A beaker of mass m resrs on scale .it contains a mass mo of oil with denisty pa.

Question

A beaker of mass m resrs on scale .it contains a mass mo of oil with denisty pa. A block of iron of mass m i and denisty pi is suspended from spring scale and partially immersed in oil , such that one third of the block is in air

The reading of the top scale Ft and bottom Fb, scale are given, respectively, by Ft = mig - B, Fb = (m + mo)g+ B. Ft = mig, Fb = (m + mo) g - B . Ft = mig + B, Fb = (m + mo)g - B. Ft = mig, Fb = (m + mo)g + B . Ft = 0, Fb = (mi + m + mo)g. Ft =mig, Fb = (m + mo)g. Ft = mig - B, Fb = (m + mo)g. Ft = mig + B, Fb =(m + mo)g + B. Ft = mig + B, Fb = (m + mo)g. Ft = mig - B, Fb= (m + m0)g - B. Neglecting the buoyancy of the air, the buoyant force B acting on the block is B =rho i/rho o mog. B = 2/3 rho i/rho o mog. B = 2/3 rho o/rho i mig. B = 1/3 rho i/rho o mog. B = 1/3 rho o/rho i mig. B = mig. B = rho i/rho o mi g. B = rho o/rho i mi g . B = 1/3 rho i/rho o mi g. B = 2/3 rho i/rho o mi g.

Explanation / Answer

It is option 7.

since free body diagram of mass mi means buoyancy and Ft upwards and weight of mass downwards

Thus Ft= mig -B

Similarly Free body of whole system implies Fb= (m+mo)*g

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