AP EAMCET Mock Test - 4 - JEE MCQ

For steel wire: Y = 20 × 10¹⁰ Pa; L = 1.5 m ; d = 0.3 cm = 0.3 × 10⁻²m
Therefore, area of cross-section of the steel wire,

The stretching force for the steel wire,
F = 4.0 + 6.0 = 10 kgf = 10 × 9.8 N
If l is extension in the steel wire, then

For brass wire: Y = 9.0 × 10¹⁰ Pa ; L = 1m;

The stretching force for the brass wire,

F = 6 kgf = 6 × 9.8 N

Weight of the hanging portion of the chain = mg/3
Weight of the part of the chain on the table = 2mg/3

For the hanging part of the chain,

For the part of the chain on the table, we have

and we have

From 1, 2, 3 and 4, we have

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,

For A, T = f = 2mg

2mg − mg = ma₁

∴ a₁ = g

For B,

From the force diagram shown in the figure,

2mg − mg = 3ma₂

a² = g/3

For C,

∴ 3mg − mg = 2ma₃

⇒ a₃ = g

So, a₁ = a₃ > a₂

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).

Angle of prism = 30^o = A
Angle of incidence = i₁ = 60^o
Angle of emergence = i₂ = ?
Angle of devation = D = 30^o
∴ i₂ = A + D - i₂ = 30 + 30 - 60 = 0^o
⇒ r₂ = 0^o
∴ r₁ = A-r₂ = 30 - 0 = 30^o
∴ μ = sin i 1 sin r 1 = sin 60^o /sin 30^o = √ 3

Given that P(A′) = 0⋅3, P(B) = 0⋅4, P(A∩B′) = 0⋅5

P(B′) = 1 − P(B) = 1 − 0⋅4 = 0⋅6

P(A) = 1 − P(A′) = 1 − 0⋅3 = 0⋅7

P(A∪B′) = P(A) + (B′) − P(A∩B′)

= 0⋅7 + 0⋅6 − 0⋅5 = 0⋅8

⇒ 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

Projection of 

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

For the case when the number n is not a positive integer the binomial theorem becomes, for -1 < x < 1,

l = lower boundary point of median class
c.f. = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
n = total number of values or observations
h = class length of median class

3P(X = 0) = P(X = 1)

⇒ 1 - P = 11P
⇒ P = 1/12

⇒ 11³ - 11 = 1320

⇒

⇒

On integrating both sides, we get

log(x/y) = x ⇒ x/y = e^x

⇒ x = ye^x

Since, we know that cos²θ + sin²θ = 1

Put cosx = t, −sinx dx = dtcos,

* M₁ * M₂ * M₃ * M₄ * M₅ *
Five men can be seated in 5! ways.
The women can occupy any of the places marked *.
Thus, the women can be seated in ⁶P₄ = 360 ways.
 Total number of ways = 5! × 360 = 43200

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

The area of the region ODC is

So, the area of the given region OABC is equal to
Area of OABD – Area of OCD

First term = a
Common difference = (b - a)
Last term = 2a
=> Tn = 2a = a + (n - 1)(b - a)

Therefore, maximum value of xy is