UGEE SUPR Mock Test- 5 - JEE MCQ

where L = length of an organ pipe and that of closed organ pipe is

So, the ratio of p^th overtone of open to closed.

where q = charge of the cyclotron,

B = intensity of the magnetic field,

R = radius of orbit

And the m = mass of the particle,

Hence, E_K is independent of frequency of revolution.

where, m = magnification,

f_ο = focal length of objective

and f_e = focal length of eyepiece

⇒ f_ο = 9f_e …(i)

Also, distance between objective and eyepiece

⇒ f_ο = f_e = (given)

⇒ 9f_e + f_e = 20 ⇒ f_e = 2 cm

⇒ f_ο = 9f_e = 18 cm

μ_e = i x A

where i = current and A = area.

where T = time period of revolution and r = radius orbit

where, v = velocity of revolving electron

⇒ μ_e = evr/2

Hence, the magnetic moment of a revolving electron is evr/2.

This is because, more the frequency of incident light, more will be the energy of the photons incident photons. (E = hv)

Thus, more will be the energy (kinetic) acquired from the electrons.

The more the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential).

T = mg - mω² cos λ

When the body is suspended at the equator, then λ = 0 and r = R

∴ From Eq. (i), we have

where, T = tension in string

m = mass/length

and l = length of string.

New frequency,

⇒ 13v = 10v + 90

v = 30Hz

Increased length of wire

Since the cell is the same, therefore the new balancing length = 1/3 (new increased length)

= L/2

Therefore, for the same cell the balance point will be as length L/2 from positive end.

linear velocity = v

Length of string = Radius = r

If string is halved, r’ = r/2

Angular momentum. L = r x p

where , r = radius vector

and p = linear momentum

= r x mv

= mvt

As L remain constant,

mvr = constant

⇒ r' → r/2 , v' = 2v such that

mv' r' = m 2v x (r/2)

= m v r

Hence, linear velocity will be 2 v.

⇒

Since, fifth dark fringe is formed opposite to one of the slit.

∴ 2x₅ = d (distance between slits]

Put n = 5 in Eq. (i). we get

Current at output circuit of transformer, i₂ = 2A

V₂ = 440V

Hence, the current drawn by the primary windings is 5A.

speed of electron of the hydrogen atom in the ground state,

v₁ = 2.2 x 10⁶ m/s

Since

∴

[for third excited state. n₂ = 4]

⇒

5.5 x 10⁵ m/s

(A + B) · (A - B) = 0

A·A + B·A - A·B - B·B = 0

|A| + BAcosθ - AB cosθ - |B|² = 0

|A|² = |B|²

Hence. |A| = |B|

Time period of simple pendulum inside a stationary lift

..(i)

where, l = length of string and

g = gravitational acceleration

Acceleration of lift in upward direction,

a = g/3

∴ Time period of the pendulum

i.e., F_d ∝ v ⇒ F_d = bv

where, b is a damping constant.

∴ Damping constant. b = F_d/v

SI unit of damping constant

Hence, the SI unit of damping constant is kg/s.

Stopping potential = V

If λ₀ be the threshold wavelength, then the maximum kinetic energy of emitted electrons.

Again, the wavelength of used light

λ' = 2λ

Stopping potential V' = V/3

Then

⇒

From Eqs (i) and (ii), we have

⇒ λ0 = 4λ

So, the threshold wavelength is 4 times of wavelength of light.

If g be the value of gravitational acceleration on the surface of the earth, then weight of body mg = 300N

If g' be the gravitational acceleration at the bottom of the hole, then

∴ Weight of the body on the bottom erf hole,

mg' = mg/2 = 300/2 = 150N

The monomers used in the preparation of dextron are lactic acid and glycolic acid. There action for the formation of dexron can be written as

The enzyme which converts maltose to glucose is maltase. During digestion, starch is partially transformed into maltose by pancreatic or salivary enzymes called amylases; maltase secreted by the intestine men converts maltose into glucose. The glucose so produced is either utilised by the body or stored in the liver as glycogen.

Atomicity of heteronuclear molecule = total number of atoms present

∴ Atomicity of AlPO₄ = 1 +1 + 4 = 6

Formula for osmotic pressure, π = CRT

considering the set given in option (d), i.e. 30 gL^-1 urea and 17.1 gL^-1 sucrose.

Given, the molecular mass of urea 60 g mol^-1 and molecular mass of sucrose 342 g mol^-1.

For urea.

conc, C = 3/60 = 1/20

Osmotic pressure π₁ = (1/20) x R x T

For sucrose conc, C = 17.1/342 = 1/20

∴ Osmotic pressure, π₂ = (1/20) RT

Thus, the set of solutions of urea and sucrose given in option (d) is isotonic.

2MnO₂ + 4KOH + O₂ → 2K₂MnO₂ + 2H₂O

α = 1-i/n'-1

where, n is the number of ions formed alter dissociation.

The relationship can be obtained as follows;

For the reaction, A ⇌ n'B

Initially 1 mole 0

After dissociation (1 - α) mole n' α

Total number of moles present in the solution

= (1 - α) + n'α = 1 + (n'-1)α

van't Hoff factor, i = 1 + (n' - 1), α > 1 if n' ≥ 2

∴ α = i-1/n'-1

Iodide ion acts a homogenous catalyst as the phase of I^- and reactant, H₂O₂ is aqueous (same).

⇒ 2 moles of sodium metal reacts to give 1 mole of ethane.

Also,

1 mole of Na metal = 23 g

∴ 2 motesol Na metal = 23x 2 = 46g

Thus, 46 g of sodium is required to prepare one mole of ethane from methyl chloride by Wurtz reaction.