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Topic Topic: Weight of submerged object is reduced by buoyant force Post ReplyNew Question
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Joined: 26 June 2005
Posts: 918
Posted: 20 July 2005 at 11:45am | IP Logged Quote System

M is a cube-shaped object with 10 cm sides. The density of M is 5 g/cm.  M is placed on a weight scale and both objects are fully submerged in water. By assuming water density as  1 g/cm  and gravitational acceleration as  10 m/s ,  what will be the weight of M as measured by the scale?
    A.   25 N
    B.   30 N
    C.   35 N
    D.   40 N
    E.   45 N
    F.   50 N
 QUIZ QUESTION #317 created by Beka on 19 July 2005 at 2:26pm

Correct answer: D    Correct clicks: 41.67%  Wrong clicks: 58.33%
    A.   [20 clicks 13.89%] 
    B.   [14 clicks 9.72%] 
    C.   [14 clicks 9.72%] 
    D.   [60 clicks 41.67%] 
    E.   [13 clicks 9.03%] 
    F.   [23 clicks 15.97%] 
Please discuss this quiz question.
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Joined: 26 June 2005
Posts: 27
Posted: 23 July 2005 at 4:37pm | IP Logged Quote Beka

Performing all calculations in SI :-

Volume of M = 0.1 m  0.1 m  0.1 m = 0.001 m

Mass of M = 0.001 m  5000 kg/m = 5 kg

Weight of M = 5 kg  10 m/s = 50 N

This is the normal weight of M. Underwater, buoyant force pushes upwards and reduces its weight. According to Bernoulli's principle, magnitude of buoyant force equals the weight of the displaced liquid. Hence :-

Volume of displaced water = Volume of M =  0.001 m

Mass of displaced water = 0.001 m  1000 kg/m = 1 kg

Weight of displaced water = 1 kg  10 m/s = 10 N

Buoyant Force = Weight of displaced water = 10 N

Therefore the normal weight of M is reduced by 10 N.

The measured weight = 50 N - 10 N = 40 N


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