Two glass spheres of equal volume are connected by a small horizontal tube containing a small amount of mercury. The spheres are sealed at 20°C with exactly 1 litre of air on each side. Cross-sectional area of tube is 5 × 10⁻⁴ m². The mercury will be displaced by 34 × 10⁻ˣ centimeter if the temperature of one sphere is raised by 0.1°C while the other is maintained at 20°C. Value of x is (neglect expansion of glass spheres and tube):
Physics · Mixed Chapter
JEE Mains

Question

Two glass spheres of equal volume are connected by a small horizontal tube containing a small amount of mercury. The spheres are sealed at 20°C with exactly 1 litre of air on each side. Cross-sectional area of tube is 5 × 10⁻⁴ m². The mercury will be displaced by 34 × 10⁻ˣ centimeter if the temperature of one sphere is raised by 0.1°C while the other is maintained at 20°C. Value of x is (neglect expansion of glass spheres and tube):

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let mercury displacement be s. At equilibrium both gas pressures are equal. If the original volume of each side is V=10⁻³ m³ and tube area A=5×10⁻⁴ m², the heated side gains volume As and the other loses As. Step 2: Ideal gas equation gives (T+ΔT)/(V+As)=T/(V−As), where T≈293 K, ΔT=0.1 K. Step 3: Solve for s=VΔT/[A(2T+ΔT)] ≈ (10⁻³×0.1)/(5×10⁻⁴×586.1)=3.41×10⁻⁴ m=3.41×10⁻² cm. Step 4: 3.41×10⁻² cm≈34×10⁻³ cm, hence x=3.