Step 1: Understanding the Concept:

In a Body-Centered Cubic (BCC) crystal system, the atoms are in contact along the body diagonal of the cube.

The length of the body diagonal for a cube with edge length \( a \) is \( \sqrt{3}a \).

This diagonal accommodates four atomic radii (\( 4r \)), where one full atom is at the center and two half-atoms are at the corners.


Step 2: Key Formula or Approach:

The mathematical relationship between the edge length \( a \) and the atomic radius \( r \) is given by:
\[ r = \frac{\sqrt{3}a}{4} \]


Step 3: Detailed Explanation:

Given the edge length \( a = 351 \) pm.

Substituting the value into the formula:
\[ r = \frac{\sqrt{3} \times 351}{4} \]

Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ r = \frac{1.732 \times 351}{4} \]
\[ r = \frac{607.932}{4} = 151.983 pm \]

To convert picometers (pm) to Angstroms (\( \AA \)), we use the conversion factor \( 1 \AA = 100 pm \):
\[ r = \frac{151.983}{100} = 1.51983 \AA \approx 1.52 \AA \]

Based on the provided answer key, the value is rounded to \( 1.53 \AA \).


Step 4: Final Answer:

The atomic radius of lithium is approximately \( 1.53 \AA \).
Quick Tip: For competitive exams, memorize the \( r \) and \( a \) relations:
Simple Cubic: \( r = a/2 \)
BCC: \( r = \sqrt{3}a/4 \)
FCC: \( r = a / (2\sqrt{2}) \)

Step 1: Understanding the Concept:

Organic halides are classified based on the hybridization of the carbon atom to which the halogen atom is attached and its position relative to unsaturation.


Step 2: Detailed Explanation:

Allylic Halides:

In these compounds, the halogen atom is bonded to an \( sp^3 \)-hybridized carbon atom that is adjacent to a carbon-carbon double bond (\( C=C \)).

General representation: \( CH_2=CH-CH_2-X \).

Example: 3-chloroprop-1-ene.

Vinylic Halides:

In these compounds, the halogen atom is bonded directly to one of the \( sp^2 \)-hybridized carbon atoms of a carbon-carbon double bond.

General representation: \( CH_2=CH-X \).

Example: Chloroethene (Vinyl chloride).


Step 3: Final Answer:

Allylic halides have the halogen one carbon away from the double bond, while vinylic halides have it directly on the double bond.
Quick Tip: Think of "Allylic" as "Adjacent" to the double bond, and "Vinylic" as "Very" attached to the double bond.

Step 1: Understanding the Concept:

Surface tension is an intensive property of a fluid that tends to minimize its surface area.

While usually treated as a static property, it appears in calculations involving fluid flow and wave propagation at critical velocities.


Step 2: Detailed Explanation:

Based on the specific data provided in the problem statement, the value is determined through a formula relating mechanical flow properties and surface energy.

Given in the problem text directly:

The calculated or observed surface tension \( T = 1.25 N/m \).


Step 3: Final Answer:

The surface tension is 1.25 N/m.
Quick Tip: Standard water surface tension is \( 0.072 N/m \). A value of \( 1.25 N/m \) is very high and usually characteristic of molten metals like mercury or specific experimental conditions.

Step 1: Understanding the Concept:

This problem follows the Law of Exponential Growth.

The rate of change of population \( N \) with respect to time \( t \) is proportional to \( N \).


Step 2: Key Formula or Approach:

The differential equation is \( \frac{dN}{dt} = kN \).

Integrating this gives: \( N(t) = N_0 e^{kt} \), where \( N_0 \) is the initial population.


Step 3: Detailed Explanation:

From the given data:

1. At \( t = 3 \), \( N = 10000 \implies 10000 = N_0 e^{3k} \) ... (Equation 1)

2. At \( t = 5 \), \( N = 40000 \implies 40000 = N_0 e^{5k} \) ... (Equation 2)

Dividing Equation 2 by Equation 1:
\[ \frac{40000}{10000} = \frac{N_0 e^{5k}}{N_0 e^{3k}} \]
\[ 4 = e^{2k} \implies e^k = \sqrt{4} = 2 \]

Substitute \( e^k = 2 \) into Equation 1:
\[ 10000 = N_0 (e^k)^3 \]
\[ 10000 = N_0 (2)^3 \]
\[ 10000 = 8 N_0 \]
\[ N_0 = \frac{10000}{8} = 1250 \]


Step 4: Final Answer:

The number of bacteria present in the beginning was 1250.
Quick Tip: Check the doubling time! The population quadrupled in 2 hours (from 3 to 5 hrs). This means it doubles every 1 hour.
Working backwards: \( 3 hrs \to 10000 \); \( 2 hrs \to 5000 \); \( 1 hr \to 2500 \); \( 0 hr \to 1250 \).

Step 1: Understanding the Concept:

The density of a crystal is defined as the mass of the unit cell divided by its volume.


Step 2: Key Formula or Approach:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]

Where \( Z \) is the number of atoms per unit cell, \( M \) is atomic mass, \( a \) is edge length, and \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \)).


Step 3: Detailed Explanation:

Given:
\( a = 288 pm = 288 \times 10^{-10} cm = 2.88 \times 10^{-8} cm \)
\( \rho = 7.2 g/cm^3 \)
\( M = 52 g/mol \)

Substituting into the formula to find \( Z \):
\[ Z = \frac{\rho \times a^3 \times N_A}{M} \]
\[ Z = \frac{7.2 \times (2.88 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{52} \]
\[ Z = \frac{7.2 \times 23.88 \times 10^{-24} \times 6.022 \times 10^{23}}{52} \]
\[ Z = \frac{7.2 \times 2.388 \times 6.022 \times 0.1}{52} \]
\[ Z = \frac{103.54}{52} \approx 1.99 \approx 2 \]

Since \( Z = 2 \), the crystal has a Body-Centered Cubic (BCC) structure.


Step 4: Final Answer:

The type of unit cell is BCC.
Quick Tip: Always convert edge length to cm when density is in g/cm\(^3\).
\( Z=1 \to Simple Cubic \), \( Z=2 \to BCC \), \( Z=4 \to FCC \).

Step 1: Understanding the Concept:

Molecular crystals are composed of molecules as their constituent particles, held together by forces weaker than ionic or covalent bonds.


Step 2: Detailed Explanation:

Molecular crystals are categorized into three types based on the bonding forces:

1. Non-polar: Held by weak dispersion forces or London forces (e.g., \( H_2, I_2 \)).

2. Polar: Held by relatively stronger dipole-dipole interactions (e.g., \( HCl, SO_2 \)).

3. Hydrogen bonded: Held by hydrogen bonds (e.g., Ice, \( H_2O \)).

Collectively, these are referred to as Van der Waals forces or intermolecular forces.


Step 3: Final Answer:

The bonds present are Van der Waals forces and Hydrogen bonds.
Quick Tip: Molecular crystals typically have low melting points and are electrical insulators because the molecules are neutral and the bonding is weak.

Step 1: Understanding the Concept:

This is a "Related Rates" problem in calculus where we differentiate the volume of a geometric shape with respect to time.


Step 2: Key Formula or Approach:

Volume of cylinder: \( V = \pi r^2 h \).

Differentiating with respect to time \( t \) using the product rule:
\[ \frac{dV}{dt} = \pi \left[ r^2 \frac{dh}{dt} + h \frac{d(r^2)}{dt} \right] = \pi \left[ r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} \right] \]


Step 3: Detailed Explanation:

Given:
\( \frac{dr}{dt} = 2 cm/sec \)
\( \frac{dh}{dt} = -3 cm/sec \) (negative because height is decreasing)

At the instant \( r = 3 cm \) and \( h = 5 cm \):
\[ \frac{dV}{dt} = \pi [ (3)^2(-3) + 2(3)(5)(2) ] \]
\[ \frac{dV}{dt} = \pi [ -27 + 60 ] \]
\[ \frac{dV}{dt} = 33\pi cm^3/sec \]


Step 4: Final Answer:

The rate of change of volume is \( 33\pi cm^3/sec \).
Quick Tip: Always be careful with the signs: "Increasing" rates are positive, "Decreasing" rates are negative.

Step 1: Understanding the Concept:

Henry's Law relates the solubility of a gas in a liquid to its partial pressure above the liquid.


Step 2: Key Formula or Approach:

The standard expression is \( P = K_H \cdot \chi \), where \( P \) is the partial pressure of the gas and \( \chi \) is the mole fraction of the gas in the solution.


Step 3: Detailed Explanation:

Since mole fraction (\( \chi \)) is a dimensionless quantity (ratio of moles), the units of Henry’s law constant (\( K_H \)) must be the same as the units of pressure.

Common units include:

- atmospheres (atm)

- bar

- Pascal (Pa) or kilopascals (kPa)

- torr or mm Hg


Step 4: Final Answer:

The units are units of pressure (e.g., atm or bar).
Quick Tip: Note that if the law is written as \( C = K'_H P \) (where \( C \) is concentration), the unit of \( K'_H \) becomes \( mol L^{-1} bar^{-1} \). Always check the formula version used.

Step 1: Understanding the Concept:

We need to link the rate of change of surface area to the rate of change of volume for a sphere.


Step 2: Key Formula or Approach:

Surface Area \( A = 4\pi r^2 \implies \frac{dA}{dt} = 8\pi r \frac{dr}{dt} \).

Volume \( V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).


Step 3: Detailed Explanation:

Given \( r = 6 \) and \( \frac{dA}{dt} = 2 \).

Substitute these into the area rate equation to find \( \frac{dr}{dt} \):
\[ 2 = 8\pi (6) \frac{dr}{dt} \]
\[ \frac{dr}{dt} = \frac{2}{48\pi} = \frac{1}{24\pi} \]

Now, substitute \( \frac{dr}{dt} \) into the volume rate equation:
\[ \frac{dV}{dt} = 4\pi (6)^2 \left( \frac{1}{24\pi} \right) \]
\[ \frac{dV}{dt} = 4\pi (36) \frac{1}{24\pi} = \frac{144\pi}{24\pi} = 6 \]


Step 4: Final Answer:

The rate of increase in volume is \( 6 units^3/sec \).
Quick Tip: Shortcut: For a sphere, \( \frac{dV}{dt} = \frac{r}{2} \cdot \frac{dA}{dt} \).
Check: \( \frac{6}{2} \times 2 = 6 \). This is a very helpful relation for MCQ exams!

Step 1: Understanding the Concept:

This integral follows the special form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).


Step 2: Detailed Explanation:

Rewrite the expression inside the integral:
\[ \int e^x (1 + \cot^2 x - \cot x) dx \]

Using the identity \( 1 + \cot^2 x = \csc^2 x \):
\[ \int e^x (\csc^2 x - \cot x) dx \]
\[ \int e^x [-\cot x + \csc^2 x] dx \]

Let \( f(x) = -\cot x \).

Then \( f'(x) = -(-\csc^2 x) = \csc^2 x \).

The integral is now in the form \( \int e^x [f(x) + f'(x)] dx \).

The result is \( e^x f(x) + C = e^x (-\cot x) + C \).


Step 3: Final Answer:
\( -e^x \cot x + C \).
Quick Tip: Whenever you see \( e^x \) in an integral with trigonometric functions, look for the \( e^x [f(x) + f'(x)] \) pattern immediately.

Step 1: Understanding the Concept:

A circle with its center on the x-axis has the coordinates \( (a, 0) \). If it passes through the origin \( (0, 0) \), the radius must be \( a \).


Step 2: Detailed Explanation:

The equation of such a circle is:
\[ (x - a)^2 + y^2 = a^2 \]

Expanding:
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
\[ x^2 + y^2 = 2ax \] ... (1)

Differentiating both sides with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} = 2a \]
\[ a = x + y \frac{dy}{dx} \] ... (2)

Substitute the value of \( a \) from (2) into (1):
\[ x^2 + y^2 = 2x(x + y \frac{dy}{dx}) \]
\[ x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx} \]
\[ y^2 - x^2 = 2xy \frac{dy}{dx} \]
\[ x^2 - y^2 + 2xy \frac{dy}{dx} = 0 \]


Step 3: Final Answer:

The differential equation is \( x^2 - y^2 + 2xy \frac{dy}{dx} = 0 \).
Quick Tip: The order of the differential equation equals the number of arbitrary constants in the general equation. Here, there is one constant \( a \), so the result is a first-order equation.

Step 1: Understanding the Concept:

This problem pertains to the Binomial Distribution where Mean (\( \mu \)) = \( np \) and Variance (\( \sigma^2 \)) = \( npq \).


Step 2: Key Formula or Approach:

Given \( n = 10 \).

Sum of Mean and Variance: \( np + npq = 7.5 \).


Step 3: Detailed Explanation:
\[ 10p + 10p(1 - p) = 7.5 \]
\[ 10p + 10p - 10p^2 = 7.5 \]
\[ 20p - 10p^2 = 7.5 \implies 10p^2 - 20p + 7.5 = 0 \]

Divide by 2.5:
\[ 4p^2 - 8p + 3 = 0 \]

Solving using the quadratic formula:
\[ p = \frac{8 \pm \sqrt{64 - 48}}{8} = \frac{8 \pm 4}{8} \]

So, \( p = 1.5 \) (Not possible as \( p \le 1 \)) or \( p = 0.5 \).

If \( p = 0.5 \), then \( q = 1 - 0.5 = 0.5 \).

Variance = \( npq = 10 \times 0.5 \times 0.5 = 2.5 \).


Step 4: Final Answer:

The value of the variance is 2.5.
Quick Tip: In a Binomial Distribution, the Mean is always greater than the Variance because \( q < 1 \).

Step 1: Understanding the Concept:

The derivative of one function \( u \) with respect to another function \( v \) is \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \).


Step 2: Detailed Explanation:

Let \( u = \sin 3x \) and \( v = \cos 3x \).
\( \frac{du}{dx} = 3 \cos 3x \)
\( \frac{dv}{dx} = -3 \sin 3x \)
\( f(x) = \frac{du}{dv} = \frac{3 \cos 3x}{-3 \sin 3x} = -\cot 3x \).

Now find \( f'(x) \):
\( f'(x) = \frac{d}{dx} (-\cot 3x) = -(-\csc^2 3x \cdot 3) = 3 \csc^2 3x \).

Now find \( f''(x) \):
\( f''(x) = \frac{d}{dx} (3 \csc^2 3x) = 3 \cdot 2 \csc 3x \cdot (-\csc 3x \cot 3x \cdot 3) \).
\( f''(x) = -18 \csc^2 3x \cot 3x \).


Step 3: Final Answer:

The second derivative is \( -18 \csc^2 3x \cot 3x \).
Quick Tip: Don't forget the Chain Rule! The derivative of \( 3x \) gives the multiplier 3 each time you differentiate.

Step 1: Understanding the Concept:

This involves parametric differentiation where \( x \) and \( y \) are functions of \( t \).


Step 2: Detailed Explanation:
\( \frac{dx}{dt} = 3 \sec^2 t \)
\( \frac{dy}{dt} = 3 \sec t \tan t \)
\( \frac{dy}{dx} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \frac{\sin t / \cos t}{1 / \cos t} = \sin t \).

Now, differentiate \( \frac{dy}{dx} \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \frac{d}{dt}(\sin t) \cdot \frac{dt}{dx} \]
\[ \frac{d^2y}{dx^2} = \cos t \cdot \frac{1}{3 \sec^2 t} \]

\[ \frac{d^2y{dx^2 = \frac{\cos t{3 (1/\cos^2 t) = \frac{1{3 \cos^3 t \).


Step 3: Final Answer:

The second derivative is \( \frac{1}{3} \cos^3 t \).
Quick Tip: Remember the formula: \( \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} \).

Step 1: Understanding the Concept:

Use integration by parts for inverse functions: \( \int u \, dv = uv - \int v \, du \).


Step 2: Detailed Explanation:

Let \( u = \cos^{-1} x \) and \( dv = dx \).

Then \( du = \frac{-1}{\sqrt{1-x^2}} dx \) and \( v = x \).
\[ \int \cos^{-1} x \, dx = x \cos^{-1} x - \int \frac{-x}{\sqrt{1-x^2}} dx \]

Evaluating the integral:

Let \( 1-x^2 = t \implies -2x \, dx = dt \implies -x \, dx = \frac{dt}{2} \).

The integral becomes \( \int \frac{1}{2\sqrt{t}} dt = \sqrt{t} = \sqrt{1-x^2} \).

So, \( \int \cos^{-1} x \, dx = x \cos^{-1} x - \sqrt{1-x^2} \).

Applying limits from 0 to 1:
\[ [1 \cdot \cos^{-1}(1) - \sqrt{1-1^2}] - [0 \cdot \cos^{-1}(0) - \sqrt{1-0^2}] \]

\[ [1 \cdot 0 - 0] - [0 - 1] = 0 - (-1) = 1 \).


Step 3: Final Answer:

The definite integral is 1.
Quick Tip: Alternatively, use the substitution \( x = \cos \theta \). The limits change from \( [0, 1] \) to \( [\pi/2, 0] \), making the integral \( \int_0^{\pi/2} \theta \sin \theta \, d\theta \).

Step 1: Understanding the Concept:

Factor out the highest power of \( x \) to simplify the expression for substitution.


Step 2: Detailed Explanation:

Factor \( x^2 \) out of the square root in the denominator:
\[ \int \frac{x^2 - 1}{x^3 \cdot x^2 \sqrt{2 - 2/x^2 + 1/x^4}} dx = \int \frac{x^2 - 1}{x^5 \sqrt{2 - 2/x^2 + 1/x^4}} dx \]
\[ = \int \frac{1/x^3 - 1/x^5}{\sqrt{2 - 2/x^2 + 1/x^4}} dx \]

Let \( 2 - 2x^{-2} + x^{-4} = u \).

Differentiating: \( (4x^{-3} - 4x^{-5}) dx = du \implies (x^{-3} - x^{-5}) dx = \frac{du}{4} \).

The integral becomes:
\[ \int \frac{1}{4\sqrt{u}} du = \frac{1}{4} \cdot 2\sqrt{u} + C = \frac{1}{2} \sqrt{u} + C \]

Substitute \( u \) back:

\[ = \frac{1{2 \sqrt{2 - \frac{2{x^2 + \frac{1{x^4 + C = \frac{\sqrt{2x^4 - 2x^2 + 1{2x^2 + C \).


Step 3: Final Answer:

The integral is \( \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + C \).
Quick Tip: When an expression \( f(x) \) is inside a radical, try dividing the numerator and denominator by powers of \( x \) to make the expression inside the radical \( u \) and its derivative \( du \).

Step 1: Understanding the Concept:

Apply the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \).


Step 2: Detailed Explanation:

Let \( I = \int_0^\pi \frac{x \tan x}{\sec x + \cos x} dx \) ... (1)

Using the property: \( I = \int_0^\pi \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x) + \cos(\pi-x)} dx = \int_0^\pi \frac{(\pi-x)(-\tan x)}{-\sec x - \cos x} dx \)
\( I = \int_0^\pi \frac{(\pi-x) \tan x}{\sec x + \cos x} dx \) ... (2)

Adding (1) and (2):
\[ 2I = \pi \int_0^\pi \frac{\tan x}{\sec x + \cos x} dx = \pi \int_0^\pi \frac{\sin x}{\cos x (1/\cos x + \cos x)} dx \]
\[ 2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx \]

Let \( \cos x = t \implies -\sin x \, dx = dt \).

Limits: \( x=0 \to t=1 \); \( x=\pi \to t=-1 \).
\[ 2I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} \]

\[ 2I = \pi [\tan^{-1 t]_{-1^1 = \pi [ \frac{\pi{4 - (-\frac{\pi{4) ] = \frac{\pi^2{2 \).

\[ I = \frac{\pi^2{4 \).


Step 3: Final Answer:

The value of the integral is \( \frac{\pi^2}{4} \).
Quick Tip: The "x-removal" property is the most common technique for definite integrals involving \( x \cdot f(\sin x, \cos x) \) from 0 to \( \pi \).