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 made to stand vertically on a plane surface `(g= 10m^(-2))`

Options

A. `100 xx 10^4` pa
B. `10 xx 10^4`Pa
C. `14.28 xx 10^4 Pa`
D. `85.72 xx 10^4 Pa`

Answer

D

Explanation

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)