As weight of liquid in capillary is balanced by surface tension then $T \times 2 \pi r= \pi r^2 h_1 \rho g$
$h_1= \large\frac{2T}{r\rho g}$
But weight of liquid in tapered table as shown is more than uniform tube of radius the r then in order to balance the weight $h < h_1$
$\therefore h < \large\frac{2T}{r \rho g}$
Hence c is the correct answer.