KIITEE Mock Test - 8 - JEE MCQ

Inside the shell, there is no mass. So, gravitational field is zero. Outside the shell, gravitational field is inversely proportional to the square of the distance.

x = a sin t ...(i)
y = a(1 - cos t) ...(ii)
x² = a² sin² ωt = a²(1 - cos² ωt)
x² = a² - a² cos2 t
a² cos² ωt = a² - x² ...(iii)
y = a(1 - cos t) = a - a cos t
a - y = a cos t
(a - y)² = (a cos t)²
a² + y² - 2ay = a² cos² ωt ...(iv)
Put the value from equation (iii),
a² + y² - 2ay = a² - x²
x² + y² - 2ay = 0
This is the equation of a circle.

If F > μ_s mg, the block m will start moving on block M. It will, therefore, exert a force on block M due to kinetic friction m and M.
This force f_k = μ_k mg.
Thus, the acceleration of block

Hence, the correct choice is (1).

Their final relative velocity = 0
Relative velocity of A w.r.t B is u = v₁ - v₂
Relative acceleration = -a

Magnitude of the Impulse received | ΔI| = 4mcu

V₁ and V₂ are velocities after collision.
Magnitude of impulse is the same on both A and B.
For body A:
2mv₁ - 6mu = -4mcu
2mv₁ = 6mu - 4cum
V₁ = 3u - 2cu(→)
For body B:
mv2 - (-mu) = 4mcu
mv2 = 4mcu - mu
v2 = 4cu - u(→) - speed of B

Scale reading is due to the force acting on its pan. This force is due to the rate of change of momentum given to it because of colliding particles.
Velocity with which the particle collides with the pan, v =  =  = _^2.8√5 ms^-1
= 12.52 x 10^-3 kg ms^-1
 Momentum imparted to pan per second = 12.52 x 10^-3 x 100
= 1.252 kg = 1252 g (weight)

For the block with mass m, the force experienced is the spring force.
∴ Spring force = Mass × acceleration = ma
For block with mass M, forces are spring force and the external F

When two vectors are perpendicular to each other, then their dot product is zero.

Average force = Change in momentum/Time interval

According to de-Broglie relation,

Acid hydrolysis of an ester gives two different organic compounds.

3-chloro prop-1-ene (allyl halide) undergoes SN¹ reactions through the formation of an intermediate carbocation, which is resonance stabilised as follows:

Therefore, it will undergo nucleophilic substitution most readily.

Yes, it is the correct answer.

I. The compound which has the least solubility will be precipitated first.
(a) BaSO₄
Given, Ksp = 10^-11
Let the solubility be x mol/L.

Since BaSO₄ has the least solubility, therefore it will be precipitated first.

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.

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.

SN² mechanism means bimolecular nucleophilic substitution mechanism. In this, rate of reaction is determined by concentration of two molecules. It is characteristic of 1⁰ alkyl halides.
CH₃ -CH₂ -Cl
1⁰ alkyl halide

H_β line is formed due to the transition of electrons from 4 to 2.

According to Gibb's equation:

The reaction becomes spontaneous at a temperature at which  is a negative value.
At 298K:

Since ΔGº is positive, the reaction is not spontaneous at this temperature.



Hence, the reaction is spontaneous above 1100 K or 827^oC.

The slope of the first line is 4/3 and the slope of the second line is 3/4.
Clearly, these slopes are neither negative reciprocal of each other nor equal.
Hence, they are neither parallel nor perpendicular.

cos^-1
= cos^-1
= cos^-1 {cos(π - x) = cos(π + x) = -cos x}
= cos^-1
( 0 ≤ cos^-1 x ≤ π)

The equation of a tangent of slope m to the parabola 

If it passes through (4, 10), then