Question

A 4.5 m long rod has a mass of 2.8 kg and is stretched 0.21 m by force. What is the strain on the rod?

Select one:

a) 21.43     b) 0.047     c) 0.13    d) 6.1

Assuming atmospheric pressure to be 1.01 x 10⁵ Pa and the density of sea water to be 1025 kg/m³ , What is the absolute pressure at a depth of 15.0 m below the surface of the ocean?

Select one:

a. 1.508 x 10⁵ Pa

b. 1.16 x 10⁵ Pa

c. 4.98 x 10⁴ Pa

d. 2.52 x 10⁵ Pa

Answer 1

Answer: (i)

Length of the rod $L=4.5~m$

Change in lenght, $\Delta L=0.21~m$

Strain, $\epsilon =\frac{\Delta L}{L}=\frac{0.21}{4.5}=0.047$

(ii) Atmospheric pressure, $P_A=1.01\times10^5~Pa$

density of the water, $\rho =1025~kg/m^3$

depth, $h=15.0~m$

absolute pressure at a depth h below the surface of the ocean,

$P=P_A+\rho gh=(1.01\times10^5~Pa)+(1025\times9.8\times15.0)~Pa$

$=(1.01\times10^5~Pa)+(1.51\times10^5~Pa)=2.52\times10^5~Pa$