JEE Questions & Answers

Q001

Problem

Let $a, b, c$ be three non-zero real numbers such that the equation \[ \sqrt{3} a \cos x + 2 b \sin x = c, \, x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \] has two distinct real roots $\alpha$ and $\beta$ with $\alpha + \beta = \frac{\pi}{3}$. Then, the value of $\frac{b}{a}$ is

Solution

Step 1: Rewrite the equation $\sqrt{3} a \cos x + 2 b \sin x = c$ as a linear combination of sine and cosine, and compare it to the standard form $R\cos(x - \theta)$ where $\tan \theta = \frac{2b}{\sqrt{3}a}$.

Step 2: Use the identity for the sum of roots of a trigonometric equation, which is $2\alpha = \frac{\pi}{3}$, in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so finding the roots requires $\cos^{-1}\left(\frac{c}{R}\right)$ and $-\cos^{-1}\left(\frac{c}{R}\right)$.

Step 3: Given that $\alpha + \beta = \frac{\pi}{3}$, solve for $\tan\left(\frac{\pi}{3}\right) = \frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$. Thus, $\frac{2b}{\sqrt{3}a} = \sqrt{3}$, which simplifies to $b = a$.

Final Answer: $\frac{b}{a} = 1.00$

Q002

JEE PHYSICS Capilarity

Problem

A uniform capillary tube of inner radius $ r $ is dipped vertically into a beaker filled with water. The water rises to a height $ h $ in the capillary tube above the water surface in the beaker. The surface tension of water is $ \sigma $. The angle of contact between water and the wall of the capillary tube is $ \theta $. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?
(A) For a given material of the capillary tube, $ h $ decreases with increase in $ r $
(B) For a given material of the capillary tube, $ h $ is independent of $ \sigma $
(C) If this experiment is performed in a lift going up with a constant acceleration, then $ h $ decreases
(D) $ h $ is proportional to contact angle $ \theta $

Solution

Step 1: The height $ h $ to which the liquid rises in a capillary tube is given by the formula:

$$ h = \frac{2\sigma \cos \theta}{\rho g r} $$ where $ \rho $ is the density of the liquid and $ g $ is the acceleration due to gravity.

Step 2: Analyzing each statement:

(A) $ h $ is inversely proportional to $ r $, so $ h $ decreases with an increase in $ r $. This statement is true.

(B) $ h $ is directly proportional to $ \sigma $, so it is not independent of $ \sigma $. This statement is false.

(C) If the lift is going up with an acceleration $ a $, effective gravity becomes $ g_{\text{eff}} = g + a $. Thus, $ h $ is given by:

$$ h = \frac{2\sigma \cos \theta}{\rho (g + a) r} $$ Hence, $ h $ decreases. This statement is true.

(D) $ h $ is proportional to $ \cos \theta $, not directly to $ \theta $. Therefore, this statement is false.

Final Answer: The true statements are (A) and (C).

Q003

Chemistry Advanced Paper 1

Problem

The calculated spin only magnetic moments of [Cr(NH3)6]3+ and [CuF6]3- in BM, respectively, are (Atomic numbers of Cr and Cu are 24 and 29, respectively) (A) 3.87 and 2.84 (B) 4.90 and 1.73 (C) 3.87 and 1.73 (D) 4.90 and 2.84.

Solution

Step 1: Determine the oxidation state and electron configuration of Cr in [Cr(NH3)6]3+.

Cr in [Cr(NH3)6]3+ has an oxidation state of +3, so its electronic configuration is [Ar] 3d³. This gives 3 unpaired electrons.

Step 2: Calculate the spin-only magnetic moment for [Cr(NH3)6]3+.

The spin-only magnetic moment $\mu$ is calculated using the formula:

$$ \mu = \sqrt{n(n+2)} \, \text{BM} $$

where $n$ is the number of unpaired electrons.

For 3 unpaired electrons:

$$ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM} $$

Step 3: Determine the oxidation state and electron configuration of Cu in [CuF6]3-.

Cu in [CuF6]3- has an oxidation state of +3, so the electronic configuration is [Ar] 3d⁸. This gives 2 unpaired electrons.

Step 4: Calculate the spin-only magnetic moment for [CuF6]3-.

$$ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \, \text{BM} $$

Final Answer: (A) 3.87 and 2.84

Q004

Physics Electricity and Capacitance Advanced Paper 1

Problem

Three identical capacitors $C_1$, $C_2$, and $C_3$ have a capacitance of 1.0 µF each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_r$. The cell electromotive force (emf) $V_0 = 8 \, \text{V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C_3$ is found to be 5 µC. The value of $\varepsilon_r =$ __________.

Solution

Step 1: Calculate the initial charge on capacitor $C_3$ when $S_1$ is closed and $S_2$ is open. The capacitance between $C_3$ and the series combination of $C_1$ and $C_2$ is $C_{eq} = 1 \, \mu F + \frac{1}{\frac{1}{1} + \frac{1}{1}} \, \mu F = 1 \, \mu F + 0.5 \, \mu F = 1.5 \, \mu F$. The charge is $Q = C_{eq} \cdot V_0 = 1.5 \, \mu F \times 8 \, \text{V} = 12 \, \mu C$.

Step 2: When $S_1$ is opened and $S_2$ is closed, the charge redistributes between $C_1$, which has a dielectric, and $C_3$. Let the effective capacitance of $C_1$ be $C_{1}' = \varepsilon_r \times 1\, \mu F$. The total charge $12 \, \mu C$ is now shared between $C_1'$ and $C_3$.

Step 3: Set up the equation for charge distribution: $Q_3 = \frac{C_3}{C_1' + C_3} \times 12 \, \mu C = 5 \, \mu C$. Given $C_3 = 1 \, \mu F$, solve $ \frac{1}{\varepsilon_r + 1} \times 12 = 5$.

Step 4: Solve $\frac{12}{\varepsilon_r + 1} = 5$. Multiply to clear the fraction: $12 = 5(\varepsilon_r + 1)$, leading to $\varepsilon_r = \frac{12}{5} - 1 = \frac{7}{5} = 1.4$.

Final Answer: $\varepsilon_r = 1.40$.

Q005

Maths Number theory Advanced Paper 1

Problem

Let $ S = \{ a + b\sqrt{2} : a, b \in \mathbb{Z} \} $, $ T_1 = \{ (-1 + \sqrt{2})^n : n \in \mathbb{N} \} $, and $ T_2 = \{ (1 + \sqrt{2})^n : n \in \mathbb{N} \} $. Then which of the following statements is (are) TRUE?
(A) $ \mathbb{Z} \cup T_1 \cup T_2 = S $
(B) $ T_1 \cap \left(0, \frac{1}{2024}\right) = \emptyset $, where $ \emptyset $ denotes the empty set.
(C) $ T_2 \cap (2024, \infty) eq \emptyset $
(D) For any given $ a, b \in \mathbb{Z} $, $ \cos\left(\pi(a + b\sqrt{2})\right) + i\sin\left(\pi(a + b\sqrt{2})\right) \in \mathbb{Z} $ if and only if $ b = 0 $, where $ i = \sqrt{-1} $.

Solution

Step 1: Analyze statement (A). $ S = \{ a + b\sqrt{2} : a, b \in \mathbb{Z} \} $ denotes all numbers of the form $ a + b\sqrt{2} $. Since $ T_1 $ and $ T_2 $ are subsets of $ S $, the union $ \mathbb{Z} \cup T_1 \cup T_2 \subset S $, and not equal to $ S $. Hence, (A) is False.

Step 2: Analyze statement (B). $ T_1 = \{ (-1 + \sqrt{2})^n : n \in \mathbb{N} \} $ and its terms are of the form $ x^n $ where $ -1 < (-1 + \sqrt{2}) < 0 $. Hence none of the elements are positive, confirming $ T_1 \cap \left(0, \frac{1}{2024}\right) = \emptyset $. Thus, (B) is True.

Step 3: Analyze statement (C). Since $ T_2 = \{ (1 + \sqrt{2})^n : n \in \mathbb{N} \} $ and $ 1 + \sqrt{2} > 2 $, for large $ n $, elements grow exponentially. So indeed, there exist elements in $ (2024, \infty) $. Hence, (C) is True.

Step 4: Analyze statement (D). The expression $ \cos(\pi(a + b\sqrt{2})) + i\sin(\pi(a + b\sqrt{2})) $ simplifies to an integer form, equivalent to Euler's formula $ e^{i\theta} $ being real if and only if $ \theta = n\pi $ (an integer multiple of $ \pi $), implying $ a + b\sqrt{2} \in \mathbb{Z} $. This implies $ b = 0 $. Hence, (D) is True.

Final Answer: B, C, and D are True.