Question

Q.4
The initial temperature of a liquid is
(80.0 + 0.1)°C. After it has been
cooled, its temperature is
(10.0 + 0.1)°C. The fall in
temperature in degree centigrade is
70.0
70.0 +0.3
70.0 +0.2
70.0 +0.1​

Answer 1

Answer:

(70.0 +0.2)°C

Explanation:

We know that If x=a−b

x±Δx=(a±Δa)−(b±Δb)

⇒±Δx=±(Δa+Δb)

∴  (70±0.1)°C−(10.0±0.1)°C     Error gets added up always

=([70-10]±[0.1+0.1])°C

=(70.0 ±0.2)°C

Answer 2

The fall in temperature in degree centigrade is 70.0 ± 0.2.

We have,

x ± Δx = (a ± Δa) − (b ± Δb)

⇒ ± Δx= (a ± Δa) − (b ± Δb) - x

⇒ ± Δx= (a ± Δa) − (b ± Δb) - a − b  ,  ∵ x = a − b

⇒ ± Δx= ± (Δa + Δb)

We know that the error is always added up.

∴  (70 ± 0.1)° C − (10.0 ± 0.1)° C = ( [70-10] ± [0.1+0.1] )° C  

                                                  =(70.0 ±0.2)° C

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