# DETERMINATION OF MODULUS OF RIGIDITY OF A WIRE BY STATICAL METHOD USING HORIZONTAL PATTERN OF APPARATUS

• Theory: The Modulus of Rigidity

η= $$180MgD(L1-L2)/π^2r^4(θ1 – θ2 )$$     …………….(1)

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.
• Scale
• 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:
 Sl No Reading along any direction (X) Reading along perpendicular to the previous direction (Y) Uncorrected  diameter (X+Y)/2 (cm) Mean uncorrected diameter (cm) Mean corrected diameter (cm) Mean radius (r) (cm) msr (cm) csr total   (cm) msr (cm) csr total (cm)
• 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) 1 0 2 500 3 1000 … …
• Calculation:

•  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

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