-->

Mechanical Properties Of Fluids

Question
CBSEENPH11018116

One meter long elastic wire is fixed at one end and at the other end a weight of 3kg is attached. The mass is pulled away from vertical line (passing through point of suspension) to a distance 0.6m and made to revolve with horizontal circle of radius 0-6m. Find the increase in length of the wire. Given that area of cross-section of wire 2x10–7m2 and Young's modulus of material of wire 2.5x1010 N/m2.

Solution

The given question is illustrated in the figure. below.


From right angled ΔSOA,
s i n theta equals fraction numerator 0.6 over denominator 1 end fraction equals 0.6
therefore space space space space space space c o s theta equals square root of 1 minus s i n squared straight theta end root equals 0.8
The different forces acting on mass are:

(i) Weight mg in vertically downward direction.
(ii) Tension T in the string along AS.
Now resolving the components of T as shown in fig. above.
The component Tcosθ balances the weight mg of the mass and component Tsinθ provides the necessary centripetal force to whirl the mass.
straight i. straight e. space space space space space space space straight T space c o s theta equals m g space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis

rightwards double arrow space space space space space space straight T equals fraction numerator m g over denominator c o s theta end fraction equals fraction numerator 3 cross times 9.8 over denominator 0.8 end fraction equals 36.75 straight N
N o w comma
space space space space space space space space space space space straight Y equals fraction numerator straight T divided by straight A over denominator capital delta l divided by straight l end fraction

rightwards double arrow space space space space space capital delta l equals fraction numerator Y l over denominator Y A end fraction
space space space space space space space space space space space space space equals fraction numerator 36.75 cross times 1 over denominator 2.5 cross times 10 to the power of 10 cross times 2 cross times 10 to the power of negative 7 end exponent end fraction

space space space space space space space space space space space space space equals 7.35 cross times 10 to the power of negative 3 end exponent straight m equals 7.35 m m