KIITEE Mock Test - 1 - JEE MCQ

Total flux linked with the coil at time t = 0 is

Flux linked with the coil at time t = 1 is

Now, induced emf = Change in flux in unit time

The incident ray OA is refracted along BD by the lens. Singe the image coincides with the object. the refracted ray BD must retrace its path after reflection from the mirror, i.e., it falls normally on the mirror.
Let C be the point on the axis where BD produced meets the axis. It is clear that C is the centre of curvature of the mirror. Also is the imago of the object O formed by the lens alone. Thus
f = +10 cm and u = - 12 cm
The distance v of the image is obtained from the lens formula

which gives v = - 60 cm
Since x = 10 cm. R = v - x = 60 - 10 = 50 cm.

Hence, the focal length of the mirror = R/2 = 25cm

MAss number of daughter nucleus (M) = 208- 4 = 204
Now, total energy of distintegration = energy of daughter nucleus + energy of α-particle

PE = -2 KE

if ω is the angular speed acquired by the coil when a charge Q is passed through it for very short time Δt, then 

Actual molecular weight of naphthoic acid (C₁₁H₈O₂) = 172
Molecular mass (calculated) 


= 0.5

= 1.10 - 0.0295
= 1.0705 V

Possible excited states for an electron in a Bohr orbit of a hydrogen atom can be found using the formula for the energy levels of the hydrogen atom:

En = -13.6 eV / n^2

where En is the energy of the nth level, n is the principal quantum number (n=1, 2, 3, ...), and -13.6 eV is the ground state energy (n=1).

For an excited state, n > 1. Let's find the energies for the next few levels:

- For n = 2 (first excited state):
E2 = -13.6 eV / 2^2 = -13.6 eV / 4 = -3.4 eV

- For n = 3 (second excited state):
E3 = -13.6 eV / 3^2 = -13.6 eV / 9 ≈ -1.51 eV

- For n = 4 (third excited state):
E4 = -13.6 eV / 4^2 = -13.6 eV / 16 = -0.85 eV

These are the possible excited states for an electron in a Bohr orbit of a hydrogen atom. Note that as the principal quantum number, n, increases, the energy levels get closer together and approach zero but never become positive. This is because positive energy would correspond to an unbound electron, which is not part of the hydrogen atom's energy levels.

In zero-valent metal complexes, central metal atom has zero oxidation state.
(a) [Cu(NH₃)₄]SO₄
Oxidation number of Cu:
x + (4 x 0) - 2 = 0
x - 2 = 0
x = +2
(b) Oxidation number of Pt in [Pt(NH₃)₂Cl₂]:
x + (2 x 0) + 2 x (-1) = 0
x - 2 = 0
x = +2
(c) Oxidation number of Ni in [Ni(CO)₄]]:
x + (4 x 0) = 0
x = 0
(d) Oxidation number of Fe in K₃[Fe(CN)₆]:
3 x (+1) + x + 6 x (-1) = 0
3 + x - 6 = 0
x = +3
 [Ni(CO)4] is a zero-valent compound.

V.P does not depend on the amount of water.
The volume occupied by water molecules in vapour phase is 1 × 10^-4 dm³ which is approximately equal to 1 x 10^-3 m³(volume of the flask)
Ideal Gas Law -
PV = nRT
where
P - Pressure
V - Volume
n - No. of Moles
R - Gas Constant
T - Temperature

PV = nRT

Diethyl ether, when heated with CO at 150^oC and 500 atm pressure in the presence of BF3, forms ethyl propionate.

Mn in [Mn(H₂O)₆]²⁺ is present as Mn²⁺(3d⁵).

sp³d² hybridisation occurs in [Mn(H₂O)₆]²⁺
Due to the presence of a weak field ligand, pairing of electrons do not take place. Hence, the number of unpaired electrons is five.

Nylon is a polymer which contains nitrogen. It contains amide linkage as shown below.

Put x = 1
Then, 

l = lower boundary point of median class
c.f. = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
n = total number of values or observations
h = class length of median class

When number at unit place is 3, then other 3 numbers can be arranged in 3! ways.
∴ The sum of the digits in unit place when 3 is their at unit place = 3!x3...(1)
Similarly.
the sum of the digits in.... when

from (1), (2), (3), (4)
∴ The sum of the digits in the unit place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time is

Given: ∑x = 18 and ∑x² = 30
n = 12
Now,

= = e⁴
as x  
 0 and

For the case when the number n is not a positive integer the binomial theorem becomes, for -1 < x < 1,

According to the question, L₁ and L₃ are not parallel.

So the required values of k are given by

or 5k² + 51k + 54 = 0