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GS Paper IUPSC Prelims 1998
Health & Medicine
Question No. 234Interactive Test Mode
00:00
+2-0.66

Q234. A ball is dropped from the top of a high building with a constant acceleration of 9.8 m/s2. What will be its velocity after 2 seconds?

A ball is dropped from the top of a high building with a constant acceleration of 9.8 m/s2. What will be its velocity after 2 seconds?

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Official Answer Key & Detailed Solution

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Official UPSC Answer KeyUPSC Prelims 1998 • Science and Technology

Correct Answer: Option (B) — 19.6 m/s

Concept & Option Breakdown

Standard Reference Verified

Using the first kinematic equation of motion: $v = u + at$ Given: Initial velocity $u = 0$ (dropped from rest) Acceleration $a = 9.8\text{ m/s}^2$ Time $t = 2\text{ seconds}$ $v = 0 + (9.8 \times 2) = 19.6\text{ m/s}$.

Elimination Heuristic & Exam Approach

Velocity after time $t$ under gravity is simply $v = gt = 9.8 \times 2 = 19.6\text{ m/s}$.

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Question Specifications & Syllabus Mapping

Examination & YearUPSC Civil Services Preliminary Examination 1998 (Q234)
Paper & Marking SchemeGS Paper I • 2 Marks (-0.66 Negative for incorrect response)
Syllabus Subject & TopicScience and Technology → Health & Medicine
Official Answer KeyOption (B) — 19.6 m/s

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Frequently Asked Questions — UPSC Prelims 1998 Q234

What is the official answer for UPSC Prelims 1998 GS Paper I Question 234 (Science and Technology)?

The official UPSC answer key for this 1998 Science and Technology (Health & Medicine) question is Option (B): "19.6 m/s". Correct attempts carry 2 Marks with -0.66 Negative penalty for wrong answers.

Which UPSC Prelims syllabus topic does this 1998 question belong to?

This question belongs to GS Paper I under the subject "Science and Technology" and micro-topic "Health & Medicine". You can explore all repeated questions from Health & Medicine in Acme AI's 3D Prelims PYQ Realm or practice 50 full-length and sectional tests in the APT Prelims Test Series.

How can I solve this Health & Medicine UPSC Prelims question using option elimination?

Velocity after time $t$ under gravity is simply $v = gt = 9.8 \times 2 = 19.6\text{ m/s}$.

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