- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
medium
It is estimated that per minute each $cm^2$ of earth receives about $2\ cal (1\ cal = 4.18\ J)$ of heat energy from the sun. This is called Solar constant. In $SI$ units the value is :-
A$1060$
B$178.4$
C$17.84$
D$1393.33$
Solution
Solar constant $=2 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}$
$2 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}=\mathrm{X} \mathrm{J} / \mathrm{m}^{2}-\mathrm{sec}$
$X=2$
$\left(\frac{c a l}{J}\right)\left(\frac{\sec }{\min }\right)\left(\frac{m^{2}}{c m^{2}}\right)$
$=2(4.18)\left(\frac{1}{60}\right)(10000)=1393.33$
$2 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}=\mathrm{X} \mathrm{J} / \mathrm{m}^{2}-\mathrm{sec}$
$X=2$
$\left(\frac{c a l}{J}\right)\left(\frac{\sec }{\min }\right)\left(\frac{m^{2}}{c m^{2}}\right)$
$=2(4.18)\left(\frac{1}{60}\right)(10000)=1393.33$
Standard 11
Physics
Similar Questions
Match List $I$ with List $II$ and select the correct answer using the codes given below the lists :
| List $I$ | List $II$ |
| $P.$ Boltzmann constant | $1.$ $\left[ ML ^2 T ^{-1}\right]$ |
| $Q.$ Coefficient of viscosity | $2.$ $\left[ ML ^{-1} T ^{-1}\right]$ |
| $R.$ Planck constant | $3.$ $\left[ MLT ^{-3} K ^{-1}\right]$ |
| $S.$ Thermal conductivity | $4.$ $\left[ ML ^2 T ^{-2} K ^{-1}\right]$ |
Codes: $ \quad \quad P \quad Q \quad R \quad S $
