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An iron cylindar of base area 7 cm^2 and...

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))`

A

`100xx10^(4)` Pa

B

`10xx10^(4)` Pa

C

`14.28xx10^(4)` Pa

D

`85.72xx10^(4)` Pa

Text Solution

Verified by Experts

The correct Answer is:
D
Given, mass of cylinder, `m_(c)=10" kg"`
Base area of the cylinder,
`A_(c)=" 7 cm"^(2)=7xx(10^(-2))^(2)=7xx10^(-4)m^(2)`
Pressure due to cylinder.
`P_(c)=(F_(c))/(A_(c))=(m_(c)g)/(A_(c))=(10xx10)/(7xx10^(-4))=14.28xx10^(4)" Pa"`
Mass of the rod, `m_(r)=10" kg"`
Base area of the rod,
`A_(r)1" cm"^(2)=1xx(10^(-2))^(2)=10^(-4)m^(2)`
Pressure due to rod.
`p_(r)=(F_(r))/(A_(r))=(m_(r)g)/(A_(r))=(10xx10)/(10^(-4))=100xx100^(4)" Pa"`
Increase in pressure,
`P_(r)-P_(c)=(100-14.28)xx10^(4)" Pa"`
Hence, the correct option is (d).
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