AP EAMCET Mock Test - 10 - JEE MCQ

Charge will flow from A to B until their potentials become equal. If charge q flows from A to B, then 
or, Q - q = a/b, which gives q = 
Hence, charge left on A = Q - q = Q - 
Hence, the correct choice is (d).

Let T = temperature of the junction J
L = length and A = area of each rod
Heat conducted in first rod = heat conducted by second rod + heat conducted by third rod

Ratio of masses is 1:5.
Ratio of acceleration is 5:1.
Ratio of distances covered is  5:1.
Distance to be covered  =(12R−R−2R)=9R
Distance moved by smaller sphere

Since, from Rutherford - Soddy law, the number of atoms left after half-lives is given by

Where, N₀ is the original number of atoms.
The number of half-lives, 

Relation between effective disintegration constant (λ) and half life (T)


Effective half life,

dv/dx = x³ - x = 0
⇒ x(x² - 1) = 0
⇒ x = 0, ±1
 = (3x² - 1) = Positive for x = ±1
i.e. V is minimum at x = ±1
⇒ V_max = - J
⇒ K_max = 2 -  =  J
 × 1 × v² = 
⇒ v = 3/√2 m/sec

Given,
velocity of the particle is 
time instance is 1 sec.
The acceleration of the particle is the rate of change of velocity with respect to time thus, acceleration is given by,

Acceleration of the particle remains constant at all times.

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. 

Acid hydrolysis of an ester gives two different organic compounds.

Cu from sulphide ore is obtained in a Bessemer's convertor as follows:


Copper pyrite ore CuFeS₂ also contains FeS which is removed as follows:

Roasting converts FeS to FeO, which combines with silica flux to form slag.

From the given equations, we have
K₁ = 
K₂ = 
K₃ = 
For the given reaction,
K = 

K = 

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.

For the reaction

ΔG^o for the reaction
2ZnS + 3O₂ → 2ZnO + 2SO₂
Can be obtained by adding eqn. (1), (2) and (3)
⇒ ΔG^o = 293 − 480 − 544 = −731 kJ

The given equation reduces to (−ω)^m    = (−ω²)^m 

⇒ ω^m  = 1 

⇒ m = 3

When two sides and angle opposite to one of then be given. In this case, the triangle is not always uniquely determined. It is quite possible to have no triangle, one triangle and two triangles with this type of data.
So, it is called an ambiguous case.
Let b, c and B are given parts.
When B is acute and b < c sin B, then there is no triangle
By Cosine Law
b² = c² + a² - 2ac cos B
⇒ a² - 2ac cos B = b² - c²
⇒ a² - 2ac cos B + c² cos² B
= b² - c² + c² cos² B
⇒ (a - c cos B)² = b² - c² (1 - cos² B)
= b² - c² sin² B

This equation helps us to determine a, when b, c and B are being given. We observe that:
When b < c sin B, then a is imaginary and so there is no solution.

Statements are logically equivalent to their contrapositives.

To form the CONVERSE of a statement, the "If" and the "then" sections of the statement switch places.

The determinant of skew-symmetric matrix of odd order is always zero
Therefore the inverse of skew-symmetric matrix does not exist