When we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure different the soft drink rises in the straw and we are able to take the soft drink easily.

When we cut vegetables with a sharp knife the area of contact is less. Hence the stress applied becomes more \([Stress =\frac{\text { Force }}{\text { Area }}=\frac{F}{\Delta A}]\) the vegetables would be cut easily. So, it is so easy to cut vegetables with a sharp knife rather than a blunt knife.

The sugar lump dissolves more rapidly at some points than at others. Where it dissolves, the surface tension of the liquid is reduced. As the force of surface tension reduces by different amounts in different points of the sugar lump, coffee runs up into a sugar lumb when one corner of it is held in the liquid.

\(\text {Length of a cylinder } \quad l=1.5 \mathrm{~m} \)
\(\text {Diameter of the cylinder } \quad d=4 \mathrm{~cm} \)
\(\therefore \text { Radius of the cylinder } \quad \mathrm{r}=\frac{4}{2}=2 \times 10^{-2} \mathrm{~m} \)
\(\text {Tangential force } \quad \mathrm{F}_{\mathrm{t}}=4 \times 10^{5} \mathrm{~N} \)
\(\text {Rigidity modulus of the cylinder }=6 \times 10^{10} \mathrm{Nm}^{-2} \)
\(\text {Rigidity modulus } \quad \eta_{R}=\frac{F_{t}}{\Delta A \theta} \)
\(\Delta A=\pi r^{2}=3.14 \times 2 \times 2 \times 10^{-4} \)
\(=12.56 \times 10^{-4} \mathrm{~m}^{2} \)
\(\eta_{R} =6 \times 10^{10}=\frac{4 \times 10^{5}}{12.56 \times 10^{-4} \theta}  \)
\(\text {First produced } \theta=\frac{4 \times 10^{+5}}{6 \times 10^{10} \times 12.56 \times 10^{-4}}  \)
\(=\frac{4 \times 10^{5-10+4}}{75.36} \)
\(=\frac{0.4}{75.36}\)
= 0.005307

Let the mass of Ag block be x kg.
Tension in the string T = 37.12 N
Relative density of liquid R_e =0.72.
Relative density of silver R_Ag =10
According to the principle of floatation,
\(V \rho_{g}=m g \)
\(\therefore V \rho=\mathrm{m} \)    ....(1)
\(\rho \quad- density\)
Weight of Ag block \(=mg =V \rho_{g}\)
\(=\mathrm{R}_{\mathrm{Ag}} \mathrm{V}_{\mathrm{g}}\)
Weight of Ag block \(W=10 V_{g}\)
Force of buoyancy \(F_{B}=R_{\text {liq }} V_{g}\)
\(=0.72 \mathrm{~V}_{\mathrm{g}}\)
Apparent weight \(\mathrm{W}_{\mathrm{app}}=10 \mathrm{~V}_{\mathrm{g}}-0.72 \mathrm{~V}_{\mathrm{g}}\)
\(=9.28 \mathrm{~V}_{\mathrm{g}}\)
Tension in the string = Apparent weight/Volume
\(37.12=9.28 \mathrm{~g}\)
= mg
\(\therefore Mass \ x=\mathrm{m}=\frac{37.12}{9.28}\)
Mass of Ag block = 4 kg