SRMJEE Mock Test - 5 (Engineering) - JEE MCQ

• High input resistance
• Low output resistance
• More efficiency
• More current amplification
• More voltage amplification

K.E. of emitted electrons = radiation of energy - work function
= 6.2 eV - 4.2 eV = 2 eV = 2 x 1.6 x 10^-19 J = 3 x 10^-19 J (approx.)

Photoelectricity ∝ Intensity of light.
The intensity of light increases the number of rays striking the metal per time and thus increases the photoelectric effect.

n = 2 [Number of stereogenic area]
Total number of stereoisomers = 2ⁿ
[When symm. is/are absent]
Total number of stereoisomers = 2² = 4

Teflon is a polymer of tetrafluorothylene. It is used for coating articles and cookware to make them non-sticky.

Nylon−66 is a polymer of adipic acid and hexamethylenediamine. Glyptal is a polymer of ethylene glycol and phthalic acid. Buna−S is a polymer of butadiene and styrene.

F₂​C = CF₂​ (tetrafluoroethylene) is the monomer for Teflon. Nylon, Glyptal, and Buna-S have different monomers. Answer A is correct.

Ptyalin is present in human saliva. The optimum pH for Ptyalin activity is 5.6 − 6.9. But normally stomach pH is 1.5 − 3.5. So activity of Ptyalin decreases in stomach.

Ptyalin is active in saliva (pH  6.8) but denatures in stomach acid, decreasing its activity.

In alkali metals, the reactivity increases down the group due to decrease in IE₁ . But in case of halogens, the reactivity decrease down the group due to decrease in their electrode potentials.

The compound of a metal M, Mx​(NO₃​)y​, when treated with hydrogen sulphide in neutral or dilute acidic medium forms a black precipitate of M and its sulphide.
A black sulphide is formed by the action of H₂​S on CuCl₂​.

CdCl₂​ forms yellow ppt and ZnCl₂​ forms white ppt.

Physical adsorption increases with an increase in pressure. It also increases with the surface area. As adsorption is exothermic process so as temperature decreases, adsorption increases. It is a multilayer adsorption.

Halogens (ns²np⁵) after getting one electron occupy ns²np⁶ configuration, thus have EA₂ zero

α + β =
αβ = (-1)/1 
So, α² + β² = (α + β)² - 2αβ = (2)² - 2(-1) = 4 + 2 = 6

The probability mass function of Poisson distribution is given by:

Here, mean m = np = 10 × (1/50) = (1/5) = 0.2
Probability of number of packets containing not more than 2 defective pens = [P(0) + P(1) + P(2)] =
The number of packets containing not more than 2 defective pens = × 10,000 = 9989
Hence, the number of packets containing more than 2 defective pens in a consignment of 10,000 packets = 10,000 - 9,989 = 11

According to De Moivre 's theorem we get, (cosθ + i sinθ)ⁿ = cos nθ + isin nθ
Let z = √3 + i ...........(1)
r = |z| =
z =
Let z = r(cosθ + i sinθ) .........(2)
By comparing (1) and (2), we get
cos θ =
sin θ =
Therefore,
z¹⁰ = 2¹⁰
= 1024
= 1024
= 512 - 512 √3i
a = 512 and b = -512√3

Probability of hitting a target (p) = 1/5
∴ Probability of not hitting a target (q) =
Probability that none will be hit in 10 shots = (4/5)10
∴ Required probability = 

Given the quadratic equation:
ax² + 2bx + c = 0

Since b² = ac, we can rewrite the equation as:
ax² + 2√(ac)x + c = 0

Factoring, we get:
(√a * x + √c)² = 0

So, x = -√c / √a , -√c / √a

Thus, aα = aβ

Now, consider the equation:
cx² + 2bx + a = 0

Rewriting using b² = ac:
cx² + 2√(ac)x + a = 0

Factoring, we get:
(√c * x + √a)² = 0

So, x = -√a / √c , -√a / √c

Thus, cγ = cδ

Therefore, aα = aβ = cγ = cδ

Let one side of the quadrilateral be x and another side be y.
So, 2(x + y) = 34
or (x + y) = 17 …(i)

Let the area of the quadrilateral be A(x) = x(17 - x).
⇒ A'(x) = (17 - x) - x
⇒ A'(x) = 17 - 2x
Setting A'(x) = 0, we get x = 17/2.

A''(x) = -2, which is less than 0.

Since A''(x) is negative, the area will be maximum at x = y = 17/2.
Thus, the area = x ⋅ y = (17/2) × (17/2) = 289/4 = 72.25 cm².

Given, x = 2t − |t|, y = t³ + t²|t|

Since, 

So when t ≥ 0,

x = 2t - t
⇒ x = t

And y = t³ + t² × t
⇒ y = 2t³

When t < 0,

x = 2t - |t|
⇒ x = 2t - (-t)
⇒ x = 2t + t
⇒ x = 3t

And y = t³ + t² |t|
⇒ y = t³ + t² × (-t)
⇒ y = 0

Hence,

Differentiate both sides w.r.t. x,

At t = 0 (i.e. x = 0),

LHD = 6(0)² = 0 and RHD = 0
∴ f′(0) = 0.

∴ 

Let C.P. of the car = Rs. x
Amount spent on repairs = 10% of x = Rs. x/10
⇒ Total C.P. of the car for John = Rs.(x + x/10) = Rs. 11x/10
Profit = Rs. 11,000
⇒ S.P. of Car = C.P of Car + Profit
= Rs. 11x/10 + Rs. 11,000 = Rs.
Given:
Profit percentage = 20%
Now,
C.P. =
⇒
⇒
⇒ 132x = 110x + 11,000 100
⇒ 132x – 110x = 11,000 100 … (i)
⇒ 22x = 11,000 100
⇒ x = Rs. (11,000 x 100)/22 = Rs. 50,000
Amount spent on repairs = Rs. x/10  = Rs.50,000/10 = Rs. 5000

It is given that,

Ganesh's position from left = 12

Rajan's position from right = 15

After interchange,

Rajan's position from right = 20

Total boys in the class = Ganesh's position from left + Rajan's position from right -1
Boys = 12 + 20 - 1 = 32 - 1 = 31.

Thus, there are 31 boys in the class.

Hence, option D is the correct answer.

Let the cost price of an article be ₹x.

Selling price of the article at a gain of 12.5% is:
SP = Cost Price + Gain % of Cost Price
⇒ x + 12.5% of x
⇒ x + (1/8) × x
⇒ ₹(9/8) x

Selling price of the article at a gain of 25% is:
SP = Cost Price + Gain % of Cost Price
⇒ x + 25% of x
⇒ x + (1/4) × x
⇒ ₹(5/4) x

According to the question, the difference between the selling prices at 25% and 12.5% gain is ₹22.50:
(5/4)x - (9/8)x = ₹22.50

Converting to a common denominator:
(10x/8) - (9x/8) = ₹22.50
(x/8) = ₹22.50
x = ₹22.50 × 8
x = ₹180

Thus, the cost price of the article is ₹180.

Above condition can be followed in two cases.
Case I: Five 1s and one 2 and 3.
No of seven-digit numbers = 7! / 5! = 42
Case II: Four 1s and three 2s
No of seven-digit numbers = 7! / 4! × 3! = 35
Total ways = 42 + 35 = 77
Hence, the correct answer is 77.