Question

Review Problem. A light string with a mass per unit length of 8.20 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (Fig. P16.30). An object of mass m is suspended from the center of the string, putting a tension in the string. (A) (mg)^1/2(B) 1/4(mg x 1000)^1/2 (C) 1/8(5mg x 1000)^1/2(D) None

Answer 1

Answer:

The answer is 'D' none

Explanation:

In the figure shown we have

$2Tsin(\theta )=mg\\\\\therefore T=\frac{mg}{2sin(\theta )}$

From the figure we can see that

$cos(\theta )=\frac{\frac{3L}{8}}{\frac{L}{2}}\\\\\therefore \theta =cos^{-1}(\frac{3}{4})\\\\sin(\theta )=\frac{\sqrt{7}}{4}$

Thus value of tension be will be

$T=\frac{2mg}{\sqrt{7}}$