Ice point and steam point on BYJUS scale reads 8- B and 72- B respectively- The temperature on - C scale is 25- C- find reading on BYJUS scale at same temperatureA- 24- BB- 32- BC- 18- BD- 26- B

The correct option is A 24^o BB F.P/B.P F.P = C 0/100 0 B 8/64 = C/100 B = 64 . C/100 + 8 = 64 × 25/100 + 8 nbsp; nbsp; nbsp;= 16 + 8 = 24^o ...

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Standard XII

Physics

Scale Conversions

Ice point and...

Question

Ice point and steam point on BYJUS scale reads $8^{o} B$ and $72^{o} B$ respectively. The temperature on $^{o} C$ scale is $25^{o} C,$ find reading on BYJUS scale at same temperature

^{o} B

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^{o} B

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^{o} B

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^{o} B

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Solution

The correct option is A 24 $^{o} B$
$\frac{B - F . P}{B . P - F . P} = \frac{C - 0}{100 - 0}$
$\frac{B - 8}{64} = \frac{C}{100}$
$B = \frac{64 . C}{100} + 8$
$= \frac{64 \times 25}{100} + 8$
= 16 + 8 = 24 $^{o} B$

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Ice point and steam point on BYJUS scale reads 8oB and 72oB respectively. The temperature on oC scale is 25oC, find reading on BYJUS scale at same temperature

The correct option is A 24oBB−F.PB.P−F.P=C−0100−0 B−864=C100 B=64.C100+8 =64×25100+8 = 16 + 8 = 24oB

Ice point and steam point on BYJUS scale reads $8^{o} B$ and $72^{o} B$ respectively. The temperature on $^{o} C$ scale is $25^{o} C,$ find reading on BYJUS scale at same temperature

The correct option is A 24 $^{o} B$
$\frac{B - F . P}{B . P - F . P} = \frac{C - 0}{100 - 0}$
$\frac{B - 8}{64} = \frac{C}{100}$
$B = \frac{64 . C}{100} + 8$
$= \frac{64 \times 25}{100} + 8$
= 16 + 8 = 24 $^{o} B$