JEE Main 2021 · 17 March · evening · Single correct · +4 / −1
An object is located at 2 km beneath the surface of the water. If the fractional compression (Δ V)/V is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ____________. [Given : density of water is 1000 kgm⁻³ and g = 9.8 ms⁻²]
- (a) 1.44 × 10⁷ Nm⁻²
- (b) 1.44 × 10⁹ Nm⁻²
- (c) 1.96 × 10⁷ Nm⁻²
- (d) 2.26 × 10⁹ Nm⁻²
Answer and solution
Answer (b) 1.44 × 10⁹ Nm⁻²
Solution
Physics sir · 3 steps
Hydraulic stress is the water pressure at that depth; divide it by the volume strain to get the bulk modulus.
- Hydraulic stress. The extra pressure on the object 2 km down is p = ρ g h = 1000 × 9.8 × 2000 = 1.96 × 10⁷ N m⁻².
- Hydraulic strain. The strain is the fractional compression (Δ V)/V. Change the percentage to a fraction: 1.36% = 0.0136.
B = p/(Δ V / V) = (1.96 × 10⁷)/0.0136 = 1.44 × 10⁹ N m⁻²
- So. The ratio of hydraulic stress to hydraulic strain, which is the bulk modulus, is 1.44 × 10⁹ N m⁻². That is option (b).
The idea to keep. Stress over strain for a change in volume is the bulk modulus. Pressure at depth is ρ g h, and a percentage strain must be divided by 100 before you use it.
Answer from the NTA final answer key. Solution by PrepWiser.