UGEE SUPR Mock Test- 3 - JEE MCQ

Let ρ the density of weight.

σ be the density of liquid.

and V be the volume of the weights.

From (1) and (2)

From (1) and (3)

where T₁ and T₂ are the temperatures of a source and sink respectively.

When T₁ and T₂ both are decreased by 100 K each, (T₁ - T₂) stays constant. T₁ decreases.

∴ η increases.

Poles are not actually at the end of the magnet they are little inside from the end this makes the magnetic length of the magnet little less than the Geometric length.

Geometric length is the actual length of the magnet.

So when the relationship between these two measured it comes to be,

Magnetic length = 0.8× Geometric length

If the effective decay constant is λ, then

Now,

So if the velocity of the electron increases, the de Broglie wavelength decreases.

∴ The electric flux, through any face

∴ μ = (1/sinθ) …(i)

When a ray of light incident from the air on the glass with the angle of incidence θ. If r be the angle of refraction, then by Snell’s law,

μ = sinθ/sinr

⇒ [fromEq. (i)]

⇒ sinr = 1/μ²r = sin^-1(1/μ²)

According to Stefan's law, the energy emitted by a body per unit area per second is proportional to the fourth power of the absolute temperature.

E = σT⁴where E = energy emitted area/second,

T = absolute temperature in kelvin

and σ = Stefan's constant ⇒ σ = E/T⁴

Unit of Stefan's constant

⇒ J/m²s/K⁴ = J/m²sK⁴Dimensions of Stefan’s constant,

Option (c) characteristics of the solar cell.

Option (d) → reverse bias characteristics 1 drawn in the first quadrant can be photodiode or Zener diode.

I_net = 10 - 13 + 10 + 7 - 12 + 18 = 20A

We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by

Where, i = current in wire and r = perpendicular distance.

B = 40μT

Where, F = force and q = electric charge

Dimensions of E or [E] = [MLT^-2]/[AT] = [MLT^-0A^-1]

Power of mass = 1

Electric dipole moment, m = l x A

Where, l = current and A = area

Dimensions of m = [M] = [A L²] = [M⁰L²T⁰A]

Hence, power of mass = 0

The Poisson’s ratio for homogeneous isotropic material must be between - 10 to + 05 because of the requirement. For Young’s modulus, the shear modulus and bulk modulus have positive values. Hence, 0.8 cannot be the value of Poisson’s ratio for homogeneous isotropic material.

∴ Area,

∴ Magnetic dipole moment associated with circular coil,

μ_c = il²/4π …(i)

When metal wire is bent into a square coil then side of square a = l/4

∴ Area, A = a² = l²/16

∴ Magnetic dipole moment associated with square coil, μ_c = iA = i(l²/16)

∴

Hnce, the ratio of magnetic dipole moment of circular coil and square coil is 4 : π.

Magnetic induction at the centre O due to straight wire,

Net magnetic induction at O,

A battery depolarizer absorbs electrons during discharge of a cell, so it acts as an oxidising agent.

In dry cells, manganese dioxide act as a depolariser.

Thermal stability is the stability of the molecule towards heat. Thermal stability of hydrides decreases down the group, as on going down the group, the size of the element E (side element with hydrogen) increases. So, the E−H bond length increases, hence- bond strength becomes weak and breaks easily to liberate hydrogen gas.

Thus, the order of thermal stability of hydrides is NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

he reagent metal/Hydrogen is used in the catalytic hydrogenation. It is a syn addition to the pi-bond.

Hence, the reactivity depends on the steric hindrance of the groups present in the molecule.

The double bond present in

is less sterically hindered than

Lime treatment removes bicarbonate hardness by forming insoluble CaCO₃ as,

2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO^2-₄ + 4H⁺

When SO₂ reacts with H₂S, it produces sulphur and water. In this reaction, ‘SO₂’ act as an oxidising agent.

Following electrode reactions are possible:

In this electrolysis, H₂ at cathode and Cl₂, at anode are given off.

Given, f(xy) =

Also, f(500) = 3

⇒ f(100 × 5) = = 3

⇒ f(100) = 15

∴ f(600) = f(100 × 6)

=

= 2.5

Let

Then

∴ Im(z) = 0, Re(z) < />

Let n, n + 1, n + 2 be three consecutive numbers.

Then, P(n) = n³ + (n + 1)³ + (n + 2)³

We prove by induction on n.

P(1) = 1 + 8 + 27 = 36, which is divisible by 9.

Let the result be true for k, i.e. P(k) = k³ + (k + 1)³ + (k + 2)³ is divisible by 9.

For k + 1, we have

P(k + 1) = (k + 1)³ + (k + 2)³ + (k + 3)³

= (k + 1)³ + (k + 2)³ + k³ + 9k² + 27k + 27

= k³ + (k +1)³ + (k + 2)³ + 9(k² + 3k + 3) = P(k) + 9(k² + 3k + 3), which is divisible by 9.

Hence, P(n) is divisible by 9 for all natural numbers.

nth term = n(n + 1)(n + 2) = n(n² + 3n + 2) = n³ + 3n² + 2n. Let the sum be S, then

S =

S = [{n(n + 1)/(2)}² + 3n(n + 1)(2n + 1)/6 + 2n(n + 1)/2]

S = n(n + 1)[n(n + 1)/4 + (2n + 1)/2 + (1)]

S = n(n + 1)[(n² + n + 4n + 2 + 4)/4]

S = n(n + 1)(n + 2)(n + 3)/4

= -1/2

Let PQ be the focal chord of the parabola. Let the coordinates of P and Q be (at₁², 2at₁) and (at₂², 2at₂), respectively.

Then, t₁t₂ = -1

Let (h, k) be coordinates of the mid-point of PQ. Then,