The Total solution for NCERT class 6-12
Let us consider F1┬а=86o┬аF
And F2┬а=95o
We know, F = 9/5C + 32
F1┬а=9/5 C1┬а+ 32
F1┬атАУ32 = 9/5 C1
C1┬а=5/9 (F1┬атАУ 32)
= 5/9 (86 тАУ 32)
= 5/9 (54)
= 5 ├Ч 6
= 30o┬аC
Now,
F2┬а=9/5 C2┬а+ 32
F2┬атАУ32 = 9/5 C2
C2┬а=5/9 (F2┬атАУ 32)
= 5/9 (95 тАУ 32)
= 5/9 (63)
= 5 ├Ч 7
= 35o┬аC
тИ┤ The range oftemperature of the solution in degree Celsius is 30┬░ C and 35┬░ C.