Answer:
(1) 1×10⁻⁴
Explanation:
From the question,
α = (ΔL/L)/(ΔT)............. Equation 1
Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.
Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C
Substitute these values into equation 1
α = 1×10⁻⁴/10
α = 1×10⁻⁵ °C⁻¹ .
β = (ΔA/A)/ΔT................... Equation 2
Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.
make ΔA/A the subject of the equation
ΔA/A = β×ΔT.......................... Equation 3
But,
β = 2α.......................... Equation 4
Substitute equation 4 into equation 3
ΔA/A = 2α×ΔT................ Equation 5
Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹
Substitute into equation 5
ΔA/A = ( 2)×(1×10⁻⁵)×(5)
ΔA/A = 10×10⁻⁵
ΔA/A = 1×10⁻⁴
Hence the right option is (1) 1×10⁻⁴
If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.
Calculation:
Step-1:
It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.
The gauge pressure at a particular depth of ocean water is calculated as:
Here 
is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.
Step-2:
Now we are substituting the values to calculate the pressure at the depth of 2452-m.
Learn more about gauge pressure here,
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