JEE Main 22 January 2025 Shift 1 (Previous Year Paper)

Concept Used:

The number of equivalence relations on a set is equivalent to the number of ways the set can be partitioned. For a set containing n elements, this number is represented by the Bell number Bn.

Calculation:

⇒ Partitions of A = {1, 2, 3} are:

⇒ 1. {{1}, {2}, {3}}

⇒ 2. {{1, 2}, {3}}

⇒ 3. {{1, 3}, {2}}

⇒ 4. {{2, 3}, {1}}

⇒ 5. {{1, 2, 3}}

⇒ Number of partitions = 5

⇒ Bell number B3 = 5

⇒ Number of equivalence relations = 5

The empty relation is not included

∴ The maximum number of equivalence relations on A = {1, 2, 3} is 5

Hence option 3 is correct

Calculation 

f(x + y) = f(x).f(y) 

⇒ f(x) = e^λx f′(0) = 4a 

⇒ f′(x) = λe^λx ⇒ λ = 4a

So, f(x) = e^4ax

f′′(x) – 3af′(x) – f (x) = 0 

⇒ λ² – 3aλ – 1 = 0 

⇒ 16a² – 12a² – 1 = 0 ⇒ 4a² = 1 ⇒ (\mathrm{a}=\frac{1}{2})

F(x) = e^2x

Area = (\int_{0}^{2} \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{2}-1)

Hence option 1 is correct

Calculation 

Let ‘G’ be the centroid of Δ formed by (1, 3) (3, 1) & (2, 4)

(\mathrm{G} \cong\left(2, \frac{8}{3}\right))

Image of G w.r.t. x + 2y – 2 = 0

(\frac{α-2}{1}=\frac{β-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4})

= (\frac{-2}{5}\left(\frac{16}{3}\right))

⇒ (α=\frac{-32}{15}+2=\frac{-2}{15}, β=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15})

15(α – β) = – 2 + 24 = 22

Hence option 4 is correct

Calculation

Given:

( Z_{1}=e^{-i\pi/4}, Z_{2}=1, Z_{3}=e^{i\pi/4} )

( |z_{1}\overline{z}{2}+z{2}\overline{z}{3}+z{3}\overline{z}_{1}|^{2} = |e^{-i\frac{\pi}{4}}\times1+1\times e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}\times e^{i\frac{\pi}{4}}|^{2} )

( = |e^{-i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}|^{2} )

( = |2e^{-i\frac{\pi}{4}}+1|^{2} )

( = |\sqrt{2}-\sqrt{2}i+i|^{2} )

( = (\sqrt{2})^{2}+(1-\sqrt{2})^{2} = 2+1+2-2\sqrt{2} = 5-2\sqrt{2} )

( \alpha=5, \beta=-2 )

( \Rightarrow \alpha^{2}+\beta^{2}=29 )

Hence option (4) 29 is correct

Calculation 

16(sec^–1 x)² + (cosec^–1 x)²

(\operatorname{Sec}^{-1} \mathrm{x}=\mathrm{a} \in[0, π]-\left{\frac{π}{2}\right})

(\operatorname{cosec}^{-1} \mathrm{x}=\frac{π}{2}-\mathrm{a})

= (16\left[\mathrm{a}^{2}+\left(\frac{π}{2}-\mathrm{a}\right)^{2}\right]=16\left[2 \mathrm{a}^{2}-π \mathrm{a}+\frac{π^{2}}{4}\right])

(\max ]_{a=π}=16\left[2 π^{2}-π^{2}+π \frac{2}{4}\right]=20 π^{2})

(\min ]_{a=\frac{π}{4}}=16\left[\frac{2 \times π^{2}}{16}-\frac{π^{2}}{4}+\frac{π^{2}}{4}\right]=2 π^{2})

Sum = 22π² 

Hence option 4 is correct

Calculation 

HHH → 0

HHT → 0

HTH → 1

HTT → 0

THH → 1

THT → 1

TTH → 1

TTT → 0

Probability distribution

(\begin{array}{c|c|c} \mathrm{x}{\mathrm{i}} & 0 & 1 \ \hline \mathrm{P}\left(\mathrm{x}{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array})

(\mu=\sum \mathrm{x}{\mathrm{i}} \mathrm{p}{\mathrm{i}}=\frac{1}{2})

(\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2})

= (\frac{1}{2}-\frac{1}{4}=\frac{1}{4})

(64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48)

Hence option 2 is correct

Calculation

a₁ .a₅ = 28 ⇒ a.ar⁴ = 28 ⇒ a² r⁴ = 28 …(1)

a₂ + a₄ = 29 ⇒ ar + ar³ = 29 

⇒ ar(1 + r²) = 29 

a² r² (1 + r²)² = (29)² …(2) 

By Eq. (1) & (2)

(\frac{\mathrm{r}^{2}}{\left(1+\mathrm{r}^{2}\right)^{2}}=\frac{28}{29 \times 29})

⇒ (\frac{\mathrm{r}}{1+\mathrm{r}^{2}}=\frac{\sqrt{28}}{29} \Rightarrow \mathrm{r}=\sqrt{28})

∵ a² r⁴ = 28 ⇒ a² × (28)² = 28 

⇒ (a=\frac{1}{\sqrt{28}})

∴ (a_{6}=a r^{5}=\frac{1}{\sqrt{28}} \times(28)^{2} \sqrt{28}=784)

Hence option 3 is correct

Calculation

P(2λ + 1, 3λ + 2, 4λ + 3) on L₁

Q(3µ + 2, 4µ + 4, 5µ + 5) on L₂

Dr’s of PQ = 3µ – 2λ + 1, 4µ – 3λ + 2, 5µ – 4λ + 2

PQ ⊥ L₁ 

⇒ (3µ – 2λ + 1)2 + (4µ – 3λ + 2)3 + (5µ – 4λ + 2)4 = 0

38µ – 29λ + 16 = 0 …(1)

PQ ⊥ L₂ 

⇒ (3µ – 2λ + 1)3 + (4µ – 3λ + 2)4 + (5µ – 4λ + 2)5 = 0

50µ – 38λ + 21 = 0 …(2)

By (1) & (2) 

(λ=\frac{1}{3} ; \mu=\frac{-1}{6})

∴ (\mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right) & \ \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right))

Line PQ 

(\begin{array}{ccc} \frac{x-\frac{5}{3}}{\frac{1}{6}} & \frac{y-3}{\frac{-1}{3}} & \frac{z-\frac{13}{3}}{\frac{1}{6}} \end{array})

(\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1})

(\text { Point }\left(\frac{14}{3},-3, \frac{22}{3}\right))

lies on the line PQ 

Hence option 4 is correct

Calculation

(\rm \mathrm{e}^{5(ℓ n x)^{2}+3}=\mathrm{x}^{8})

⇒ (\ln \mathrm{e}^{5(ℓ \mathrm{n} x)^{2}+3}=\ln \mathrm{x}^{8})

⇒ 5(ℓnx)² + 3 = 8ℓnx

(ℓnx = t)

⇒ 5t² – 8t + 3 = 0

(\mathrm{t}{1}+\mathrm{t}{2}=\frac{8}{5})

(\ell \operatorname{nx}{1} x{2}=\frac{8}{5})

x₁ x₂ = e^8/5

Hence option 1 is correct

Calculation

T_n = S_n – S_n–1

⇒ (\mathrm{T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3))

⇒ (\frac{1}{T_{n}}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)})

(\lim {n \rightarrow \infty} \sum{r=1}^{n} \frac{1}{T_{r}}=\lim {n \rightarrow \infty} 8 \sum{r=1}^{n} \frac{1}{(2 n-1)(2 n+1)(2 n+3)})

= (\lim _{n \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n+1)(2 n+3)}\right))

= (\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right])

= (\frac{2}{3})

Hence option 3 is correct

Calculation 

= (\underbrace{12 \mathrm{C}{2}}{\substack{\text { Sclection of two } \ \text { letters before } \mathrm{M}}} \times \underbrace{{ }^{13} \mathrm{C}{2}}{\substack{\text { Selection of two } \ \text { letters after } \mathrm{M}}}=5148)

Hence option 4 is correct

Calculation

(\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{1}{\mathrm{y}^{2}}\right) \mathrm{x}=\frac{1}{\mathrm{y}^{3}})

(\text { I.F. }=\mathrm{e}^{\int \frac{1}{{y}^{2}} \mathrm{dy}}=\mathrm{e}^{-\frac{1}{y}})

⇒ (x \cdot e^{-\frac{1}{y}}=\int\left(e^{-\frac{1}{y}}\right) \cdot \frac{1}{y^{3}} d y)

Put (-\frac{1}{y}=t)

(+\frac{1}{\mathrm{y}^{2}} \mathrm{dy}=\mathrm{dt})

(x \cdot e^{-\frac{1}{y}}=-\int t \cdot e^{t} d t)

(x \cdot e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C)

(x \cdot e^{-\frac{1}{y}}=\frac{+1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C)

x = 1, y = 1 

(\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{e}}+\frac{1}{\mathrm{e}}+\mathrm{C})

⇒ (\mathrm{C}=-\frac{1}{\mathrm{e}})

Put y = (\frac{1}{2})

(\frac{\mathrm{x}}{\mathrm{e}^{2}}=\frac{2}{\mathrm{e}^{2}}+\frac{1}{\mathrm{e}^{2}}-\frac{1}{\mathrm{e}})

x = 3 – e

Hence option 3 is correct

Calculation

y = x² + px – 3

Let P(α, 0), Q(β, 0), R(0, –3)

Circle with centre (–1, –1) is (x + 1)² + (y + 1)² = r²

Passes through (0, –3)

1² + (–2)² = r²]

r² = 5

(x + 1)² + (y + 1)² = 5

Put y = 0

(x + 1)² = 5 – 1

(x + 1)² = 4

x + 1 = ±2

x = 1 or x = –3 

P(1, 0) and Q(–3, 0)

Area of ΔPQR = (\frac{1}{2}\left|\begin{array}{ccc} 1 & 0 & 1 \ -3 & 0 & 1 \ 0 & -3 & 1 \end{array}\right|=6)

Hence option 2 is correct

Calculation. 

S₁ : (x + 2)² + (y – 2)² = 2²

S₂ : (x – 2)² + (y – 5)² = r²

Both circle intersect at two points 

∴ |r₁ – r₂| < c₁ c₂ < r₁ + r₂

|r – 2| < 5 < 2 + r 

⇒ 3 < r < 7

r ∈ (3, 7)

α = 3, β = 7

3β – 2α = 15

Hence option 1 is correct

Calculation 

f(x) = (7tan⁶ x – 3tan² x)(sec² x)

(I_{1}=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right) d x)

Put tanx = t

(I_{1}=\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0)

(I_{2}=\int_{0}^{\pi / 4} x \underbrace{\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)}_{\mathrm{II}} d x)

= (\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]{0}^{\pi / 4}-\int{0}^{\pi / 4}\left(\tan ^{7} x-\tan ^{3} x\right) d x)

= (0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right)\left(1+\tan ^{2} x\right) d x)

Put tanx = t

= (-\int_{0}^{1}\left(t^{5}-t^{3}\right) d t=-\left[\frac{t^{6}}{6}-\frac{t^{4}}{4}\right]=\frac{1}{12})

7I₁ + 12I₂ = 1 

Hence option 3 is correct

Calculation 

f(x + y) = f(x) f′(y) + f′(x) f(x)

Put = x = y = 0

f(0) = f(0)f′(0) + f′(0)f(0) 

f′(0) = (\frac{1}{2})

Put y = 0

f(x) = f(x) f′(0) + f′(x)f(0)

(f(x)=\frac{1}{2} f(x)+f^{\prime}(x))

(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(\mathrm{x})}{2})

(\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2})

⇒ (\ell n y=\frac{x}{2}+c)

∵ f (0) = 1 ⇒ C = 0 

(\ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2})

(\ell n \ \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2})

(\sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2})

= 2525

Hence option 2 is correct

Concept:

Counting reduced proper fractions:

• Set A: A = {1, 2, 3, …, 10}.
• Set B: B = {m/n : m,n ∈ A, m < n, gcd(m,n)=1}.
• Key idea: For a fixed denominator n, the number of integers
• 1 ≤ m < n with gcd(m,n)=1 equals φ(n).
• Important terms: Euler's totient function φ(n),
• reduced fraction, distinct fractions.
• Therefore: n(B) = Σ_n=2¹⁰ φ(n).

 

Calculation:

A = {1, 2, ….10}

(\mathrm{B}\left{\frac{\mathrm{~m}}{\mathrm{n}}=\mathrm{m}, \mathrm{n} \in \mathrm{~A}, \mathrm{~m}<\mathrm{n}, \operatorname{gcd}(\mathrm{~m}, \mathrm{n})=1\right})

n(B)

(\mathrm{n}=2 \quad\left{\frac{1}{2}\right})

⇒ φ(2) = 1

(\mathrm{n}=3 \quad\left{\frac{1}{3}, \frac{2}{3}\right})

⇒ φ(3) = 2

(\mathrm{n}=4 \quad\left{\frac{1}{4}, \frac{3}{4}\right})

⇒ φ(4) = 2

(\mathrm{n}=5 \quad\left{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right})

⇒ φ(5) = 4

(\mathrm{n}=6 \quad\left{\frac{1}{6}, \frac{5}{6}\right})

⇒ φ(6) = 2

(\mathrm{n}=7 \quad\left{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right})

⇒ φ(7) = 6

(\mathrm{n}=8 \quad\left{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right})

⇒ φ(8) = 4

(\mathrm{n}=9 \quad\left{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right})  

⇒ φ(9) = 6 

(\mathrm{n}=10 \quad\left{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right})

⇒ φ(10) = 4

Hence n(B) = 31

∴ n(B) = 31.

Calculation

(y^{2}=2 \sqrt{3} x)

((x-2 \sqrt{3})^{2}+y^{2}=(2 \sqrt{3})^{2})

(A=\frac{π r^{2}}{2}-2 \int_{0}^{2 \sqrt{3}} \sqrt{2 \sqrt{3} x} d x)

(\frac{π(12)}{2}-2 \sqrt{2 \sqrt{3}} \frac{\left(\mathrm{x}^{3 / 2}\right)_{0}^{2 / 3}}{3 / 2})

= 6π – 16

Hence option 3 is correct

Calculation

(P=\frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9}+\frac{6}{10} \times \frac{5}{9}}=\frac{5}{9})

m = 5, n = 9

m + n = 14

Calculation

 

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

(\ell(\mathrm{LR})=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5})

Hence option 3 is correct

CONCEPT:

Faraday's Laws of Electrolysis

• Faraday's first law of electrolysis states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
• The relationship can be expressed as:

m = (ItM)/(nF)

where m is the mass of the substance deposited, I is the current, t is the time, M is the molar mass of the substance, n is the number of electrons transferred per ion, and F is the Faraday constant.

EXPLANATION:

• For the electrolysis of aluminium chloride (AlCl₃), the reaction at the cathode is:

Al³⁺ + 3e^- → Al

• Given data:
• Current (I) = 2A
• Time (t) = 30 minutes = 30 × 60 = 1800 seconds
• Molar mass of Aluminium (M) = 27 g mol^-1
• Faraday constant (F) = 96500 C mol^-1
• In the reaction, the number of electrons (n) transferred per ion is 3.
• Using the formula:

m = (ItM)/(nF)

m = (2 × 1800 × 27)/(3 × 96500)

m = (97200)/(289500)

m = 0.336 g

Therefore, the amount of aluminium deposited at the cathode is 0.336 g.

CONCEPT:

Radioactive Decay

• Radioactive decay is a process by which the nucleus of an unstable atom loses energy by emitting radiation. The decay constant (λ) is a probability rate constant that quantifies the likelihood of decay of an atom per unit time.
• Key characteristics of radioactive decay:
• The decay constant (λ) is independent of temperature.
• The half-life (t_1/2) of a radioactive substance is the time required for half of the radioactive atoms to decay. The relationship between half-life and decay constant is given by:

t_1/2 = ln(2) / λ

EXPLANATION:

• Statement 1: Amount of radioactive substance remained after three half-lives is 1/8 of the original amount.
• This is true. After one half-life, half of the original amount remains. After three half-lives, (1/2)³ = 1/8 of the original amount remains.
• Statement 2: Decay constant does not depend upon temperature.
• This is true. The decay constant is a fundamental property of the radioactive substance and is not affected by temperature.
• Statement 3: Decay constant increases with increase in temperature.
• This is not true. The decay constant is independent of temperature.
• Statement 4: Half life is ln(2) times of 1/rate constant.
• This is true. The half-life (t_1/2) is given by t_1/2 = ln(2) / λ, where λ is the decay constant.

Therefore, the statement that is not true for radioactive decay is: Decay constant increases with increase in temperature.

CONCEPT:

Stereoisomers and Geometrical Isomerism

• Stereoisomers are compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of atoms.
• Geometrical isomerism occurs in compounds with restricted rotation around a double bond or ring, resulting in cis-trans isomers.
• Each chiral center or double bond contributes to possible isomerism.

EXPLANATION:

• 

It has 4 stereoisomers (\left[\begin{array}{ll} \mathrm{R} \text { cis } & \mathrm{R} \text { trans } \ \mathrm{Scis} & \mathrm{Strans} \end{array}\right])

• Two factors contribute to stereoisomerism here:
• The double bond between the third and fourth carbon introduces cis and trans isomers.
• The chiral carbon (second carbon with -CH₃, -CH=CH-CH₃, -OH, and -H as substituents) introduces R and S configurations.
• Combining these, the molecule can exhibit the following configurations:
• Cis with R configuration
• Cis with S configuration
• Trans with R configuration
• Trans with S configuration
• Hence, the molecule has 4 possible stereoisomers.

Therefore, the total number of stereoisomers for the given molecule is 4.

EXPLANATION:​.

Electronegativity

• Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used scale for electronegativity.
• The electronegativity values (on the Pauling scale) for some elements are as follows:
• | | Li | Be | B | C | N | O | F | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = On pauling scale | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4.0 |

For elements in the second period and third period:

• | | Na | Mg | Al | Si | P | S | Cl | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = | 0.9 | 1.2 | 1.5 | 1.8 | 2.1 | 2.5 | 3.0
  |

 

Therefore, the electronegativity order that is incorrect is: Al < Mg < B < N

CONCEPT:

Electronic Configuration of Lanthanoids

• Lanthanoids are a series of 15 elements from lanthanum (La) to lutetium (Lu) in the periodic table, characterized by the filling of the 4f orbitals.
• The general electron configuration of lanthanoids is [Xe] 4fⁿ 5d^x 6s².
• The common oxidation states of lanthanoids include +2, +3, and sometimes +4.
• Key electron configurations:
• Europium (Eu): [Xe] 4f⁷ 6s²
• Gadolinium (Gd): [Xe] 4f⁷ 5d¹ 6s²
• Terbium (Tb): [Xe] 4f⁹ 6s²
• Samarium (Sm): [Xe] 4f⁶ 6s²

EXPLANATION:

• To determine which ions have a 4f⁷ configuration, we must consider the electron loss when forming the given ions:
• Europium (Eu):
• Eu²⁺: [Xe] 4f⁷ (Loses 2 electrons from 6s²)
• Eu³⁺: [Xe] 4f⁶ (Loses 2 electrons from 6s² and 1 electron from 4f⁷)
• Gadolinium (Gd):
• Gd³⁺: [Xe] 4f⁷ (Loses 2 electrons from 6s² and 1 electron from 5d¹)
• Terbium (Tb):
• Tb³⁺: [Xe] 4f⁸ (Loses 2 electrons from 6s² and 1 electron from 4f⁹)
• Samarium (Sm):
• Sm²⁺: [Xe] 4f⁶ (Loses 2 electrons from 6s²)

Therefore, the lanthanoid ions with a 4f⁷ configuration are Eu²⁺ and Gd³⁺.

EXPLANATION:

Trends in the Periodic Table

• Ionic Radii: The size of the ionic radii decreases as the positive charge of the ion increases.
• Order: Al³⁺ < Mg²⁺ < Na⁺ < F^-
• Ionisation Enthalpy: The energy required to remove an electron from an atom in its gaseous state. It generally increases across a period from left to right.
• Order: B < C < O < N
• Metallic Character: The degree to which an element exhibits metallic properties, which decreases across a period and increases down a group.
• Order: B < Al < Mg < K
• Electronegativity: The tendency of an atom to attract electrons in a chemical bond. It generally increases across a period from left to right.
• Order: Si < P < S < Cl

The correct matching is  A-IV, B-I, C-II, D-III

CONCEPT:

Vitamins and their Chemical Names

• Vitamins are essential nutrients required by the body to function properly. Each vitamin has a specific chemical name.
• Ascorbic acid is the chemical name for Vitamin C.

EXPLANATION:

• Adipic acid: This is not a vitamin, but a dicarboxylic acid commonly used in the production of nylon.
• Aspartic acid: This is not a vitamin, but an amino acid involved in the synthesis of proteins.
• Ascorbic acid: This is Vitamin C, an essential nutrient for the repair of tissues and the enzymatic production of certain neurotransmitters.
• Saccharic acid: This is not a vitamin, but an oxidized derivative of glucose.

Therefore, the acid that is a vitamin is ascorbic acid (Vitamin C).

The correct answer is: (3) Ascorbic acid

CONCEPT:

First Law of Thermodynamics

• The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (q) plus the work done on the system (w):
• ΔU = q + w
• In a thermally insulated (adiabatic) system, there is no heat exchange with the surroundings, so:
• q = 0
• When work is done on the system (for example, by mechanically stirring the liquid), the work done (w) is positive, which increases the internal energy (ΔU) of the system.

EXPLANATION:

• Since the vessel is thermally insulated, there is no heat exchange with the surroundings:
• q = 0
• The work is done on the liquid by mechanically stirring it, so:
• w > 0
• According to the first law of thermodynamics:
• ΔU = q + w
• Since q = 0, we have ΔU = w
• When w > 0, ΔU > 0

Therefore, the correct option for the thermodynamic parameters is  ΔU > 0, q = 0, w > 0​

CONCEPT:

Bohr's Model of the Atom

• In Bohr's model, the radius of the ( n )th orbit of an atom is given by:
• (r_n = \frac{n^2 a_0}{Z})
• where:
• ( rn ) = radius of the ( n )th orbit
• ( a0 ) = Bohr radius (radius of the first stationary state of the hydrogen atom)
• ( n ) = principal quantum number
• ( Z ) = atomic number of the element
• For the first excited state, ( n = 2 ).

APPLICATION TO HELIUM ION (He⁺):

• For the Helium ion (He⁺), the atomic number ( Z ) is 2.
• For the first excited state, ( n = 2 ).
• Substituting these values into the formula for the radius:
• (r = \frac{n^2 a_0}{Z} \ = \frac{(2)^2 a_0}{2} \ = \frac{4 a_0}{2} \ = 2 a_0)

Therefore, the radius of the first excited state of the Helium ion is r = 2a0

CONCEPT:

SN₁ and SN₂ Reactions

• SN₁ Mechanism: This is a two-step mechanism where the rate-determining step involves the formation of a carbocation intermediate. It is favored by:
• Stable carbocations (tertiary > secondary > primary).
• Polar protic solvents.
• SN₂ Mechanism: This is a single-step mechanism involving a backside attack by the nucleophile. It is favored by:
• Less steric hindrance (methyl > primary > secondary; tertiary is unfavorable).
• Polar aprotic solvents.

EXPLANATION:

CH3 – O – CH2 – Cl will undergo SN1 mechanism because (\mathrm{CH}{3}-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H}{2})  is highly stable. 

• Statement I: CH₃–O–CH₂–Cl undergoes the SN₁ mechanism even though it is a primary halide because the intermediate carbocation (CH₃–O–CH₂⁺) formed is stabilized by resonance with the oxygen atom.
• Statement II: CH₃–C(CH₃)₂–CH₂–Cl (neopentyl chloride) does not undergo the SN₂ mechanism easily due to significant steric hindrance around the reactive center, which obstructs the nucleophilic backside attack.
• Both statements are correct based on the mechanisms and the structural properties of the molecules.

Therefore, both Statement I and Statement II are correct.

CONCEPT:

Reactions of Propyne with Sodium and Sodium Amide

• When alkynes react with metallic sodium in the presence of liquid ammonia, they release hydrogen gas. The number of moles of H₂ gas liberated depends on the available acidic hydrogen atoms in the molecule.
• Reaction with sodium amide (NaNH₂) leads to the formation of a sodium salt of the alkyne and releases ammonia (NH₃) gas. This reaction is used to identify terminal alkynes.

EXPLANATION:

• Statement I: One mole of propyne (CH₃–C≡CH) reacts with an excess of sodium (Na) to liberate 0.5 moles of hydrogen gas (H₂), as each terminal hydrogen in the alkyne contributes to the reaction. This statement is correct.
• Statement II: Four grams of propyne contain 0.1 moles of propyne (molecular weight = 40 g/mol). When reacted with NaNH₂, this amount of propyne will release 0.1 moles of NH₃ gas, which occupies 2240 mL at STP (not 224 mL as stated). Hence, this statement is incorrect.
• (\frac{4}{40}=0.1 \text { mole } \quad \frac{0.1 \mathrm{~mole}}{2240 \mathrm{~mole}})
• Statement I is correct but Statement II is incorrect

Therefore, Statement I is correct, but Statement II is incorrect.

CONCEPT:

Equilibrium Constant (K_P) for Gas Reactions

• The equilibrium constant, K_P, is calculated using the partial pressures of the gases involved in the reaction at equilibrium.
• For a reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K_P = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b

where P represents the partial pressure of the gases at equilibrium.

• Pure solids and liquids do not appear in the expression for K_P.

EXPLANATION:

• The given reaction is:

CO₂(g) + C(s) ⇌ 2CO(g)

• CO₂ is the reactant gas.
• Graphite (C) is a solid and does not appear in K_P expression.
• CO is the product gas.
• Initial pressure of CO₂ is 0.5 atm.
• Let 'x' be the pressure change due to the conversion of CO₂ to CO at equilibrium:
• Pressure of CO₂ at equilibrium: (0.5 - x) atm
• Pressure of CO at equilibrium: 2x atm
• (\begin{array}{lc} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{~s}) & \rightleftharpoons 2 \mathrm{CO}(\mathrm{~g}) \ 0.5 & - \ 0.5-\mathrm{x} & 2 \mathrm{x} \end{array})
• Given: Total pressure at equilibrium is 0.8 atm:
• (0.5 - x) + 2x = 0.8
• Solving for x: 0.5 + x = 0.8
• x = 0.3
• Equilibrium pressures:
• P_CO2 = 0.5 - 0.3 = 0.2 atm
• P_CO = 2(0.3) = 0.6 atm
• Plugging these values into the K_P expression:
• K_P = (P_CO)² / (P_CO2)
• K_P = (0.6)]² / 0.2
• K_P = 0.36 / 0.2
• K_P = 1.8 atm

Therefore, the correct answer is 1.8 atm

CONCEPT:

IUPAC Nomenclature of Organic Compounds

• The IUPAC name of a compound is determined based on the longest carbon chain containing the principal functional group.
• Priority is given to functional groups in the following order: carboxylic acid (-COOH) > ester (-COOR) > ketone (-C=O) > alcohol (-OH) > amine (-NH₂) > alkene > alkyne > alkane.
• Substituents are named and numbered to give the lowest possible position numbers to the functional groups and substituents.

EXPLANATION:

• The given compound contains two functional groups:
• A carboxylic acid (-COOH) group, which has the highest priority.
• An ester group (-COOCH₃) as a substituent.
• The longest carbon chain containing the carboxylic acid group has six carbons (hexane). The main chain is numbered from the carboxylic acid carbon to give substituents the lowest possible numbers.
• The substituents on the chain are:
• A methoxycarbonyl (-COOCH₃) group at position 6.
• Methyl groups at positions 2 and 5.
• The compound's name is written as: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

Therefore, the IUPAC name of the compound is: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

CONCEPT:

Electrolysis of Sulfuric Acid

• During electrolysis, an electrolyte is decomposed into its constituent ions through the application of an electrical current.
• H₂S₂O₈ (peroxodisulfuric acid) can be produced by the electrolysis of concentrated sulfuric acid solutions.
• At the anode, the half-reaction is:

2 HSO₄^- → H₂S₂O₈ + 2e^-

EXPLANATION:

• To efficiently produce H₂S₂O₈, a concentrated solution of sulfuric acid is required because:
• In a concentrated solution, the concentration of HSO₄^- ions is high enough to facilitate the formation of peroxodisulfuric acid at the anode.
• Dilute solutions of sulfuric acid or sodium sulfate do not provide the necessary ion concentrations for this reaction.

Therefore, the correct answer is: (3) Concentrated solution of sulfuric acid.

CONCEPT:

Fehling's Test

• Fehling's test is used to detect the presence of reducing sugars and aldehydes.
• Compounds with an aldehyde group (-CHO) or α-hydroxy ketones give a positive Fehling's test, resulting in the formation of a red precipitate of cuprous oxide (Cu₂O).
• Ketones and aromatic aldehydes (like benzaldehyde) generally do not give a positive Fehling's test.

EXPLANATION:

• Compound (A): Benzaldehyde (C₆H₅CHO) is an aromatic aldehyde and does not give a positive Fehling's test.
• Compound (B): Acetophenone (C₆H₅COCH₃) is a ketone and does not give a positive Fehling's test.
• Compound (C): Glucose derivative (HOCH₂–CO–(CHOH)₃–CH₂OH) contains an α-hydroxy aldehyde functional group and gives a positive Fehling's test.
• Compound (D): Propionaldehyde (CH₃CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.
• Compound (E): Phenylacetaldehyde (C₆H₅CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.

Therefore, the compounds (C), (D), and (E) give a positive Fehling's test.

CONCEPT:

Crystal Field Stabilization Energy (CFSE)

• CFSE is the energy difference between the energy of an electron configuration in the ligand field and in an isotropic field.
• It depends on:
• The splitting of d-orbitals (t_2g and e_g) in the crystal field.
• The number of electrons in these orbitals.
• CFSE is calculated as:

CFSE = [(number of electrons in t_2g) × (-0.4Δ₀)] + [(number of electrons in e_g) × (+0.6Δ₀)]

EXPLANATION:

For complex K3 [Fe(SCN)6] 

Calculation of CFSE

= (–0.4 × 3 + 0.6 × 2) Δ0

= 0 Δ0

• [Fe(NH₃)₆]Br₂:
• Fe²⁺ (d⁶) forms a high-spin complex due to NH₃ being a weak field ligand. This results in CFSE ≠ 0.
• [Fe(en)₃]Cl₃:
• Fe³⁺ (d⁵) forms a low-spin complex due to en (ethylenediamine) being a strong field ligand. CFSE ≠ 0.
• K₄[Fe(CN)₆]:
• Fe²⁺ (d⁶) forms a low-spin complex due to CN^- being a strong field ligand. CFSE ≠ 0.
• K₃[Fe(SCN)₆]:
• SCN^- is a weak field ligand, and Fe³⁺ (d⁵) forms a high-spin complex. The electron configuration is t_2g³ e_g², resulting in:
• CFSE = [-0.4(3) + 0.6(2)]Δ₀
• CFSE = 0

Therefore, in K₃[Fe(SCN)₆], the CFSE (Δ₀) is equal to zero.

CONCEPT:

Boiling Point Elevation

• The boiling point elevation (ΔT_b) of a solution is directly proportional to the molal concentration (m) of the solute and the van't Hoff factor (i).
• The formula is given by:

ΔT_b = i × K_b × m

where K_b is the ebullioscopic constant of the solvent.

• For ionic compounds, the van't Hoff factor (i) represents the number of particles the compound dissociates into in solution.
• For non-electrolytes like urea, i = 1.

EXPLANATION:

ΔTb = i Kb . m . ∝ i.C.

where C = concentration 

 

B.P. order (ii) < (i) < (iii) < (iv)

Therefore, the correct answer is (ii) < (i) < (iii) < (iv).

CONCEPT:

Aromatic Substitution and Diazotization

• Bromination in AcOH: Bromine in acetic acid selectively introduces a bromine atom at the para position relative to the nitro group (-NO₂) due to its electron-withdrawing nature.
• Reduction: The nitro group (-NO₂) is reduced to an amino group (-NH₂) using Sn and HCl.
• Diazotization: Amino groups react with NaNO₂ and HCl at low temperatures (273 K) to form diazonium salts (-N₂⁺).
• Reaction with Ethanol: Diazonium salts react with ethanol (EtOH) to form aldehydes, specifically acetaldehyde (CH₃CHO), as a product of the Etard reaction mechanism.

EXPLANATION:

• Step 1: Bromination
• The nitrobenzene reacts with Br₂ in acetic acid (AcOH) to produce para-bromo nitrobenzene.
• Step 2: Reduction
• Para-bromo nitrobenzene is reduced with Sn and HCl to form para-bromo aniline (p-bromoaniline).
• Step 3: Diazotization
• p-Bromoaniline reacts with NaNO₂ and HCl at 273 K to form para-bromo benzene diazonium chloride.
• Step 4: Reaction with Ethanol
• The diazonium salt reacts with ethanol (EtOH), forming para-bromo benzene and acetaldehyde (CH₃CHO).

Therefore, the products formed are para-bromo benzene and acetaldehyde (CH₃CHO).

CONCEPT:

Magnetic Behaviour and Geometries

• The magnetic behaviour of a complex can often be used to infer its geometry and the oxidation state of the central metal ion.
• [NiCl₄]²⁻:
• Ni²⁺ has the electronic configuration [Ar] 3d⁸.
• In a tetrahedral field, the 3d orbitals do not experience significant splitting, retaining unpaired electrons and leading to a paramagnetic nature.
• This suggests a tetrahedral geometry.
• [Ni(CO)₄]:
• Ni in zero oxidation state (Ni(0)) has the electronic configuration [Ar] 3d¹⁰.
• In a tetrahedral field, the d orbitals become fully paired in the presence of strong field ligands (like CO), leading to a diamagnetic nature.
• This suggests a tetrahedral geometry.

EXPLANATION:

• [NiCl₄]²⁻:
• Oxidation state of Ni is +2.
• Ni²⁺ electronic configuration: [Ar] 3d⁸.
• If the geometry is tetrahedral, the hybridization is sp³.
• With 2 unpaired electrons, it exhibits paramagnetism.
• [Ni(CO)₄]:
• Oxidation state of Ni is 0.
• Ni(0) electronic configuration: [Ar] 3d¹⁰.
• If the geometry is tetrahedral, the hybridization is sp³.
• With no unpaired electrons, it exhibits diamagnetism.

Therefore, the correct answer is [NiCl₄]²⁻: Ni²⁺, tetrahedral, [Ni(CO)₄]: Ni(0), tetrahedral

CONCEPT:

Geometrical Isomerism

• Geometrical isomerism (cis-trans isomerism) occurs in compounds with restricted rotation around a double bond or in cyclic structures.
• Cis-isomer: Identical groups are on the same side of the double bond.
• Trans-isomer: Identical groups are on opposite sides of the double bond.
• Properties such as dipole moment and melting point differ for cis and trans isomers:
• Cis-isomers usually have higher dipole moments due to group orientation.
• Trans-isomers generally have higher melting points due to better packing in the solid state.

EXPLANATION:

• (A) Propene does not show geometrical isomerism:
• Geometrical isomerism requires two different groups on each carbon of the double bond, which propene (CH₃–CH=CH₂) lacks.

Incorrect

• (B) Trans-isomer has identical groups on opposite sides:
• This is true; geometrical isomers differ based on the position of groups around the double bond.

Correct

• (C) Cis-but-2-ene has a higher dipole moment than trans-but-2-ene:
• 

(dipole moment only)

• In cis-isomer, the dipoles of substituent groups add up, while in trans-isomer, the dipoles cancel out.

Correct

• (D) 2-Methylbut-2-ene shows two geometrical isomers:
• 

 (does not show GI) 

• This is incorrect as 2-methylbut-2-ene does not have two different groups on each carbon of the double bond.

Incorrect

• (E) Trans-isomer has a lower melting point than cis-isomer:
• 

 (Melting point) 

• Trans-isomers generally have a higher melting point due to better packing in the solid state.

Incorrect

Therefore, the incorrect statements are: (A), (D), and (E).

Concept:

Vernier Caliper

• Vernier Caliper is a precision instrument that can be used to measure internal and external distances accurately.
• A vernier caliper consists of two main parts:
• the main scale engraved on a solid L-shaped frame and the vernier scale that can slide along the main scale.
• It works on the principle of vernier and can measure the dimensions to an accuracy of 0.02 mm.

Parts of a Vernier caliper:

• Outside jaws: Used to measure the external diameter or width of an object
• Inside jaws: Used to measure the internal diameter of an object
• Locking screw: to lock the jaws
• Adjusting screw: to take an accurate measurement of the workpiece.

 

Explanation:

In general one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct.

Also least count is given by one main scale division / number of vernier scale division for normal vernier calliper.

Calculation:

Total charge inside the cube

= (\frac{\lambda \frac{\mathrm{a}}{2}}{4}=\frac{\lambda \mathrm{a}}{8})

∴ (\phi=\frac{\mathrm{q}{\mathrm{in}}}{\varepsilon{0}}=\frac{\lambda \mathrm{a}}{8 \varepsilon_{0}})

Calculation:

The circuit can be considered as 

∴ R_eq = 0.5 + (\frac{0.5 \times 2}{2+0.5}=\left(\frac{5}{10}+\frac{10}{25}\right) \Omega)

= (\frac{45}{50}=\frac{9}{10}=0.9)

∴ i = (\frac{0.9}{0.9}=1 \mathrm{~A})

CONCEPT:

Young's double-slit experiment

• Young’s double-slit experiment helped in understanding the wave nature of light.
• The original Young’s double-slit experiment used diffracted light from a single monochromatic source of light.
• The light that comes from the monochromatic source is passed into two slits to be used as two coherent sources.
• At any point on the screen at a distance ‘y’ from the center, the waves travel distances l1 and l2 to create a path difference of Δl at that point.
• If there is a constructive interference on the point then the bright fringe occurs.
• If there is a destructive interference on the point then the dark fringe occurs.
• The distance of the nth bright fringe from the central fringe is given as,

(⇒ y=\frac{nλ D}{d})

Where d = distance between slits, D = distance between slits and screen, and λ = wavelength

​

Explanation:

β (fringe width) = (\frac{λ \mathrm{D}}{\mathrm{~d}})

In denser medium, λ ↓ ⇒ β ↓

⇒ fringe come closer 

Also, µ = (\frac{c}{V} ) ⇒ V = (\frac{c}{μ})

Frequency remains same, 

⇒ μ = (\frac{\lambda_{\text {vac. }} \mathrm{f}}{\lambda_{\text {med }} \mathrm{f}}) ⇒ (\lambda_{\text {med }}) = (\frac{\lambda_{\mathrm{vac} .}}{\mu})

Calculation:

We know that

(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4} \Rightarrow \mathrm{P} \propto \mathrm{AT}^{4})

(\frac{\mathrm{P}{\text {smaller }}}{\mathrm{P}{\text {larger }}}=\frac{(0.2)^{2} \times 800^{4}}{(0.8)^{2} \times 400^{4}})

(\frac{1}{16} \times 16=1)

∴ P_larger = P_smaller 

Calculation:

(\rm\Delta Q_{1}=m \times S_{1} \times \Delta T=10^{-3} \times 2100 \times 10=21 \mathrm{~J})

(\rm\Delta \mathrm{Q}{2}=\mathrm{m} \times \mathrm{L}{\mathrm{f}}=10^{-3} \times 3.35 \times 10^{5}=335 \mathrm{~J})

(\rm\Delta Q_{3}=m \times S_{w} \times \Delta T=10^{-3} \times 4180 \times 100=418 \mathrm{~J})

(\rm \Delta \mathrm{Q}{4}=\mathrm{m} \times \mathrm{L}{\mathrm{v}}=10^{-3} \times 2.25 \times 10^{6}=2250 \mathrm{~J})

(\rm\Delta Q_{5}=\mathrm{m} \times \mathrm{S}_{\mathrm{v}} \times \Delta \mathrm{T}=10^{-3} \times 1920 \times 10=19.2 \mathrm{~J})

(\rm \Delta Q_{\mathrm{net}}=3043.2 \mathrm{~J})

Concept:

de Broglie wavelength:

• Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
• The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength can be calculated from Planks constant h divided by the momentum of the particle.

(\lambda = \frac {h}{mv})

where λ is de Broglie wavelength, h is Plank's constant, and mv is the momentum.

• If different particles have the same velocity, then the wavelength is inversely proportional to the mass of that particle.

Calculation:

(\lambda=\frac{\mathrm{h}}{\mathrm{mv}})

(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi})

(\mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}})

(\lambda=\frac{2 \pi \mathrm{rh}}{\mathrm{nh}})

(\lambda \propto \frac{\mathrm{r}}{\mathrm{n}})

(\frac{\lambda_{1}}{\lambda_{4}}=\frac{r_{1} n_{4}}{n_{1} r_{4}}=\frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}})

(\frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4})

Calculation:

⇒ p_eq = p₁ + p₂ + p₃  

⇒ p1 = (\left(\frac{1}{3\left|\mathrm{R}_{1}\right|}\right))

⇒ p2 = (\left(\frac{1}{2}\right)\left(\frac{1}{-\left|R_{1}\right|}-\frac{1}{-\left|R_{2}\right|}\right))

⇒ p2 = (\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}{2}\right|}-\frac{1}{\left|\mathrm{R}{1}\right|}\right))

⇒ p3 = (\left(\frac{1}{3}\right)\left(\frac{1}{-\left|\mathrm{R}{2}\right|}-\frac{1}{\infty}\right)=-\frac{1}{3\left|\mathrm{R}{2}\right|})

⇒ peq = (\frac{1}{3}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right)-\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right))

=  (-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right))

Calculation:

⇒ t = (\frac{\ell}{V_{x}}=\frac{10 × 10^{-2}}{10^{6}}=10^{-7})

⇒ V_y = u_y + a_yt

⇒ Vy = 0 + (\frac{\mathrm{eE}}{\mathrm{~m}} × 10^{-7})

⇒ Vy = (\frac{1.6 × 10^{-19}}{9.1 × 10^{-31}} × 9.1 × 10^{-2} × 10^{-7})

⇒ Vy =  16 × 10⁴

Explanation:

Resistivity is independent of temperature for wire-bound resistors

Calculation:

9^th harmonic of closed pipe = (\frac{9 V_{1}}{4 \ell_{1}})

4^th harmonic of open pipe = (\frac{2 \mathrm{~V}{2}}{\ell{2}})

∴ (\frac{9 \mathrm{~V}{1}}{4 \ell{1}}=\frac{2 \mathrm{~V}{2}}{\ell{2}})

∴ (\frac{9}{4 \ell_{1}} \sqrt{\frac{\mathrm{~B}}{\rho_{1}}}=\frac{2}{\ell_{2}} \sqrt{\frac{\mathrm{~B}}{\rho_{2}}} \Rightarrow \frac{\ell_{2}}{\ell_{1}}=\frac{8}{9} \sqrt{\frac{\rho_{1}}{\rho_{2}}})

(\ell_{2}=\ell_{1} \times \frac{8}{9} \times \frac{1}{4}=\frac{20}{9} \mathrm{~cm})

Calculation:

(\mathrm{I}=\frac{\mathrm{MR}^{2}}{2}-\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^{2}}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^{2}\right])

(\mathrm{I}=\frac{13}{32} \mathrm{MR}^{2})

Calculation: 

(\mathrm{F}_{1}=\frac{\mathrm{GMm}}{(2 \mathrm{R})^{2}}\quad\quad...(1))

(\mathrm{F}_{2}=\frac{\mathrm{GMm}}{(2 \mathrm{R})^{2}}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^{2}}\right))

(\mathrm{F}_{2}=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}\quad\quad...(2))

F₁ : F₂ = 12 : 11

CONCEPT:

• Photoelectric effect: When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.

Photo-electric cell

• The photo-electric cell works on the principle of the photoelectric effect.
• The photo-electric cell consists of an evacuated glass tube containing two electrodes emitter (C) and Collector (A).
• The emitter is always kept at a negative potential.
• The collector is made of a metal rod and is always kept at a positive potential.
• When the light of frequency more than the threshold frequency of material of emitter is made incident on the emitter, photo-emission takes place.
• Then the photo­electrons are attracted to the collector which is positive with respect to the emitter. Thus current flows in the circuit. If the intensity of incident radiation is increased the photoelectric current increases.

Calculation:

The energy of incident light is E = (\frac{1240}{\lambda}=\frac{1240}{550} \simeq 2.25) eV

Thus it is less than 2.5 eV (Li) and greater than 1.9 eV (Cs).

Photoelectric effect only can be possible for Cs.

Explanation:

We know that 

B = μ₀ni

(\left[\frac{\mathrm{B}}{\mu_{0}}\right]=[\mathrm{ni}]=\mathrm{L}^{-1} \mathrm{~A}^{1})

Cakculation:

(\frac{1}{2} \mathrm{mv}{\mathrm{A}}^{2}=\frac{1}{2} \mathrm{mv}{\mathrm{B}}^{2}+\mathrm{mgh})

⇒ (\frac{1}{2} m(5 g \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^{2}+\mathrm{mg} \frac{\ell}{2})

⇒ (\frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}})

⇒ (\mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell)

(\frac{1}{2} \mathrm{mv}{\mathrm{C}}^{2}=\frac{1}{2} \mathrm{mv}{\mathrm{D}}^{2}+\mathrm{mg} \frac{\ell}{2})

⇒ (\mathrm{KE}{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \Rightarrow \frac{\mathrm{KE}{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2)

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:

1. Primary cell: This type of cell cannot be recharged.
2. Secondary cell: This type of cell can be recharged.

• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

• If the number of cells are connected end to end that the positive terminal of one cell is connected to the negative terminal of the succeeding cell then it is called a series arrangement of cells.
• The equivalent emf of cells in series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

• The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

Cells in parallel:

• If the number of cells is connected such that the positive terminals are connected together at one point and the negative terminals of these cells are connected together at another point then it is called a parallel arrangement of cells.
• The equivalent emf of cells in a parallel arrangement is given as,

(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n})

• The equivalent internal resistance of cells in a parallel arrangement is given as,

(⇒ \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...+\frac{1}{r_n})

Calculation:

= (\rm \frac{\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\varepsilon)

Explanation:

Forward Biased:

• The diode is said to be forward-biased when the P region is connected to a more positive terminal and N region is connected to the negative terminal of the battery.
• A forward-biased diode offers very little resistance to the flow of current and hence a large current flow through the diode.
• When forward biased, the diode acts as an ON switch.

Reversed Biased:

• A diode is said to be reversed biased when the P terminal is connected to lower potential than the N terminal
• A reverse-biased diode offers very large resistance to the flow of current and hence a small minority of current flows through the diode

How to identify forward and reversed biased diode:

1. If the input voltage is greater than the other end voltage, then it is forward-biased and vice-versa.
2. If the voltage at the p-section is greater than the n-section, then is forward biased and vice-versa.

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
• In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​Parallel plate capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​
• The electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,

(\Rightarrow E=\frac{σ}{\epsilon_o}=\frac{Q}{A\epsilon_o})

• The potential difference between the plates is given as,

(\Rightarrow V=\frac{Qd}{A\epsilon_o})

• The capacitance C of the parallel plate capacitor is given as,

(\Rightarrow C=\frac{Q}{V}=\frac{A\epsilon_o}{d})

Where A = area of the plates, d = distance between the plates, Q = charge on the plates, and σ = surface charge density

Calculation:

Δq = (KC - C)V

= 40 × 10^-6 × 100

= 4000 × 10^-3 = 4mC

(\Delta \mathrm{U}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{2}-\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2}(\mathrm{~K}-1) \mathrm{CV}^{2})

= (\frac{1}{2} \mathrm{CV}^{2}(2-1))

= (\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 40 \times 10^{-6} \times 10000)

= 0.2 J

Calculation:

P_eq = 2P_ℓ + P_m

(-\frac{1}{\mathrm{f}{\mathrm{Q}}}=\frac{2}{\mathrm{f}{\ell}}-\frac{1}{\mathrm{f}_{\mathrm{m}}})

= (\frac{4(\mu-1)}{R}-\frac{2}{-R}=\frac{1}{R}(4 \mu-4+2))

(-\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}(4 \mu-2))

⇒ (\frac{1}{f_{e q}}=\frac{-1}{R}(4 \mu-2))

(\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{R}}{2})

R = (2 f_{\mathrm{eq}}=-2\left(\frac{\mathrm{R}}{4 \mu-2}\right)=\frac{-\mathrm{R}}{(2 \mu-1)})