- Theory: The Modulus of Rigidity
η= \( 180MgD(L1-L2)/π^2r^4(θ1 – θ2 ) \) …………….(1)
where M = load suspended.
D = Diameter of the pulley.
R = Radius of the experimental rod.
G = Acceleration due to gravity.
L1 – L2 = distance between two pointers.
θ1 – θ2 = Difference of the angular twist produced in the rod.
- Apparatus used:
- Horizontal pattern of apparatus.
- Set of ½ kg weights.
- Screw gauge.
- Experimental Results:
- Table for least count of the screw gauge:
|Pitch of the screw(p) (cm)||No. of circular scale division||Least count = p/n (cm)|
Zero error of the screw gauge = …………. c.m.
- Table for determination of the radius (r) of the rod:
|Reading along any direction (X)||Reading along perpendicular to the previous direction (Y)||Uncorrected diameter (X+Y)/2
|Mean uncorrected diameter
|Mean corrected diameter
|Mean radius (r)
- Table for the Determination of the diameter (D) of the pully:
|Circumference of the pully (L) in cm||Radius of the pully (R) in cm||Diameter of the pully (D) in cm|
- Difference in the length between the two pointer (L1 – L2) = ……… cm
- Table for the determination of the Twist:
|No of obs||Weight placed on the pan (M) in gm||Twist of the first pointer in degree (θ1)||Twist of the second pointer in degree (θ2)||(θ1~θ2) in degree|
|Load increasing||Load decreasing||Mean value of angle of twist (θ1)||Load increasing||Load decreasing||Mean value of angle of twist (θ2)|
- Percentage Error :
\( dη/η=dM/M+dg/g+dD/D+d(L1-L2)/(L1-L2)+4dr/r+d(θ1 – θ2 )/(θ1 – θ2 ) \)
\( dM \) =0, \( dg \) =0, \( dD \) =0.1cm, \( d(L1-L2) \) =2 *0.1 cm, dr=least count of the screw gauge,
d(θ1 – θ2 )=2*10.
Percentage error = \( dη/η \) *100%
- Result: The modulus of rigidity ( η) of the rod is ….. dyne/cm2