JEE Main Mock Test - 1 - JEE MCQ

On removing an electron, the Helium atom becomes a single electron atom.
Applying Bohr's model, the energy of the single electron is given by 
(where Z is atomic number and n is number of the shell in which the electron resides).
Thus, the energy required is:

Energy required to remove the electron from singly ionized Helium atom = 54.4 eV

Energy required to remove the electron from Helium atom = x eV

So, 54.4 eV = 2.2x

x = 24.73 eV

Total energy required to ionize the Helium atom completely = 54.4 + 24.73 = 79.13 eV

Hence, the nearest option is 'C'.

Length , l = 4,234 m
Breadth, b = 1.005 m
Thickness , t = 2.01 x 10^-2 m
Volume = l x b x t
⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m³ (significant figure = 3)

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL² / 3 α (∵ I_rod = ML² / 3)
Hence, angular acceleration, α = 3g / 2L

Explanation:Uniform motion is the kind of motion in which a body covers equal displacement in equal intervals of time. It does not matter how small the time intervals are, as long as the displacements covered are equal.

If a body is involved in rectilinear motion and the motion is uniform, then the acceleration of the body must be zero.

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get 
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s₃ = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s₄= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7

B = B₀sin(π × 10⁷C)t + B₀sin (2π × 10⁷C)t
Since there are two EM waves with different frequency, to get maximum kinetic energy, we take the photon with higher frequency.
B₁ = B₀sin(π × 10⁷C)t  v₁ = 10⁷/2 x c 
B₂ = B₀sin(2π × 10⁷C)t v₂ = 10⁷C
where C is speed of light C = 3 × 10⁸ m/s
v₂ > v₁
so KE of photoelectron will be maximum for photon of higher energy.
v₂ = 10⁷C Hz
hv = φ + KE_max
energy of photon
E_ph = hv = 6.6 × 10^-34 × 10⁷ × 3 × 10⁹
E_ph = 6.6 × 3 × 10^-19J

KE_max = E_ph - φ
= 12.375 - 4.7 = 7.675 eV ^_≈ 7.72 eV

Work done in isothermal process on the gas,

For the prism, angle of deviation is given by:

Hence, emergent ray will be perpendicular to the second face or we say that the angle made by the emergent ray is 90°.

Energy of the emitted photon,

where E_A and E_N are energies of photons from atom and nucleus, respectively.

E_A is order of eV and E_N is order of MeV

By law of distribution of Boolean Algebra, we have A+ (B. C) = (A+B). (A + C)
Step 1: Write outputs for the OR gates with inputs W and X, and W and Y.
Input for first OR gate are W and X. The output for this OR gate is, Y₁ = W + X
Input for second OR gate are W and Y. The output for this OR gate is, Y₂ = W + Y
Step 2: Write output for AND gate whose inputs are the output of OR gate.
F = Y₁ • Y₂ ⇒ F = (W + X). (W + Y)
By using Law of distribution of Boolean Algebra, A + (B. C) = (A+B). (A + C)
Therefore, F = W + (X. Y)

Let emf of the cell be E. Current through the voltmeter (when connected between 1-2) is 
Current through R₁ = 4/R₁
∴ Current through 
∴ Potential difference across 
∴
When the voltmeter is connected between 2 - 3
Current through voltmeter = 6/200 A
Current through R₂ = 6/100 A
∴ Current through

∴
From (1) and (2) 
Put this in 
When connected across 1-3, the voltmeter will read E = 12 V.

Given, Z 50, A - Z = 120 - 50 = 70
Δm = Z.m_p + (A - Z) m_n - M= [50 × 1.007825 +70 × 1.008665 - 119.902199] = 1.095601 MeV
E = 1.095601 × 931.478 MeV = 1020.53 MeV ≈ 1021 MeV

Let radius of circular orbit of the Earth and Venus be r_e and r_v respectively (r_c/r_v) = (365/220)² [Kepler's third law] 

From the drawing given in the problem M'N'/MN = N'V/NV

Explanation: 

• From the given graph we can see that the potential energy is minimum at point B and D
• Whereas it has maximum potential energy at point A and C respectively
• Hence point A and C will be an unstable state since they have maximum potential energy compared to B and D.

Force on the particle is 
This force is always perpendicular to the velocity. Since deflection is small, the force is nearly in (↑) direction always.
Impulse is: 

By Bernoulli's equation: mgd = 1/2 mv²
v = √2gd ........... (1)
By Projectile motion equation: 1/2 gt² = h

Therefore, 
So, d = R²/4h

= 

= 22 or 22 %

Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance x from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
[ m(1.27 – x) + 35.5mx ] / (m + 35.5m)  =  0
m(1.27 – x) + 35.5mx =  0
1.27 - x = -35.5x
∴ x = -1.27 / (35.5 - 1)  =  -0.037 Å
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.

Let t be the time taken to cover the horizontal distance of 500 m from the gun to the target ,then
t = 500 / 5000 = .1 s
During this time,bullet will fall down vertically due to acceleration due to gravity.
Vertical distance covered
h = 1/2gt² = .005 m
= 5cm.
Hence the gun should be aimed at 5cm above the target

The amplitude of the magnetic field of an electromagnetic wave in a vacuum, B₀ = 510nT = 510 x 10^-9 T

Speed of light in a vacuum, c = 3 x 10⁸ m/s

The relation between the amplitude of the electric field and the magnetic field is c = E₀/B₀

E = cB_o = 3 x 10⁸ x 510 x 10^-9 = 153 N/C

A radio can tune to minimum frequency, v₁ = 7.5 MHz = 7.5 x 10⁶ Hz

Maximum frequency, v₂ = 12MHz = 12 x 10⁶ Hz

Speed of light, c = 3 x 10⁸ m/s

Corresponding wavelength for v₁ can be calculated as: 
Corresponding wavelength for v₂ can be calculated as: 

V² = 9.56 × 10¹²
V = 3.09 × 10⁶ m/sec.

Electronic configuration of Cr atom (z = 24) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5

sp² hybridisation gives three sp² hybrid orbitals which are planar triangular forming an angle of 120° with each other.
The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls²2s²2p⁶
B ls²2s²2p⁶3s²3p³
C ls²2s²2p⁶3s²3p^s 

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show S_N reactions; Cl directly attached (C=C) bond, i.e. vinyl group.