4 g of C3H4 = 0.1 mol
From 0.1 mol of P, 0.01 mol of R will be produced
\(⇒ 1.62 \) g of R is produced
From 0.1 mol of P, 0.032 mol of U is produced
\(= 3.2 \)g of U is produced
If 0.01 mol of $\mathrm{P_4O_{10}}$ is removed from 0.1 mol, then the remaining molecules of $\mathrm{P_4O_{10}}$ will be:
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is: