AP EAMCET Mock Test - 5 - JEE MCQ

E.g.

The hydroboration oxidation of an alkene will give the alcohol as the product and the OH group will attach to the carbon which is less hindered. Thus this reaction follows Anti-Markownikoff's rule.

Hence isobutylene on hydroboration oxidation will isobutyl alcohol.

Dettol is mixture of

Hence compound ‘B’ is Terpineol.

The equation for the reaction is as follows:

For 1st order reactions, R → P

Rate = -d(R)/dt = k[R]

On integrating this elation, we get

In [R] = -kt + l …(i)

At t = 0, R = [R]₀, In[R]₀ = -k x 0 + l

In[R]₀ = l

Substituting the value of I in (i) and on rearrangement, we get

In[R] = -kt + InR₀And, k = 1/t in [R₀]/[R]

∴ 2(+1) + 2(x) + 7(-2) = 0

2 + 2x - 14 = 0

2x = 12 x = +6

2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO^2-₄ + 4H⁺

When SO₂ reacts with H₂S, it produces sulphur and water. In this reaction, ‘SO₂’ act as an oxidising agent.

x² + px + q = 0
Let, α, α² are the roots of the given equation.
⇒ α + α² = −p
and α⋅α² = q
⇒ α3 = q
∵ α + α² = −p
⇒ (α + α²)³ = −p³
⇒ α⁶ + α³ + 3α.α²(α + α²) = −p³
⇒ q² + q + 3α³(α + α²) = −p³
⇒ q² + q + 3q(−p) = −p³
⇒ p³ + q² + q − 3pq = 0
⇒ p³ + q² + q(1 − 3p) = 0

Let P(A) and P(B) be the probabilities that a man and a woman live for another 10 years respectively.

Both the given curves are rectangular hyperbolas.

And, we know that the eccentricity of a rectangular hyperbola is e₁ = e₂ = √2.

Hence, e₁ − e₂ = 0.

Given:

then,

Let E₁, E₂ and A be the events defined as follows
E₁ = six occurs
E₂= six does not occurs
A = the man reports that it is a six. We have.

Now, P(A/E₁)= Probability that the man reports that there is a six on the die given then 6 has occured on die
= Probability the man speaks truth

=3/4

P(A/E₂) = Probability that the man reports that there is six is on die given that six has not occured on die
= Probability that the man does not speak truth

We have to find P(E₁/A) i.e. the probability that there is six on the die given that the man has reported that there is six. By Baye's rule, we have

Consider the Chairman and the Vice-Chairman as one individual.
So, instead of 15 members to be arranged, we now have only 14.
These 14 members can be arranged around a round table in 13! ways.
Also, the Chairman and the Vice-Chairman can be arranged in 2! ways.
∴ The total number of ways is 2 × 13!.
If the Chairman and the Vice-Chairman are never to be seated together, then the total number of ways is = 14! - 2 × 13! = 12 × 13!

X__X__X__X__X.

The four digits 3, 3, 5, 5 can be arranged at 4 places in 
The five digits 2, 2, 8, 8, 8, 8 can be arranged at X places in 
∴ Total number of arrangements = 6 × 10 = 60 ways

Total number of ways of selecting a pair of numbers which are different from set (0,1,2,3,...,9)

Now, correct pair is only one. Hence, probability that the number dialed is correct = 1/90

As the straight line and circle intersect so x = −y satisfies the equation of circle x² + y² + 4y = 0

(−y)² + y² + 4y = 0 ⇒ 2y² + 4y = 0 ⇒ 2y(y + 2) = 0 ⇒ y = 0, −2

∴ x = 0, 2

Thus, the interesetion of line and circle is (0,0) and (2, −2).

Since, options are of form y² = 4ax, so putting (2,−2) in y² = 4ax

Putting back in standard equation of parabola,

which also satisfies point (0, 0)

Thus, y² = 2x is the equation of parabola.