AP EAMCET Mock Test - 1 - JEE MCQ

N = F = 10N
mg = f = μN = 0.2 x 10
Weight = mg = 2 N

For the block with mass m, the force experienced is the spring force.
∴ Spring force = Mass × acceleration = ma
For block with mass M, forces are spring force and the external F

Referring to the figure, the forces acting on arms BC and AD are equal and opposite. The force on arm AB is given by

which is directed towards the wire. The force on arm CD is given by

which is directed away from the wire. Since F₁ > F₂, the loop will move towards the wire. Hence the correct choice is (3).

(i) Effect due to rotation: Acceleration due to gravity is maximum at poles because there is no effect of earth's rotation at poles which are situated on the axis of rotation. Also, the earth is flattened at the poles due to which the distance from centre of the earth decreases and cause an increase in gravity. 

Acceleration due to gravity is minimum at the equator because there is maximum possible effect of the earth's rotation.

So, as we move from the equator to the poles, the acceleration due to gravity g increases. 

(ii) Effect due to radial distance from centre: The acceleration due to gravity is maximum at earth's surface. It decreases on moving towards the centre and above the surface.

Of the given options, option (1) satisfies the above condition.

SbCl₅ metal halide is more covalent than other given options. It can be explained on the basis of Fajan’s rule. The higher the charge on the cation, the ,,greater is its polarising power. In SbCt_6, Antimony (Sb) is in +5 oxidation state, it has maximum polarising power. Hence, possess a the higher covalent character.

The enzyme which converts maltose to glucose is maltase. During digestion, starch is partially transformed into maltose by pancreatic or salivary enzymes called amylases; maltase secreted by the intestine men converts maltose into glucose. The glucose so produced is either utilised by the body or stored in the liver as glycogen.

All molecules or ions with free pair of electrons or at least one rt-bood can act as nucleophiles.

Among given species CN^-, H₂O, R—OH are nucleophiles.

At constant pressure, q_p = ΔU -W.

q_p = ΔU - pΔV

q_p = U₂ - U₁ - (-p(V₂ - V₁))

q_p = (U₂ + pV₂) - (U₁ + pV₁)

q_p = H₂ - H₁

q_p = ΔH

As we know, ΔH = ΔU + pΔV

ΔH = ΔU + W

Putting the value of ΔH in (1)

q_p = ΔU + W

Other corrected options are as follows:

• For adiabatic process ΔU = W

• For an isochoric process ΔU = q_v

• For an isothermal process q = -W

Given

Using Kohlrausch law, to find molar conductivity of acetic acid.

taking log on both sides, we get

y log x = (x - y) log e = (x - y)

when x = 1, then y (log 1) = (1 - y)

⇒ y = 1

On differentiating both sides,

when x = 1 the

⇒ dx/dt = 27 - 3t²

Since, the direction will be reverse,

∴ dx/dt = 0

⇒ 27 - 3t² = 0

⇒ t = 3

∴ Distance x = 2 + 27t - t³

= 2 + 27 x 3 - (3)³

= 2 + 81 - 27 = 56 units

We have equation of hyperbola

25x² - 9y² = 225

On comparing with we get

A² = 9, b² = 25

Therefore:

Let X be a random variable that denotes the amount received by the player.

Then.X can take values 5, 2, and 1.

Now, P(X = 5) = 1/4, P(X = 2) = 2/4 = 1/2 and P(X = 1) = 1/4

Thus, the probability distribution of X is

Put sin^-1 (0.4) = θ ⇒ sin θ = 0.4

∴ E = sin(3θ) = 3 sinθ - 4sin³θ

= 3(0.4) - 4(0.4)³

= 1.2 -4(0.064)

= 1.2 - 0.256

= 0.944

We know that, a_n = S_n - S_n-1

= 5(2ⁿ - 1) - 5(2^n-1 - 1)

= 5(2ⁿ - 2^n-1)

= 5(2^n-1)

Given, pair of lines, x² - 4pxy + 8y² = 0

Which is in the form of ax² + 2hxy + by² = 0

According to the question,

Sum of the slopes = 3 x product of the slopes

⇒ m₁ + m₂ = 3 x (m₁ m₂)

⇒

⇒ 4p = 3 ⇒ p = 3/4

Clearly, f(x)will be defined when

3 - x > 0 ⇒ x < />

and x - 2 ≥ 0 ⇒ x ≥ 2

∴ Domain of f(x) is [2 3).

Option (a) (q ∧ r) v (∼p ∧ s) = (T ∧ F) v (F ∧ F) = F v F = F

Option (b) (-P → q) ↔ (r ∧ s) = (p v q) → (r v s) (∵ p → q = -p v q)

Option (d) is similar to option (a).

We have

∴ The volume of the parallelepiped with AB, AC, and AD on the coterminous edges

= |2 (2 + 10) + 9(7 + 4) -1(35 - 4)|

= |2 (12) + 9(11) - 1(31)|

= |(24 + 99 - 31)|

= |92| = 92 cubic units

We have polar coordinates of P are (2, π/6). If O is the image of P about the X-axis

∴ Q = (2, π/6)

⇒ Q = (2, 11π/6)

It is given that P (X ≥ 1) ≥ 0.9375

= 1 - P(X = 0) ≥ 0.9375

= 1 - ⁿC_o(p^o)(g)^n-o ≥ 0.9375

= 1 - (¼)ⁿ ≥ 0.9375

= 1 - 0.9375 ≥ (¼)ⁿ

= 0.0625 ≥ (¼)ⁿ

= 625/10000 ≥ (¼)ⁿ

= 1/16 ≥ (¼)ⁿ

= 16 ≤ 4^th

⇒ n = 2

= a.[(b x a ) + (b x c) + (c x a) + (c x b)]

= a[(b x a) + ( b x c) + (c x a) - ( b x c)]

= a[(b x a) + (c x a)]

=[a b a] + [a c a] = 0 + 0 = 0