Q106 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
A force of $450$ N acts on a surface area of $15$ cm$^2$. Find the pressure exerted.
- A$3.0 \times 10^4$ Pa
- Correct answerB$3.0 \times 10^5$ Pa
- C$4.5 \times 10^4$ Pa
- D$4.5 \times 10^5$ Pa
Solution
Given, force $F = 450$ N
Area $A = 15$ cm$^2 = 15 \times 10^{-4}$ m$^2$
Pressure $P = \dfrac{F}{A}$
$P = \dfrac{450}{15 \times 10^{-4}}$
$P = 30 \times 10^4$ Pa
$P = 3.0 \times 10^5$ Pa
Answer: $3.0 \times 10^5$ Pa
Area $A = 15$ cm$^2 = 15 \times 10^{-4}$ m$^2$
Pressure $P = \dfrac{F}{A}$
$P = \dfrac{450}{15 \times 10^{-4}}$
$P = 30 \times 10^4$ Pa
$P = 3.0 \times 10^5$ Pa
Answer: $3.0 \times 10^5$ Pa
Hence, the correct answer is option: 'B'.