Two masses, M1 = 3 kg and M2 = 2 kg, are connected by a light inextensible string passing over a frictionless pulley (Atwood machine). The acceleration of the system is (Take g = 10 m/s^2): MCQ with Answer and Explanation
Two masses, M1 = 3 kg and M2 = 2 kg, are connected by a light inextensible string passing over a frictionless pulley (Atwood machine). The acceleration of the system is (Take g = 10 m/s^2):
A. 2 m/s^2
B. 4 m/s^2
C. 5 m/s^2
D. 10 m/s^2
Answer: Option A
Solution (By JKSSB Mock Tests)
For an Atwood machine, the common acceleration of both masses is given by the formula a = [(M1 - M2) / (M1 + M2)] * g. Substituting the given values: a = [(3 - 2) / (3 + 2)] * 10 = (1 / 5) * 10 = 2 m/s^2. The heavier mass accelerates downwards, the lighter one upwards.
Explanation:
Volume V = a³. For error propagation: % error in V = 3 × % error in a. Given 2% error in side, volume error = 3×2% = 6%. Power rule: if Q = xⁿ, %ΔQ = |n|×%Δx. Memory tip: Volume depends on cube of length ⇒ errors triple. Common in experimental error analysis questions.
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