BITSAT Mock Test - 8 - 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 = , which gives q =
Hence, charge left on A = Q - q = Q -
Hence, the correct choice is (4).

Let P be the position of the mark. Q is the position of its image.
Since the incident ray PA lies in a medium of refractive index μ₂, and is refracted into a medium of refractive index μ₁, our formula becomes


where u = -2 R = -10 cm,
R = -5 cm, μ₂ = 1.5 and μ₁ = 1
Putting these values in the above formula, we have

which gives v = -20 cm
∴ Distance of image Q from O = 20 - 5 = 15 cm

Shunt resistance

For first order reaction,

Here, [A_o] = Initial concentration of reactant and [A] = Final concentration of reactant
When initial concentration drops to 3/4, then


But, t_1/4 = t_3/4 (given)
Therefore,


=

According to anti-Markovnikov's rule, X is 1-bromobutane and Y is 2-bromobutane.

In this reaction, organic peroxide dissociates butene into free radicals. There is the formation of free radicals as intermediates.

More stable free radicals give major products according to anti-Markovnikov's rule.

Secondary amines, on reaction with nitrous acid at low temperature, give an oily nitrosamine.

Enthalpy of neutralisation is defined as the amount of heat evolved when 1 equivalent of H⁺ ions from an acid is completely neutralised by 1 equivalent of HO^- ions from a base.
For a strong base and a strong acid, which are strong electrolytes, this value is fixed under standard conditions at -57.62 kJ mol^-1.
But for a weak acid like HCN or a weak base, the value is less as some of the heat liberated is utilised to ionise the weak acid or the weak base.

t-butyl methyl ether is prepared by Williamson synthesis process. It involves the reaction of an alkoxide ion with a primary alkyl halide via an S_N2 reaction.

Solutions of alkali metals in liquid ammonia are blue coloured due to ammoniated electrons.

These ammoniated electrons absorb red colour from the visible light and so the transmitted light is blue.
Hence, option (4) is correct.
Option (1) is incorrect because the reducing power of the alkali metals increases from sodium to cesium. However, Li has the highest reducing power because of its highest heat of hydration.
Option (2) is incorrect because Li⁺, being highly hydrated, has the least ion mobility and the least value of ionic velocity.
Option (3) is incorrect because the metal ions are extensively hydrated in their aqueous solutions and the trend of the hydrated ionic radius is as follows.
Li⁺(aq) > Na⁺(aq) > K⁺(aq) > Rb⁺(aq) > Cs⁺(aq)
[Note: In solid ionic crystals, the trend of the radii of metal ions is the same as that for the atomic radii, which is Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺]

2S S
Given: [] = s = 2 x 10^-4
[Ag⁺] = 2s = 2 x 2 x 10^-4
= 4 x 10^-4
K_sp = [Ag⁺]²
= [4 x 10^-4]²[2 x 10^-4]
= 32 x 10^-12

The AB₂ type compounds have a fluorite type structure in which the cations and anions are present in ratio of 1 : 2. Among the given options, CaF₂ has AB₂ type structure.
The Ca²⁺ are arranged in ccp type arrangement (Ca²⁺ ions present at all corners as well as the centre of each face) and the fluoride ions are present in the tetrahedral voids.

Hence, the formula is CaF₂.

Standard heat of formation  of methane is represented by
C (graphite) + 2H₂(g) → CH₄(g)

The paragraph accounts for government measures or intervention at both producer end, i.e. farmer, and consumer end. Only then can farming continue to be profitable or sustainable else lands would be put to alternative uses, i.e. farming will come to an end. This essence is captured only by option 1.

In each step, one element is replaced by a new one alternately on either end, while the whole set of elements rotates by 45° in anti-clockwise direction. Options (2), (3) and (4) have a new third element at the same end again and are ruled out. Hence, the correct option is (1).

The first two letters of the first group are each moved two steps forward, and the last two letters are each moved one step forward to obtain the corresponding letters of the second group. The pattern is as follows:
Q + 2 = S
D + 2 = F
X + 1 = Y
M + 1 = N
Similarly,
U + 2 = W
I + 2 = K
O + 1 = P
Z + 1 = A

According to the passage, the meaning of the sentence, war is the negation of truth is that wars always spread and give rise to falsehood. Therefore, (4) is correct.

According to the directions given, the arrangement from left to right is:
F D A C E B
There are 4 persons to the right of D.

The sentence is in simple past tense. We add 'was' in the passive form of sentence (was + rocked).
The doer of the action is mentioned at the end of the sentence with the preposition 'by'.
Option 4 has all the modifications in the sentence as per the rules of active/passive voice. Thus, it is the correct answer.

This is the correct answer because there are three dots on part of the total.
So, in total there will be 12 dots and in the same sequence.

(32)(32)^1/6(32)^1/36 … up to can be written as

We know

So, = 45a^{45 - 1} = 45a⁴⁴

Put = t, .e^x dx = dt
dt = e^t + C = + C