Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : A spherical body of radius (5 ± 0.1) mm having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is 4 %.
Reason R : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below

Solution

Physics sir · 4 steps

Terminal velocity goes as r², so the 2% error in radius doubles to 4% in velocity.

1. Terminal velocity. When the sphere reaches terminal velocity, weight is balanced by buoyancy plus Stokes' drag. That gives the expression below, where ρ is the body's density, σ the liquid's density and η its viscosity.

   vₜ = 2r²(ρ - σ)g/9η ⇒ vₜ ∝ r²

2. Check Assertion A. The densities, g and η are fixed, so only r carries error. For a power, the percentage error is multiplied by the power: (Δ vₜ)/vₜ = 2(Δ r)/r = 2 × 0.1/5 × 100% = 4%. A is true.
3. Check Reason R. R says vₜ ∝ 1/r, but the formula shows vₜ ∝ r². A bigger sphere falls faster, not slower. R is false.
4. So. A is true and R is false, which is option (b).

The idea to keep. For y ∝ xⁿ, the percentage error in y is |n| times that in x. Write the real formula first, read off the power, then multiply the error.

Answer from the NTA final answer key. Solution by PrepWiser.