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**Practical**Institute:
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Biju Patnaik University of Technology BPUT
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B. TECH PHYSICS LABORATORY
MANUAL
GOVERNMENT COLLEGE OF ENGINEERING,
KALAHANDI, BHAWANIPATNA
Mrs. A. Gupta
Dr. S.K. Mohanta
Department of Basic Science
©Dept. of Basic Science, GCEK, Bhawanipatna

Contents
1. Acceleration due to gravity (g) by bar (Compound) pendulum.
2. Determination of wavelength of Sodium Yellow light using Newton’s ring set-up.
3. Young’s modulus by Searl’s apparatus.
4. Determination of Grating element.
5. Hall effect.
6. Surface tension by Capillary rise method.
7. Young’s modulus by bending of beam.
8. Rigidity modulus by Static method.
9. P-N diode charactreistics.
10. Characteristics of BJT.

EXPERIMEN NO. 1
Aim:
To determine the acceleration due to gravity (g) by using a Bar pendulum and also
determine radius of gyration about an axis through the center of gravity for the bar pendulum.
Apparatus Required:
(i) A bar pendulum, (ii) a knife–edge with a platform, (iii) a sprit level, (iv) a precision stop
watch, and (v) a meter scale.
Theory:
A simple pendulum consists of a small body called a “bob” (usually a sphere) attached to
the end of a string the length of which is great compared with the dimensions of the bob and the
mass of which is negligible in comparison with that of the bob. Under these conditions the mass
of the bob may be regarded as concentrated at its center of gravity, and the length of the pendulum
is the distance of this point from the axis of suspension. When the dimensions of the suspended
body are not negligible in comparison with the distance from the axis of suspension to the center
of gravity, the pendulum is called a compound, or physical, pendulum. A rigid body mounted upon
a horizontal axis so as to vibrate under the force of gravity is a compound pendulum.
In Fig.1 a body of irregular shape is pivoted about a horizontal frictionless axis through P and is
displaced from its equilibrium position by an angle 𝜃. In the equilibrium position the center of
gravity G of the body is vertically below P. The distance GP is l and the mass of the body is m.
The restoring torque for an angular displacement 𝜃 is
1

For small amplitudes (𝜃 ≈ 0),
𝜏 = − 𝑔 sin 𝜃
𝐼
2𝜃
2
……………..
= − 𝑔𝜃
……………..
where I is the moment of inertia of the body through the axis P. Eq. (2) represents a simple
harmonic motion and hence the time period of oscillation is given by
𝐼
𝑔
𝑇 = 𝜋√
.….…………
Now 𝐼 = 𝐼𝐺 +
, where 𝐼𝐺 is the moment of inertia of the body about an axis parallel with axis
of oscillation and passing through the center of gravity G.
……………..(4)
𝐼𝐺 =
where K is the radius of gyration about the axis passing through G. Thus,
𝑇 = 𝜋√
2
2
+
𝑔
𝐾2
+
𝑙
𝜋√
=
…………….
𝑔
The time period of a simple pendulum of length L, is given by
𝑇 = 𝜋√
Comparing with Eq. (5) we get
=
+
……………(6)
𝑔
2
……………..
This is the length of “equivalent simple pendulum”. If all the mass of the body were concentrated
2
at a point O (See Fig.1) such that 𝑃 = +
, we would have a simple pendulum with the same
time period. The point O is called the ‘Centre of Oscillation’. Now from Eq. (7)
−
+
=
………………
i.e. a quadratic equation in . Equation (8) has two roots
And
=
+
=
and
such that
……………(9)
Thus both 𝑎
are positive. This means that on one side of C.G there are two positions of the
centre of suspension about which the time periods are the same. Similarly, there will be a pair of
positions of the centre of suspension on the other side of the C.G about which the time periods will
be the same. Thus there are four positions of the centers of suspension, two on either side of the
C.G, about which the time periods of the pendulum would be the same. The distance between two
such positions of the centers of suspension, asymmetrically located on either side of C.G, is the
length L of the simple equivalent pendulum. Thus, if the body was supported on a parallel axis
through the point O (see Fig. 1), it would oscillate with the same time period T as when
2

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