MHT CET Mock Test - 5 - JEE MCQ

Distance between the slits, d=0.28×10−3 m

Distance between the slits and the screen, D=1.4m

Distance between the central fringe and the fourth (n=4) fringe, 

u=1.2×10−2 m

In case of a constructive interference, we have the relation for the distance between the two fringes as:

u=nλD/d

⇒λ=ud/nD=6×10−7m=600nm

Escape velocity is the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.

Angular acceleration. Angular acceleration is the rate of change of angular velocity. In three dimensions, it is a pseudovector. In SI units, it is measured in radians per second squared (rad/s²), and is usually denoted by the Greek letter alpha (α).

Explanation:Alternating Current is the current in which the polarity of source continuously changes on a fixed frequency. So the positive and the negative terminals ‘alternate’

 Answer :- 

Solution :- torque when =180°

= 1.5 × 0.22 × sin 180°

= 0.33 × 0 = 0 Nm

Ionic behavior implies dipole moment.

HCl is an ionic compound (H⁺Cl^-)
CO₂, CH₄ and BF₃  is a covalent compound

We know that,
eV=(hc/λ)-w
V=(hc/eλ)-(w/e)
Arithmetic progression =>V₂=(V₁+V₂)/2
Now,
(hc/eλ₂)-w/e=1/2[(hc/eλ₁)-(w/e) +(hc/eλ₃) -(w/e)]
=>1/ λ₂=1/2[(1/ λ₁)+(1/λ₃)]
=>2/ λ₂=1/ λ₁ + 1/λ₃
Hence the correct answer is harmonic Progression.

Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R

In the below reaction , no bond with chiral carbon is disturbed, hence, configuration is retained in product.

Carbon dioxide CO₂ has polar bonds but is a nonpolar molecule. The structure of CO₂ is linear. The individual bond dipoles cancel each other as they point in opposite direction and are equal in magnitude. Hence, the molecule is non polar with zero dipole moment.

Greater the number of hydroxy groups , greater its a bility to form H-bonds with water, hence greater solubility.

Sucrose is a non-reducing sugar because the two monosaccharide units are held together by a glycosidic linkage between C₁ of α-glucose and C₂ of β-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.

It gives racemic mixture of oxiranes.

On heating glycerol with excess amount of HI, the product formed is Isopropyl iodide.

The correct answer is Option D.
 

(a)

Nearest neighbours are along the diagonal at the half distance apart, Diagonal AB = √3a

∴ Distance of the nearest neighbour

Next nearest neighbours are along the cell edge = 520 pm

K-atom at the centre is surrounded by eight K-atoms as nearest neighbours.

In bcc, coordination number is six as next nearest neighbours

The correct answer is option C
CH_3​CH₂​CONH₂​ (A)⟶ CH₃​ − CH₂ ​− NH₂​ (B)⟶ ​CH_3​ − CH₂ ​− OH
In the above sequence A & B respectively are Br₂​/KOH and HNO₂​
The first step is Hoffmann bromamide degradation reaction in which an amide (propanamide) is converted to an amine  (ethylamine) containing one carbon atom less. The reagent A is bromine in presence of KOH. In the second step, aliphatic primary amine (ethyl amine) reacts with nitrous acid (reagent B) to form aliphatic primary alcohol (ethyl alcohol).
 

We want to find the indefinite integral of |x| with respect to x:

∫ |x| dx

1. Consider x ≥ 0:

   • Here, |x| = x.
   • So, ∫ |x| dx = ∫ x dx = (x²)/2 + C₁

2. Consider x < 0:

   • Here, |x| = -x.
   • So, ∫ |x| dx = ∫ (-x) dx = - (x²)/2 + C₂

To combine these results into a single expression, we note that for an indefinite integral, constants C₁ and C₂ can be absorbed into a single arbitrary constant C. A concise way to write this is:

∫ |x| dx = (x · |x|)/2 + C

• For x ≥ 0, this becomes x·x/2 = x²/2.
• For x < 0, this becomes x·(-x)/2 = -x²/2.
• Combining both the results , we get x|x|/2 + C

Find the shortest distance between the lines 

On comparing them with :

we get : 

Since f ‘(x) = 2x > 0 for x > 0,and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing. Infact , f is strict increasing on [0 , ∞) and strict decreasing on (- ∞,0].

 Therefore , f is strictly increasing on R.

Since AB is the diameter of the circle. 
From figure, 
Also equation of the circle is 

equation of tangent to this circle at (0, 0) is ax + by = 0 ……(1)
∴ m = length of perpendicular from


and n = length of perpendicular from 


Hence (D) is the correct answer.

A.B = [(-1(-1) + 2(-2) + 3(-3)   -1(-3) + 2(1) + 3(2)]

A.B = [1 - 4 - 9     3 + 2 + 6]

A.B = [-12   11]

Minimum value   

and maximum value =