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The path taken by Dhiraj can be analyzed by tracking the direction and distance at each turn, starting from Point A as the origin (0, 0) on a coordinate plane:

Given:

1. Dhiraj starts from Point A and drives 11 km towards west. 2. He then takes a right turn, drives 8 km, turns right and drives 14 km. 3. He then takes a right turn and drives 11 km. 4. He takes a final right turn, drives 3 km and stops at Point P.

Therefore, Dhiraj should drive 3 km to the north to reach Point A again.

Hence, the correct answer is "Option 1".

Logic followed here is:

The letters in the series follow a mathematical pattern where each subsequent term is derived by adding or subtracting a constant value from the positional values of the letters in the preceding term.

Analysis of the sequence:

Therefore, the missing term in the given series based on the English alphabetical order is FWEF.

Hence, the correct answer is "Option 4".

The logic followed in this pattern is based on alphabetical shifts where the letters are incremented by specific values in an alternating sequence.

For 'BIRD' related to 'CLSG':

For 'RAIN' related to 'SDJQ':

Applying the same logic to 'STAR':

Thus, STAR is related to TWBU.

Hence, the correct answer is 'Option 1'.

The logic followed here is:

Logic: The difference between the positional values of the first letters of the two clusters is 6, and the difference between the positional values of the second letters of the two clusters is 6.

Option 1) TL - NF

Option 2) ME - HA

This pair does not follow the rule as the differences are 5 and 4 instead of 6.

Option 3) WO - QI

Option 4) YQ - SK

Therefore, the letter-cluster pair that is different from the others is ME - HA.

Hence, the correct answer is 'Option 2'.

1. Since all persons face the centre: right = counterclockwise, left = clockwise. Fix Ay at pos(1). Cv sits to the immediate right of Ay (counterclockwise): Ay(1) → Cv(7).

2. Du sits third to the left of Ay (clockwise): Ay(1) → pos(2) → pos(3) → Du(4). Thus, Du is at pos(4).

3. Bk is the immediate neighbour of Fo and Cv. Since Cv is at pos(7) and Ay is at pos(1), Bk must be at pos(6). Fo then must be at pos(5).

4. Gw is not an immediate neighbour of Du. The remaining positions are pos(2) and pos(3). Neighbors of Du(4) are pos(3) and pos(5). Since Gw cannot be at pos(3), Gw must be at pos(2).

5. The final person, Es, occupies the last remaining position, pos(3). The full arrangement is: Ay(1), Gw(2), Es(3), Du(4), Fo(5), Bk(6), Cv(7).

Position of Es with respect to Fo: From Fo(5), moving counterclockwise (right) → Du(4) (1^st right) → Es(3) (2^nd right). Therefore, Es sits second to the right of Fo.

Hence, the correct answer is 'Option 1'.

The correct answer is -30 cm.

Key Points

• According to the New Cartesian Sign Convention, the focal length (f) of a concave lens is always taken as negative; hence, f = −15 cm.
• Since a concave lens always produces a virtual and erect image on the same side as the object, the image distance (v) is also negative; hence, v = −10 cm.
• The relationship between the object distance (u), image distance (v), and focal length (f) is given by the Lens Formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \).
• To find the object distance (u), we rearrange the formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \).
• By substituting the values: 1/u = 1/(−10) − 1/(−15). This simplifies to 1/u = −1/10 + 1/15.
• Finding a common denominator (30), the equation becomes: 1/u = (−3 + 2) / 30 = −1/30.
• Therefore, u = −30 cm, indicating the object is placed 30 cm in front of the lens.

 Additional Information

• Concave Lens Characteristics:
• It is known as a diverging lens because it spreads out light rays that travel parallel to its principal axis.
• The images formed are consistently virtual, erect, and diminished in size, regardless of the object's position.
• The magnification (m) for a concave lens is always positive and less than 1, calculated as m = v/u.
• Sign Convention Rules:
• All distances are measured from the optical center of the lens.
• Distances measured in the direction of incident light are positive, while those measured against it are negative.
• The object distance (u) is almost always negative in standard optical setups as objects are placed to the left of the lens.

The correct answer is 250 J.

Key Points

• Work Done is defined in physics as the product of the Force applied to an object and the Displacement of that object in the direction of the force.
• When a porter lifts luggage vertically against gravity, the force required is equal to the Weight of the luggage, which is calculated as Mass (m) × Acceleration due to gravity (g).
• The specific values provided in the problem are: Mass (m) = 10 kg, Height (h) = 2.5 m, and Acceleration due to gravity (g) = 10 m/s².
• The formula used for calculating the work done against gravity is W = m × g × h, which also represents the Gravitational Potential Energy gained by the object.
• By substituting the values into the equation: W = 10 kg × 10 m/s² × 2.5 m.
• Performing the multiplication: 10 × 10 = 100; then 100 × 2.5 = 250. Thus, the total work performed is 250 Joules.

 Additional Information

• Work Done (W):
• It is a Scalar Quantity, meaning it has magnitude but no specific direction in space.
• The SI Unit of work is the Joule (J), where 1 J = 1 Newton-meter (N·m).
• Work can be positive, negative, or zero depending on the Angle (θ) between the force vector and the displacement vector (W = Fs cosθ).
• Acceleration due to Gravity (g):
• On Earth, the standard value is approximately 9.8 m/s², though for simplified competitive exam calculations, it is often taken as 10 m/s².
• It represents the constant acceleration experienced by an object in Free Fall near the Earth's surface.
• Potential Energy:
• Lifting an object increases its Stored Energy based on its position. This is specifically termed Gravitational Potential Energy.
• The work done by the porter is effectively transferred into the luggage as this form of energy.

The correct answer is Because each method exploits specific differences in physical and chemical properties between the ore and the gangue.

Key Points

• Concentration of Ore, also known as ore dressing or beneficiation, is the process of removing unwanted earthy and rocky impurities called gangue from the mined ore.
• The selection of a specific concentration method is strictly governed by the physical and chemical properties of the metal compound and the gangue particles.
• Hydraulic Washing (Gravity Separation) utilizes the difference in specific gravities between the ore and the gangue; typically, the ore is heavier than the impurities.
• Magnetic Separation is applied when either the ore or the gangue possesses magnetic properties, allowing separation via electromagnetic rollers.
• Froth Flotation exploits differences in surface wettability; sulphide ores are preferentially wetted by oils (like pine oil) while gangue is wetted by water.
• Chemical methods, such as Leaching, are used when the ore is soluble in a particular solvent while the gangue remains insoluble.

 Additional Information

• Hydraulic Washing:
• Commonly used for oxide ores like Haematite (Fe₂O₃) and Cassiterite (SnO₂).
• Involves an upward stream of running water to wash away lighter gangue.
• Froth Flotation Details:
• Collectors: Substances like pine oils or xanthates that enhance the non-wettability of mineral particles.
• Froth Stabilizers: Compounds like cresols and aniline which stabilize the froth.
• Depressants: Used to separate two sulphide ores (e.g., using NaCN to separate ZnS from PbS).
• Leaching Examples:
• Bauxite (Aluminium ore) is concentrated using hot concentrated NaOH (Bayer’s process).
• Gold and Silver are leached using dilute solutions of NaCN or KCN in the presence of air.

The correct answer is The screen should be placed anywhere between 40 cm and 80 cm from the lens.

Key Points

• According to the Cartesian sign convention, the focal length (f) of a convex lens is positive (+40 cm), and the object distance (u) is negative (−100 cm) as it is placed on the left side.
• Applying the Lens Formula, 1/f = 1/v − 1/u, we substitute the values: 1/40 = 1/v − 1/(−100), which simplifies to 1/v = 1/40 − 1/100. Calculating the Least Common Multiple (LCM) of 40 and 100 gives 200, resulting in 1/v = (5 − 2) / 200 = 3/200, thus v = 66.67 cm.
• For a convex lens, the point 2F is located at twice the focal length, which is 2 × 40 = 80 cm. Since the object is placed at 100 cm (which is beyond 2F), the laws of optics dictate that the image must form between F (40 cm) and 2F (80 cm) on the opposite side.
• The linear magnification (m) is defined as the ratio of image distance (v) to object distance (u). Here, m = (200/3) / (−100) = −2/3 ≈ −0.67. The negative sign confirms the image is real and inverted, while a magnitude less than 1 confirms it is diminished (smaller).
• Since the image is real, it can only be captured on a screen placed exactly at the point of convergence, which in this scenario is within the 40 cm to 80 cm range from the optical center.

 Additional Information

• Convex Lens Image Characteristics:
• Object at Infinity: Image forms at the focus (F), is real, inverted, and highly diminished.
• Object at 2F: Image forms at 2F on the other side, is real, inverted, and same size as the object.
• Object between F and 2F: Image forms beyond 2F, is real, inverted, and magnified.
• Object at F: Image forms at infinity, is real, inverted, and highly magnified.
• Object between F and Optical Center: Image is virtual, erect, and magnified, appearing on the same side as the object.
• Power of a Lens: It is the reciprocal of the focal length in meters (P = 1/f). For this lens, the power would be +2.5 Dioptres.

The correct answer is H2.

Key Points

• A molecule is the smallest unit of a chemical element or compound that can exist freely; it is formed when two or more atoms are chemically joined by covalent bonds.
• H2 (Hydrogen) is a homonuclear diatomic molecule, meaning it is composed of exactly two atoms of the same element.
• In an H2 molecule, two hydrogen atoms share their single valence electrons to achieve a stable duplet configuration, forming a single covalent bond.
• Under standard temperature and pressure (STP), hydrogen exists as a colorless, odorless, and highly flammable diatomic gas.
• This structure is distinct from polyatomic molecules or heteronuclear molecules, which involve either more than two atoms or atoms from different elements.

 Additional Information

• O₃ (Ozone):
• It is an allotrope of oxygen consisting of three oxygen atoms, making it a homonuclear triatomic molecule.
• It is primarily found in the Earth's stratosphere, where it protects the biosphere from harmful ultraviolet (UV) radiation.
• CO (Carbon Monoxide):
• It is a heteronuclear diatomic molecule because it consists of two atoms, but they are of different elements: carbon and oxygen.
• It is a toxic gas produced during the incomplete combustion of hydrocarbons and binds strongly to hemoglobin in the blood.
• H₂O (Water):
• It is a heteronuclear triatomic molecule composed of two hydrogen atoms and one oxygen atom.
• The atoms are held together by polar covalent bonds, giving water its unique properties as a universal solvent.