AP EAMCET Mock Test - 4 - JEE MCQ

x = a sin t ...(i)
y = a(1 - cos t) ...(ii)
x² = a² sin² ωt = a²(1 - cos² ωt)
x² = a² - a² cos2 t
a² cos² ωt = a² - x² ...(iii)
y = a(1 - cos t) = a - a cos t
a - y = a cos t
(a - y)² = (a cos t)²
a² + y² - 2ay = a² cos² ωt ...(iv)
Put the value from equation (iii),
a² + y² - 2ay = a² - x²
x² + y² - 2ay = 0
This is the equation of a circle.

Loss in energy when two capacitors are connected in parallel is

Let a and α be the linear and angular accelerations of the sphere respectively.
For translational motion,
F + f = Ma ...(1)
The magnitude of the net torque acting on the sphere = FR - fR.
Hence, for rotational motion the equation is

• FR - fR = Iα = Ia/R (∵ a = αR)

For a hollow sphere, I = 2/3 MR². Hence
FR - fR = 2/3 MR² ×  MRa
F - f = 2/3 Ma ...(2)
From Eqs. (1) and (2) we get f = .
Hence the correct choice is (4).

Let the velocity of ball I and ball III after collision be v₁ and v₂.
v₂ − v₁ = 0.5 x 10  ...(i)
mv₂ + mv₁ = m x 10   ...(ii)
⇒   v₂ + v₁=10
Solving equations (i) and (ii),

v₁=2.5 m s⁻¹ 
v₂=7.5 m s⁻¹

Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken between two consecutive collisions,

Charged particle can be considered at centre of a cube of side a, and given surface represents its one face. So, flux 

Mass of water in first tube is

Now, surface tension 
where h' is the height to which water rises in the second tube and r' its radius.
Since r' = 2r, h' = h/2.
Therefore, the mass of water in the second capillary tube is

= 

At t = 1 sec,

The electric flux is given by the surface integral. Here, the electric field E is due to charge inside the Gaussian surface only. Hence, the correct option is (d).

⇒ 3 + 5 + 7 … + (2n + 1) = 440

⇒ n/2[2 x 3 + (n-1)(2) = 440

⇒ n(3 + n - 1) = 440

⇒ n(n + 2) = 440

⇒ n = 20

The distance between 4x + 2y + 4z- 16 = 0 and 4x + 2y + 4z + 5 = 0 is 

Option (1) represents set A.
Option (2) represents set A - B.
∴ Only option (3) represents set A ∩ B.

Subject to contracts, 3x₁ + 2x₂ ≥ 9, x₁ - x₂ ≤ 3, x₁ ≥ 0, x₂ ≥ 0

On taking given constraints as equation, we get the following graphs

Here, we get feasible region is unbounded.

Let the remaining two observations be x and y. Since mean of 7 observations is 8.

⇒ 42 + x + y = 56

⇒ x + y = 14

Also variance (σ)2 = 16

⇒  (4 + 16 + 100 + 144 + 196 + x² + y²) = 80 × 7

⇒  x² + y² = 560 − 460 = 100

But   (x + y)² + (x − y)² = 2(x² + y²)

⇒ 196 + (x − y)² = 2 × 100

⇒ (x − y)2 = 4

⇒ (x − y) = ±2

(x + y) = 14 and (x − y) = 2

∴ x = 8, y = 6

If (x + y) = 14 (x − y) = −2

∴ x = 6, y = 8

So, the remaining two observations are 6 and 8.

Since, we know that cos²θ + sin²θ = 1

Put cosx = t, −sinx dx = dtcos,

The given equation is tan² 2x + 2tan2x tan3x = 1

So the given equation becomes

To find the solutions in the given interval, put n=0,1,2,3

Hence, four values in the interval 

Given, 2sec2α = tanβ + cotβ


Taking +ve sign, we have

3 pre-images from A can be selected in ⁶C₃ ways.

The rest of the 3 elements can be mapped in 3 x 3 x 3 ways.

The required number of functions = ⁶C₃ x 3³ =  540

Let C1: (x − 1)² + (y − 1)² = 2

C2: (x + 3)² + (y − 4)² = 52
Both the circle pass through the origin.

Hence, tangents at the origin (using T = 0) to C₁ and C₂ are x + y = 0 and 3x − 4y = 0, respectively.

The slope of the tangents are −1 and 3/4 respectively.

Thus, if  −1 < m < 3/4, then origin divides AB internally.

We know that

= 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

Let the roots be a, a²

Also a ⋅ a² = −k ⇒ k = −a³ = 2 ± √5

Since, the semi latus rectum of a parabola is the HM of segments of a focal chord.

∴ Latus rectum of the parabola =24/5