AP EAMCET Mock Test - 1 - JEE MCQ

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

Since there is no friction between the sphere and the horizontal surface and also between the spheres themselves, there will be no transfer of angular momentum from sphere A to sphere B due to the collision. Since the collision is elastic and the spheres have the same mass, the sphere A only transfers its linear velocity v to sphere B. Sphere A will continue to rotate with the same angular speed ω at a fixed location. Hence the correct choice is (3).

Total work done by the force F is the area under the force-displacement curve. This work is responsible to increase kinetic energy of the toy car.
Total work done by the force,

Hence, the kinetic energy of toy car at x = 11 m is,
Work done = Change in kinetic energy = 550 J
By energy conservation,
Potential energy gain by toy car at y_max = Kinetic energy at x = 11 m

(a) [Fe(H₂O)₆]Cl₃- cationic complexes

(b) [Ni(NH₃)₆]Cl₂ - cationic complexes

(d) K[Ag(CN)₂] - anionic complexes

CO(g) + (1/2)O₂(g) → CO₂(g), ΔH₂ = -Y, ….(ii)

For the formation of CO subtract Eqs. (ii) from (i), i.e,

∴ Δ_rH for formation of CO = ΔH₁ - ΔH₂

= -x+y or y-x

α = 1-i/n'-1

where, n is the number of ions formed alter dissociation.

The relationship can be obtained as follows;

For the reaction, A ⇌ n'B

Initially 1 mole 0

After dissociation (1 - α) mole n' α

Total number of moles present in the solution

= (1 - α) + n'α = 1 + (n'-1)α

van't Hoff factor, i = 1 + (n' - 1), α > 1 if n' ≥ 2

∴ α = i-1/n'-1

Let

put x = a sin² θ

⇒ dx = a(2 sinθ cos θ)dθ

when, x = 0, θ = 0 and x = a , θ = π/2

∴ k = πa

dθ/dt = -k(θ - θ₀), where k is constant

⇒ (dθ/dt) + kθ = kθ₀

Which is linear differential equation in the form of

(dy/dx) + Py = Q

IF = e^∫kt = e^kt

before the required solution,

(θ)(e^kt) = ∫(e^kt x kθ₀)dt

⇒ θe^kt = e^kt θ₀ + a

⇒ θ = θ₀ + ae^-kt

On adding Eqs. (i) and (ii), we get

Centre of circle (i) te (3, 2)

Equation of circle concentric with the circle (i) and touching the Y- axis is

(x - 3)² + (y - 2)² = (3)²

⇒ x² + 9 - 6x + y² + 4 - 4y = 9

⇒ x² + y² - 6x - 4y + 4 = 0

X is f (x) = 3(1- 2x²), 0 < x="" />< />

= 0, otherwise

= 179/864

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

On adding Eqs. (i) and (ii). we get

⇒ I = 7π/18

Put t = tan θ ⇒ θ = tan^-1 t

sin^-1 (sin θ) = θ = tan^-1 t

= cos^-1(cos θ)

= θ = tan^-1t

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)

Key Idea: Firstly find then Use the slope of normal

The slope of normal at

dy = e^xdx

Integrating on both sides, we get ∫dy = ∫e^xdx

⇒ y = e^x + C …(i)

On putting x = 0, y = 1 is Eq. (i), we get

1 = e⁰ + C ⇒ C = 0

Now, particular solution of the given differential is y = e^x

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

Let tan^-1x = t 
cot^-1x = π/2 - t


Max will occur around t = π/2
Range of f(t) = 

Plane p is ⊥ to line.

It passes through point (2, 3, -1).
Equation of plane p:
2(x - 2) + 1(y - 3) + 1(z + 1) = 0
2x + y + z - 6 = 0
Point (1, 2, 2) satisfies the above equation.

f(x) + f'(x) + f'' (x) = x⁵ + 64
f'(x) + f''(x) + f'''(x) = 5x⁴
f''(x) + f'''(x) + f^iv(x) = 20x³
f'''(x) + f^iv(x) + f^v(x) = 60x²
∴ f^v(x) - f''(x) = 60x² - 20x³
⇒ 120 - f''(1) = 40
⇒ f''(1) = 80
Also, f(1) + f'(1) + f''(1) = 65
⇒ f'(1) = -15

= 2(sinx)^x/2₀= 2(1 - 0)

= 2sq. Units

Let I = ∫logx[log(ex)]^-2 dx

Put logx = t ⇒ x = e^f

⇒ dx = e^fdt

= (e^f/1+t) + C

= (x/1+logx)+C

It is given that stapes of the lines passing through origin are m₁(let) = 1 + √2) and m₂(let) =   -1     = -(√2 - 1)
                                                                                                                                                  1+√2  

∴ Required joint equation of lines passing through origin is

[y-(1+√2)x][y+(√2-1)x] = 0

⇒y² + (√2 - 1)xy-(1 +√2)xy - (2 - 1)x² = 0

⇒Y² - 2xy - x² = 0

⇒X² + 2xy - y² = 0

⇒ a - b = 2/3