All India JEE Group

Problem Overview
A particle is projected at an angle of 30 degrees to the horizontal. After 2 seconds, it moves in a direction given by tan^(-1)(1/4). We need to find its initial speed, given that the acceleration due to gravity (g) is 10 m/s².
Key Concepts
- Projectile Motion: The motion can be analyzed in horizontal and vertical components.
- Velocity ... moreComponents: Initial speed (u) can be split into horizontal (u_x) and vertical (u_y) components.
Initial Velocity Components
- The initial velocity can be expressed as:
- u_x = u * cos(30°)
- u_y = u * sin(30°)
- Using the values:
- cos(30°) = √3/2
- sin(30°) = 1/2
Vertical Motion Analysis
- The vertical position after time (t) is given by:
- y = u_y * t - (1/2) * g * t²
- Substituting for t = 2 seconds:
- y = (u/2) * 2 - 5 * 2²
- y = u - 20
Horizontal Motion and Final Velocity
- The horizontal position after 2 seconds:
- x = u_x * t = (u * √3/2) * 2 = √3 * u
- The final vertical velocity (v_y) after 2 seconds:
- v_y = u_y - g * t = (u/2) - 20
Direction of Velocity
- The tangent of the angle θ of the velocity vector after 2 seconds is given by:
- tan(θ) = v_y / u_x
- For θ = tan^(-1)(1/4):
- 1/4 = [(u/2) - 20] / (u * √3/2)
Solving the Equation
- Rearranging gives:
- u/2 - 20 = (1/4)(u * √3/2)
- Multiplying through by 8 to eliminate fractions and solving for u leads to:
- u = 40/√3 m/s
Final Result
- The initial speed of the particle is approximately 23.09 m/s.

Reaction Overview
The reaction between manganese carbonyl (Mn(CO)5) and methyllithium (MeLi) typically results in the formation of a methyl manganese complex.
Main Steps of the Reaction
- Nucleophilic Attack:
- Methyllithium acts as a strong nucleophile. The methyl group (Me-) attacks the manganese center in Mn(CO)5.
- Formation of a Methyl Compl... moreex:
- The methyl group replaces one of the CO ligands in the manganese complex, resulting in the formation of methyl manganese carbonyl, typically denoted as [Mn(CO)4(Me)].
Final Product
- Structural Composition:
- The final product is [Mn(CO)4(Me)], which consists of a manganese atom bonded to four carbonyl (CO) ligands and one methyl (Me) group.
Key Features of the Product
- Coordination:
- The manganese retains a coordination number of five, which is common for transition metal complexes.
- Stability:
- The presence of carbonyl ligands contributes to the stability of the complex due to strong π-backbonding.
- Reactivity:
- The methyl group introduces new reactivity, making the complex useful in various organic transformations.
Conclusion
The reaction between Mn(CO)5 and MeLi leads to the formation of [Mn(CO)4(Me)], an important intermediate in organometallic chemistry that showcases the versatility of transition metal complexes.

Introduction
To find the area bounded by the curve y² = x, the line y = 4, and the y-axis, we need to first understand the shapes formed by these equations.
Step 1: Understanding the Curves
- The equation y² = x represents a rightward-opening parabola.
- The line y = 4 is a horizontal line.
- The y-axis is the vertical line where x = 0.
Step 2: Findin... moreg Intersection Points
- To find the points of intersection between the parabola and the line y = 4, substitute y into the parabola's equation:
- 4² = x → x = 16.
- Thus, the points of intersection are (16, 4) and (0, 0).
Step 3: Setting Up the Area Calculation
- The area can be calculated by integrating the area under the curve from the y-axis to the intersection point along the line y = 4.
- We will integrate the function derived from the parabola, which is x = y².
Step 4: Calculating the Area
- The area can be expressed as:
- Area = ∫ (from y=0 to y=4) (y²) dy.
- Evaluating this integral:
- Area = [y³/3] from 0 to 4
- Area = (4³/3 - 0) = 64/3.
Conclusion
The area bounded by the curve y² = x, the line y = 4, and the y-axis is 64/3. Since the correct answer choice in the question is option 'B', which is 16/3, it appears to be a mistake as the computed area is 64/3. Thus, the area calculation confirms that the answer is indeed 64/3, not 16/3.

Understanding Standard Heat of Formation
The heat of formation is a crucial concept in thermodynamics, representing the change in enthalpy when one mole of a substance is formed from its elements in their standard states. The standard state typically refers to a substance at 1 atmosphere of pressure and a specified temperature (usually 25°C).
Why Certain Substances Have Zero He... moreat of Formation
The heat of formation for elements in their most stable form at standard conditions is defined as zero. Hence, the substances listed can be analyzed as follows:
1. Aluminum (Al)
- Aluminum is an element and is found in its most stable form as a solid metal at standard conditions.
- Therefore, its heat of formation is zero.
2. Diamond (C)
- Although diamond is a form of carbon, it is not the most stable form at standard conditions; graphite is.
- As such, diamond does not have a heat of formation of zero.
3. Zinc (Zn)
- Zinc is also an element, existing in its most stable metallic form at standard conditions.
- Hence, the heat of formation for zinc is zero.
4. Graphite (C)
- Graphite is the most stable form of carbon at standard conditions.
- Therefore, the heat of formation for graphite is also zero.
Conclusion
Thus, the substances with a heat of formation of zero in their standard states are:
- Aluminum (A)
- Zinc (C)
- Graphite (D)
The correct options are A, C, and D.