JEE Main Mock Test - 7 - JEE MCQ

Let m be the mass of the coin. It will fly off when the centripetal force mrω² just exceeds the force of friction μmg. The minimum ω is given by

Hence the correct choice is (c).

A stationary wave has the equation of the form
y = A sin kx sin ωt
Here, for y₁, we have k₁ = π/L and ω₁ = ω
∴ v₁ = ω₁/k₁ = ωL/π
For y₂, we have k₂ = 2π/L and ω₂ = 2ω
∴ v² = ω₂/k₂ = ωL/π = v₁
Thus, the wave velocities are the same in both the cases. Also, they have the same amplitude. The frequency for y₂ is twice the frequency for y₁
Now, since energy ∝ (frequency)²  
∴ E₂ = 4E₁

When a lens is painted as given, only the intensity of the image reduces since the portion blackened does not refract the incident light to form image.
Whereas the transparent still refracts as in case of a biconvex lens to obtain a single image following,

Key Idea In nature, metals have positive
temperature coefficient and semiconductors have negative temperature coefficient.
Aluminium is a metal, so its conductivity decreases with increase in temperature, due to the excessive collision of electron. Whereas, Si is a semiconductor therefore its conductivity increases with increase in temperature.
All metals have positive temperature coefficient of resistivity and semiconductor has negative temperature coefficient of resistivity. Hence, assertion (A) is incorrect but reason (R) is correct.

We can not extract all electrons from metal, as we remove electrons the metal will get more and more positively charged.
This is so because  the number of proton will be more than no of electron.
Now, Attraction will be greater. So, It takes too much energy to extract the electrons any more.

Position vector of box and stone are 

When stone hits the block,

For ball A, The velocity of the ball before entering the water surface is 

When ball enters into the water, due to Upthrust of water the velocity of ball decrease (retarded). 

If h be the depth up to which ball sink, then v² - u² = 2as

For ball  B, The velocity of the ball before entering the water surface is 

When ball enters into the water, due to Upthrust of water the velocity of ball decrease (retarded).
So retardation,

If h be the depth up to which ball sink, then v² - u² = 2as

Ratio of the depth of ball 

Intensity of light after polarization is  Since the light is polarized at angle 60º so the intensity of polarized light is 

M0 = 200 MeV/c²; m = 1 GeV/c²
Initial velocity = v₀
Mv₀ = mv

pH is 4.05 when the indicator is 75% red.

Al₂O₃ can be converted to anhydrous AICl₃ by heating a mixture of Al₂O₃ and carbon in dry Cl₂ 

Morphine is anaesthetic administered by injection.

H.C. is the delocalisation of σ−bond electrons.
Resonance is the delocalisation of π−bond electrons.
Inductive effect is the partial displacement of σ−bond electrons.
β−H connected with Csp (C≡C) So bond order increase hence bond energy increase (α) H remove the ion is stabilised by delocalisation while γ does not
(i) Delocalisation of σ electron is hyperconjugation ✓
(ii) Delocalisation of π-election is resonance ✓
(iii) Inductive effect involves partial transfer of σ. electrons ×
(iv) Inductive effect involves partial displacement of σ electrons ✓
(v) Resonance effect is distance dependent ×
(vi) sp³, sp² and sp hybridised carbon atoms can participate in resonance ×

(i) The ionic radius of lanthanoids follow the following relation:

Atomic number (Z) for Pr = 59
Atomic number (Z) for Sm = 62
Atomic number (Z) for Dy = 66
and has ionic radii for positive ions as given below:

Thus, statement (i) is not correct.
(ii) Element Eu (Z=63) can exist in (+)2 and (+)3 oxidation state.
∴ (+) 3 oxidation state is the most common oxidation state for lanthanoids.
Thus, (+) 2 oxidation state oxides oxidises to (+) 3 easily and Eu²⁺ can act as strong reducing agent.
Hence, the statement (II) is correct.
(iii) The electronic configuation for
(Pu), (Z=94)=5f⁶6d⁰7s²
Therefore, it can use maximum of 6 + 2 = 8 electrons.
Thus, can show +7 oxidation state.
Thus, statement (III) is also correct.
Hence, option (d) is the correct answer.

In presence of UV light, free radical substitution reaction occurs in aliphatic side chain where as in positive q presence can halogen carrier electrophilic substitution reaction takes place.

pπ - pπ  back-bonding is the main factor responsible for the weak acidic nature of B-F bonds in BF₃.

Aspirin (or acetylsalicylic acid) is used as a pain reliever for minor aches and pains and to reduce fever. 

Structure of Aspirin:

⇒6  Carbon chain ⇒  Functional group -NH₂ is at C- 3. It will be added at suffix i.e amine 
⇒ Methyl substituent at C−5 . So, the IUPAC name of the compound is  5 - Methyl - 3 - hexanamine

The angle made by line AB with x  axis is 45º. The angle made by line AB with y axis is 120º. 

Let the angle made by line AB with z axis is θ. We have   

Given that θ is acute, So, we get  

|2z₁ + z₂| ≤ |2z₁| + |z₂| = 2|z₁| + |z₂| = 2 × 1 + 2 = 4
From the figure, |z₁ − z₂| is the least when 0,z₁,z₂ are collinear.

Then |z₁ − z₂| = 1

The given equation is equivalent to
7 cos²x + sin x cos x − 3(sin²x + cos²x) = 0
or 3 sin²x − sin x cos x − 4 cos² x = 0
Since cos x = 0 does not satisfy the given equation we divide throughout by cos²x and get 3 tan² x − tan x − 4 = 0
⇒ (3 tan x − 4)(tan x + 1) = 0
⇒ tan x = 4/3 or tan x = −1

It should be noted that the solutions given by (b) or (c) are not complete; the general solution is given by (d) only.

f(x) = log(x − [x]) is defined if x − [x] > 0
⇒ {x} > 0 ⇒ x ∉ I
∴ Domain = R − I

Total number of matrices = 3⁹
Number of symmetric matrices = 3⁶
Number of skew symmetric matrices = 3³
The zero matrix is both symmetric and skew-symmetric the required probability is 

For a ∈ N,a is a factor of a so R is reflexive. If a is factor b and b is factor of c then a is factor of c so R is transitive. If a is factor of b and b is factor of a then a = b so R is anti symmetric, 2 is factor of 4 but 4 is not a factor of 2.

The equation of the given curve is  x² = 4y

Differentiating w.r.t.  x, we get 

Now, slope of the tangent at (h,k) is given by, 

Thus, slope of the normal at 

Therefore, the equation of normal at 

Since it passes through the point  (1,2) we have 
now(h,k) lies on the curve  x² = 4y, so we have h² = 4k   ...(3)
Solving (2) and (3), we get ℎ=2 and k= 1
From (1), the required equation of the normal is: 

Let the five ice creams in the shop be A,B,C,D,E, F and a children buys six ice creams, so number of ways in which he can buy the ice creams is the intergral solution of the equation A+B+C+D+E+F = 6 and that is given as: 

But it is given in statement 1, that the number of ways is ¹⁰C₅ . So the statement 1 is incorrect.

The number of different ways of arranging 6 A's and 4 B's in a row is given as:

So it is same as the number of ways in which the children can buy the ice creams. So statement 2 is correct.

given

∴  The general term is

For independent of x,