AP EAMCET Mock Test - 9 Free Online Test 2026

The wavelength of line in cased Baimer series is given by where n = 3,4,5 and R = Rydberg constant.

So, for the Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum, etc. Hence, the given transition represents the second line of the Balmer series.

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

...(i)

This can be shown graphically as

For small time intervals, i,e„ δt →O, then the angle between v₁ and v₂ is very small i.e., θ ≈ 0°. So. from Eq. (i), we get ∅ = 90⁰

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

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

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].