SRMJEE Mock Test - 7 (Engineering) - JEE MCQ

Particle nature and wave nature of electromagnetic waves and electrons can be shown by photoelectricity and electron microscopy.
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. Explaining the experiments on the photoelectric effect led to the idea of light behaving as a particle of energy, called a photon.
In electron microscope, high-speed electron beam is used instead of light waves, which are used in optical microscope. Like light, the stream of electrons has a corpuscular and vibratory character. Electron microscope gives very high magnification and high resolution.
The electrons can be focused by electro-magnetic lenses, much like the light rays. Electron beam can vibrate like light rays but has very short wave length as compared to light rays. Wave length of electron beam λ = 0.005 nm as compared to 550 nm of visible light. Resolution increases with the decrease of wave length.

Work function,
ϕ = hc / λ = 12400 eV·Å / λ

For lithium:
λ_Li = 12400 eV·Å / 2.3 eV
λ_Li = 5391 Å ......(1)

For copper:
λ_cu = 12400 eV·Å / 4 eV
λ_cu= 3100 Å ......(2)

The wavelength range of visible light is:
4000 Å < λ < 7000 Å

From equations (1) and (2), we see that only lithium has a wavelength within the visible range.

Thus, lithium metal will be used for the photoelectric cell to work with visible light.

The deviation formula is:
δ = i₁ + i₂ − A

Given: δ = 45°, A = 60°

Substituting the values:
45° = i₁ + i₂ − 60°

So,
i₁ + i₂ = 105°

Also given:
i₁ − i₂ = 20°

Solving for i₁ and i₂:

Adding both equations:
(i₁ + i₂) + (i₁ − i₂) = 105° + 20°
2i₁ = 125°
i₁ = 62°30′

Substituting i₁ in the first equation:
62°30′ + i₂ = 105°
i₂ = 42°30′

Thus, the angles of incidence are i₁ = 62°30′ and i₂ = 42°30′.

We know that points connected by conducting wires have same potential. So, by rearranging the circuit, we get,

Here, we can see that capacitors C₁ and C₃ are in parallel combination. Let their equivalent capacitance be C_eq1.

Similarly, capacitors C₆ and C₇ are in parallel combination. Let their equivalent capacitance be C_eq2.

For parallel combination of capacitors, we know that,

Redrawing the circuit,

Using the concept of Wheatstone bridge,

Capacitor C₄ can be removed. Therefore, the circuit will become,

Now, capacitor C_eq1 is in series combination with C₅. Let the equivalent capacitance be C_eq3.

Capacitor C₂ is in series combination with C_eq2. Let their equivalent capacitance be C_eq4.

For series combination of capacitors,

Redrawing again,

Here, we can see that capacitors C_eq3 and C_eq4 are in parallel combination. So, the final equivalent capacitance will be given by,

 
=  Constant

Given:

• Pressure of gas: P = 4.5 × 10³ Pa
• Heat supplied to the gas: ΔQ = 800 kJ = 8 × 10⁵ J
• Initial Volume: V₁ = 0.5 m³
• Final Volume: V₂ = 2.0 m³

Now, work done by the gas in expanding:
ΔW = P(V₂ − V₁)
ΔW = 4.5 × 10⁵ (2.0 − 0.5)
ΔW = 4.5 × 10⁵ × 1.5
ΔW = 6.75 × 10⁵ J

Now, from the first law of thermodynamics:

ΔU = ΔQ − ΔW
where ΔU is the change in internal energy of the gas.

ΔU = 8 × 10⁵ − 6.75 × 10⁵ J
ΔU = 1.25 × 10⁵ J

Thus, the change in internal energy of the gas is 1.25 × 10⁵ J.

When the two way key is switched off, then The current flowing in the resistors R and X is
 I = 1 A   ...(i)
When the key between the terminals 1 and 2 is plugged in, then
(Potential difference across R) = IR= kl₁    ...(ii)
where k is the potential gradient across the potentiometer wire
When the key between the terminals 1 and 3 is plugged in, then 
Potential difference across (R + X) is

From equation (ii), we get|

From equation (iii), we get

X = kl₂ − R

=    (Using (iv))

2N₂O₅ → 4 NO₂ + O₂

Production rate of O₂ =
Production rate of NO₂ =
Disappearance rate of N₂O₅ =

[Rate of disappearance of N₂O₅ = Twice the production rate of O₂]

These are all Lewis bases as N-atom in each of them has tendency to donate an electron pair. Piperidine (c) is the strongest base among these. In pyrrole (b), the lone pair of electrons on N-atom is involved in delocalisation of π electrons (aromatic sextet). Hence, it has the least tendency to donate an electron pair. Aniline (d) and pyridine (a) are also weakly basic due to resonance. On the other hand, piperidine has no such delocalisation and it is a secondary amine. Hence, it is highly basic.

The reaction  is an example of substitution reaction

Glycerol reacts with oxalic acid to produce different products at temperature when glycerol is heated with crystalline oxalic acid at 100−110°C.. Initially, the glycerol gets converted to glyceryl monoxalate. Then glyceryl monoformate is formed by the removal of carbon dioxide from glyceryl monoxalate. This gives glycerol and formic acid.
Here, glycerol reacts with oxalic acid at around 260° C to form glycerol dioxalate as intermediate and further removes carbon dioxide to form allyl alcohol.

In the sp³d² hybridisation, the orbitals involved are s, px, py, pz, d_x²−y², dz².

Let the two roots of the given equation be α and 2α.
Sum of roots = α + 2α = -p or 3α = -p or α = -p/3 ... (1)
Product of roots = 2α² = q ... (2)
From equations (1) and (2), we get

The two circles whose centre and radius are C₁(0, 0), r₁ = 12, C₂(3, 4), r₂ = 5 and it passes through origin ie, the centre of C₁.

Now, C₁C₂ = = 5
and r₁ - r₂ = 12 - 5 = 7
∴ C₁C₂ < r₁ - r₂
Hence, circle C₂ lies inside the circle C₁.
From figure the minimum distance between, them is
AB = C₁B - C₁A = r₁ - 2r₂
= 12 - 10 = 2

Given: sin⁻¹(1 − x) − 2sin⁻¹x = π/2
Let x = sin y
∴ sin⁻¹(1 − siny) − 2y = π/2
⇒ sin⁻¹(1 − siny) = π/2 + 2y
⇒ 1 − siny = sin
⇒ 1 − siny = cos2y
⇒ 1 − siny = 1 − 2sin²y as cos2y = 1 − 2sin²y
⇒ 2sin²y − siny = 0
⇒ 2x² − x = 0
⇒ x(2x − 1) = 0
⇒ x = 0, 2x − 1 = 0
⇒ x = 0, 2x = 1
⇒ x = 0, x = 1/2; but x = 1/2 does not satisfy the given equation.
∴ x = 0 is the solution of the given equation.

For function to be continuous 

Also, f(0) = 0

Hence, the function is continuous at x = 0

For function to be differentiable

LHD = RHD = finite

Let O be the circumcentre and I be the incentre of ΔABC.

Here, OM and IN are perpendicular to the base.

Since OI ∥ BC, we have IN = OM ⇒ OM = r.

Also, OB = R and ∠BOM = ∠A.

From ΔOBM, we get cos A = OM / OB = r / R.

Thus, r = R cos A.

Given, 

Function f(x) is continuos at x = a if

Now, 

Since, the value of sin 1/x is between [−1,1]

= (0)² × (value between −1 and 1)

= 0

f(x)  is continuous at x = 0

if 

⇒ f(0) = 0

Given:
x = ω - ω² - 2, where ω and ω² are cube roots of unity.

We need to evaluate x⁴ + 3x³ + 2x² - 11x - 6.

Since ω and ω² satisfy the equation ω³ = 1 and 1 + ω + ω² = 0, we simplify x:
x = ω - ω² - 2
=> x + 2 = ω - ω²
=> Squaring both sides,
(x + 2)² = (ω - ω²)²
=> x² + 4x + 4 = ω² - 2ωω² + ω⁴
=> x² + 4x + 4 = ω² - 2(1) + 1
=> x² + 4x + 4 = ω² - 2 + 1 = ω² - 1
=> x² = -4x - 3

Using this in the given expression, we compute:
x⁴ + 3x³ + 2x² - 11x - 6 = -1

Thus, the correct answer is B) -1.

Given that :

∴ Range of f(x) is: 

= 

The given inequality is:

log₉(x + 1) * log₂(x + 1) − log₉(x + 1) − log₂(x + 1) + 1 < 0

We simplify this inequality:

log₉(x + 1) * (log₂(x + 1) − 1) − (log₂(x + 1) − 1) < 0

⇒ (log₉(x + 1) − 1)(log₂(x + 1) − 1) < 0

Next, we examine the factors:

log₉(x + 1) − 1 > 0 and log₂(x + 1) − 1 < 0, or log₉(x + 1) − 1 < 0 and log₂(x + 1) − 1 > 0.

Case 1:
log₉(x + 1) − 1 > 0 ⇒ log₉(x + 1) > 1 ⇒ x + 1 > 9 ⇒ x > 8

log₂(x + 1) − 1 < 0 ⇒ log₂(x + 1) < 1 ⇒ x + 1 < 2 ⇒ x < 1

This results in no valid solution for x, since x cannot simultaneously be greater than 8 and less than 1.

Case 2:
log₉(x + 1) − 1 < 0 ⇒ log₉(x + 1) < 1 ⇒ x + 1 < 9 ⇒ x < 8

log₂(x + 1) − 1 > 0 ⇒ log₂(x + 1) > 1 ⇒ x + 1 > 2 ⇒ x > 1

This gives us a valid solution for x in the range (1, 8).

The integral values of x in this range are 2, 3, 4, 5, 6, and 7.

Thus, the number of integral solutions is 6. The correct answer is:

C) 6

According to the given information, girls are sitting from left to right as given below.
Neha Shalini Rekha Neetu Pooja
So, Rekha sits in the middle.

The movements of Rohan are shown in the figure below:

Clearly, AD = BC = 2 km
So, required distance = AE
= (DE - AD)
= (3 - 2) km = 1 km

From the above diagram AF = 1 km.

It is given that;

Selling price (SP) = 540 Rupees

Loss = 10%

We know that,

In the case of 10% profit gain,

SP =  600 × 110 / 100 = 660 Rupees