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Andrej [43]
3 years ago
10

The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val

ue of this melting point is 173.2 oC. What is the percent error in this reading?
Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

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<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

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