A hydraulic lift has area ratio of output to input piston 10:1. Force applied 50 N on input piston lifts weight of MCQ with Answer and Explanation
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Question #1
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Raindrops fall to the ground with a uniform, constant velocity. This constant velocity is achieved when the gravitational pull is perfectly balanced by:
Correct Answer: Option A
Explanation:
As a raindrop falls, it accelerates due to gravity. The upward viscous drag force of the air increases with the drop's velocity (Stokes' Law). Eventually, the downward weight of the drop is exactly balanced by the upward buoyant force of the air plus the upward viscous drag force. The net force becomes zero, and it falls with a constant 'terminal velocity'.
As a raindrop falls, it accelerates due to gravity. The upward viscous drag force of the air increases with the drop's velocity (Stokes' Law). Eventually, the downward weight of the drop is exactly balanced by the upward buoyant force of the air plus the upward viscous drag force. The net force becomes zero, and it falls with a constant 'terminal velocity'.
Question #2
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The unit of work is
Question #3
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The time taken for a radioactive sample to decay to 1/16th of its initial activity is:
Correct Answer: Option D
Explanation:
After n half-lives, activity reduces to (1/2)ⁿ of initial. Set (1/2)ⁿ = 1/16 ⇒ (1/2)ⁿ = (1/2)⁴ ⇒ n = 4 half-lives. Thus time = 4 × half-life. Memory aid: '1/2ⁿ remaining after n half-lives; 1/16 = 1/2⁴ ⇒ 4 half-lives'. This radioactivity calculation is frequently tested in competitive exams. Always express fraction as power of 1/2 to find half-life count. This problem assesses understanding of exponential decay without heavy computation.
After n half-lives, activity reduces to (1/2)ⁿ of initial. Set (1/2)ⁿ = 1/16 ⇒ (1/2)ⁿ = (1/2)⁴ ⇒ n = 4 half-lives. Thus time = 4 × half-life. Memory aid: '1/2ⁿ remaining after n half-lives; 1/16 = 1/2⁴ ⇒ 4 half-lives'. This radioactivity calculation is frequently tested in competitive exams. Always express fraction as power of 1/2 to find half-life count. This problem assesses understanding of exponential decay without heavy computation.
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