physics ncert solutions class 11th

Diameter of the cone at the narrow end, d = 0.5 mm = 0.5 $*{10}^{-3}$ m

Radius, r = d/2 = 0.25 $*{10}^{-3}$ m

Area, A = $\pi *r^2$ = 1.96 $*{10}^{-7}$ $m^2$

Compressional force, F = 50000 N

Pressure at the tip of the anvil, p = F/A = 50000/1.96 $*{10}^{-7}$ Pa = 2.54 $*{10}^{11}$ Pa

Volume of water, V = 1 liter. If the water is compressed by 10%, then

ΔV = 0.10% of V= (0.1/100) $*1$ = 1 $*{10}^{-3}$

Bulk modulus of water, k = 2.2 $*{10}^9$ N/ $m^2$

From the relation, k = $V\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ , we get ΔP = k $*\frac{\mathrm{\Delta }\mathrm{V}}{\mathrm{V}}$ = 2.2 $*{10}^9*$ 1 $*{10}^{-3}\frac{}{1}$ = 2.2 $*{10}^6$ N/ $m^2$

Length of an edge of the solid copper cube, l = 10 cm = 0.1 m

Volume of the copper cube = 1 $*{10}^{-3}$ $m^3$

Hydraulic pressure, p = 7.0 $*{10}^6$ Pa

Bulk modulus of copper, k = 140 $*{10}^9$ Pa

From the relation k = $V\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ we get $\mathrm{\Delta }\mathrm{V}$ = $V\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{k}}$ = $\frac{1*{10}^{-3}*7.0*{10}^6}{140*{10}^9}$ = 5 $*{10}^{-8}$ $m^3$

Hydraulic pressure exerted on glass slab, p = 10 atm = 10 $*1.1013*{10}^5$ Pa

Bulk modulus of glass, k = 37 $*{10}^9$ N $m^{-2}$

From the relation k = $V\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ , we get $\frac{\mathrm{\Delta }\mathrm{V}}{V}$ = $\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{k}}$ = $\frac{10\mathrm{}*1.1013*{10}^5}{37*{10}^9}$ = 2.976 $*{10}^{-5}$

Initial volume, $V_1$ = 100 lit = 100 $*{10}^{-3}$ $m^3$

Final volume, $V_2$ = 100.5 lit = 100.5 $*{10}^{-3}$ $m^3$

Increase in volume, ΔV = $V_2-V_1$ = 0.5 $*{10}^{-3}$ $m^3$

Increase in pressure, ΔP = 100 atm = 100 $*$ 1.013 $*{10}^{5}Pa$

Bulk modulus, k = $V_1\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ = $\frac{100\mathrm{}*{10}^{-3}*100\mathrm{}*1.013*{10}^5}{0.5\mathrm{}*{10}^{-3}}$ Pa = 2.026 $*{10}^9$ Pa

Bulk modulus of air = 1.0 $*{10}^{5}Pa$

(Bulk modulus of water / Bulk modulus of air) = (2.026 $*{10}^9)/$ 1.0 $*{10}^5$ = 2.026 $*{10}^4$

This higher ratio is attributed to the higher compressibility of air than water.

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity,  $\omega$ = 2 rev/s

Cross sectional area of the wire, A = 0.065 ${cm}^2$ = 0.065 $*{10}^{-4}{m}^2$

Let Δl be the elongation of the wire

When the mass is placed at the position of the vertical circle, the total force on the mass is

F = mg + ml ${\omega }^2$ = 14.5 $*\left(9.81+1*2^2\right)$ = 200.25 N

Young's modulus for steel,  $Y_s$ = Stress / Strain = 2 $*{10}^{11}$ Pa

Stress = F/A = 200.25/0.065 $*{10}^{-4}$

Strain = (Δl/l) = (Δl/1) = Δl

Δl = (200.25/0.065 $*{10}^{-4})/$ 2 $*{10}^{11}$ = 1.54 $*{10}^{-4}$ m

It is given that the tension, F in each wire is same. Since the wire is of same length, Strain also will be same.

If $d_c$ is the diameter of copper wire and Young's modulus of copper $Y_c$ = 110 $*{10}^{9}Pa$ and strain is s, then $Y_c$ = $\frac{F}{\frac{\pi }{4}{d_c}^2}$ , $d_c=\sqrt{\frac{4F}{\pi *Y_c}}$

Similarly, if $d_i$ is the diameter of iron wire and $Y_i$ is the Young's modulus of iron = 190 $*{10}^{9}Pa$ , then $d_i$ = $\sqrt{\frac{4F}{\pi *Y_i}}$

$\frac{d_c}{d_i}$ = $\sqrt{\frac{Y_i}{Y_c}}$ = $\sqrt{\frac{190}{110}}$ = 1.314

Area of cross-section, A = $\pi {1.5}^2$ ${cm}^2$ = 7.07 $*{10}^{-4}{m}^2$

Maximum stress = Maximum load / cross sectional area

Maximum load = Maximum stress $*$ cross sectional area = 108 $*$ 7.07 $*{10}^{-4}$ = 7.07 $*{10}^4$ N