NHT0015 16. A metal rod is heated increasing its temperature
by 10°C. The fractional change in its length is
1 *10-4. If a plate of same metal is heated raising
its temperature by 5°C, the fractional change in
its area will be .-
(1) 1 x 10-4
(2) 0.5 x 10-4
(3) 2 x 10-4
(4) 3 x 10-4​

(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

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂