AP EAMCET Mock Test - 5 - JEE MCQ

At steady state, inductance is ineffective
I = V/R = 10/10 = 1A

ΔH = CpΔT

For isothermal expansion

ΔT = 0

∴ ΔH = 0

Initial rate = k[A]² [B]

Rate k = [2A]² [B] = 4k[A]² [B]

Thus if the concentration of A, [A] is doubled and that of [B] is constant, then initial rate increases by a factor of 4.

Dettol is mixture of

Hence compound ‘B’ is Terpineol.

The molecular dimensions of H₂O₂ in the gaseous phase is shown in the following figure.

Thus, the bond angle H — O — O in H₂O₂ in gaseous phase is 94.8º

Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds. According to this theory the primary valency gives the oxidation number and the secondary valency gives the coordination number. Also, the geometry of the complex is determined by the number and position of secondary valences in space as the ligand satisfying secondary valences are always Greeted towards the fixed position in space.

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]

Following electrode reactions are possible:

In this electrolysis, H₂ at cathode and Cl₂, at anode are given off.

∴ 2(+1) + 2(x) + 7(-2) = 0

2 + 2x - 14 = 0

2x = 12 x = +6

The given equation is

⇒ 2x + 3 = 7; 2y − 4 = 14

⇒  2x = 4; y = 9

⇒ x = 2; y = 9

∴ x + y = 2 + 9 = 11

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

f(x) = [x] - x
= x - {x} - x
= -{x}
Range of {x} = [0, 1)
Range of -{x} = (-1, 0]

Any parametric point on the ellipse can be taken as P(acosθ, bsinθ).

Now if midpoint of PS is (h, k) then

Now, to find the locus we need to eliminate θ,

Using sin²θ + cos²θ = 1

Which is also an ellipse 
Now,
Area of the ellipse 
Therefore, ratio = 1/4

Let

Feasible solutions are

Clearly, the solution region is the quadrilateral.

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

It is given that
P(AUB)=0.6 and P(AB) = 0.2
Therefore

A tautology is a compound statement which always results in truth value.

p ⇒ q is logically equivalent to ~q ⇒ ~p~.

∴ (p ⇒ q) ⇔ (~q ⇒ ~p) is a tautology but not a contradiction.

Given:

then,

The given family of lines can be represented by the equation,
xcosα + ysin α = 10               ...(1)
where ​α is an arbitrary constant.
Differentiating we have,
cosα + sinαy₁ = 0                  ...(2)
Multiplying (2) by x and subtracting it from (1),

⇒ (y − xy₁)sinα = 10          ...(3)
Multiplying (1) by y₁ and (2) by y and subtracting,
xy₁cosα − ycosα = 10y₁
⇒ (xy₁ − y)cosα = 10y₁...(4)
Squaring and adding (3) and (4) we get, (y − xy₁)² = 100(1 + y₁²)
Now on comparing, we get A = 100.