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+// Problem 8.8,Page no.212
+
+clc;clear;
+close;
+
+//Outer spring
+n_1=10 //number of coils
+D_1=3 //cm //Diameter of coil
+d_1=3 //mm //diameter of wire
+dell_1=2 //cm //deflection of spring
+
+//Inner spring
+n_2=8 //number of coils
+
+G=80 //GPa //Modulus of rigidity
+
+//Calculation
+
+R_1=D_1*2**-1
+P_1=G*10**9*dell_1*10**-2*(d_1*10**-3)**4*(64*(R_1*10**-2)**3*n_1)**-1 //Load carried outer spring for compression of 2 cm
+
+P_2=100-P_1 //N //Load carried by inner spring
+k_2=P_2*0.01**-1 //N/m //stiffness of inner spring
+
+//D_2=D_1*10**-2-d_1*10**-3-2*dell_1*10**-2-d_2 //Diameter of inner spring
+//Further simplifying above equation we get
+//D_2=0.023-d_2
+
+//Now from stiffness equation of inner spring
+//k=G*d_2**4*(8*D_2**3*n_2)**-1
+//Now substituting values and further simplifying we get
+//d**4=(0.023-d)**3*312500**-1
+
+//As d is small compared with 0.023,as a first appromixation
+d_2_1=(0.023**3*312500**-1)**0.25 //m
+
+//Second Approximation
+d_2_2=((0.023-d_2_1)**3*312500**-1)**0.25 //m
+
+//Final approximation
+d_2_3=((0.023-d_2_2)**3*312500**-1)**0.25*100 //cm
+
+//Result
+printf("Stiffness of inner spring is %.2f",k_2);printf(" N/m")
+printf("\n Wire Diameter of inner spring is %.3f cm",d_2_3)