A wire, suspended vertically from one of its ends, is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm . If the elastic energy (in J ) stored in the wire is U , What is the value of (1/U) ?

Question

A wire, suspended vertically from one of its ends, is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm . If the elastic energy (in J ) stored in the wire is U , What is the value of (1/U) ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Elastic potential energy, U=(1/2)× F× l=(1/2)× 200× 10- 3= text0.1J Therefore, (1/U)=10 .

Q. A wire, suspended vertically from one of its ends, is stretched by attaching a weight of $200 N$ to the lower end. The weight stretches the wire by $1 mm$ . If the elastic energy (in $J$ ) stored in the wire is $U$ , What is the value of $\frac{1}{U}$ ?

Elastic potential energy, $U = \frac{1}{2} \times F \times l = \frac{1}{2} \times 200 \times 1 0^{- 3} = 0.1 J$
Therefore, $\frac{1}{U} = 10$ .

Q. A wire, suspended vertically from one of its ends, is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm . If the elastic energy (in J ) stored in the wire is U , What is the value of U1​ ?

Elastic potential energy, U=21​×F×l=21​×200×10−3=0.1J Therefore, U1​=10 .

Q. A wire, suspended vertically from one of its ends, is stretched by attaching a weight of $200 N$ to the lower end. The weight stretches the wire by $1 mm$ . If the elastic energy (in $J$ ) stored in the wire is $U$ , What is the value of $\frac{1}{U}$ ?

Elastic potential energy, $U = \frac{1}{2} \times F \times l = \frac{1}{2} \times 200 \times 1 0^{- 3} = 0.1 J$
Therefore, $\frac{1}{U} = 10$ .