An iron cylindar of base area 7 cm^2 and mass 10 kg is melted and reshaped into a into a rod of area of cross section 1cm^2.Calculate the increase in the pressure exerted on a plane surface when both the cylinder and the rod are madeto stand vertically on a plane surface (g= 10m^(-2))

An iron cylindar of base area 7 cm^2 and mass 10 kg is melted and resh

1.	An iron cylindar of base area 7 cm^2 and mass 10 kg is melted and reshaped into a into a rod of area of cross section 1cm^2.Calculate the increase in the pressure exerted on a plane surface when both the cylinder and the rod are madeto stand vertically on a plane surface (g= 10m^(-2))
Answer» `100 xx 10^4` pa `10 xx 10^4`Pa `14.28 xx 10^4 Pa` `85.72 xx 10^4 Pa` Solution :GIVE , MASS of cylinder , `m_c = 10kg` Base area of the cylinder , `A_c = 7 cm^2 = 7xx(10^-2)^2 = 7xx10^-4m^2` Pressure due to cylinder . `P_c = F_c/A_c = (m_cg)/A_c = (10xx10)/(7xx10^-4) = 14.28xx10^4 Pa` Mass of the rod , `m_r = 10 kg` Base area of the rod , `A_r =1cm^2 = 1xx(10^-2)^2 = 10^-4m^2` Pressure due to rod `P_r = F_r/A_r = (m_rg)/A_r = (10xx10)/10^-4 = 100xx10^4 Pa` Pressure due to `P_r-P_r=(100-14.28) xx 10^4)` Pa Hence the CORRECT OPTION is (d)