Question

सही जोड़ी बनाइए -
कॉलम 'अ'
(i) स्थैतिक घर्षण गुणांक
(ii) 1 ev .
(iii) कोणीय संवेग संरक्षण नियम
(iv) आदर्श गैस समीकरण
(v) प्रतिबल
कॉलम 'ब'
(a) PV = nRT
(b) प्रत्यानयन बल/क्षेल
(c) us = Fs/R
(d) Io = नियतांक
(e) 1.6x10-19 जूल​

Answer 1

Answer:

i)$\mu_s = \frac{F_s}{R}$

ii)$1 eV = 1.6 \times 10^{-19}$

iii) Angular momentum conservation

$I\omega = constant$

iv) Ideal gas equation

PV = nRT

v) Stress = Force / Area

Explanation:

i) Static Friction coefficient

$F_s = \mu_s R$

so we have

$\mu_s = \frac{F_s}{R}$

ii) 1 eV

it is the unit of energy and equal to the work done to pass the electron through 1 volt

so we have

$U = (1.6 \times 10^{-19})(1) J$

$U = (1.6 \times 10^{-19}) J$

iii) Angular momentum conservation

When net external torque on the system is zero then its angular momentum is always conserved

so we will have

$L = I\omega = constant$

iv)Ideal gas equation is given as

$PV = nRT$

here we know

P = pressure of gas

V = volume of gas

n = number of moles

v) Stress

Stress is the ratio of force and area

so it is given as

$Stress = \frac{Force}{Area}$

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Topic : Ideal gas equation

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