UGEE SUPR Mock Test-2 - JEE MCQ

Given, angular momentum of electron in
H-atom = 3h/2π  .....(i)
From Bohr's postulate,
Angular momentum = nh/2π    ....(ii)
Comparing Eqs. (i) and (ii), we get
n = 3
The kinetic energy of an electron in hydrogen is given by

  (for H, Z = 1)
= 1.51 eV

As p-n junction diode conducts electricity during only positive half-cycle (forward biased condition). So, the potential difference across the capacitor is
V = Peak value of input AC voltage
= V₀ = V_rms × √2
Given, V_rms = 220V
∴ V = 220√2 V

The given circuit is as shown

The truth table for P can be shown as

The truth table for Q can be shown as

Thus, Q is 1 for any value of P. So, the values satisfying given condition are,
P = 0, Q = 1

At room temperature, some of the valence electrons acquire thermal energy greater than E_g (energy gap between conduction and valence band) and some of the valence electrons becomes free. Thus, we can say that a valence electron is shifted to conduction band leaving a vacancy of electron in valence.
This vacancy is known as hole and carries a positive charge.

The magnetic field in between because of each will be in opposite direction.

Let h be the height of platform, then speed of walking (v_g) = h/20
Speed of escalator (v_e) = h/30
Time taken by girl when she walk up on the movin escalator is

(a) Consider a string of length I connected to a particle as shown in figure

As the particle is moving with velocity v in uniform circular motion, so the net force must be equal to the centripetal force, which is provided by the tension in the string.
∴ Net force = Centripetal force (F_C)
= Tension in string (T)
⇒ 

Given, mass, M = 2 kg
I_x = 0.2 kg -m² 
and I_y = 0.3 kg -m²
According to the perpendicular axis theorem, the moment of inertia of the rectangular plate about an axis passing through O and perpendicular to the plane of plate is
I = I_x + l_y = 0.2 + 0.3
= 0.5kg m²
Also, 1 = Mk²
where, k = radius of gyration.
⇒ 

We have,
On putting x = tanθ
⇒ θ = tan^-1 x
Then we, get, 

On differentiating Eq. (i) both sides w.r.t. x, we get

∴  

We have, 
  ....(i)
On differentiating Eq. (i) both sides vv.r.t. x,
we get

Again differentiate w.r.t. to x, then we get


⇒ 
⇒

AM ≥ HM

For minimum value = 0.
Let and   are anti-parallel so

For (x) to be defined, we must have

The centre of the given circle is C (4, 2)
Let A ≡ (-3,2)

If (α, β) are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,

Thus, the coordinates of the other end of diameter are (11, 2)

Total number of ways = 6 × 6 × ..... to n times = 6ⁿ.
Total number of ways to show only to even number = 3 × 3 × ........ to n times = 3ⁿ.
∴ required number of ways = 6ⁿ - 3ⁿ.

We know that

Let end A is (a, 0) and end B is
(0, b) then 
a² + b² = 81  ....(1)
If centroid is (x, y) then

x = a/3, y = b/3, putting in (1) we get the locus as
x² + y² = 9

The given relation is xy + y² = tan x + y.
Differentiating both sides with respect to x, we get

Probability that the electronic component fails when first used is P(F) = 0.10. Therefore,
P(F') = 1 - P(E) = 0.90
Let E be the event that a new component will last for one year.

Fe³⁺(d⁵) has 5 unpaired electrons therefore magnetic moment
which is maximum among given options. As Sc³⁺ Ti³⁺ , Cr³⁺ and V³⁺ contain 0, 1, 3, and 2 number of unpaired electrons respectively.

The b.o inCl - O^- is 1
The b.o in O = Cl - O^- is 1.5

The bond length increases as b.o. decreases.

Cisplatin [Pt(NH₃)₂Cl₂] is used in cancer treatment.

Maleic acid and fumaric acids are geometrical isomers.

Acidic character of oxide ∝ Non-metallic nature of element.
Non-metallic character increases along the period. Hence order ofacidic character is
Cl₂O₇ > SO₂ > P₄O₁₀

More will be the electronegativity of X, lesser will be the bond length of X-O bond.

Given conditions
V₁ = 16.4 L, V₂ = 5 L
P₁ = 1.5 atm, P₂ = 4.1 atm
T₁  = 273 + 27 = 300 K,
T₂ = 273 + 227 = 500 K
V₁ = 1.5 atm, = 4.1 atm
Applying gas equation,