You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Calorimetry

On a hypothet...

Question

On a hypothetical scale A the ice point is $42^{o}$ and the steam point is $182^{o}$ For another scale B, the ice point is $- 10^{o}$ and steam point in $90^{o}$ ) If B reads $60^{o}$ , the reading of A is,

160^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

140^{o}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

120^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

110^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $140^{o}$
Given,
$B = 60^{0} => \frac{A - 42}{182 - 42} = \frac{B - (- 10)}{90 - (- 10)}$
$=> \frac{A - 42}{140} = \frac{B + 10}{100} => \frac{A - 42}{14} = \frac{60 + 10}{10}$
$=> \frac{A - 42}{14} = \frac{70}{10} => A - 42 = 98$
$A = 140^{0}$

Suggest Corrections

Similar questions

Q. In a given scale

X

, the ice point is

40^{\circ} X

and the steam point is

120^{\circ} X

. For another scale

Y

, the ice point and steam point are

- 30^{\circ} Y

and

130^{\circ} Y

respectively. If

X

reads

50^{\circ} X

temperature of a body, then

Y

would read

Q. In a given scale

X

, the ice point is

40^{\circ} C

and the steam point is

120^{\circ} C

. For another scale

Y

, the ice point and steam point are

- 30^{\circ} C

and

130^{\circ} C

respectively. If

X

reads

50^{\circ} C

temperature of a body, then

Y

would read

Q. On a certain scale of temperature x, the ice points is

40^{\circ}

and the steam point is

160^{\circ}

. On another scale of temperature y, the ice points is

- 20^{\circ}

and the steam point is

100^{\circ}

. Convert

50^{\circ}

in x to the y-scale.

Q. 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

State the temperature of ice point and steam point on the Kelvin scale

Join BYJU'S Learning Program

On a hypothetical scale A the ice point is 42o and the steam point is 182o For another scale B, the ice point is −10o and steam point in 90o) If B reads 60o, the reading of A is,

The correct option is B 140oGiven,B=600=>A−42182−42=B−(−10)90−(−10)=>A−42140=B+10100=>A−4214=60+1010=>A−4214=7010=>A−42=98A=1400

On a hypothetical scale A the ice point is $42^{o}$ and the steam point is $182^{o}$ For another scale B, the ice point is $- 10^{o}$ and steam point in $90^{o}$ ) If B reads $60^{o}$ , the reading of A is,

The correct option is B $140^{o}$
Given,
$B = 60^{0} => \frac{A - 42}{182 - 42} = \frac{B - (- 10)}{90 - (- 10)}$
$=> \frac{A - 42}{140} = \frac{B + 10}{100} => \frac{A - 42}{14} = \frac{60 + 10}{10}$
$=> \frac{A - 42}{14} = \frac{70}{10} => A - 42 = 98$
$A = 140^{0}$