedral geometry, but the presence of a lone pair on the sulphur atom comes into play (VSEPR theory). 
The geometry of a molecule with three sigma bond pairs and one lone pair, according to the VSEPR theory, can be represented as under:

Lone Pair − Bond Pair repulsion > Bond Pair − Bond Pair repulsion
Due to a greater repulsion between the lone pair and bond pair, the bond angle between the lone pair and each S−FS-F bond is more than the bond angles of the F−S−O and O−S−FF-S-O and O-S-F bonds. Thus, the molecule OSF2OSF2 has a trigonal pyramidal geometry.  

SO₂
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 2
Number of lone pair = 1
Total steric count = 2 + 1 = 3
A steric count of 3 has a hybridisation of sp². So, the molecule is expected to have a trigonal planar shape, but due to the presence of a lone pair of electrons on the sulphur atom, the geometry changes to a bent structure as follows:

BrF₃
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 3

Number of lone pairs = 2
Due to the presence of three sigma bonds, and two lone pairs of electrons, BrF3BrF3 shows a trigonal bipyramidal geometry with a distorted T−T-shaped structure.

SiO²⁻₃

The Lewis structure of the molecule can be represented as under :

Number of sigma bond pairs = 3

Numebr of lone pair = 0

Total steric number = 3 + 0 = 3

Hybridisation → sp²
Since there are no lone pairs on the silicon atom, the molecule shows a geometry of trigonal planar, as per its hybridisation. 

In case of 2Ca + O² → 2CaO calcium is being oxidized and oxygen is reduced. 

The Gibbs free energy change at a constant temperature and pressure is given as:

where, ΔH is the change in the enthalpy of the process and ΔS is the change in the entropy of the process.

For the process to be spontaneous, ΔG must be negative.

Chain isomerism is type of structural isomerism. Chain isomers differ in the branching of carbon skeleton. Possible chain isomers of C₆H₁₄ are:

Hence, five isomers are possible.

Synthesis of glucose molecule in photosynthesis involves following reaction:

Acid hydrolysis of an ester gives two different organic compounds.

A salt is precipitated only when the product of ionic concentrations is more than its solubility product.

So, AB will be precipitated only when the concentration of [B^-] is more than 10^-5 M.

Surface tension is maximum in water due to strong intermolecular forces of attraction (hydrogen bonds) between the water molecules.

∵ Line y = kx + 4 touches the parabola y = x - x².
So, kx + 4 = x - x²
⇒ x² + (k - 1) x + 4 = 0 has only one root.
(k - 1)² = 16
⇒ k = 5 or -3 but k > 0
So, k = 5
Hence, x² + 4x + 4 = 0
⇒ x = -2
So, P is (-2, -6) and V is (1/2 , 1/4).

So, maxima occurs at x = 2.
f(2) = 4.0 - 2.2² + 4.2 + 5 = 5
Clearly, f(x) = -1 has exactly 2 solutions.

So, f'(e) - f''(2) > 0
Option C is correct.

Let the equation of the required plane be:

∵ (2, 1, -2) lies on it, 2 + λ (-2) = 0
 ⇒ λ = 1
Hence, π : 3x + 2y - 8 = 0

Clearly, X and Y are on the opposite sides of plane .

Now, 3¹⁰ - 2¹⁰ = (3⁵ - 2⁵)(3⁵ + 2⁵)
= (211)(275)
= (35 × 6 + 1)(45 × 6 + 5)
= 6λ + 5
Hence, the remainder is 5.

Let the centre be (h, k).

⇒ 2h² = h² + k² + 2hk
The locus will be x² - y² = 2xy.

We have,

a²+2b=7   …(i)

b²+4c=−7   …(ii)

c²+6a=−14   …(iii)

Adding equations (i), (ii) and (iii), we have

It is possible only when a=−3, b=−1 and c=−2.

So,

Let

Feasible solutions are

Clearly, the solution region is the quadrilateral.