Physics 3 Marks Question

Breaking stress = 8 x 10⁸N/m²

$\frac{\text{C}}{\text{S}}$ Area of the upper wire A $=0.006\text{cm}^2=\frac{0.003}{10000}\text{m}^2$

$\frac{\text{C}}{\text{S}}$  area of the lower wire A' $=0.003\text{cm}^2=\frac{0.003}{10000}\text{m}^2$

1. Let the length of each wire = L and the maximum load on the hanger W

Total load on the lower wire = W + m₁

The stress on the lower wire (W + m₁)g/A'

$=(\text{W}+10)\frac{\text{g}}{\Big(\frac{0.003}{10000}\big)}\text{N}/\text{m}^2$

$=(\text{W}+10)\text{g}\times\frac{10^7}{3}\text{N}{\text{m}}^2$

$=\frac{(\text{W}+30)\text{g}}{\frac{(0.0060)}{10000}}\text{N}/\text{m}^2$

$=\frac{(\text{W}+30)\text{g}^*10^7}{6}\text{N}/\text{m}^2$

Equationg it to breaking stress,$\frac{(\text{W}+30)\text{g}^*10^7}{6} = 8\times10^8$

$\Rightarrow \text{W}=\frac{6^*80}{10-30}$

$\Rightarrow \text{W}=48-30=18\text{kg}$

Since the lower wire reaches the breaking stress with a lower weight, the maximum weight W = 14kg. The lower wire will break first.

2. If m₂ = 36kg

The stress on upper wire $=\frac{(\text{W}+10+36)\text{g}}{\text{A}}$

$=\frac{(\text{W}+46)\text{g}}{\frac{0.006}{10000}}$

$=\frac{(\text{W}+46)\text{g}^*10^7}{6}$

Equating with breaking stress, $\frac{(\text{W}+46)\text{g}^*10^7}{6}=8\times 10^8$

$\Rightarrow \text{W}=\frac{6^*80}{10-46}$

$=48-46=2\text{kg}$

Given:

Bulk modulus of water B = 2 × 10⁹Nm²

Depth d = 400m

Density of water at the surface $\rho_0=1030\text{Kg}/\text{m}^3$

We know that:

Density at surface $=\frac{\text{m}}{\text{V}_0}$

Density at depth $\rho_\text{d}=\frac{\text{m}}{\text{V}_\text{d}}$

$\Rightarrow\frac{\rho_\text{d}}{\rho_\text{0}}=\frac{\text{V}_0}{\text{V}_\text{d}}\ \cdots(1)$

Here:

$\rho_\text{d}$ = density of water at a depth

m = mass

V₀ = volume at the surface

V_d = volume at a depth

Pressure at a depth d $=\rho_0\text{gd}$

Acceleration due to gravity g = 10ms²

Volume strain $=\frac{\text{V}_0-\text{V}_\text{d}}{\text{V}_0}$

$\text{B}=\frac{\text{pressure}}{\text{volume}\ \text{strain}}$

$\Rightarrow\text{B}=\frac{\rho_0\text{gd}}{\Big(\frac{\text{V}_0-\text{V}_\text{d}}{\text{V}_0}\Big)}$

$\Rightarrow1-\frac{\text{V}_\text{d}}{\text{V}_0}=\frac{\rho_0\text{gd}}{\text{B}}$

$\Rightarrow\frac{\text{V}_\text{d}}{\text{V}_0}=\Big(1-\frac{\rho_0\text{gd}}{\text{B}}\Big)\ \cdots(2)$

Using equations (1), and (2), we get:

$\Rightarrow\frac{\rho_\text{d}}{\rho_0}=\frac{1}{\Big(1-\frac{\rho_0\text{gd}}{\text{B}}\Big)}$

$\Rightarrow\rho_\text{d}=\frac{1}{\Big(1-\frac{\rho_0\text{gd}}{\text{B}}\Big)}\rho_0$

$\Rightarrow\rho_\text{d}=\frac{1030}{\Big(1-\frac{1030\times10\times400}{2\times10^9}\Big)}\approx1032\text{Kg}/\text{m}^3$

Change in density $=\rho_\text{d}-\rho_0$

= 1032 - 1030 = 2Kg/m³

Hence, the required density at a depth of 400m below the surface is 2Kg/m³.

Given:

Original length of steel wire L = 1m

Area of cross-section A = 4.00mm² = 4 ×10^-2cm²

Load = 2.16kg

Young's modulus of steel Y = 2 × 10¹¹N/m²

Acceleration due to gravity g = 10ms^-2

Let T be the tension in the string after the load is suspended and $\theta$ be the angle made by the string with the vertical, as shown in the figure:

$\cos\theta=\frac{\text{x}}{\sqrt{\text{x}^2+\text{l}^2}}=\frac{\text{x}}{\text{l}}\Big\{1+\frac{\text{x}^2}{\text{l}^2}\Big\}^{\frac{-1}{2}}$

Expanding the above equation using the binominal theorem:

$\cos\theta=\frac{\text{x}}{\text{l}}\Big\{1-\frac{1}{2}\frac{\text{x}^2}{\text{l}^2}\Big\}$ (neglecting the higher order terms)

Since x << l, $\frac{\text{x}^2}{\text{l}^2}$ can be neglected.

$\Rightarrow\cos\theta=\frac{\text{x}}{\text{l}}$

increase in length:

$\triangle\text{l}=(\text{AC}+\text{CB})-\text{AB}$

$\text{AC}=\big(\text{l}^2+\text{x}^2\big)^{\frac{1}{2}}$

$\triangle\text{l}=2\big(\text{l}^2+\text{x}^2\big)^{\frac{1}{2}}-2\text{l}$

We know that:

$\text{Y}=\frac{\text{F}}{\text{A}}\frac{\text{L}}{\triangle\text{L}}$

$\Rightarrow2\times10^{12}=\frac{\text{T}\times100}{\big(4\times10^{-2}\big)\times\big[2\big (50^2+\text{x}^2\big)^{\frac{1}{2}}-100\Big]}$

From the free body diagram:

$2\text{T}\cos\theta=\text{mg}$

$2\text{T}\Big(\frac{\text{x}}{50}\Big)=2.16\times10^{3}\times980$

$\Rightarrow\frac{2\times\big(2\times10^{12}\big)\times\big(4\times10^{-2}\big)\times\Big[2\Big(50^2+\text{x}^2\frac{1}{2}\Big)-100\Big]\text{x}}{100\times50}=(2.16)\times10^3\times980$

On solving the above equation, we get x = 1.5cm

Hence, the required vertical depression is 1.5cm.