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Find the ratio of the weights as measure...

Find the ratio of the weights as measured by a spring balance, of a 1 kg block of iron nd a 1kg block of wood. Density of iron `=7800 kg m^-3` density of wood `=800 kg m^-3 and density of air =1.293 kgm^-3`.

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The correct Answer is:
A

Net weight of iron block
`w_1=mg-V_(1rho_(air))xxg`
`[m-(m/rho_1)rho_(air)]g`
`=[1-(1/7800)xx1.293]xx(9.8)`
Again net weight of wood
`=W_w=mg-V_w.rho_(air)g`
`=[m-(m/rho_w)rho_(air)]g`
`rarr (1-1/800xx1.293)9.8`
`rarr :. W_1/W_2=9.8((7800-1.293)/7800)/(9.8(800-1.293)/800)`
`=(7800-1.293)/(800-1.293)x8/78=1.0015`
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