Find the increase in pressure required to decrease volume of mercury by 0.001%. (Bulk modulus of mercury = 2.8 x 10¹⁰ N / m²) (Ans : 2.8 x 10⁵ N / m²)
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Answered by
19
it is given that decreases of volume of mercury = 0.001 %
% decreases in volume = change in volume/original volume
0.001 % = ∆V/V × 100
or, ∆V/V = 0.001 × 10^-2 = 10^-5
we know,
bulk modulus = P/(∆V/V)
or, P = bulk modulus × (∆V/V)
= 2.8 × 10^10 × 10^-5
= 2.8 × 10^5 N/m²
hence, the increase in pressure = 2.8 × 10^5 N/m²
% decreases in volume = change in volume/original volume
0.001 % = ∆V/V × 100
or, ∆V/V = 0.001 × 10^-2 = 10^-5
we know,
bulk modulus = P/(∆V/V)
or, P = bulk modulus × (∆V/V)
= 2.8 × 10^10 × 10^-5
= 2.8 × 10^5 N/m²
hence, the increase in pressure = 2.8 × 10^5 N/m²
Answered by
0
Explanation:
B=2.8×10
10
ΔV
2
0.001%
v×
100
0.001
=0v
∴
=0.00001v
0v
∴B=
Δv
pv
∴2.8×10
10
×10
−5
=P
∴2.8×10
5
N/m
2
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