$0.1 \, Pa = \dots \, Nm^{-2} = \dots \, \text{dyne} \, cm^{-2}$.
(0.1, 1) $1$. We know that $1 \, Pa = 1 \, N/m^2$. Therefore,$0.1 \, Pa = 0.1 \, Nm^{-2}$.
$2$. To convert $Nm^{-2}$ to $\text{dyne} \, cm^{-2}$,we use the conversion factors: $1 \, N = 10^5 \, \text{dyne}$ and $1 \, m^2 = (10^2 \, cm)^2 = 10^4 \, cm^2$.
$3$. Thus,$1 \, Nm^{-2} = \frac{10^5 \, \text{dyne}}{10^4 \, cm^2} = 10 \, \text{dyne} \, cm^{-2}$.
$4$. Therefore,$0.1 \, Nm^{-2} = 0.1 \times 10 \, \text{dyne} \, cm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
$5$. The final result is $0.1 \, Pa = 0.1 \, Nm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
$2$. To convert $Nm^{-2}$ to $\text{dyne} \, cm^{-2}$,we use the conversion factors: $1 \, N = 10^5 \, \text{dyne}$ and $1 \, m^2 = (10^2 \, cm)^2 = 10^4 \, cm^2$.
$3$. Thus,$1 \, Nm^{-2} = \frac{10^5 \, \text{dyne}}{10^4 \, cm^2} = 10 \, \text{dyne} \, cm^{-2}$.
$4$. Therefore,$0.1 \, Nm^{-2} = 0.1 \times 10 \, \text{dyne} \, cm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
$5$. The final result is $0.1 \, Pa = 0.1 \, Nm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
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Similar Questions
Some physical quantities are given in Column $I$ and some possible $SI$ units in which these quantities may be expressed are given in Column $II$. Match the physical quantities in Column $I$ with the units in Column $II$.
Column $I$ Column $II$ $(A)$ $GM_e M_s$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$M_s$: mass of the Sun) $(p)$ $(\text{volt})(\text{coulomb})(\text{metre})$ $(B)$ $\frac{3RT}{M}$ ($R$: universal gas constant,$T$: absolute temperature,$M$: molar mass) $(q)$ $(\text{kg})(\text{m})^3(\text{s})^{-2}$ $(C)$ $\frac{F^2}{q^2 B^2}$ ($F$: force,$q$: charge,$B$: magnetic field) $(r)$ $(\text{m})^2(\text{s})^{-2}$ $(D)$ $\frac{GM_e}{R_e}$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$R_e$: radius of the earth) $(s)$ $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$
| Column $I$ | Column $II$ |
| $(A)$ $GM_e M_s$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$M_s$: mass of the Sun) | $(p)$ $(\text{volt})(\text{coulomb})(\text{metre})$ |
| $(B)$ $\frac{3RT}{M}$ ($R$: universal gas constant,$T$: absolute temperature,$M$: molar mass) | $(q)$ $(\text{kg})(\text{m})^3(\text{s})^{-2}$ |
| $(C)$ $\frac{F^2}{q^2 B^2}$ ($F$: force,$q$: charge,$B$: magnetic field) | $(r)$ $(\text{m})^2(\text{s})^{-2}$ |
| $(D)$ $\frac{GM_e}{R_e}$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$R_e$: radius of the earth) | $(s)$ $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$ |
Pick out the two scalar quantities in the following list:
force,angular momentum,work,current,linear momentum,electric field,average velocity,magnetic moment,relative velocity.
force,angular momentum,work,current,linear momentum,electric field,average velocity,magnetic moment,relative velocity.
Medium
View SolutionThe related effort to derive the properties of a bigger,more complex system from the properties and interactions of its constituent simpler parts is:
State whether the following statements are true or false:
$(a)$ Weight is a fundamental physical quantity.
$(b)$ $A$ micrometer is used to measure the size of an oleic acid molecule at the molecular level.
$(c)$ The dimensions of a derived quantity obtained from any fundamental quantity can never be zero.
$(a)$ Weight is a fundamental physical quantity.
$(b)$ $A$ micrometer is used to measure the size of an oleic acid molecule at the molecular level.
$(c)$ The dimensions of a derived quantity obtained from any fundamental quantity can never be zero.
Medium
View SolutionWrite the unit and dimensional formula of the ratio of the gravitational constant $(G)$ to the acceleration due to gravity $(g)$.
Medium
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