AP EAMCET Mock Test - 3 - JEE MCQ

Hence, work done by the centripetal force is zero, when it through (2/3)rd of the circular path.

Initial frequency, f₁ = 10Hz and final frequency, f₂ = 20Hz

Angular velocity for rotational motion is given by ω = 2πf

∴ ω₁ = 2πf₁ = 2π x 10 = 20π rad/s and ω₂ = 2πf₂ = 2π x 20 = 40π rad/s

According to work-energy theorem, work done in rotation = change in rotational kinetic energy

.

When 8 small drops are combined to form a bigger drop, then volume will be same.

If r₁ be radius of small drop and r₂ be the radius of biggordrop, then v₂ = v₁

∴ Terminal velocity of the drop

i.e., v ∝ r²

⇒ v₁/v₂ = 1/4

v₂ = 4v₁ x 10 = 40 cm/s

Torque = Force x Perpendicular distance

i.e., τ = F x r

∴ Dimensional formula of τ = [M L T^-2] [L] = [ML² T^-2]

P = ?_o?E

where, E is the electric field, and x is electric susceptibility, and it is a dimensionless quantity, Dimensional formula of P = Dimensional formula of (ε₀) x Dimensional formula of E

= [M^-1 L^-3 T⁴ I²] [ML T^-3I^-1] = [M⁰ L^-2 T^-1 I¹]

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

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.

⇒

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

σ_max = Force/Area = Mg/πr²

As the ift is accelerating with acceleration a, then

σ_max = M(g ± a)/πr²

⇒ r² = M(g ± a)/πS [Given , σ_max = S]

d²/4 = M(g ± a)/πS [∴ r = d/2]

As acceleration maximum i.e., g’ = g + a, so

E_i = 2π² mr²₁f²_{1 ………..}(i)

Where, f₁ = frequency of revolution of the body.

When, r₂ = 2r₁ and f₂ = and 2f₁ , then

Change in kinetic energy,

ΔE = E_r -E_i

= 16E_i - E_i = 15 E_i

Optically active compounds rotate the plane of polarised light either towards the right or left side. And the compound which has at least one chiral centre will be optically active.

Since only the given compound contains a chiral carbon, it will rotate the plane of polarised light.

Option D also has chiral carbons but it also contain plane of symmetry. So, it will be meso compound.

For Heterogeneous  Equilibrium

 for a strong acid and a strong base = 
Let 
NaOH, CH₃COONa are strong electrolytes and are fully dissociated in the aqueous solutions.
∴ -57.1 - x = -55.9
x = -1.2 kJ mol^-1
Therefore, 

The Wolff-Kishner reduction is an organic reaction that is used to convert an aldehyde or ketone to an alkane, using hydrazine, base and thermal conditions. The mechanism begins with the attack of hydrazine on the aldehyde or ketone to form an imine. Proton transfer steps then result in the formation of an N=N bond. Deprotonation of nitrogen and a rearrangement reaction results in the formation of a carbanion and the release of nitrogen gas. The carbanion then picks up a proton from water to regenerate the base catalyst and provides the final alkane product.

In OF₂, fluorine has negative oxidation state due to high electronegativity. Therefore, OF₂ dissolves in water but does not give any oxyacid solution because unlike other halogens, fluorine cannot acquire a positive oxidation state. In oxyacids of halogens, all halogen atoms have positive oxidation states.

SO₂, SCI₄ and SO₄ give oxyacid solutions in water and acquire positive oxidation states as shown below.

When dispersed phase is gas and dispersion medium is solid the colloidal sol obtained is termed as solid foam

Every whole number is either odd or even.
A whole number is even if it can be divided evenly by two; the numbers 2, 4, 6, 8, 10, 12, . . . are even.
In this question we are given the information that n is an odd number.
Because n is odd, it must be one of the numbers 1, 3, 5, 7, 9, 11, 13, 15, and so on.
Now we have to reason mathematically.
If n is a member of this “odd” list, then what can we say for sure about n +1?
If we add 1 to each number in the odd list, we get the “ n +1 ” list: 2, 4, 6, 8, 10, 12, 14, 16, and so on, which are even numbers.