The photograph of a house occupies an area of 1.75 cm²on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement.

Random errors involved in a set of 100 measurements are very less as compared to the set of 5 measurements. Therefore, a set of 100 measurements is more reliable than a set of 5 measurements.

Area of the sheet = 2 (l × 0 + b × t + t × l)

                       = 2 (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

                       = 2 (4.3604739)

                       = 8.7209478 m²

Area can contain a maximum of three significant digits, therefore, rounding off, we get

Area = 8.72 m²

Also, volume = l × b × t

               V = 4.234 × 1.005 × 0.0201

                  = 0.0855289

                  = 0.0855 m³ (Significant Figures = 3)

a) 1

The given quantity is 0.007 m²

If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. 

Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all the digits are significant figures.