AP EAMCET Mock Test - 9 - JEE MCQ

The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, i.e.,

where T₂ is temperature of sink,
and T₁ is temperature of hot reservoir.

Using Kirchoff's law for the loop BADB, we get

And we have, I₁ = I₂ + I₃ ---- 3

From 1, 2 and 3, we have

I3 = -1/13 A, which means B is at high potential as compared to A

Hence, potential difference across B and D = 2/13 V

The Boolean expression for the given combination is

output Y=(A+B).C
 
The truth table is

Hence A=1, B=0, C=1

Decrease in GPE of B₁ = Increase in Gravitational PE of B₂ + Increase in elastic PE of spring. 

Again applying law of conservation of mechanical energy at the mean position we get 

Change of momentum of one bullet = m(v - u)
= 0.03 x {50 - (-30)}
= 2.4 kg ms^-1
Average force = rate of change of momentum of 200 bullets
= 200 x 2.4 = 480 N,
which is choice (4).

We have βγ + γα + αβ = 0. We can write Δ as:
Δ = 
= 
[Using C₁  C₁ + C₂ + C₃]
=  = 0 [all zero property]

2 sin² x + 3 cos x = 0
⇒ 2(1 - cos² x) + 3 cos x = 0
⇒ 2 - 2 cos² x + 3 cos x = 0
⇒ 2 cos² x - 3 cos x - 2 = 0
⇒ 2 cos² x - 4 cos x + cos x - 2 = 0
⇒ 2 cos x(cos x - 2) + 1(cos x - 2) = 0
⇒ (cos x - 2)(2 cos x + 1) = 0
Either cos x - 2 = 0 or 2 cos x + 1 = 0
But cos x - 2 = 0
i.e. cos x = 2, which is not possible.
Now, from 2 cos x + 1 = 0, we get:
cos x = -1/2
⇒ cos x = cos ()

Therefore, general solution of the equation is:

Given, |Z−3−4i|² + |Z+2−7i|² + |Z−5+2i|²
Put, Z=x+iy
=|(x−3)+(y−4)i|² + |(x+2) + (y−7)i|² + |(x−5)+(y+2)i|2
=(x−3)2+(y−4)² + (x+2)² + (y−7)² + (x−5)2+(y+2)2
=3[x² − 4x + 4 + y² − 6y + 9] + 68
=3[(x−2)² + (y−3)²] + 68
The minimum value of 3[(x−2)²+(y−3)²] + 68 is possible at x = 2 & y = 3
∴a = 2, b = 3, λ = 68
⇒ a + b + λ = 73

Area of parallelogram = 
On comparing with given equations,
a = 1, b = 2, c = 3 and c' = 5/2
a₁ = 3, b₁ = 4, d = -10 and d' = - 5
Thus, area of parallelogram =  sq. units
=  sq. units = 5/4 sq. units

Number should be greater than 7000 and divisible by 5.
Digits that has to be used are 3,5,7,8 &9,  no digit being repeated.

The number to be divisible by 5 it should end with 5 in this case.

Therefore, four-digit numbers =3⋅3⋅2⋅1=18

Five-digit numbers =4⋅3⋅2⋅1⋅1=24
Hence, the total required numbers are 42.

Given,

Put,

x + y = v

⇒ dx + dy = dv

Then, (i) becomes

This is a linear differential equation 
Hence, required solution is

Now,

Putting in (ii), we get

Substituting the values of A and B, we get

According to the question, L₁ and L₃ are not parallel.

So the required values of k are given by

or 5k² + 51k + 54 = 0

The equation has imaginary roots when b² - 4ac < 0.
Substituting a = 2 and b = 3 in b² - 4ac < 0,
3² - 4(2)c < 0
or 9 - 8c < 0
or 8c > 9
or c > 9/8
Thus, the roots are imaginary when c > 9/8

Comparing it with the general equation of hyperbola, we get
a² = 6 and b² = 3

Eccentricity for hyperbola is given by

Intersection point P of lines 

Inclination of reflected ray is 180°−60°=120°

⇒  Equation of reflected ray is

f(x) = x³ + 2x² + 2x − 1
f′(x) = 3x² + 4x + 2
D = (4)²−4(3)(2) < 0
⇒ f′(x) > 0 ∀x ∈ R
⇒ f(x) is increasing function in x∈[0,4]
⇒ Minimum of f(x) = f(0) = −1
& maximum of f(x) = f(4)
⇒ f(4) = 64 + 2(16) + 2(4) − 1
= 64 + 32 + 7
= 103
⇒ f(4) + f(0) = 102

Here μ₁ ​=30, μ₂​ = 25

and C_V1​​ = 50, C_{V2 ​​}= 60

⇒ μ₁​σ₁ ​​× 100 = 50, μ₂​σ_2​​ × 100 = 60

⇒ σ₁​=15,σ₂​ = 15 ∴ σ₁​−σ₂​=0

In sec x is increasing and it has value between 11 and √2.
Hence range is [1, √2].