$0.1 \, Pa = \dots \, Nm^{-2} = \dots \, \text{dyne} \, cm^{-2}$.
(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 $0.1 \, Pa = 0.1 \, Nm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
Explore More
Similar Questions
The coefficients of thermal expansion of steel and a metal $X$ are respectively $12 \times 10^{-6} /^{\circ}C$ and $2 \times 10^{-6} /^{\circ}C$. At $40^{\circ}C$,the side of a cube of metal $X$ was measured using a steel vernier callipers. The reading was $100 \, mm$. Assuming that the calibration of the vernier was done at $0^{\circ}C$,then the actual length of the side of the cube at $0^{\circ}C$ will be:
Match List-$I$ with List-$II$:
$LIST$-$I$ $LIST$-$II$ $(A)$ Surface tension $(I)$ $kg m^{-1} s^{-1}$ $(B)$ Pressure $(II)$ $kg m s^{-1}$ $(C)$ Viscosity $(III)$ $kg m^{-1} s^{-2}$ $(D)$ Impulse $(IV)$ $kg s^{-2}$
Choose the correct answer from the options given below:
| $LIST$-$I$ | $LIST$-$II$ |
| $(A)$ Surface tension | $(I)$ $kg m^{-1} s^{-1}$ |
| $(B)$ Pressure | $(II)$ $kg m s^{-1}$ |
| $(C)$ Viscosity | $(III)$ $kg m^{-1} s^{-2}$ |
| $(D)$ Impulse | $(IV)$ $kg s^{-2}$ |
Choose the correct answer from the options given below:
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge,whose pitch is $0.5 \ mm$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5 \ mm$ and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2 \%$,the relative percentage error in the density is (in $\%$)
DifficultIIT 2011
View SolutionOut of the following,the only pair that does not have identical dimensions is
Medium
View SolutionPick 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 SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo