D. Solve the following numericals.
1. Wax melts at 335 K. Express this temperature in the Fahrenheit and Celsius scales.
2. The temperature of milk in a glass is 60° C. Express this temperature in the Fahrenheit and
Kelvin scales.
3. What should be the time period of a wave with a frequency of 300 Hz?
4. Find the frequency of a wave whose time period is 0.2 second.
5. Aditi claps and hears the echo after reflection from a cliff which is 990 m away. If the velocity
of sound is 330 m/s, calculate the time taken for hearing the echo.
Answer:
Given,
Temperature at which Wax Melts is 335
For any temperature scale,
\boxed{ \dfrac{ x – LFP }{UFP – LFP} = \sf \: constant}
UFP−LFP
x−LFP
=constant
For Kelvin scale,
\frac{k – 273}{373 – 273} = \frac{k – 273}{100} = \sf \: constant
373−273
k−273
=
100
k−273
=constant
For Celsius scale,
\frac{c – 0}{100} = \frac{c}{100} = \sf \: constant \:
100
c−0
=
100
c
=constant
For Fahrenheit scale,
\frac{f – 32}{212 – 32} = \sf \: constant
212−32
f−32
=constant
Here, k, c, f are temperatures on respective scales.
According to given question,
k =335
Conversion to Celsius scale
\begin{gathered} \frac{k – 273}{100} = \frac{c}{100} \\ \\ k – 273 = c \\ \\ 335 – 273 = c \\ \\ c = 62\end{gathered}
100
k−273
=
100
c
k−273=c
335−273=c
c=62
Conversion to Fahrenheit scale
\begin{gathered} \frac{k – 273}{100} = \frac{f – 32}{180} \\ \\ \frac{k – 273}{5} = \frac{f – 32}{9} \\ \\ \frac{335 – 273}{5} = \frac{f – 32}{9} \\ \\ \frac{62}{5} = \frac{f – 32}{9} \\ \\ f – 32 = \frac{62 \times 9}{5} \\ \\ f – 32 = 111.6 \\ \\ f = 111.6 + 32 \\ \\ f = 143. 6\end{gathered}
100
k−273
=
180
f−32
5
k−273
=
9
f−32
5
335−273
=
9
f−32
5
62
=
9
f−32
f−32=
5
62×9
f−32=111.6
f=111.6+32
f=143.6
Therefore,
Temperature on Kelvin = 335 K
Temperature on Celsius = 62°C
Temperature on Fahrenheit = 143.6°F.