Question

Q10. A mass of 3 kg is placed on the massless piston
of area 10 cm2 of a cylindrical vessel containing a
gas. Find the fractional change in surface area of
sphere if its bulk modulus is 2 x 10000 N/m2 and
placed in the given vessel
(1) 0.5
(2) 1.5
(4) Zero
(3) 1
​

Answer 1

answer : option (3) 1

explanation : first find pressure exerted on piston.

so, pressure, P = weight of object placed on piston/base area

= m × g/A

here, m = 3Kg, g = 10m/s²

and A = 10cm² = 10 × 10^-4 m² = 10^-3 m²

P = 3 kg × 10 m/s²/(10^-3 m²)

= 3 × 10⁴ N/m²

now, Bulk modulus = P/(∆V/V)

given, bulk modulus = 2 × 10⁴ N/m²

so, 2 × 10⁴ = 3 × 10⁴/(∆V/V)

or, ∆V/V = 3/2........(1)

now, volume of sphere, V = 4/3 πr³

differentiating with respect to r,

dV = 4πr² dr

or, dV/V = (4πr²)dr/(4/3 πr³)

or, dV/V = 3 dr/r

or, ∆V/V = 3 ∆r/r .....(2)

similarly, surface area of sphere, A = 4πr²

after differentiating with respect to r,

or, dA = 8πr dr

or, dA/A = 2 dr/r

or, ∆A/A = 2 ∆r/r ......(3)

from equations (2) and (3),

∆A/A = (2/3) ∆V/V

from equations (1),

∆A/A = (2/3) × (3/2) = 1

hence, change in surface area = 1

Answer 2

explanation : first find pressure exerted on piston.

so, pressure, P = weight of object placed on piston/base area

= m × g/A

here, m = 3Kg, g = 10m/s²

and A = 10cm² = 10 × 10^-4 m² = 10^-3 m²

P = 3 kg × 10 m/s²/(10^-3 m²)

= 3 × 10⁴ N/m²

now, Bulk modulus = P/(∆V/V)

given, bulk modulus = 2 × 10⁴ N/m²

so, 2 × 10⁴ = 3 × 10⁴/(∆V/V)

or, ∆V/V = 3/2........(1)

now, volume of sphere, V = 4/3 πr³

differentiating with respect to r,

dV = 4πr² dr

or, dV/V = (4πr²)dr/(4/3 πr³)

or, dV/V = 3 dr/r

or, ∆V/V = 3 ∆r/r .....(2)

similarly, surface area of sphere, A = 4πr²

after differentiating with respect to r,

or, dA = 8πr dr

or, dA/A = 2 dr/r

or, ∆A/A = 2 ∆r/r ......(3)

from equations (2) and (3),

∆A/A = (2/3) ∆V/V

from equations (1),

∆A/A = (2/3) × (3/2) = 1

hence, change in surface area = 1