WAEC Physics Past Questions & Answers - Page 83

The formula for the frequency in a stringed instrument : \(f = \frac{1}{2} \sqrt{\frac{T}{m}}\)

f = frequency; T = tension in the string; m = mass per unit lenth of the string.

f\(_1\) = 14 = \(\frac{1}{2} \sqrt{\frac{T}{m}}\).

When T is quadrupled, we have

f\(_2\) = new frequency = \(\frac{1}{2} \sqrt{\frac{4T}{m}}\)

= 2(\(\frac{1}{2} \sqrt{\frac{T}{m}}\))

= 2 f\(_1\)

= 2 x 14 

= 28 Hz

No explanation has been provided for this answer.

No explanation has been provided for this answer.

Lens formula:

\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

where f, d\(_o\) and d\(_i\) are the focal length, object distance and image distance respectively.

The distance from the lens to the focal point is called the focal length. For converging lenses, the focal length is always positive, while diverging lenses always have negative focal lengths.

⇒ \(\frac{-1}{30} = \frac{1}{d_o} + \frac{1}{20}\)

\(\frac{1}{d_o} = \frac{-1}{30} - \frac{1}{20}\)

\(\frac{1}{d_o} = \frac{-2 + -3}{60}\) 

→ \frac{-5}{60}\) = \frac{-1}{12}\)

\(d_o = - 12 cm\)

The object distance -12cm = 12cm on the other side of the lens.

Sin C = \(\frac{1}{\eta} = \frac{1}{1.5} = 0.6667\)

C = sin\(^{-1}\) 0.6667

= 42°