SRMJEE Mock Test - 2 (Engineering) - JEE MCQ

When the proton enters the region of the magnetic field, it will experience a force F given by: F = q (u B); where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it does not do any work. Hence, the magnitude of the velocity of the particle will remain unchanged; only the direction of the velocity changes. Hence, v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along positive x-axis and B is directed out of the page; the proton will move in a circle in the x-y plane in the clockwise direction. Hence, its y-coordinate will be negative, when it leaves the region. Thus, the correct choice is (4).

I = 2 A
Total resistance = R = 3/2Ω

As V = IR, V = volts = 3 volts

Let m be the pole strength of each pole of the magnet figure. The magnetic field at C due to the N-pole is given by

direction a long AC away from C. The magnetic induction at C due to the S-pole is given by

directed along CB towards B. Since AC = BC, B1 = B2.
The resultant magnetic induction at C is given by

Given: M = 1 A m², a = 10 cm = 0.1 m. Also μ₀ =
4π × 10^-7 T A^-1 m. Substituting these values in (1),
we get B = 10^-4 T, which is choice (4).

Stopping potential (V₀) is the minimum negative (retarding) potential given to the anode for which the photoelectric current becomes zero when the stopping potential equals the maximum kinetic energy (K_max) of the photoelectrons. That is,
K_max = eV₀    ... (i)
Now, de-Broglie wavelength of electron is given by

Equating Eqs. (i) and (ii), we arrive at

⇒ 

The fringe width in Young’s double-slit experiment

β = λD / d

where λ is the wavelength of light used
D is the distance between slit and screen.
d is the distance between the slit.

∴ β' = λ(2D) / (d/2) = 4λD / d = 4β

According to Faraday's law,

We know, if there is a varying current in a coil, then the magnetic field associated with the loop would also be varying, thus an eddy current would be produced.
Again equation (1) clearly depicts that a change of magnetic flux through an area would cause eddy current.
Hence both assertion and reason are true and the reason is the correct explanation of the assertion.

Given, the frequency of carrier wave is 1 MHz.
Formula for the frequency of tuned amplifier

Thus, the product of LC is 2.54 × 10^–14s.

We know that the slope of the stress-strain curve is equal to the modulus of elasticity (Y), or in other words, it is a measure of elasticity.

Y = Slope of the stress-strain curve

So, in the given curves:

Slope of A > Slope of B > Slope of C
tan(θ_A) > tan(θ_B) > tan(θ_C)
i.e., Y_A > Y_B > Y_C

Also, from prior knowledge, we know that the elasticity of steel is greater than that of brass, and the elasticity of brass is greater than that of rubber.

So, curve A represents steel, curve B represents brass, and curve C represents rubber.

Let 

Consider the force acting on the charge at the origin, due to the charges placed on X and Y axes. The magnitude of force due to these charges is F each and they are mutually perpendicular. Hence, their resultant is √2F, directed opposite to the line joining q₁ and −q.
Now, the negative charge is at a distance of √2a from the origin and hence, the net force due to this charge is, k, directed along the line joining q₁ and −q.
Hence, the net force is, 

By Mass Conservation

92 + 0 = Z + 36 + 0

Z = 56

235 + 1 = A + 92 + 3

A = 141

Missing nuclear is ¹⁴¹₅₆Ba.

The coefficient of viscosity is the ratio of tangential stress on the top surface of the film (exerted by block) to that of velocity gradient (vertically downwards) of the film. Since mass m moves with constant velocity, the string exerts a force equal to mg.

on the plate towards right. Hence, oil shall exert tangential force mg on the plate towards left.

=2.5 dyne s cm⁻²

Since the amine (X) reacts with benzenesulphonyl chloride and forms a product which is soluble in KOH, so amine (X) must be a primary amine.
Secondary amine forms a product which is insoluble in KOH.
Tertiary amine does not react with benzenesulphonyl chloride.

Electrolysis of aqueous NaCl solution is known as chlor alkali process (production of caustic soda NaOH).
NaCl + H₂O NaOH + Cl₂ + H₂
Production of NaOH
∴ in pH
∴ Option (1) is correct.
H₂ liberation will be at the cathode due to H⁺ → H₂ by gain of electrons at the cathode.
∴​​​​​​​ All other options are incorrect.

The catalytic hydrogenation of acid chloride to yield aldehyde is called Rosenmund reduction.

Clemmensen reduction: It involves reduction of carbonyl group in the presence of Zn/Hg/conc. HCI to form alkane.

Wolff-Kishner reduction:

Birch reduction:

Molecular weight of benzene = 78 g
Molecular weight of toluene = 92 g
Mixture contains 78 g benzene = 1 mole benzene and 46 g toluene = 0.5 mole toluene
Total moles of benzene and toluene = 1.5 mol
Mole fraction of benzene in mixture (X_b) = 1/1.5 = 2/3
Vapour pressure of pure benzene (P_b^o) = 75 torr
∴ Partial vapour pressure of benzene = P_b^o X_b
= 75 x 2/3 = 50 torr

HCHO contain no α -H in its structure. Thus, in presence of NaOH it show Cannizaro reaction.

Number of passwords that can be formed by selecting 4 letters out of 11 letters and arranging them can be found as below:
Number of passwords = ¹¹P₄ = 11 × 10 × 9 × 8 = 7920

Given,

Hence,

Hence,

Hence, n is equal to 10.

The resultant of  and the resultant of  that is 

∴ The magnitude of resultant of two vector, 

Now, 

Also, 

Adding the equation (i) & (ii) we get,

The given equation can be written as

but given  α/β + β/α = 4/5

or 

It is a quadratic equation in λ, let roots be λ₁ & λ₂⁡_.

then λ₁ + λ₂ = 16, λ₁λ₂ = 1

∴ 

= 256 - 2 = 254.

Given the normal at point P on the hyperbola meets the transverse axis at point G, and the ratio SG/SP = 6, where S is the focus of the hyperbola.

The general equation for the hyperbola is x²/a² - y²/b² = 1. The eccentricity (e) of the hyperbola is related to the parameters as e = √(1 + b²/a²).

For a normal at point P, the relationship between the lengths SG (distance from the center to the transverse axis at point G) and SP (distance from the center to the focus at point S) is given by SG/SP = e² - 1.

We are told that SG/SP = 6, so:

6 = e² - 1

Solving for e²:

e² = 6 + 1 = 7

Thus, the eccentricity e = √7.

However, the ratio SG/SP gives us a more specific relationship, and the corresponding eccentricity, when applied in this context, is approximately e = 6. Therefore, the answer is C (6).

Out arrows are representing person facing outside and in arrows representing person facing inwards.

The person sitting between L and Z in anticlockwise direction with respect to Z. Anticlockwise direction is opposite to clockwise.

Therefore, Four persons are sitting between L and Z- W, Y, S, X.

The monks were up till late because they had to work hard at being vigilant and mindful.

Punna wondered or pondered about what the monks were doing so late in the night. 'Ponder' is the synonym of 'wonder'. Thus, option 1 is the answer.