JEE Main Mock Test - 12 - JEE MCQ

 will not effect electron and  will reduce electron velocity as electric field on it is in direction opposite to field.
So (I) is correct


So, (II) is correct
No net force acts on rectangular loop placed in magnetic field.
So (III) is correct.

The pressure due to surface tension = 4T / R

The pressure due to electrostatic forces = σ² / 2ε₀

Just before the bubble bursts,

4T / R = σ² / 2ε₀ or R = 8Tε₀ / σ²

Escape speed is given by

Now given for a planet (lets say A )

and for planet B,

Hence, v₂ = 48√2 m/s = v√2 ms⁻¹(given)
∴v = 48

rom work energy theorem
for upward motion,
1/2m (16)² = mgh + W
for downward motion
1/2m (8)² = mgh – W
h = 8 m

Magnetic field at origin due to part of wire parallel to Y−axis is only useful.
Magnetic field due to semi-infinite wire, B = μ₀I / 4πd
Here, 

W_g =  −ΔU
Work done by gravity = decrease in potential energy

= 

In the forward biased mode, we apply voltage in a direction opposite to that of barrier potential, that is, p-side to the positive terminal and n-side to the negative terminal of the battery. Thus, electrons in the n-side, holes in the p-side are pushed towards the junction which results in increased diffusion.
In reverse biased mode, we apply voltage in a direction of that of barrier potential. The electrons in n-side, holes in p-side pushed away from the junction that results in the drift current.

We know that,
p₁V = nRT₁
⇒ 1 × V = nR(35 + 273)
V = nR × 308 ...(i)
and
p₂V = nRT₂
⇒ 3 × V = nR T₂ ...(ii)
On dividing Equation (ii) by Equation (i), we get
3 = T₂ / 308
⇒ T₂ = 924 K
T₂ = 924 - 273
= 651 °C

Initial angular moment about 'O' is zero and also no torque about 'O'.
So ω = 0
∴   Not possible

If the point is at a distance x from water at 100 °C, heat conducted to ice in time t,


So ice melted by this heat

Similarly heat conducted by the rod to the water at 100 °C in time t,

So steam formed by this heat

According to the given problem, m_ice = m_steam , i.e.,

i.e., 200 °C temperature must be maintained at a distance 10.34 cm  from water at 100 °C.

If we consider break down in Zener diode, then potential across R₃ will be 10 V and R₁ will be 2 V.
So current in R₃ will be i₃ = 10 / 1500 = 2 / 300 A
and current in R₁ will be i₁ = 2 / 500 A
⇒ i₁ < i₃, which is not possible
⇒ The potential difference across the Zener diode does not reach to break down voltage. So no current will flow through the reverse-biased Zener diode.

Induced electric field,

Torque due to field about centre of ring,

The ring starts rotating when,
τ due to electric field = τ due to friction.
τ₁ = (μmg)r
Solving,

= 4 s

(A) No precipitate with K₂CrO₄ in acetic acid as its K_sp is high.
(B) Ca²⁺ + 2 K⁺ + [Fe(CN)₆]⁴⁻ ⟶ K₂Ca[Fe(CN)₆]↓ (white)
(C) It imparts a brick-red colour to Bunsen flame.
(D) Ca(HCO₃)₂ is formed, which is water-soluble.

Volatility is the tendency of a substance to vaporise. Volatile substances: Volatile substances have the capability to go into the vapour phase. 
The boiling point is defined as the temperature at which a liquid's saturated vapour pressure equals the atmospheric pressure surrounding it.
Volatile nature ∝ 1 / Boiling point
(CH3)3N→ There is no H-bonding, so the Boiling point will be less.
Hence, it is most volatile.

X_A′ = 0.675

Given that the atomic masses of He and Ne are 4 amu and 20 amu, respectively.
Given Temperatures:
T_He = −73°C = 200 K
T_Ne = 727°C = 1000 K
De Broglie Wavelength Formula:
λ = h / (mv)
From the kinetic theory of gases:
(1/2) mv² = (3/2) k_B T
where:
m = mass of the gas molecule
k_B = Boltzmann constant
Now,
mv = √(3mk_B T)
According to the question:
λ_He = M λ_Ne
Substituting values:

Solving for M:
M = √(25) = 5

We have, 

⇒ A = 2(π + 1)

Hence, f(x) = (π + 1) sin x + 2(π + 1)

∴ fₘₐₓ = 3(π + 1) = M

and fₘᵢₙ = (π + 1) = m

⇒ M / m = 3

Let, ∠ABC = ∠ACB = θ and the slope of AC = m.

Slope of BC = -1/2

Slope of AB = 3/4

Now, tan(∠ABC) = tan(∠ACB)

⇒ Equation of AC is x = 2a

Perimeter = 2(2α + 2 cos 2α)

P = 4(α + cos 2α)

dP/dα = 4(1 - 2 sin 2α) = 0

sin 2α = 1/2

2α = π/6 , 5π/6

d²P/dα² = -4 cos 2α

For maximum α = π/12

Area = (2α)(2 cos 2α)

= (π/6) × 2 × (√3/2) = π / (2√3)

x² + y² + 4x + 6y − 3 = 0
has centre C(−2,−3) and radius =4
∴r₂ = 4 + OC / 2 & r₁ = 4 − OC / 2

∴  Coefficient of x⁵ in the given expression
= Coefficient of x⁵ in 
= Coefficient of x⁵ in 
= ³¹C₆ − ²¹C₆

Given equation of hyperbola is

Equation of directrix will be x = ±a / e,
Where, 
i.e. k≥  −1
 is the equation of directrix


⇒ k = 2 as k = −3 is rejected
So equation of hyperbola will be 
∴  (√5,−2  satisfy the given hyperbola

Let

Let, 


Adding (1) and (2), we get

Exhaustive number of cases is 6³ = 216.
Now if a larger number appears than the previous number each time, all the three numbers are distinct.
Now three numbers can be selected from six numbers in ⁶C₃ ways and there is only one way in which three selected numbers can appear.
Hence, probability is 20 / 216 = 5 / 54

Clearly, f(x) = x − [x] = {x}
 which has period 1
Let sin(1/x) be periodic with period
then, 

Now for a variable x and constant T, the given relation cannot hold ∀ allowable x. Hence, sin1/x is not periodic.
Similarly, for f(x) = xcosx, let T be the period.
Hence, (x + T) cos(x + T) = xcosx

Note that LHS is a constant while RHS varies as x varies for allowable values of x. Hence, no such T is possible, so xcosx also non-periodic.

p : probability of getting head
q : probability of getting tail

Such that, p + q = 1

According to question we can get head in 2nd, 4th, 6th ... tail.

Hence required probability = qp(1 + q² + q⁴ + ... ) = 2/5

⇒ qp / (1 - q²) = 2/5

⇒ 5pq = 2 - 2q²

⇒ 5p(1 - p) = 2 - 2(1 - p)²

⇒ 5p - 5p² = 2 - 2(1 + p² - 2p)

⇒ 5p - 5p² = 2 - 2 - 2p² + 4p

⇒ 3p² = p

⇒ p(3p - 1) = 0

⇒ p = 0, p = 1/3

∵ p ≠ 0

∴ p = 1/3

⇒ 9p = 3

⇒ p = 3

⇒I = π² / 2
10I = 5π² = 49

Given, 
and 
Let 
So, 

On comparing we get,

On solving equation (i), (ii) and (iii) we get,

So,  = 29