Two metallic wires P$P$ and Q$Q$ have same volume and are made up of same material. If their area of cross sections are in the ratio 4:1$4:1$ and force F₁$F_1$ is applied to P$P$, an extension of Δ l$\Delta l$ is produced. The force which is required to produce same extension in Q$Q$ is F₂$F_2$. The value of (F₁)/(F₂)$\frac{F_1}{F_2}$ is:

Related Formula

Core Logic

The change in pressure Δ P$\Delta P$ is the hydrostatic pressure at depth h$h$.

Step 2: Calculation

Given values: ρ = 10³ kg/m³$\rho = 10^3\mathrm{\,kg/m^3}$ g = 10 m/s²$g = 10\mathrm{\,m/s^2}$ β = 9 × 10⁸ N/m²$\beta = 9 \times 10^8\mathrm{\,N/m^2}$ (Δ V)/(V) = -0.02% = -(0.02)/(100)$\frac{\Delta V}{V} = -0.02\% = -\frac{0.02}{100}$

Substitute into the equation:

Chapter Mix

Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids