A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cross section 1cm find the energy stored in rod .Y=3.5 ×10^10 Nm2​

Energy stored in the rod, W = 4.55 × 10⁻⁴ J

Explanation:

From the question, the following values are given:

Mass of the brass rod, m = 10kg

Length of rod, l = 1 m

Radius of brass rod, r  =1 cm

Change in length of rod, e = ?

Energy stored, W = ?

Young's modulus of brass, Y = 3.5 × 10¹⁰ N/m²

Using the formula for energy stored in an elastic material;

W = 1/2 F × e

Force due to the load, F = mg =  10 × 10 = 100N

To determine the change in length, e, we use the formula for Young's modulus of elasticity;

Y = (F/A)/(e/l) = F × l / A × e  

e = (F × l) / (Y × A)

where A = πr²

e = 100 ×  1 / (3.5 × 10¹⁰ × π × (0.01)²

e = 9.09 × 10⁻⁶ m

Therefore, energy stored in the rod, W =  1/2 × 100 × 9.09 × 10⁻⁶

Energy stored in the rod, W = 4.55 × 10⁻⁴ J

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.