Medium
Express:
$(a)$ The energy required to break one bond in $DNA$ in $eV$.
$(b)$ The kinetic energy of an air molecule $(10^{-21} \; J)$ in $eV$.
$(c)$ The daily intake of a human adult in kilocalories.
$(a)$ The energy required to break one bond in $DNA$ in $eV$.
$(b)$ The kinetic energy of an air molecule $(10^{-21} \; J)$ in $eV$.
$(c)$ The daily intake of a human adult in kilocalories.
(N/A) The energy required to break one bond of $DNA$ is given by:
$\frac{10^{-20} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.06 \; eV$.
Note: $0.1 \; eV = 100 \; meV$ ($100$ millielectron volt).
$(b)$ The kinetic energy of an air molecule is given by:
$\frac{10^{-21} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.0062 \; eV$.
This is equivalent to $6.2 \; meV$.
$(c)$ The average daily energy intake of a human adult is approximately $10^7 \; J$. Converting this to kilocalories:
$\frac{10^7 \; J}{4.2 \times 10^3 \; J/kcal} \simeq 2400 \; kcal$.
$\frac{10^{-20} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.06 \; eV$.
Note: $0.1 \; eV = 100 \; meV$ ($100$ millielectron volt).
$(b)$ The kinetic energy of an air molecule is given by:
$\frac{10^{-21} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.0062 \; eV$.
This is equivalent to $6.2 \; meV$.
$(c)$ The average daily energy intake of a human adult is approximately $10^7 \; J$. Converting this to kilocalories:
$\frac{10^7 \; J}{4.2 \times 10^3 \; J/kcal} \simeq 2400 \; kcal$.
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Similar Questions
Match List-$I$ with List-$II$.
List-$I$ List-$II$ $(A)$ Coefficient of viscosity $(I)$ $[ML^0T^{-3}]$ $(B)$ Intensity of wave $(II)$ $[ML^{-2}T^{-2}]$ $(C)$ Pressure gradient $(III)$ $[M^{-1}LT^2]$ $(D)$ Compressibility $(IV)$ $[ML^{-1}T^{-1}]$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(A)$ Coefficient of viscosity | $(I)$ $[ML^0T^{-3}]$ |
| $(B)$ Intensity of wave | $(II)$ $[ML^{-2}T^{-2}]$ |
| $(C)$ Pressure gradient | $(III)$ $[M^{-1}LT^2]$ |
| $(D)$ Compressibility | $(IV)$ $[ML^{-1}T^{-1}]$ |
The phenomenon of physics that deals with the constitution and structure of matter at the minute scales of atoms and nuclei is
In a new system,the unit of mass is $10 \ kg$,the unit of length is $5 \ m$,and the unit of time is $10 \ s$. Then,in this new system,$20 \ g/cm^3$ is equal to:
Medium
View SolutionChoose the physical quantity pair with the same dimensions.
Match the physical quantities given in List-$I$ with their dimensions in List-$II$.
List-$I$ List-$II$ $A$. Gravitational potential $I$. $M^0 L^2 T^{-2} K^{-1}$ $B$. Stefan's constant $II$. $M^0 L^2 T^{-2}$ $C$. Permittivity $III$. $M L^0 T^{-3} K^{-4}$ $D$. Specific heat capacity $IV$. $M^{-1} L^{-3} T^4 I^2$
$(The dimensions of mass, length, time, temperature and current are $M$, $L$, $T$, $K$ and $I$ respectively.)$
| List-$I$ | List-$II$ |
| $A$. Gravitational potential | $I$. $M^0 L^2 T^{-2} K^{-1}$ |
| $B$. Stefan's constant | $II$. $M^0 L^2 T^{-2}$ |
| $C$. Permittivity | $III$. $M L^0 T^{-3} K^{-4}$ |
| $D$. Specific heat capacity | $IV$. $M^{-1} L^{-3} T^4 I^2$ |
$(The dimensions of mass, length, time, temperature and current are $M$, $L$, $T$, $K$ and $I$ respectively.)$
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