An ungraduated thermometer of uniform bore is attached to a centimeter scale and is found to read 10 cm in melting ice- 30 cm in boiling water and 16 cm in a liquid- Find temperature of the liquid-

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Toppr Admin · Accepted Answer

Temperature at 10 cm is 00C-Temperature at 30cm si 1000C-So at L-16 cm- let say it measure T-xA0-So-xA0-L-x2212-1030-x2212-10-T-x2212-0100-x2212-016-x2212-1030-x2212-10-T-x2212-0100-x2212-0T-300C

An ungraduated thermometer of uniform bore is attached to a centimeter scale and is found to read $10 c m$ in melting ice, $30 c m$ in boiling water and $16 c m$ in a liquid. Find temperature of the liquid.

Temperature at 10 cm is $0^{0} C$ .
Temperature at $30 c m$ si $100^{0} C$ .
So at L=16 cm, let say it measure T.
So
$\frac{L - 10}{30 - 10} = \frac{T - 0}{100 - 0}$

$\frac{16 - 10}{30 - 10} = \frac{T - 0}{100 - 0}$
$T = 30^{0} C$

An ungraduated thermometer of uniform bore is attached to a centimeter scale and is found to read 10 cm in melting ice, 30 cm in boiling water and 16 cm in a liquid. Find temperature of the liquid.

Temperature at 10 cm is 00C.Temperature at 30cm si 1000C.So at L=16 cm, let say it measure T. So L−1030−10=T−0100−016−1030−10=T−0100−0T=300C

An ungraduated thermometer of uniform bore is attached to a centimeter scale and is found to read $10 c m$ in melting ice, $30 c m$ in boiling water and $16 c m$ in a liquid. Find temperature of the liquid.

Temperature at 10 cm is $0^{0} C$ .
Temperature at $30 c m$ si $100^{0} C$ .
So at L=16 cm, let say it measure T.
So
$\frac{L - 10}{30 - 10} = \frac{T - 0}{100 - 0}$

$\frac{16 - 10}{30 - 10} = \frac{T - 0}{100 - 0}$
$T = 30^{0} C$