Question

An object of 5kg mass is placed on a cylindrical stand having circular cross section. The radius of the base of the stand is 4 cm . If the same object is placed on cylindrical stand with radius of the base 2cm, the pressure becomes

(1) One fourth
(2) Half
(3) Two times
(4) Four times

Answer 1

two times because pressure is inversely proportional to area

Answer 2

Answer: (4) Four times

Explanation:

Case 1:

Mass of object, m = 5 kg

Radius of base of cylindrical stand, r = 4 cm = 0.04 m

$\because \text { Pressure }=\frac{\text { Force }}{\text { Base area }}$

$\text{Force} =\text { Body weight }=m g=5 \times 10=50 \ N$

$\text{Base area} =\pi r^{2}=\pi(0.04)^{2} \mathrm{m}^{2}$

$\therefore \text { Initial Pressure }=\frac{50 N}{\pi(0.04)^{2}} \rightarrow(1)$

Case 2:

Radius if base of cylindrical stand, r' = 2 cm = 0.02 m

$\because \text { Pressure }=\frac{\text { Force }}{\text { Base area }}$

Force = weight of body = mg = 50 N

$\text{Base area} =\pi r^{\prime 2}=\pi(0.02)^{2} \mathrm{m}^{2}$

$\therefore \text { Final pressure }=\frac{50 N}{\pi(0.02)^{2}} \rightarrow(2)$

From equations (1) and (2),

$\frac{\text {Initial pressure}}{\text {Final pressure}}=\frac{\frac{50 N}{\pi(0.04)^{2}}}{\frac{50 N}{\pi(0.02)^{2}}}$

$=\frac{1}{4}$

$\text{Initial pressure} =\frac{\text {Final pressure}}{4}$

$\text { Final pressure }=4 \times \text { Initial pressure }$