Question

(D) उपर्युक्त में कोई नहीं
Draw a graph showing variation of spring force (F) with extension
ideal spring. Write S.I. unit of spring constant.
[2]
आदर्श स्प्रिग के लिए स्प्रिग-बल (F) में लम्बाई वृद्धि (X) के साथ होने वाले परिवर्तन का ग्राफ बनाए । स्प्रिर
नियतांक का S.I. मात्रक लिखिए ।
Using dimensional analysis, derive Stoke's law for the viscous force acting or
spherical body of radius (r), falling freely with a velocity (v) through a fluid
viscosity n.
वेमीय विश्लेषण का उपयोग करके, n श्यानता के तरल में, v वेग से स्वतंत्रतापूर्वक गिरते हुए, 1 त्रि
​

Answer 1

from Hook's law,

spring force is directly proportional to elongation or compression of spring.

if we assume F is spring force acts on it, then spring elongates or compresses x from its equilibrium position.

from Hook's law, F ∝ x

⟹ F=kx

here, k is spring constant.

hence, graph between force and extension of spring is a straight line as shown in figure.

from graph, it is clear that force will be negative when spring is compressed and force will be positive when spring is elongated.

si unit of spring constant, k = si unit of force/si unit of extension (length)

= N/m

dimension of viscous force, F = [MLT^-2]

dimension of radius , r = [L]

dimension of velocity, v = [LT^-1]

dimension of viscosity , n = [ML^-1T^-1]

from dimensional analysis,

F ∝ r^x v^y n^z

[MLT^-2] ∝ [L]^x [LT^-1]^y [ML^-1T^-1]^z

[MLT^-2] ∝ [M^z] [L]^(x + y - z) [T]^(-y-z)

comparing both sides,

z = 1,

x + y - z = 1 => x + y = 2

-y - z = -2 => y + z = 2 => y = 1

so, x = 1 , y = 1 and z = 1

then, F ∝ rvn

hence, viscous force is directly proportional to radius, velocity and viscosity.