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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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Answer» Correct Answer - 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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