Answer :

Correct answer is : 6

$h_{app}=\frac{t_1}{{\mathrm{\mu <!--\; \mu \; -->}}_1}+\frac{t_2}{{\mathrm{\mu <!--\; \mu \; -->}}_2}=\frac{6}{7}\times <!--\; \times \; -->3+\frac{6}{5}\times <!--\; \times \; -->3$

$\mathrm{=}\frac{18}{7}+\frac{18}{5}$

= 2.57 + 3.60 = 6.17 cm ⇒ N = 6