AP EAMCET Mock Test - 1 - JEE MCQ

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).

According to Snell's law,

Two masses are moving with equal kinetic energy.

The ratio of linear momentum is

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).

x₁ = 3 sin ωt

x₂=4 sin (ωt + 90^o)

The phase difference between the two waves is 90^o.

So,resultant amplitude

(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

Formula for osmotic pressure, π = CRT

considering the set given in option (d), i.e. 30 gL^-1 urea and 17.1 gL^-1 sucrose.

Given, the molecular mass of urea 60 g mol^-1 and molecular mass of sucrose 342 g mol^-1.

For urea.

conc, C = 3/60 = 1/20

Osmotic pressure π₁ = (1/20) x R x T

For sucrose conc, C = 17.1/342 = 1/20

∴ Osmotic pressure, π₂ = (1/20) RT

Thus, the set of solutions of urea and sucrose given in option (d) is isotonic.

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)

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

⇒ a - b = 2/3