- Home
- Standard 11
- Physics
10-1.Thermometry, Thermal Expansion and Calorimetry
hard
$0°C$ તાપમાને એક ગોળો ${\omega _0}$ કોણીય વેગથી ભ્રમણ કરે છે. હવે તાપમાન $100°C$ થતા નવી કોણીય ઝડપ કેટલી થાય? ( ${\alpha _B} = 2.0 \times {10^{ - 5}}{\rm{\,per}}°C^{-1}$ )
A
$1.1{\omega _0}$
B
$1.01{\omega _0}$
C
$0.996{\omega _0}$
D
$0.824{\omega _0}$
Solution
${R_{100}} = {R_0}[1 + \alpha \times 100]$
$R_{100}^2 = R_0^2[1 + 2\alpha \times 100]$
${I_1}{\omega _1} = {I_2}{\omega _2}$
$\Rightarrow$ $\frac{2}{5}MR_0^2{\omega _1} = \frac{2}{5}MR_{100}^2{\omega _2}$
$\Rightarrow$ $R_0^2{\omega _1} = R_0^2[1 + 2 \times 2 \times {10^{ – 5}} \times 100]{\omega _2}$
$\Rightarrow$ ${\omega _2} = \frac{{{\omega _1}}}{{[1 + 4 \times {{10}^{ – 3}}]}} = \frac{{{\omega _0}}}{{1.004}} = 0.996{\omega _0}$
Standard 11
Physics
