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Byju's Answer

Standard XII

Physics

Area and Volume Expansion

A manufacture...

Question

A manufacture marks the thermometer wrongly. At 0 $^{o}$ C it reads -10 $^{o}$ C, at 100 $^{o}$ C it reads 85 $^{o}$ C. Then the reading at 50 $^{o}$ C will be :

^{o}

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32.5

^{o}

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37.5

^{o}

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42.5

^{o}

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Solution

The correct option is C 37.5 $^{o}$ C
From the interpolation we can get the value of x.

$\frac{100^{o} - 50^{o}}{85^{o} - x^{o}} = \frac{50^{o} - 0^{o}}{x^{o} - 10^{o}}$

$x^{o} = {37.5}^{o}$

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A manufacture marks the thermometer wrongly. At 0o C it reads -10o C, at 100o C it reads 85oC. Then the reading at 50o C will be :

The correct option is C 37.5o CFrom the interpolation we can get the value of x.100o−50o85o−xo=50o−0oxo−10oxo=37.5o

A manufacture marks the thermometer wrongly. At 0 $^{o}$ C it reads -10 $^{o}$ C, at 100 $^{o}$ C it reads 85 $^{o}$ C. Then the reading at 50 $^{o}$ C will be :

The correct option is C 37.5 $^{o}$ C
From the interpolation we can get the value of x.

$\frac{100^{o} - 50^{o}}{85^{o} - x^{o}} = \frac{50^{o} - 0^{o}}{x^{o} - 10^{o}}$

$x^{o} = {37.5}^{o}$