Major Competitive Exams

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Major Competitive Exams

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Major Competitive Exams MCQ & Objective Questions

Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.

What You Will Practise Here

Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams

Exam Relevance

The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.

Common Mistakes Students Make

Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam

FAQs

Question: How can I improve my performance in Major Competitive Exams?
Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.

Question: What types of questions should I focus on for these exams?
Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.

Question: Are there specific strategies for tackling objective questions?
Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.

Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!

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Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum if the mass remains the same?

A. 2ω
B. 4ω
C. ω
D. ω/2

Solution

Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.

Correct Answer: B — 4ω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled while keeping the mass constant, what will be the new moment of inertia?

A. 2I
B. 4I
C. I
D. I/2

Solution

The moment of inertia of a disk is I = (1/2)MR^2. If the radius is doubled, the new moment of inertia becomes I' = (1/2)M(2R)^2 = 4I.

Correct Answer: B — 4I

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled while keeping the mass constant, what will be the new angular momentum?

A. 2Iω
B. 4Iω
C. Iω
D. I(2ω)

Solution

The moment of inertia I of a disk is proportional to r^2, so if the radius is doubled, I becomes 4I. Thus, angular momentum L = Iω becomes 4Iω.

Correct Answer: B — 4Iω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular velocity to conserve angular momentum?

A. ω
B. 2ω
C. ω/2
D. ω/4

Solution

To conserve angular momentum, if the radius is doubled, the angular velocity must be halved.

Correct Answer: C — ω/2

Q. A distance is measured as 100 m with an error of 2 m. What is the absolute error?

A. 2 m
B. 0.02 m
C. 0.2 m
D. 20 m

Solution

The absolute error is given as 2 m.

Correct Answer: A — 2 m

Q. A diverging lens has a focal length of -10 cm. What is the image distance when the object is placed at 5 cm?

A. -3.33 cm
B. -5 cm
C. 3.33 cm
D. 5 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we have 1/(-10) = 1/v - 1/5. Solving gives v = -3.33 cm.

Correct Answer: A — -3.33 cm

Q. A diverging lens has a focal length of -15 cm. What is the nature of the image formed when an object is placed at 10 cm from the lens? (2023)

A. Real and inverted
B. Virtual and upright
C. Real and upright
D. Virtual and inverted

Solution

A diverging lens always forms a virtual and upright image when the object is placed in front of it.

Correct Answer: B — Virtual and upright

Q. A doctor is faced with a patient who refuses a life-saving treatment due to personal beliefs. Should the doctor respect the patient's wishes or insist on treatment?

A. Respect the patient's wishes
B. Insist on treatment
C. Seek a second opinion
D. Consult legal advice

Solution

The doctor should respect the patient's wishes, as autonomy is a key ethical principle in healthcare.

Correct Answer: A — Respect the patient's wishes

Q. A doctor is to a hospital as a teacher is to a _____.

A. school
B. classroom
C. university
D. library

Solution

A teacher works in a school, just as a doctor works in a hospital.

Correct Answer: A — school

Q. A doctor is to a patient as a mechanic is to a _____.

A. car
B. tool
C. engine
D. garage

Solution

A doctor treats patients, while a mechanic repairs cars.

Correct Answer: A — car

Q. A doctor prescribes a medication that costs $15 per dose. If a patient needs to take it twice a day for 10 days, what will be the total cost?

A. $300
B. $150
C. $450
D. $200

Solution

Total doses = 2 doses/day * 10 days = 20 doses. Total cost = 20 doses * $15/dose = $300.

Correct Answer: A — $300

Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 0.8 m, what is the torque about the hinges?

A. 8 Nm
B. 10 Nm
C. 16 Nm
D. 20 Nm

Solution

Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.

Correct Answer: C — 16 Nm

Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 1 m, what is the torque about the hinges?

A. 10 Nm
B. 20 Nm
C. 30 Nm
D. 40 Nm

Solution

Torque (τ) = F × d = 20 N × 1 m = 20 Nm.

Correct Answer: B — 20 Nm

Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1 m, what is the torque about the hinges?

A. 25 Nm
B. 50 Nm
C. 75 Nm
D. 100 Nm

Solution

Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.

Correct Answer: B — 50 Nm

Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1.2 m, what is the torque about the hinges?

A. 60 Nm
B. 50 Nm
C. 70 Nm
D. 40 Nm

Solution

Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.

Correct Answer: A — 60 Nm

Q. A double convex lens has a focal length of 10 cm. If it is made of a material with a refractive index of 1.5, what is the radius of curvature of each surface assuming they are equal?

A. 15 cm
B. 20 cm
C. 25 cm
D. 30 cm

Solution

Using the lens maker's formula, R = 2f(n-1) = 2*10*(1.5-1) = 20 cm.

Correct Answer: B — 20 cm

Q. A double convex lens has a radius of curvature of 20 cm on both sides. What is its focal length if the refractive index is 1.5?

A. 10 cm
B. 15 cm
C. 20 cm
D. 25 cm

Solution

Using the lens maker's formula, f = R/(n-1), we find f = 20/(1.5-1) = 40 cm.

Correct Answer: A — 10 cm

Q. A dress is marked at $120. If a discount of 15% is applied, what is the selling price?

A. $102
B. $100
C. $105
D. $110

Solution

Selling Price = Marked Price - (Discount % of Marked Price) = 120 - (15/100 * 120) = 120 - 18 = $102.

Correct Answer: A — $102

Q. A dress is marked at $200. If it has a 30% discount followed by an additional 10% discount, what is the final price?

A. $140
B. $130
C. $150
D. $120

Solution

First discount: 200 - (30/100 * 200) = 200 - 60 = $140. Second discount: 140 - (10/100 * 140) = 140 - 14 = $126.

Correct Answer: A — $140

Q. A dress is marked at $250 and has a discount of 15% followed by another discount of 10%. What is the final price?

A. $212.50
B. $215.00
C. $220.00
D. $225.00

Solution

First discount: 250 - (15/100 * 250) = 250 - 37.5 = $212.5. Second discount: 212.5 - (10/100 * 212.5) = 212.5 - 21.25 = $191.25.

Correct Answer: A — $212.50

Q. A dress is marked at $250 and has a discount of 15%. What is the selling price?

A. $212.50
B. $225
C. $237.50
D. $240

Solution

Selling Price = Marked Price - (Discount % of Marked Price) = 250 - (15/100 * 250) = 250 - 37.5 = $212.50.

Correct Answer: A — $212.50

Q. A dress is marked at $250 and has a discount of 20% followed by an additional discount of 10%. What is the final selling price?

A. $200
B. $225
C. $210
D. $215

Solution

First discount = 250 * 20/100 = 50. New price = 250 - 50 = 200. Second discount = 200 * 10/100 = 20. Final price = 200 - 20 = $180.

Correct Answer: C — $210

Q. A dress is marked at $250 and has two successive discounts of 10% and 20%. What is the final selling price?

A. $200
B. $180
C. $190
D. $210

Solution

First discount: 250 - (10/100 * 250) = 250 - 25 = $225. Second discount: 225 - (20/100 * 225) = 225 - 45 = $180.

Correct Answer: C — $190

Q. A dress is marked at $250. If it is sold for $200 after two successive discounts of 10% and 20%, what is the selling price after the first discount?

A. $225
B. $240
C. $200
D. $250

Solution

First discount: 250 - (0.10 * 250) = 250 - 25 = $225.

Correct Answer: A — $225

Q. A drop of liquid is in equilibrium on a surface. What is the condition for the drop to remain in equilibrium?

A. Weight equals surface tension
B. Weight equals gravitational force
C. Surface tension equals gravitational force
D. Surface tension equals buoyant force

Solution

For a drop to remain in equilibrium, the upward force due to surface tension must balance the downward gravitational force.

Correct Answer: C — Surface tension equals gravitational force

Q. A drop of liquid is spherical in shape. This is due to which of the following?

A. Gravity
B. Surface tension
C. Viscosity
D. Pressure

Solution

The spherical shape of a liquid drop is due to surface tension, which minimizes the surface area for a given volume.

Correct Answer: B — Surface tension

Q. A drop of oil spreads on water instead of sinking. This is due to:

A. Higher density of oil
B. Lower surface tension of oil
C. Higher surface tension of water
D. Lower viscosity of oil

Solution

The drop of oil spreads on water because the surface tension of water is higher than that of oil, causing the oil to spread out.

Correct Answer: C — Higher surface tension of water

Q. A drop of water is spherical in shape due to which property?

A. Viscosity
B. Surface tension
C. Capillarity
D. Density

Solution

A drop of water is spherical due to surface tension, which minimizes the surface area.

Correct Answer: B — Surface tension

Q. A factory produces 1000 units in 5 days. How many units will it produce in 3 days if the production rate remains constant?

A. 500 units
B. 600 units
C. 700 units
D. 800 units

Solution

The factory produces 1000/5 = 200 units per day. In 3 days, it will produce 200 * 3 = 600 units.

Correct Answer: B — 600 units

Q. A factory produces 1200 units of a product in 8 hours. How many units can it produce in 5 hours at the same rate?

A. 750
B. 600
C. 900
D. 800

Solution

The rate of production is 1200 units / 8 hours = 150 units/hour. In 5 hours, it can produce 150 * 5 = 750 units.

Correct Answer: D — 800

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